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Pythagorean RightAngled Triangles
Rightangled triangles with whole number sides have fascinated mathematicians
and number enthusiasts since well before 300 BC when Pythagoras wrote about his famous "theorem".
The oldest mathematical document in the world, a little slab of clay that would fit in your hand,
is a list of such triangles. So what is so fascinating about them? This page starts from scratch
and has lots of facts and figures with several online calculators to help with your own
investigations.
Contents of this page
The icon means there is a
Things to do section of questions to start your own investigations.
The calculator icon
indicates that there is a live interactive calculator in that section.
Rightangled Triangles and Pythagoras' Theorem
Pythagoras and Pythagoras' Theorem
Pythagoras was a mathematician born in Greece in about 570 BC. He was interested in
mathematics, science and philosophy. He is known to most people because of
the Pythagoras Theorem that is about a property of all
triangles with a rightangle (an angle of 90°):
If a triangle has one angle which is a rightangle (i.e. 90°)
then there is a special relationship between the lengths of its three sides:
If the longest side (called the hypotenuse) is h
and the other two sides (next to the
right angle) are called a and b, then: a^{2} + b^{2} = h^{2}Pythagoras' Theorem
or,
the square of the longest side is the same as the sum of the squares of
the other two sides. h^{2} = a^{2} + b^{2}
is only true for rightangled triangles.
If all the angles of a triangle are less than 90° then
h^{2} < a^{2} + b^{2}
For example, in an equilateral triangle with sides 1 1 1 and all angles 60°
1^{2} = 1 < 1^{2} + 1^{2} = 2
If one the angles of a triangle is greater than 90° then
h^{2} > a^{2} + b^{2}
Note that in any triangle, the longest side h cannot be longer than the sum of the other two sides.
so h < a + b.
If it equalled the sum of the other two then the triangle is just a line of length a + b = h!
For example, if the two shorter sides of a rightangled triangle are 2 cm and 3 cm, what is
the length of the longest side?
If the longest side is h, then, by Pythagoras' Theorem, we have:
h^{2}
=
2^{2} + 3^{2} = 13
h
=
√13 = 3·60555
Some visual proofs of Pythagoras' Theorem
My favourite proof of the lookandsee variety is on the right.
Both diagrams are of the same size square of side a + b.
Both squares contain the same four identical rightangled triangles in white
(so it is whiteangled )
with sides a, b, c.
The left square also has two blue squares
with areas a^{2} and b^{2}
whereas the right hand one replaces them with one red square of area c^{2}.
This does not depend on the lengths a, b, c; only that they are the sides of a rightangled triangle.
So the two blue squares
are equal in area to the red square, for any rightangled triangle:
a^{2} + b^{2} = c^{2}
This makes an effective visual aid by pushing the squares from their locations on the left to
where they are shown on the right. Don't turn them or flip them, just move them
to their respective corners.
There is a very nice illustration of A Device That Illustrates Pythagoras' Theorem
that is a Mathematica Demonstration.
Click on the image on the right here to see an animation in a new window
or to download the active controls version usable with the free Mathematica player.
Bill Richardson has a nice animation of Bhaskara's proof
The 345 Triangle
In the example above, we chose two wholenumber sides and found the longest
side, which was not a whole number.
It is perhaps surprising that there are some rightangled triangles where
all three sides are whole numbers called Pythagorean Triangles.
The three whole number sidelengths are called
a Pythagorean triple or triad.
An example is a = 3, b = 4 and
h = 5, called "the 345 triangle". We can check it as follows:
3^{2}+4^{2} = 9 + 16 = 25 = 5^{2} so
a^{2} + b^{2} = h^{2}.
This triple was known to the Babylonians (who lived in the area of presentday Iraq and Iran)
even as long as 5000 years ago! Perhaps they used it to make a rightangled triangle so they could
make true rightangles when constructing buildings  we do not know for certain.
It is very easy to use this to get a rightangle using equallyspaced knots
in a piece of rope, with the help of two friends.
If you hold the ends of the rope together together and one friend holds the fourth knot and
and the other the seventh knot and you all then tug to stretch the rope into a triangle,
you will have the 345 triangle
that has a true rightangle in it.
The rope can be as long as you like
so you could lay out an accurate rightangle of any size.
The sum of the sides of a triangle is called its perimeter.
We can also easily draw a 3 4 5 triangle as follows:
The
is a 3 4 5 triangle
But all Pythagorean triangles are even easier to draw on squared paper because all their sides are
whole number lengths.
Measure the lengths
of the two smaller sides (those around the rightangle) as lengths along and up
from the same point and then join the two
endpoints together.
So Pythagorean triangles also tell us which pairs of points with wholenumber coordinates
are a wholenumber distance apart not
in a horizontal or vertical direction.
Test a Triangle  is it Pythagorean?
Here is a little calculator that, given any two sides of a rightangled triangle
will compute the third, or you can give it all three sides to check.
It will check that it is rightangled and, if so, if it is Pythagorean
(all the sides are integers).
Here a and b are the
two legs, the sides surrounding the rightangle, and h is the longest side,
the hypotenuse:
Is the 345 the only Pythagorean Triple?
No, because we can double the length of the sides of the 345 triangle
and still have a rightangled triangle:
its sides will be 6810 and we can check that
10^{2} = 6^{2} + 8^{2}.
Continuing this process by tripling 345 and quadrupling
and so on we have an infinite number of Pythagorean triples:
3
4
5
6
8
10
12
16
20
15
20
25
18
24
30
...
or we can take the series of multiples 1, 11, 111, 1111, etc. and get a pattern:
3
4
5
33
44
55
333
444
555
3333
4444
5555
...
All of these will have the same shape (have the same angles)
but differ in size  the mathematical term is that they are all
similar triangles. If they were the same size but in different positions
or orientations, the triangles are called congruent.
Are there any other differentlyshaped rightangled triangles with whole
number sides?
Yes; one is 5, 12, 13 and another is 7, 24, 25.
We can check that they have right angles by using Pythagoras's Theorem that the squares of the two smaller sides sum to the square
of the longest side. For example 5^{2} + 12^{2} = 25 + 144 = 169 =
13^{2} 7^{2} + 24^{2} = 49 + 576 = 625 =
25^{2}
Graphs of the numbers of Pythagorean Triangles
The number of Pythagorean Triangles with a hypotenuse up to a given limit is remarkable consistent as the limit increases
as is shown by these graphs:
Graphs of the number of Pythagorean Triangles with Hypotenuse<H:
H
Only Primitives
All
Primitives+All
100
300
1000
5000
H
Later we will look more closely at these graphs and find a surprising number appears in connection
with these straight lines.
But for now, how do we find these Pythagorean triangles? Is there a systematic method?
Methods of Generating Pythagorean Triangles
The simplest method of finding all Pythagorean triples
Richard du Croo de Jongh wrote to me in July 2019 pointing out that what is surely
the simplest method of generating all Pythagorean triangles!
The method is mentioned in
Kraitchik's Mathematical Recreations on page 97: see the reference below.
If a^{2} + b^{2} = h^{2} then a^{2} = h^{2}  b^{2} which factorises: a^{2} = (h + b)(h  b)
So find two factors of a^{2}, say P and Q and P>Q. Then P = h + b, Q = h  b which means that h = (P + Q)/2, b = (P  Q)/2.
In order that P and Q are whole numbers, P and Q must be both odd or both even and P > Q (or else b is 0 or negative).
Let's try an example with a = 12: a^{2} = 144. Possible P and Q are:
P=144,Q=1 but one is even and one odd (giving a fractional value for b and h)
P=72, Q=2 and both are even
(P+Q)/2 = h = 37; (PQ)/2 = b = 35: the triple is a=12, b=35, h=37
P=48, Q=3 but one is even and one odd (giving a fractional value for b and h)
P=36, Q=4 and both are even
(P+Q)/2 = h = 20; (PQ)/2 = b = 16: the triple is a=12, b=16, h=20
P=24, Q=6 and both are even
(P+Q)/2 = h = 15; (PQ)/2 = b = 9: the triple is a=12, b=9, h=15
P=18, Q=8 and both are even
(P+Q)/2 = h = 13; (PQ)/2 = b = 5: the triple is a=12, b=5, h=13
P=16, Q=9 but one is even and one odd (giving a fractional value for b and h)
So there are just 4 Pythagorean triangles with a side of 12.
A similar method involving factors is the HypotenuseLeg difference method see below.
A simple twounitfraction method of generating PTs
This is a very simple method of generating Pythagorean triangles. It is based on forming the sum of
two unit fractions for either consecutive odd numbers or consecutive even numbers.
odd A
next B=A+2
^{1}/_{A}+^{1}/_{B}
Hyp
1
3
4/3
5
3
5
8/15
17
5
7
12/35
37
7
9
16/63
65
...
Two consecutive odd unit fractions
Take two odd numbers that differ by two, such as 3 and 5.
Make them into unit fractions: ^{1}/_{3} and ^{1}/_{5}
and add them: ^{1}/_{3} + ^{1}/_{5} = ^{8}/_{15}
The two numbers in the sum are always two sides of a primitive Pythagorean triangle!
Here the Pythagorean triangle is 8, 15, 17.
even A
next B=A+2
^{1}/_{A}+^{1}/_{B}
Hyp
2
4
3/4
5
4
6
5/12
13
6
8
7/24
25
8
10
9/40
41
...
Two consecutive even unit fractions
Take two even numbers that differ by two, such as 2 and 4.
Make them into unit fractions: ^{1}/_{2} and ^{1}/_{4}
and add them: ^{1}/_{2} + ^{1}/_{4} = ^{3}/_{4}
The two numbers in the reduced sum are always two sides of a primitive Pythagorean triangle!
Here it is the 3, 4, 5
triangle.
However, neither using the two odds nor the two evens
generate all the primitive Pythagorean triangles, but there is a method using
two fractions that does.
The TwoFractions method of generating Pythagorean Triples
Here is a very simple way of generating as many Pythagorean triangle as you want:
Method:
Example 1:
Example 2:
Take any two fractions (or whole numbers) whose product is 2
Notice that the fractions do not have to be in their lowest form:
1/3
6
4/2
2/2
Add 2 to each fraction:
7/3
8
8/2
6/2
Cross multiply to turn both into whole numbers
7
24
16
12
These are two sides of a Pythagorean triangle:
7
24
16
12
To find the third, add the squares of these two numbers:
7^{2}+24^{2} = 49 + 576 = 625
16^{2}+12^{2} = 256 + 144 = 400
... and take the squareroot to find the hypotenuse:
√625 = 25
√400 = 20
to get the Pythagorean triangle:
7 24 25
16 12 20
This works for any two fractions whose product is
2 and always generates a Pythagorean triangle:
Start with
to get:
1
2
3 4 5
2/2 1
2 4/2
6 8 10
1/2 2/4
8/2 4
5 12 13
3/3 1
2 6/3
9 12 15
2/3
3
8 15 17
4/4 2/2 1
2 4/2 8/4
12 16 20
1/3
6
7 24 25
3/2
4/3
20 21 29
1/4
8
9 40 41
In fact, all primitive Pythagorean triples are generated from two fractions in lowest form and
all nonprimitive Pythagorean triples are generated when at least one of the fractions is not in its
lowest form.
(See later on this page).
Here is a calculator for your experiments and some questions to investigate below it.
Generate PTs using Two Fractions Calculator
C A L C U L A T O R
Give two fractions whose product is 2
: they do not have to be in lowest form
and
with fractions
ka/kb, 2b/a
with k= and
a=
up to
b= up to
Factorize nonprimitive triples:
Use 2b/a in its lowest form:
Find starting fractions that generate the multiples of 3 4 5:
6 8 10, 9 12 15, 12 16,20, ...
Are some multiples impossible to generate by this method?
Which multiples of 5 12 13 can you generate from two fractions?
10 24 26, 15 36 39, 20 48 52, ...?
Use algebra to prove the twofractions method always produces a Pythagorean triangle from any two
initial numbers, as follows,
starting with
a
and
c
2 b
b
a
where a, b and c are whole numbers.
Hint: expand ( a^{2} + 2ab + 2 b^{2} )^{2}
a/b + 2 = (a+2b)/b
(2b)/a + 2 = 2(a+b)/a crossmultiplying:
(a+2b) a, 2(a+b) b; squaring each and adding gives:
(a^{2} + 2ab + 2 b^{2})^{2}
Find a triple that needs two fractions (not a fraction and a whole number) to generate it.
3/2 and 4/3 give 20 21 29
How many pairs of numbers generate 6 8 10?
4
1
2
2
2
2
I do not have a reference for this method, so if anyone can help, please do email me (details on the link that is my name
the foot of this page). Thanks!
But rather than a method ....
... is there is a formula for generating Pythagorean triples?
The m,n formula for generating Pythagorean Triples
2mn
m^{2} + n^{2}
m^{2} – n^{2}
Yes  we can generate Pythagorean Triples by supplying two different positive integer
values for m and n in this diagram.
You can multiply out these terms and check that
Once we have found one triple, we have seen that we can generate many others by just
scaling up all the sides by the same factor.
A Pythagorean triple which is not a multiple of another is called a primitive Pythagorean triple.
So 3,4,5 and 5,12,13 are primitive Pythagorean triples
but 6,8,10 and 333,444,555
and 50,120,130 are not.
Are all the Pythagorean triples generated by m,n?
The bad news is that the answer is "No", but the good news is
that all primitive Pythagorean triples are generated by some m,n values
in the formula above!
The formula using m,n will not give
all triples since it misses some of the nonprimitive ones, such as 9,12,15. This is a
Pythagorean triple since, as a triangle, is it just three times the 3,4,5
triangle (by which we mean
that we just MULTIPLY the lengths of each side of a 3,4,5 triangle by 3,
which we already know is rightangled).
But 9, 12, 15 is missed by our m,n formula because:
Our formula said m and n were positive whole numbers
and the Pythagorean triple was
m^{2} – n^{2} , 2mn , m^{2} + n^{2}
and, since we want (positive) whole number values in our triple, then m > n
(otherwise the first number in the triple is negative).
The 2mn value is one of the sides and the only even side in 9, 12, 15 is 12,
so 12 = 2 m n.
Hence m n = 6. But m > n, so we can only have two cases:
m = 6 with n = 1 OR
m = 3 and n = 2
The first case gives the triple 35, 12, 37 and the second case
gives 5, 12, 13, neither of which is
the 9, 12, 15 triple.
There are two values which generate 9, 12, 15 and they are
m = 2√2, n=√2.
In fact there are always m,n values for all PTs
but they are not always integers:
If m,n generates a, b, h then
g×a, g×b, g×h is generated by
√g m, √g n.
m^{2} – n^{2} , 2mn , m^{2} + n^{2}
All the primitive Pythagorean triangles are each generated once if and only if)
one of m,n pair is odd and the other is even.
This is often described as
m and n have opposite parity.
Since all the sides of the triangle are positive then we also need m>n.
For more on this and a proof, see the Hardy and Wright book "Introduction to the Theory of numbers" in the
References at the foot of this page.
Here is a table of Pythagorean triangles with a smaller side up to 40:
Triple
primitive?
m,n
3, 4, 5
primitive
2,1
5, 12, 13
primitive
3,2
6, 8, 10
2× 3, 4, 5
3,1
7, 24, 25
primitive
4,3
8, 15, 17
primitive
4,1
9, 12, 15
3× 3, 4, 5
–
9, 40, 41
primitive
5,4
10, 24, 26
2× 5, 12, 13
5,1
11, 60, 61
primitive
6,5
12, 16, 20
4× 3, 4, 5
4,2
12, 35, 37
primitive
6,1
13, 84, 85
primitive
7,6
14, 48, 50
2× 7, 24, 25
7,1
15, 20, 25
5× 3, 4, 5
–
15, 36, 39
3× 5, 12, 13
–
15, 112, 113
primitive
8,7
16, 30, 34
2× 8, 15, 17
5,3
16, 63, 65
primitive
8,1
17, 144, 145
primitive
9,8
18, 24, 30
6× 3, 4, 5
–
18, 80, 82
2× 9, 40, 41
9,1
19, 180, 181
primitive
10,9
20, 48, 52
4× 5, 12, 13
6,4
20, 99, 101
primitive
10,1
20, 21, 29
primitive
5,2
Triple
primitive?
m,n
21, 28, 35
7× 3, 4, 5
–
21, 72, 75
3× 7, 24, 25
–
21, 220, 221
primitive
11,10
22, 120, 122
2× 11, 60, 61
11,1
23, 264, 265
primitive
12,11
24, 32, 40
8× 3, 4, 5
6,2
24, 45, 51
3× 8, 15, 17
–
24, 70, 74
2× 12, 35, 37
7,5
24, 143, 145
primitive
12,1
25, 60, 65
5× 5, 12, 13
–
25, 312, 313
primitive
13,12
26, 168, 170
2× 13, 84, 85
13,1
27, 36, 45
9× 3, 4, 5
6,3
27, 120, 123
3× 9, 40, 41
–
27, 364, 365
primitive
14,13
28, 96, 100
4× 7, 24, 25
8,6
28, 45, 53
primitive
7,2
28, 195, 197
primitive
14,1
29, 420, 421
primitive
15,14
30, 40, 50
10× 3, 4, 5
–
30, 72, 78
6× 5, 12, 13
–
30, 224, 226
2× 15, 112, 113
15,1
31, 480, 481
primitive
16,15
32, 60, 68
4× 8, 15, 17
8,2
32, 126, 130
2× 16, 63, 65
9,7
32, 255, 257
primitive
16,1
Triple
primitive?
m,n
33, 44, 55
11× 3, 4, 5
–
33, 180, 183
3× 11, 60, 61
–
33, 544, 545
primitive
17,16
33, 56, 65
primitive
7,4
34, 288, 290
2× 17, 144, 145
17,1
35, 84, 91
7× 5, 12, 13
–
35, 120, 125
5× 7, 24, 25
–
35, 612, 613
primitive
18,17
36, 48, 60
12× 3, 4, 5
–
36, 160, 164
4× 9, 40, 41
10,8
36, 105, 111
3× 12, 35, 37
–
36, 323, 325
primitive
18,1
36, 77, 85
primitive
9,2
37, 684, 685
primitive
19,18
38, 360, 362
2× 19, 180, 181
19,1
39, 52, 65
13× 3, 4, 5
–
39, 252, 255
3× 13, 84, 85
–
39, 760, 761
primitive
20,19
39, 80, 89
primitive
8,5
40, 96, 104
8× 5, 12, 13
10,2
40, 75, 85
5× 8, 15, 17
–
40, 198, 202
2× 20, 99, 101
11,9
40, 399, 401
primitive
20,1
40, 42, 58
2× 20, 21, 29
7,3
Generate PTs using the m,n formula Calculator
2mn
m^{2} + n^{2}
m^{2} – n^{2}
Here is a calculator to compute the sides of a triangle using the formula above  just type
in the values for m and n.
Remember that the formula will find all the Primitive Triples but it will not
find all the nonprimitives. The calculator will tell you if your values of
m and n generate a Primitive
triple or a multiple of a primitive.
Alternatively, you can give it a single m value or a range of
m values and it will show you all the triangles it can generate.
Finally, give it a Pythagorean triangle and it will
test if it has generators m and n or not.
The twofraction method and the m,n generators method
We can now show that the twofractions method that we saw earlier on the page
generates all primitive Pythagorean triples by identifying n with b
and m with a + b
to find two starting fractions.
This is equivalent to choosing a = m – n and
b = n for fractions
a
and
2b
b
a
to get the primitive triple with m,n generators.
If the triangle is nonprimitive, say it is k times a primitive triangle, then the
substitutions above give the primitive triangle and we need only multiply one of the
fractions by k on the top and on the bottom to get the nonprimitive one.
The Fibonacci method
The Fibonacci numbers are generated by starting from 1 and 2 and then
using the method add the latest two to get the next. We use this method here to generate PTs.
Take any 2 numbers to start your Fibonacci series, such as 1 and 3.
In a Fibonaccilike way, add them to produce the next: 1, 3, 4 and extend your series once more by
the same rule: 1, 3, 4, 7.
Now you can make a Pythagorean Triple as follows:
Using the Fibonaccitype series of 4 numbers: 1, 3, 4, 7:
First leg:
Multiply the middle two numbers and double the result:
here 3 times 4
is 12 which we double to get 24 : the first side of our Pythagorean Triangle
Second leg:
Multiply the two outer numbers:
here 1 times 7
gives 7 : the second side of the Pythagorean triangle
Hypotenuse:
EITHER add the squares of the middle two numbers:
here 3^{2} + 4^{2} = 25
OR from the product of the last two subtract the product of the first two:
here 4×7 – 1×3 = 25
Both give 25 : the hypotenuse of the Pythagorean triangle
So we have found the primitive Pythagorean triangle 7, 24, 25.
You can start with any two numbers and use the
Fibonacci Rule: add the latest two to get the next to generate two more.
The four numbers will always generate a Pythagorean Triangle. Try it for yourself or using the
Calculator below!
The explanation of why the Fibonacci method works is simple and is related to the m,n formula method
above:
The two middle values are the m and
n values for the m,n formula for generating Pythagorean Triples
that we looked at earlier on this page:
m – n, n, m, m + n
Using the method above for this Fibonaccitype series of 4 numbers, the sides of the Pythagorean triangle are :
twice the product of the middle two: 2 m n
the product of the outer two values: (m – n) (m + n) = m^{2} – n^{2}
the sum of squares of the inner two values: m^{2} + n^{2}
We saw earlier that not all Pythagorean triples can be generated by the mn method.
We have just shown that the Fibonaccimethod is equivalent to that method and so
the Fourterm Fibonacci method will also fail to produce all triples.
It can produce all primitive triples but not all of the composite ones.
Let's call our four number in a Fibonaccitype progression a, b, c, d so that
a + b = c, b + c = d.
Then we have:
the legs of the triangle are 2 b c and a d
the hypotenuse is
b^{2} + c^{2} = d^{2} – 2 b c = a^{2} + 2 b c
the area is a b c d
the perimeter is 2 c d
the inradius is a b
the exradii (that is, the three circles outside the
triangle that are tangential to all three sides of the triangle) are
a c, b d and c d
so that, in many ways, the lengths of significant constructs in a Pythagorean triangle are
more easily and more naturally described using the four Fibonacci sequence valuesa b c d than
the with the m,n generators.
Pythagorean triangles from the Fibonacci Series C W Raine Scripta Mathematica
vol 14 (1948) page 164
seems to be the earliest reference to this method but only for the Fibonacci numbers.
Fibonacci Number Triples A F Horadam American Mathematical Monthly vol 68 (1961) pages 751753
gives the generalisation to any two starting numbers.
A Note on the ramifications concerning the construction of Pythagorean Triples from recursive sequences
H T Freitag in Applications of Fibonacci Numbers, vol 3 G E Bergum, A N Philippou, A F Horadam (eds), (Kluwer Academic 1990), pages 101106.
HypotenuseLeg difference
Hassan Ouramdane emailed me (3 November 2014) with this alternative method of generating PTs in terms of the difference between the
hypotenuse and one side.
For PT a, b, h suppose the difference
between one side, b say,
and the hypotenuse h is d then we have b + d = h.
Using Pythagoras' Theorem we have:
a^{2} + b^{2} = h^{2} and we can replace h by b + d
a^{2} + b^{2} = (b + d)^{2}Now expand the brackets:
a^{2} + b^{2} = b^{2} + 2 b d + d^{2}and we see that we can subtract b^{2}from both sides:
a^{2} = 2 b d + d^{2}and the righthand side will now factorize
a^{2} = d (2 b + d)
This tells us that
the difference between the leg b
and the hypotenuse h must be a factor of the square of the other leg a
which gives us an easy way of generating such triangles, given one side of a PT a
and a difference d:
find all the factors of a^{2} that are less than a
because the other factor is (2 b + d) which is bigger than d
All of these factors can be a value for d.
To find the PTs, one of the sides being a do the following for each value of d.
Since a^{2} = d (2 b + d)
then
a^{2}
– d = 2 b
d
:
Divide the given side squared (a^{2}) by d
From the result subtract d
IF the result is even, divide it by 2 to find the second leg b
For some factors we will not get an even number and therefore such factors cannot be values for d.
If the value was a whole number, then add d to it to find the hypotenuse h
in the PT a, b, b+d
For instance, find PTs with one side 12:
12^{2} = 144 = 2^{4} 3^{2}
The factors of 144 that are less than 12 are: 1, 2, 3, 4, 6, 8 and 9 which are possible values for the difference h  b = d:
Taking each in turn:
1?
144/1  1 = 143 which we cannot divide by 2
2?
144/2  2 = 72  2 = 70 which is even so b = 35: 12, 35, 37=35+2
3?
144/3  3 = 48  3 = 45 which is not even.
4?
144/4  4 = 36  4 = 32 which is even so b = 16: 12, 16, 20=16+4
6?
144/6  6 = 24  6 = 18 which is even so b = 9: 12, 9, 15=9+6
8?
144/8  8 = 18  8 = 10 which is even so b = 5: 12, 5, 13=5+8
9?
144/9  9 = 16  9 = 7 which is not divisible by 2
An easy method of writing down a series of Triples
Look at the following series of Pythagorean triples. It is easy to spot the pattern and to
remember it. If you then write it down to show your friends it will
look as if you have impressive calculating skills!
21
220
221
201
20200
20201
2001
2002000
2002001
20001
200020000
200020001
It uses n+1 for m
in the formula and then lets n be powers of 10. This simplifies the triples
to be 2n+1, 2n(n+1), 2n^{2}+2n+1.
Solution to Problem: Find a Scheme for writing mechanically an unlimited number of Pythagorean Triangles
M Willey, E C Kennedy, American Mathematical Monthly vol 41 (1934) page 330.
There is a Calculator below that you can use to generate
many more simple patterns like this one.
Can you find another simple method like those above that always produces Pythagorean Triangles?
Patterns in Pythagorean Triples
Let's call the two sides of the triangle that form the rightangle, its legs and use the letters
a and b. The hypotenuse is the longest side opposite the rightangle and we will often use
h for it. The two legs and the hypotenuse are the three sides of the triangle,
triple or triad a ,b, h.
Different authors use different ways of writing triads such as abh but we will use
a, b, h on this page.
The series of lengths of the hypotenuse of primitive Pythagorean triangles begins 5, 13, 17, 25, 29, 37, 41 and is
A020882 in Sloane's
Online Encyclopedia of Integer Sequences. It will
contain 65 twice  the smallest number that can be the hypotenuse of more than one primitive Pythagorean triangle.
The series of numbers that are the hypotenuse of more than one primitive Pythagorean triangle is
65, 85, 145, 185, 205, 221, 265, 305,... A024409
There are lots of patterns in the list of Pythagorean Triples above. To start off your investigations here are
a few.
Hypotenuse and Longest side are consecutive
3,
4,
5
5,
12,
13
7,
24,
25
9,
40,
41
11,
60,
61
The first and simplest Pythagorean triangle is the 3, 4, 5
triangle. Also near the top of the list is 5, 12, 13.
In both of these the longest side and the hypotenuse are consecutive integers.
The list here shows there are more.
Can you spot the pattern?
Chris Evans when a student of Diss High School found the following pattern:
1
1 3
=
4 3
→
3 4 (5)
2
2 5
=
12 5
→
5 12 (13)
3
3 7
=
24 7
→
7 24 (25)
4
4 9
=
40 9
→
9 40 (41)
...
where the fractions give the two sides and the hypotenuse is the numerator + 1
You will have noticed that the smallest sides
are the odd numbers 3, 5, 7, 9,... So the smallest sides are of the form 2i+1.
The other sides, as a series are 4, 12, 24, 40, 60... Can we find a formula here?
We notice they are all multiples of 4: 4×1, 4×3, 4×6, 4×10, 4×15,.. . The series of multiples:
1, 3, 6, 10, 15,... are the
Triangle Numbers
with a formula ^{i(i+1)}/_{2}.
So our second sides are 4 times each of these, or, simply, just 2i(i+1).
The third side is just one more than the second side: 2i(i+1)+1,
so our formula is as follows:
shortest side = 2i+1;
longest side = 2i(i+1);
hypotenuse = 1+2i(i+1)
Check now that the sum of the squares of the two sides is the same as the square of the hypotenuse (Pythagoras's Theorem).
i
a:2i+1
b:2i(i+1)
h=b+1
1
3
2×1×2=4
5
2
5
2×2×3=12
13
3
7
2×3×4=24
25
4
9
2×4×5=40
41
5
11
2×5×6=60
61
6
13
2×6×7=84
85
7
15
2×7×8=112
113
Bill Batchelor points out that the sum of the two consecutive sides is 4 i^{2} + 4 i + 1
which is just the square of the smallest side.
This gives us an alternative method of generating these triples:
Take an odd number as the smallest side: e.g. 9
square it (81)  which will be another odd number
split the square into two halves (40·5),
rounding one down (40)
and the other rounded up (41), to form the other two sides (9, 40, 41)
Alternatively, let's look at the m,n values for each of these triples. Since the
hypotenuse is one more than a leg, the 3 sides have no common factor so are primitive and therefore
all of them do have m,n values:
Triple
m
n
3, 4, 5
2
1
5, 12, 13
3
2
7, 24, 25
4
3
9, 40, 41
5
4
11, 60, 61
6
5
It is easy to see that m = n + 1.
The m,n formula in this case gives
a = m^{2} – n^{2} = (n+1)^{2} – n^{2} = 2 n + 1
b = 2 m n = 2 (n+1) n = 2 n^{2} + 2 n
h = m^{2} + n^{2} = (n+1)^{2} + n^{2} = 2 n^{2} + 2 n + 1
So h is b+1 and the pattern is always true:
if m = n+1 in the m,n formula
then it generates a (primitive) triple with hypotenuse = 1 + the longest leg.
Are these all there are? Perhaps there are other m,n values with a leg and hypotenuse consecutive numbers.
In fact, they are all given by the formula above because:
The triangles must be primitive so we know they have an m,n form. h = m^{2} + n^{2} could be either one more than
either m^{2} – n^{2}
or 2 m n:
If h = m^{2} + n^{2} = (m^{2} – n^{2}) + 1
then, taking m^{2} from both sides:
n^{2} = – n^{2} + 1
2 n^{2} = 1 or n^{2} = ^{1}/_{2}
and this is not possible for a whole number n
So h is never a + 1.
If h = m^{2} + n^{2} = 2 m n + 1 then m^{2} – 2 m n + n^{2} = 1 which is the same as (m – n)^{2} = 1. Therefore m – n = 1 or m – n = –1
But the hypotenuse is a positive number and is m^{2} – n^{2}
so we must have so m>n and so m – n cannot be –1
The only condition we can have is that m – n = 1, that is m = n + 1.
There are no other triangles with hypotenuse one more than a leg except those generated by consecutive
n+1, n as values in the m,n formula.
The series 4, 12, 24, 40, 60,... of longest legs [ numbers given by the formula 2 n(n+1) ]
and 5, 13, 25, 41, 61, ... [ numbers of the form 2 n(n+1) + 1 ] are also
connected by this unusual pattern of square number sums:
Proof without Words: Pythagorean Runs Michael Boardman
Mathematics Magazine 73 (2000) page 59.
The two legs are consecutive
Also in the 3, 4, 5 triple, the two legs of the triangle a and
b are
consecutive, b = a+1. Are there any more like this?
Yes! 20, 21, 29.
Although the list above does not contain any more, there are larger examples:
3, 4, 5 20, 21, 29 119, 120, 169 696, 697, 985
Because the two legs are consecutive numbers, they will have no common factor so all of these will be
primitive. We can therefore find certain special values for m and n
in the m,n formula above. Here is the same list with their m,n values:
m
n
a=m^{2}n^{2}
b=a+1=2mn
h=m^{2}+n^{2}
2
1
3
4
5
5
2
21
20
29
12
5
119
120
169
29
12
697
696
985
This already suggests a way that we can use the generators
m and n
for one triple to find generators for the next.
See if you can find how and also find a formula for this pattern of triples.
Kayne Johnston aged 13 has also found the neat pattern to compute this table without using
the m and n generators, each row being computed simply from the two rows before:
the next row in the table has a smallest side
that is
6 times the previous smallest side
minus the smallest side before that (the penultimate in the list)
plus 2.
For instance, after the first two rows, where the smallest sides are 3 and 20,
the next is 6 20 – 3
+ 2 = 120 –3 + 2 = 119.
The other side is just one more than the smallest, so here is it 120.
If the smallest side of the n^{th} triple is s(n)
then a formula for s(n) is:
Can you find a similar method to compute the hypotenuse but without using Pythagoras Theorem on the
two sides?
The series here are:
The shortest sides are 3, 20, 119, 696,... A001652 formula: 6×last – penultimate + 2
The second legs are 4, 21, 120, 697,... A046090 formula: 6×last – penultimate – 2
The hypotenuses are 5, 29, 169, 985,... A001653 formula: 6×last – penultimate
The recursion formulae are often written more precisely in terms of the index number (n) of the term, so if we call the
series a, the general term is a(n) or a_{n} and
therefore the previous term in the series is a(n1) or a_{n1}.
The formula next term is 6×last term – penultimate term + 2 can be written
a_{n} = 6 a_{n1} – a_{n2} + 2.
We also have:
The odd length legs are 3, 21, 119, 697, ... A046727 formula: a_{n} = 6 a_{n1} – a_{n2} – 4 (1)^{n}
The even length legs are 4, 20, 120, 696, ... A046729 formula: a_{n} = 6 a_{n1} – a_{n2} + 4 (1)^{n}
Beiler (see reference at the foot of this page) gives a formula for these triples
so that the r^{th} triple in
this list is given directly in terms of r.
Dan Sikorski points out that the ratio of successive hypotenuses tends to 3 + 2 √2.
This is also borne out by the continued fraction
for this value, which is [5; 1,4] = 5.82842712474619
and its convergents are
5
,
6
,
29
,
35
,
169
,
204
,
985
, ...
1
1
5
6
29
35
169
The m,n values for consecutivelegs triangles
m
n
a=m^{2}n^{2}
b=a+1=2mn
h=m^{2}+n^{2}
2
1
3
4
5
5
2
21
20
29
12
5
119
120
169
29
12
697
696
985
In the previous section we found a recursion pattern in the series of smallest sides in the consecutive leg triangles.
Now let's look again at these triangles but concentrate on their m,n values
as shown in the table here:
The m,n values are successive terms of a single series: 1, 2, 5, 12, 29, ....
Can you guess the next number in this series?
It is a similar kind of series to the smallest sides, where only the previous two numbers are needed to compute the next.
This time the rule is
2 times the previous one plus the one before that
For example, after 1, 2 the next would be
2×2 + 1 = 5.
and after 1, 2, 5 the next would be
2×5 + 2 = 12,
and so on. Extending the series beyond 29
we have 1, 2, 5, 12, 29, 70, 169, 408, 985, ... .
This series of numbers is called the Pell numbers
(A000129).
Any pair of neighbouring numbers used as m,nvalues generates all of the
Pythagorean triangles with consecutive sides and only those triangles.
You Do The Maths...
With the Fibonacci numbers, the ratio of one Fibonacci number to the previous one
gets closer and closer to Phi. What value does this ratio approach for the Pell Numbers:
1, 2, 5, 12, 29, ...?
We can also classify PTs by the difference in their two legs: b  a;
or the difference between one leg and the hypotnuse: h  a and h  b.
There is a further difference that has some nice properties: the excess.
In every triangle we must have every pair of sides having a total which is bigger than the third side.
If not, the pair of sides will not meet to make a triangle because the third side is too big.
In our Pythagorean triangles, the two legs must together excess the hypotenuse and the difference is
called the excess = h  (a + b).
The excess tells us how much further we have to walk if we went from A to B following the two sides of the
rightangled triangle (a+b)
as opposed to the direct route along the hypotenuse A to B (h).
excess = a + b − h
We can illustrate the excess a+b−h geometrically also:
Measure off side b along h (from point A) Show me
Take the remainder of h: (hb), away from side a Show me
What is left of side a is the excess:a(hb)=a+bh=OP Show me
We can do the same starting with side a:
Measure off side a along h (from point B) Show me
Take the remainder of h: (ha), away from side b Show me
What is left of side a is the excess:b(ha)=a+bh=OQ Show me
This excess is the diameter of the inCircle Show me
Start again
The excess is the diameter of the incircle  the largest circle that fits inside the triangle touching the three sides.
This is because
the diameter parallel to OB in the diagram meets line OA at a distance ha + (a+bh)/2 = (h+ba)/2 from A
the diameter parallel to OA in the diagram meets line OB at a distance hb + (a+bh)/2 = (hb+a)/2 from B
Since these two distances sum to h, then they define a unique point on AB which is (h+ba)/2 from A and (hb+a)/2 from B.
More patterns
There are many more patterns in the Pythagorean Triples!
For instance
there are primitive triangles whose longest side and hypotenuse
differ by 2, such as 8, 15, 17 and 12, 35, 37
and many more.
What is the mathematical pattern in these triples?
Another idea is to take the formula and find special cases,
remembering that the formula does not generate all
Pythagorean triples.
For instance, let n=1. We then have triples
m^{2}–1, 2m, m^{2}+1, although we have
to make the restriction that m>1 for the hypotenuse to be a positive number:
n
m
m^{2}–1
2m
m^{2}+1
1
2
3
4
5
1
3
8
6
10
1
4
15
8
17
1
5
24
10
26
1
6
35
12
37
Notice that not all of these are primitive and also that there are other triples with a side two less than the hypotenuse that
are not in this list.
Can any number be a side in some Pythagorean Triangle?
Note that here we use the terms leg, side, hypotenuse as follows:
there are two legs and a hypotenuse making the 3 sides of each
Pythagorean triangle.
The Number of Pythagorean Triangles having a side n
These sequences are the counts for n=1, 2, 3,... in each of the 6 categories:
so that 0, 0, 1, 1, 2,.. means
0 triangles for n=1, 0 for n=2, 1 for n=3, 1 for n=4, 2 for n=5 etc.
It looks like every fourth number after 2, namely 6, 10, 14, 18,... cannot be the side of a primitive triangle.
Also you might guess that there are no gaps in the list of numbers that can be a side of at least one triple.
These are indeed true.
To prove any number, X, can be the side of some Pythagorean triangle, we use the m,n generating
method on two cases: X even and X odd.
if X is even:
let n=1 and m=X/2 then the side generated by 2 m n is X
If X is odd:
then we can halve it and let m = (X+1)/2 and n = (X–1)/2
so that the side generated by
m^{2} – n^{2} = (m – n)(m + n) = 1×X is X
So the answer to our question Can any number be a side in some Pythagorean Triangle? is Yes!
From the table above, we can make an ordered list of the numbers themselves that can appear as sides in each category.
If we look for triples with a side that is a power of 2, we find:
for side 2^{1}=2 there are none
for side 2^{2}=4: 3 4 5 there is only 1
for side 2^{3}=8: 6 8 10, 8 15 17 so there are 2
for side 2^{4}=16: 12 16 20, 16 30 34, 16 63 64 so there are 3
...
for side 2^{n+1} there are exactly n
E2460
D Meyers, C F Pinzka, W R Westphal, H M Marston, R B Eggleton
The American Mathematical Monthly vol 82 (1975) pages 303304.
The Possible Sides of Pythagorean Triangles
Here the actual sides are listed.
If there is more than one possible Pythagorean triangle with a given side, the side is repeated in these sequences.
The sequence of possible side lengths without repetitions is given in brackets.
Primitive
All
Legs
3, 4, 5, 7, 8, 9, 11, 12, 12, 13, 15, 15, 16,... A024355
(A042965)
these are all the numbers except those
of the form 4n + 2 = 2, 6, 10, 14, ...
5, 13, 17, 25, 29, 37, 41, 53, 61, 65, 65,... A020882
(A008846=A002144)
these are the integers that have all their prime factors
of the form 4n + 1
5, 10, 13, 15, 17, 20, 25, 25, 26,... A009000
(A009003)
these are the integers that have at least one prime factor
of the form 4n + 1
Sides
3, 4, 5, 5, 7, 8, 9, 11, 12, 12, 13, 13,... A024357
(A042965)
this is every number except those of the form 4n + 2 = 2, 6, 10, 14, ...
3, 4, 5, 5, 6, 7, 8, 8, 9, 9, 10, 10, 11, 12, 12, 12, 12, 13, 13,... A009070
this series includes every integer>2
It is fairly easy to show that prime numbers are the sides a unique primitive Pythagorean triangle
(with thanks to Alf Gunnarsson for pointing this out to me).
All odd numbers are the difference of two squares since (n+1)^{2}  n^{2} = 2n + 1.
All prime numbers after 2 are odd so they are also the difference of two squares, however this is
only possible in one way since
m^{2} − n^{2} = (m + n)(m − n).
Since prime numbers have only two factors namely 1 and the number itself, then
m − n must be 1 and this solution is unique. So m = n+1 and m+n is 2n+1, the prime number.
All primitive Pythagorean triangles fit the pattern m^{2}n^{2}, 2mn, m^{2}+n^{2}
All primes except 2 are odd so cannot be the side 2mn and are therefore the difference of two squares,
which, we have just shown, is possible in only one way.
The conditions on m and n to generate a PPT are that they should be of opposite parity
as we saw earlier. The m,n pairs for PPT with a prime leg are of the form n+1,n.
One of these must be odd and the other even so the m,n pairs n+1,n where 2n+1 is prime.
prime= 2n+1
n
m= primen
PPT
3
1
2
3,4,5
5
2
3
5,12,13
7
3
4
7,24,25
11
5
6
11,60,61
13
6
7
13,84,85
17
8
9
17,144,145
Since n+1,n generate these PPTs, the second leg is 2mn = 2(n+1)n
which will always be larger than (n+1)^{2}−n^{2}=2n+1 for n≥0 so the prime
side will be the smallest and the second leg and hypotenuse will be consecutive numbers.
If m=n+1 and 2n+1 is prime, m,n generate the PPT 2n+1 = prime, 2(n+1)n, 2n^{2}+2n+1
There are also nonprime numbers that are the sides of just one PPT too:
4 in 3, 4, 5 and 8 in 8, 15, 17.
There is more about finding squares that sum of any number later on this page.
Number Series in Pythagorean Triangles
Other series of interest here are:
A020883Longest legs in primitive triples:
4, 12, 15, 21, 24, 35, 40, 45, 55, 56,... (or
A024354 without duplicates or
A024360
as a count of the number of triangles with longest side n or
A024410
longest legs in more than one primitive triangle)
A020884Shortest legs in primitive triples:
3, 5, 7, 8, 9, 11, 12, 13, 15, 16, 17, 19, 20, 20,...
(A024352 without repetitions
or A024359
as a list of counts of the number of triangles with each shortest side n or
A0244411
shortest legs in more than one primitive triangle)
Except for 0 and 1, these are all the numbers that are the difference of two squares or,
equivalently, every number that when divided by 4 has a remainder of 0, 1 or 3
A042965).
A024409Hypotenuse of more than one primitive triangle.
A002144 Prime Hypotenuse: 5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97,...
or primes whose square is the sum of two nonzero squares.
A024406Area of primitive triangles:
6, 30, 60, 84, 180, 210, 210, 330,... (
A020885 is areas divided by 6,
A024407
areas of more than one primitive triangle
since all such areas are a multiple of 6)
A024364 The perimeters of primitive triangles:
12, 30, 40, 56, 70, 84, 90,... (A024408
perimeters of more than one primitive triangle); also
A099829 Smallest perimeter of n Pythagorean triangles.
A120090 12, 30, 56, 90, 132, 154, 182, ...
are numbers whose squares are perimeters of primitive Pythagorean triangles.
A006593 the smallest number that
is a side in n triads for n=1,2,3,4,5,...: 3, 5, 16, 12, 15, 125, 24,...
so this series states that 3 is the smallest side in just 1 triad,
5 is the smallest that occurs in exactly 2 triads, 16 in 3 triads, etc.)
In terms of side differences, since 3,4,5 has a hypotenuseleg difference of 1 then
by expanding the triangle by a factor of D will produce a triangle with a hypotenuseside difference of D too.
For primitive triangles, the hypotenuseside differences are limited to
1, 2, 8, 9, 18, 25, 32, 49, 50, .. (A096033) which are the
odd squares together with half the even squares.
When is n a member of a Pythagorean triple? Dominic and Alfred Vella,
Mathematical Gazette Note 87.04, pages 102105, vol 87 (2003).
This article
and others on Pythagorean triples
are available in PDF format from Dominic Vella's
mathematics page.
Graphs of the Primitive Triples
We can plot the Pythagorean triangles on a graph in a natural way
as xy coordinates
if we use the two legs are the coordinates.
The hypotenuse is then the distance of the point from the origin.
Plotting all Pythagorean Triangles
If we plot all Pythagorean triangles with a leg up to size 100
we get the graph shown here:
Since each triple a b h is the same triple as its 'reflection'
b a h, each triple is plotted twice,
the reflections of the black points being in red.
The prominent straight lines are the multiples of the smaller Pythagorean triangles 3 4 5
in black and 4 3 5 in red.
There are more straight lines through the origin if we plot more triples:
The lines of points are densest for the multiples of the smaller (primitive) Pythagorean triples, so next we see
5 12 13 and so on:
Plotting the Primitive Pythagorean triangles
If we plot just the primitive Pythagorean triangles, the straight lines disappear and something else interesting appears.
The curved lines are even clearer on this
The UAD Tree of Primitive Pythagorean Triangles
The UAD Tree growing to the right!
3 4 5
U 5 12 13
U 7 24 25
A 55 48 73
D 45 28 53
A 21 20 29
U 39 80 89
A 119 120 169
D 77 36 85
D 15 8 17
U 33 56 65
A 65 72 97
D 35 12 37
We can organise all the primitive triangles into a kind of genealogical tree starting from 3 4 5.
The "tree" grows to the right and each node in the tree has exactly 3 descendants, called Up, Along and Down,
or U,A and D for short.
Each "descendant" triple is generated by a different transformation of
the "parent" triple a b h on its left
as follows:
E.g: 3 4 5
Starting from a, b, h
go
Up
to:
a – 2b + 2h,
2a – b + 2h,
2a – 2b + 3h
5 12 13
go
Along
to:
a + 2b + 2h,
2a + b + 2h,
2a + 2b + 3h
21 20 29
go
Down
to:
–a + 2b + 2h,
–2a + b + 2h,
–2a + 2b + 3h
15 8 17
The article by Hall (see end of this section) proves that every primitive triple is in this tree and that the tree contains
only primitive triples.
4 3 5
U 8 15 17
A 20 21 29
D 12 5 13
If we swop a and b in any triple
then the three U, A and D transformations still produce the same three triples
but the Up and Down triples have swopped places and in all three
the two legs a and b have swopped places too.
The m,n generators in the UAD tree
The m,n values
2,1
U 3,2
U 4,3
A 8,3
D 7,2
A 5,2
U 8,5
A 12,5
D 9,2
D 4,1
U 7,4
A 9,4
D 6,1
Twenty years after Hall's article about the UAD tree, other investigators
found there is a set of simpler transformations
that produces the same tree as Hall's but using the m,n generators for each triple
as follows:
Starting from the triple with generators m,n on the left we have:
m,n
go
Up
to
2m – n,
m
go
Along
to
2m + n,
m
go
Down
to
m + 2n,
n
The UAD tree Calculator
For any primitive triple, this calculator will
show the three primitive U, A and D triples following it
the UAD path to a triple from 3,4,5
find the triple at the end of given a path in the tree as a string of U, A and D directions
for example AU is the path to the triple 39 80 89
Find the paths to the triples with consecutive legs:
20 21 29, 119 120 169
, ...
What paths take us to the triples with hypotenuse one greater than a leg:
3 4 5, 5 12 13, 7 24 25, ...?
What property have all the triples with paths D, DD, DDD, DDDD, ....?
From any triple in the tree, on which branch is another
with the same difference between
hypotenuse h and the side b?
hypotenuse h and side a?
sides a and b?
Classroom Note 232. Genealogy of Pythagorean Triads A Hall
The Mathematical Gazette vol 54 (1970) pages 377379.
The family tree of Pythagorean triples A R Kanga IMA Bulletin vol 26 (1990) pages 1527.
78.12 The family tree of Pythagorean triplets revisited R Saunders, T Randall
Mathematical Gazette vol 78 (1994) pages 190193.
Over Pythagorese en bijnaPytharoese driehoeken en een generatieproces met behulp van unimodulaire matrices
(On Pythagorean and quasiPythagorean triangles and a generation process with the help of unimodular matrices)
F J M Barning Math. Centrum Amsterdam Afd Zuivere Wisk ZW001 (1963) in Dutch,
PDF
Barning preceded Hall in the discovery of these matrices, seemingly independently discovering the same tree:
so you may see them referred to as Hall Matrices or Barning's Pythagorean Tree or the BarningHall method.
(Cited in:
Matrix Generation of Pythagorean nTuples D Cass, P J Arpaia
Proc. American Math Soc. Vol 109 (1990) page 70.)
Pytagoreiska trianglar (in Swedish), B. Berggren, Tidskrift för elementär matematik, fysik och kemi 17 (1934), 129139.
This is an even earlier article about the UAD tree (private email from H Lee Price)
The Pythagorean Tree: A New Species H. Lee Price (2008) arXiv:0809.4324 (pdf)
A very interesting paper on another type of tree generating all primitive Pythagorean triples using three matrices with the same
values in each position but differing in sign:
2
1
–1
–2
2
2
–2
1
3
2
1
1
2
–2
2
2
–1
3
2
–1
1
2
2
2
2
1
3
Other triangle properties
Perimeter
If I have a piece of string of length n, how many rightangled triangles can I make from it if the sides have integer lengths?
The diagram here shows the only configuration if the rope has length 3+4+5=12. The length of the rope is the perimeter of the
triangle.
The perimeter of a triangle a b h
is the sum of the lengths of the sides a+b+h.
Since the sides of Pythagorean triangles are integers, so is the perimeter.
the perimeter is the distance your pencil has to travel to draw it accurately
if it was a triangular field, it is the length of the fencing needed to enclose the field
You can use the Calculator (opens in a new window)
to find all Pythagorean triangles with longest side up to 100, together with
the length of the perimeter P and its area A:
All Pythagorean Triples with sides up to 100
arranged in order of hypotenuse (longest side):
with Perimeter (a+b+h)
Triad
Primitive?
Perimeter
3, 4, 5
primitive
12
6, 8, 10
2× 3, 4, 5
24
5, 12, 13
primitive
30
9, 12, 15
3× 3, 4, 5
36
8, 15, 17
primitive
40
12, 16, 20
4× 3, 4, 5
48
15, 20, 25
5× 3, 4, 5
60
7, 24, 25
primitive
56
10, 24, 26
2× 5, 12, 13
60
20, 21, 29
primitive
70
18, 24, 30
6× 3, 4, 5
72
16, 30, 34
2× 8, 15, 17
80
21, 28, 35
7× 3, 4, 5
84
12, 35, 37
primitive
84
15, 36, 39
3× 5, 12, 13
90
24, 32, 40
8× 3, 4, 5
96
9, 40, 41
primitive
90
Triad
Primitive?
Perimeter
27, 36, 45
9× 3, 4, 5
108
30, 40, 50
10× 3, 4, 5
120
14, 48, 50
2× 7, 24, 25
112
24, 45, 51
3× 8, 15, 17
120
20, 48, 52
4× 5, 12, 13
120
28, 45, 53
primitive
126
33, 44, 55
11× 3, 4, 5
132
40, 42, 58
2× 20, 21, 29
140
36, 48, 60
12× 3, 4, 5
144
11, 60, 61
primitive
132
39, 52, 65
13× 3, 4, 5
156
25, 60, 65
5× 5, 12, 13
150
33, 56, 65
primitive
154
16, 63, 65
primitive
144
32, 60, 68
4× 8, 15, 17
160
42, 56, 70
14× 3, 4, 5
168
48, 55, 73
primitive
176
Triad
Primitive?
Perimeter
24, 70, 74
2× 12, 35, 37
168
45, 60, 75
15× 3, 4, 5
180
21, 72, 75
3× 7, 24, 25
168
30, 72, 78
6× 5, 12, 13
180
48, 64, 80
16× 3, 4, 5
192
18, 80, 82
2× 9, 40, 41
180
51, 68, 85
17× 3, 4, 5
204
40, 75, 85
5× 8, 15, 17
200
36, 77, 85
primitive
198
13, 84, 85
primitive
182
60, 63, 87
3× 20, 21, 29
210
39, 80, 89
primitive
208
54, 72, 90
18× 3, 4, 5
216
35, 84, 91
7× 5, 12, 13
210
57, 76, 95
19× 3, 4, 5
228
65, 72, 97
primitive
234
60, 80, 100
20× 3, 4, 5
240
28, 96, 100
4× 7, 24, 25
224
Pythagorean Triangles with Equal Perimeters
A009096 lists the perimeters of all Pythagorean
triangles in order: 12, 24, 30, 36, 40, 48, 56, 60, 60, 70,... A009096
(A010814 without repetitions).
For primitive triangles only, the perimeters are 12, 30, 40, 56, 70, 84, 90, 126, ... (A024364)
If we want the smallest length of string that makes exactly n different rightangled triangles, then
The smallest perimeter is 12 that is the perimeter of only 1 triangle: 3, 4, 5
a perimter of 60 is the smallest for two 2 different triangles:
15, 20, 25 = 5× 3, 4, 5
and 10, 24, 26 = 2× 5, 12, 13
but it has to be 120 for 3 triangles: 10x 3, 4, 5, 4x 5, 12, 13,
24, 45, 5, so the series is 12, 60, 120, ...A099830
If we collect the list of smallest perimeters of exactly 2, 3, 4, ... PTs into one list, we have
the first perimeter of exactly 2 triangles is 60, of exactly 3 triangles is 120,
of 4 is 240 and the series
continues somewhat erratically as 420, 720, 1320, 840, 2640, 1680,
... A099830.
A099831 Perimeters of exactly 2 Pythagorean triangles:
The smallest is 60 but are what other perimeters of exactly 2 PTs are there?
The next is 84 being the perimeter of 7x 3, 4, 5 and of
12, 35, 37.
The next is then 90 and so on so the list begins:
60, 84, 90, 132, 144, 210, ... A099831
Other numbers which are the perimeters of exactly 3 Pythagorean triangles: 120 is the smallest
(perimeters of 20, 48, 52, 24, 45, 51 and 30, 40, 50)
and then in order we have 120, 168, 180, 252, 280, ...A099832
Other numberss that are the perimeters of exactly 4 Pythagorean triangles:
240, 360, 480, 504, 630, ...A099833
Perimeters common to exactly 5 PTs are 420, 660, 924, 1008, 1200, 1584 ... A156687
Almost all the triangles which share a perimeter with others are nonprimitive. For Primitive Pythagorean Triangles we have quite different results.
The first primitive Pythagorean triangles with the same perimeter are
195, 748, 773 and 364, 627, 725
with a perimeter of 1716.
The next such perimeters are 2652, 3876, 3960, ... A024408
and we have to go up to a perimeter of 14280 before we find three primitive triangless with the same perimeter:
119, 7080, 7081 and 168, 7055, 7057 and 3255, 5032, 5993.
with the next perimeter being 72930 for 2992, 34905, 35033 and
7905, 32032, 32993 and 18480, 24089, 30361.
See also
The area of Pythagorean triangle a, b, h is just half the product of the two legs
(the sides that make the rightangle) ^{ab}/_{2}.
The area of the triangle is the amount of paint you would need to colour it in
The area determines how much grass seed you would need to fill a triangular field
Since we halve the product of the two legs of the triangle, we can ask:
Is the area always an integer for Pythagorean triangles?
If we can have two odd legs in a Pythagorean triangle then the answer is no.
The true answer is always Yes because:
From the m,n formula above, one side is 2mn if primitive
or else a multiuple of this if not primitive so there is always one side which is even.
In the table above 3 4 5 is the only triangle with an area smaller than its perimeter.
We can find two triples with a perimeter equal to its area: 6 8 10 has P=24 and A=24 and
5 12 13 has P=30 and A=30 (primitive)
Triangles with an area twice their perimeter are: 12 16 20 has P=48 and A=96 10 24 36 has P=60 and A=120 9 40 41 has P=90 and A=180 (primitive)
All Pythagorean Triples with sides up to 100
arranged in order of hypotenuse (longest side):
with Area ^{ab}/_{2}
Triad
Primitive?
Area
3, 4, 5
primitive
6
6, 8, 10
2× 3, 4, 5
24
5, 12, 13
primitive
30
9, 12, 15
3× 3, 4, 5
54
8, 15, 17
primitive
60
12, 16, 20
4× 3, 4, 5
96
15, 20, 25
5× 3, 4, 5
150
7, 24, 25
primitive
84
10, 24, 26
2× 5, 12, 13
120
20, 21, 29
primitive
210
18, 24, 30
6× 3, 4, 5
216
16, 30, 34
2× 8, 15, 17
240
21, 28, 35
7× 3, 4, 5
294
12, 35, 37
primitive
210
15, 36, 39
3× 5, 12, 13
270
24, 32, 40
8× 3, 4, 5
384
9, 40, 41
primitive
180
Triad
Primitive?
Area
27, 36, 45
9× 3, 4, 5
486
30, 40, 50
10× 3, 4, 5
600
14, 48, 50
2× 7, 24, 25
336
24, 45, 51
3× 8, 15, 17
540
20, 48, 52
4× 5, 12, 13
480
28, 45, 53
primitive
630
33, 44, 55
11× 3, 4, 5
726
40, 42, 58
2× 20, 21, 29
840
36, 48, 60
12× 3, 4, 5
864
11, 60, 61
primitive
330
39, 52, 65
13× 3, 4, 5
1014
25, 60, 65
5× 5, 12, 13
750
33, 56, 65
primitive
924
16, 63, 65
primitive
504
32, 60, 68
4× 8, 15, 17
960
42, 56, 70
14× 3, 4, 5
1176
48, 55, 73
primitive
1320
Triad
Primitive?
Area
24, 70, 74
2× 12, 35, 37
840
45, 60, 75
15× 3, 4, 5
1350
21, 72, 75
3× 7, 24, 25
756
30, 72, 78
6× 5, 12, 13
1080
48, 64, 80
16× 3, 4, 5
1536
18, 80, 82
2× 9, 40, 41
720
51, 68, 85
17× 3, 4, 5
1734
40, 75, 85
5× 8, 15, 17
1500
36, 77, 85
primitive
1386
13, 84, 85
primitive
546
60, 63, 87
3× 20, 21, 29
1890
39, 80, 89
primitive
1560
54, 72, 90
18× 3, 4, 5
1944
35, 84, 91
7× 5, 12, 13
1470
57, 76, 95
19× 3, 4, 5
2166
65, 72, 97
primitive
2340
60, 80, 100
20× 3, 4, 5
2400
28, 96, 100
4× 7, 24, 25
1344
The possible areas of a Pythagorean triangle are
6,24,30,54,60,84,96,... (A009112)
which Mohanty and Mohanty (see reference in the next paragraph) call Pythagorean numbers.
The areas of primitive triangles are
6,30,60,84,180,... (A024365)
which they call primitive Pythagorean numbers.
They also prove that
Every Pythaqorean triangle's area is a multiple of 6 (Theorem 5):
The areas are 6,24,30,54,60,84,96,...A009112)
and these are 6 times 1,4,5,9,10,14,16,... (A177887)
all the areas end with 0, 4 or 6 as the unit's digit and there are an infinite number of each.
Although the areas seem to get further and further apart for larger triangles:
they are in fact reasonably regular in that
There is always a Pythagorean number between n and 2n
once we get beyond n = 12 (Theorem 6)
The Four Factor property All Pythagorean numbers, A,
have 4 different factors, a, b, c, d where
a b = c d = A and a + b = c – d
and also any number which has this 4factor property must also be
a Pythagorean number (Theorem 1):
Area A
=a
×b
=c
×d
a+b=cd
Triangle
6
2
3
6
1
5
3 4 5
24
4
6
12
2
10
6 8 10
30
10
3
15
2
13
5 12 13
54
6
9
18
3
15
9 12 15
60
5
12
20
3
17
8 15 17
from which we notice that a + b = c – d = hypotenuse
If integers r, s = r + m, t = r + 2m so that r, s and
t are in arithmetic progression
with common difference m then n = r s t m is a Pythagorean number (Theorem 2):
Further, if gcd(s,m)=1 and one of s and m is even and the other odd,
then n is a primitive Pythagorean number
The product of 3 consecutive numbers is a Pythagorean number ( Theorem 2 Corollary ).
There are an infinite number or twin Pythagorean numbers, that is, pairs of integers A and
A + 6 that are both Pythagorean numbers (Theorem 9), for instance:
Area A
triangle
A+6
triangle
24
6 8 10
30
5 12 13
54
9 12 15
60
8 15 17
210
12 35 37 20 21 29
216
18 24 30
330
11 60 61
336
14 48 50
...
No square number is a Pythagorean number (Theorem 11), which was first proved
by
Pierre de Fermat (16011665).
Pythagorean Numbers S Mohanty and S P Mohanty Fibonacci Quarterly 28 (1990)
pages 3142
proves many interesting results about the area of Pythagorean Triangles
Nasty Numbers B Miller The Mathematics Teacher 73 (1980) page 649
is an early paper on the sequence of Pythagorean numbers but defined by the FourFactor property
^{*} For which n is there always a Pythagorean triangle with an area n times its perimeter?
^{*} For which n is there always a primitive Pythagorean triangle with an area n times its perimeter?
Find two Pythagorean triangles with the same area.
[Check your answers with A009127.]
Find the first few areas that are common to exactly two Pythagorean triangles.
[Hint: The numbers in A009127
will help with this and the next two Problems.]
Find three Pythagorean triangles with the same area.
What about four?
What is the smallest area common to 2, 3, 4, ... Pythagorean triangles?
[ Check your answer with A094805 ]
If I have a piece of string of length 12 I can only make one Pythagorean triangle from it, namely 3,4,5.
There is also just one Pythagorean triangle if the string was of length 24.
How does the series of string lengths that begins 12, 24, continue if there
is a unique Pythagorean triangle with that perimeter?
A string of length 60 gives me two
possible triangles with perimeter 60. What are they?
We have just seen that the smallest length of string that can be the perimeter of just one Pythagorean triangle is 12.
What is the smallest length that makes just two triangles? What about three? And four?
[Check your answer with A098714]
Find the Pythagorean triangles with areas 2×3×4, 3×4×5, 4×5×6, ... i(i+1)(i+2),
using the Calculator later on this page. You can enter the area as, for example, 3*4*5.
What do you notice about the m,n generating values for each? Prove your conjecture is true for the general case.
^{*} The answers to 1 and 2 are found in:
Right Triangles with Perimeter and Area Equal W Parsons
The College Mathematics Journal vol 15 (1984) page 429.
The ratio of Area to Perimeter
In the Table above we list some Pythagorean triangles together with their Perimeter and Area.
For 3, 4, 5 if we expand it by a factor k we have 3k, 4k, 5k, Perimeter P = 12 k and Area A = 6 k^{2} so
that A/P = k/2. Hence we can find a triangle with any whole number (and half any odd number) as the ratio
A/P .
The situation is more easily explained by seeing that
In the Pythagorean triangle a, b, h the perimiter a+b+h
is always a factor of a b.
Alfred and Dominic Vella prove this in their online article
More Properties of Pythagorean Triangles. Since the area is a b/2 then we will always find whole numbers and
some halves in the ratios of Area to Perimeter.
There is always at least one pythagorean triangle with any given wholenumber ratio. We will demonstrate this below.
Paul Cleary of Oxford (UK) wrote to me in October 2013 to say that there are only 6 pythagorean triangles that have
a given prime number as the ratio A/P. It seems there are always more than 6
when the ratio is a composite number for a ratio of 5 or more. For example,
for ratios up to 8, we only have the following PTs:
All Pythagorean Triangles with a ratio A/P from 1 to 8
PT=g×PPT
g
PPT
m,n of PPT
A
P
A/P
5, 12, 13
1
5, 12, 13
3, 2
30
30
1
6, 8, 10
2
3, 4, 5
2, 1
24
24
9, 40, 41
1
9, 40, 41
5, 4
180
90
2
10, 24, 26
2
5, 12, 13
3, 2
120
60
12, 16, 20
4
3, 4, 5
2, 1
96
48
13, 84, 85
1
13, 84, 85
7, 6
546
182
3
14, 48, 50
2
7, 24, 25
4, 3
336
112
15, 36, 39
3
5, 12, 13
3, 2
270
90
16, 30, 34
2
8, 15, 17
4, 1
240
80
18, 24, 30
6
3, 4, 5
2, 1
216
72
20, 21, 29
1
20, 21, 29
5, 2
210
70
17, 144, 145
1
17, 144, 145
9, 8
1224
306
4
18, 80, 82
2
9, 40, 41
5, 4
720
180
20, 48, 52
4
5, 12, 13
3, 2
480
120
24, 32, 40
8
3, 4, 5
2, 1
384
96
21, 220, 221
1
21, 220, 221
11, 10
2310
462
5
22, 120, 122
2
11, 60, 61
6, 5
1320
264
24, 70, 74
2
12, 35, 37
6, 1
840
168
25, 60, 65
5
5, 12, 13
3, 2
750
150
28, 45, 53
1
28, 45, 53
7, 2
630
126
30, 40, 50
10
3, 4, 5
2, 1
600
120
25, 312, 313
1
25, 312, 313
13, 12
3900
650
6
26, 168, 170
2
13, 84, 85
7, 6
2184
364
27, 120, 123
3
9, 40, 41
5, 4
1620
270
28, 96, 100
4
7, 24, 25
4, 3
1344
224
30, 72, 78
6
5, 12, 13
3, 2
1080
180
32, 60, 68
4
8, 15, 17
4, 1
960
160
33, 56, 65
1
33, 56, 65
7, 4
924
154
36, 48, 60
12
3, 4, 5
2, 1
864
144
60, 32, 68
4
8, 15, 17
4, 1
960
160
29, 420, 421
1
29, 420, 421
15, 14
6090
870
7
30, 224, 226
2
15, 112, 113
8, 7
3360
480
32, 126, 130
2
16, 63, 65
8, 1
2016
288
35, 84, 91
7
5, 12, 13
3, 2
1470
210
36, 77, 85
1
36, 77, 85
9, 2
1386
198
42, 56, 70
14
3, 4, 5
2, 1
1176
168
33, 544, 545
1
33, 544, 545
17, 16
8976
1122
8
34, 288, 290
2
17, 144, 145
9, 8
4896
612
36, 160, 164
4
9, 40, 41
5, 4
2880
360
40, 96, 104
8
5, 12, 13
3, 2
1920
240
48, 64, 80
16
3, 4, 5
2, 1
1536
192
The counts of the number of PTs with an Area/Peri ratio of k for k from 1 is:
2, 3, 6, 4, 6, 9, 6, 5, 10, 9, 6, ... A156688
For the ratio k=8 there are only 5 PTs but for all larger ratios there are
a minimum of 6 PTs.
A little algebra will show you that all the following 6 pythagorean triangles have a ratio A/P = k
for any value of k and they are all distinct PTs if k>8:
These numbers also hide another pattern. If we let A/P = k
and we saw above that a general
PT is d times a PPT which can be generated using the m,n formula
so that:
A general PT is a = d (m^{2} – n^{2}),
b = d 2 m n,
h = d (m^{2} + n^{2})
Area
= k =
a b
=
d
(m – n)n
Perimeter
2(a+b+h)
2
A bit of algebra will show that (a  4 k) (b  4 k) = 8 k^{2}
which leads to a nice algorithm that Paul Cleary found
for computing all the PTs for a given ratio k:
Find all pairs of factor A × B = 8 k^{2}
Let a = A +4 k and b = B + 4 k
The legs of a PT with A/P = k are
a, b, √
a^{2} + b^{2}
All of the PTs with A/P = k are generated by this method and only those with A/P = k
For example, let's find those PTs with k = 5
The pairs of factors of 8 k^{2} = 8×25 = 200 are
1,200
2, 100
4, 50
5, 40
8, 25
10, 20
Add 4k=20 to each factor:
which are the legs of all the 6 PTs with k = 5
21,220
22,120
24,70
25,60
28,45
30,40
Hypotenuses
221
122
74
65
53
50
Now we can see that, as Paul Cleary found, that
the minimum number of PTs is 6 when the ratio k is prime.
The factors of 8k^{2} are then only the 6 pairs:
1×8k^{2},
2×4k^{2},
4×2k^{2},
8×k^{2},
k×8k,
2k×4k
If k = 1, 2, 4, or 8 then some of them will be duplicates.
The number of primitive pythagorean triangles with A/P ratio from 1 are given by:
1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, ..... A068068
In this rightangled triangle with sides a, b, h,
we have drawn in a line from the rightangle perpendicular to the
hypotenuse, of length H. If the triangle's base is the hypotenuse then this is its height
so that the area of the triangle is hH/2.
The two smaller triangles are therefore similar to the original triangle (same shape but different sizes)
because the three angles of the original are the same three angles in the two smaller triangles.
So the ratio of the hypotenuse to the longer side will also be identical in all these triangles.
In the smallest triangle, this is a/H and in the
original (largest) triangle this is h/b so we have:
a/H = h/b
and, dividing by a we have:
1/H = h/(a b)
Squaring this we have:
1/H^{2} = h^{2}/(a^{2} b^{2})
but Pythagoras' Theorem tells us that a^{2} + b^{2} = h^{2} so we have
1/H^{2} = (a^{2}+b^{2})/(a^{2}b^{2})
or
1
=
1
+
1
H^{2}
a^{2}
b^{2}
which we can call the Reciprocal Pythagorean Theorem!
We have also shown that
H =
2 Area
=
ab
h
h
H will not be a whole number generally but it will be for the nonprimitive Pythagorean triangle
h×(a,b,h) = (ha, hb, h^{2})
when H = ab.
The product of the 3 sides
The perimeter is the sum of the three sides a+b+h.
What about their product a×b×h?
This number does
not seem to have any geometrical or practical significance in terms of the triangle
but is of interest mathematically.
The products of the three sides of some small Pythagorean triangles are:
3 4 5 has product 60
6 8 10 has product 480
5 12 13 has product 780
9 12 15 has product 1620
8 15 17 has product 2040
7 24 25 has product 4200
20 21 29 has product 12180
The list of sideproducts in order begins as follows: 60, 480, 780, 1620, 2040, 3840, 4200, ...
(A057096).
In every Pythagorean triangle the following six facts are always true:
one side is a multiple of 3
one side is a multiple of 4
so the product of the two legs is always a multiple of 12
and the area is therefore always a multiple of 6
one side is a multiple of 5
so the product of all three sides is always a multiple of 3×4×5=60
So if we divide the sidesproducts by 60 we have
1,8,13,27,34,64,70,104,125,...(A057097).
This series contains all the cubes 1^{3}=1,
2^{3}=8, 3^{3}=27, ...
If we restrict ourselves to only primitive triangles, the ordered list of sideproducts is: 60, 780, 2040, 4200, 12180, 14760, 15540,... (A063011)
and the series as multiples of 60 is 1, 13, 34, 70, 203, 246, ... (A081752)
It is an "open question" (we do not know if the answer is Yes or No) whether
the ordered series of products for all Pythagorean triangles has a repeated item in it
or if all the products are in fact unique:
Unsolved Problems in Number Theory, R K Guy (2nd ed.) SpringerVerlag
(1994),
Problem D21 "Triangles with Integer Sides, Medians, and Area" pages 188190.
The Incircle and Inradius
First we find three simple formula for the inradius, r, of a rightangled triangle. Then we will
see how to find Pythagorean triangles with a given inradius and to count the number of triangles too.
Three formulae for the Inradius
We can draw a circle touching all 3 sides of any triangle, called the incircle
with radius the inradius usually denoted by r and centre the incentre.
From the symmetry of the circle, a line from its centre to each vertex of the triangle will halve each of the
angles in the triangle.
Lines from the incentre to the vertices (shown in
gray here) divide the triangle into three smaller ones, each having the same
height, r on a base of one side of the whole triangle.
The area of a triangle is one half of the base of the triangle times its height.
So the three separate areas sum to the whole area:
area =
a r
+
b r
+
c r
= r
a + b + c
2
2
2
2
The sum of the sides of a triangle is the length of its perimeter and,
since we have the perimeter halved in the formula
we often find this expressed using the semiperimeter s:
area = inradius × semiperimeter = r s
We now have a simple formula for the inradius, r, of any triangle:
r =
2 area
=
area
perimeter
semiperimeter
In a rightangled triangle, the area is just half the product of the two legs,
^{a b} / _{2} or
2 area = a b so the formula for r is even simpler:
r =
a b
a + b + h
Since all primitive rightangled Pythagorean triangles can be derived using the m,n
formula above, then, in terms of m and n we have
r =
(m^{2}–n^{2}) 2mn
=
(m^{2}–n^{2}) 2mn
=
(m–n)(m+n)2mn
= (m–n)n
m^{2}–n^{2} + 2mn + m^{2}+n^{2}
2m^{2} + 2mn
2m(m+n)
r = (m – n) n
So we have, in primitive rightangled triangles, the inradius, r
is therefore always an integer because m and n are.
Nonprimitive triangles are just multiples of primitives, so their inradius is an integer too.
Therefore:
In all Pythagorean triangles, the inradius is a whole number
There is an even simpler formula for r:
Since we have a rightangled triangle, we can split the two legs
into a – r and r on side a and b – r and r on side b.
These lengths a – r and b – r
are duplicated on the two sections of the hypotenuse AB and indeed together make up the whole
of AB. So we have
h = (a – r) + (b – r)
which we can rearrange to find a new expression for r:
r =
a + b – h
2
This also shows that the excess = (a+b)  h, which we met earlier, is 2r
so the excess (a+b)−h is the diameter of the incircle.
It is a simple exercise now to check that r = (m – n) n by substituting the values for
a, b, h from the m,n generator formula in the formula just found for
r.
The two formulae for r in terms of the sides a, b, h
were known to the Chinese mathematician,
Liu Hui
(approx dates: 220  280) and he writes about them in his commentary of the year 263
on an even older mathematical work called
Nine Chapters on the Mathematical Arts
which may even date back to 1000 BC!
Putting Pythagoras in the frame D G Rogers,
Mathematics Today vol 44 (June 2008), pages 123125.
Finding Pythagorean Triangles with a given Inradius
Here are a few Pythagorean triangles with small inradius r:
Primitive
Triple
Perimeter
Area
r
3,4,5
12
6
1
5,12,13
30
30
2
7,24,25
56
84
3
8,15,17
40
60
3
9,40,41
90
180
4
11,60,61
132
330
5
12,35,37
84
210
5
13,84,85
182
546
6
20,21,29
70
210
6
Nonprimitive
Triple
multiple of
Perimeter
Area
r
6,8,10
2× 3,4,5
24
24
2
9,12,15
3× 3,4,5
36
54
3
12,16,20
4× 3,4,5
48
96
4
10,24,26
2× 5,12,13
60
120
4
15,20,25
5× 3,4,5
60
150
5
18,24,30
6× 3,4,5
72
216
6
15,36,39
3× 5,12,13
90
270
6
14,48,50
2× 7,24,25
112
336
6
16,30,34
2× 15,8,17
80
240
6
From the table you can see that the number of Pythagorean triangles having an inradius of
r=1 is 1, for r=2 it is 2,
r=3 has 3 as do r=4 and r=5
but r=6 has 6 triples, and so on. This series of counts:
1, 2, 3, 3, 3, 6, ... is A078644.
For just the primitive triangles we have counts
1, 1, 2, 1, 2, 2, ... and the whole series is A068068.
The 3 4 5 triangle has an inradius of 1. So if we scale it up
by a factor r,
its inradius will become r also.
We have just shown that there is a nonprimitive Pythagorean triangle with
inradius r
for all whole numbers
r ≥ 2.
It is less obvious that there is a primitive Pythagorean triangle for each
inradius r.
However, if you look carefully at the table above, you will find there is a particular
primitive Pythagorean triangle
with a specific pattern for each inradius from r= 1 to 6,
and this should give you the clue you need to prove
that there is always a primitive Pythagorean triangle with inradius=k for every whole number k.
[Hint: you have already seen the pattern]
So we have our answer:
There is both a primitive and a nonprimitive Pythagorean triangle with inradius r for any
whole number r ≥ 2
On the Number of Primitive Pythagorean Triangles with a Given Inradius Neville Robbins, Fibonacci Quarterly
(2006) 44, pages 368369.
Another curious fact is that there are exactly two
primitive
Pythagorean triangles with an inradius which is a given prime number >2.
For instance, and here the odd numbers have been included to help point out the pattern...
Can you spot the two patterns here? What is the formula for each column of triples?
You have therefore shown that there at least two Pythagorean triples
for each odd r.
Find the m,n generators for each of the two primitive triangles in the table above
Can you show that the two triangles above for each primer are both primitive?
Find odd numbers of inradius for which there are more than 2 primitive triangles
For which odd numbers r are primitive Pythagorean triangles above the only two?
Do both patterns apply for even inradius r?
The full answer to all these questions is revealed in the next section but some of the
fun of maths is trying to prove these things for yourself.
E380 W F Cheney, C W Trigg
The American Mathematical Monthly vol 47 (1940) pages 240241.
How many Primitive Pythagorean Triangles are there with a given Inradius?
We have seen in the Puzzles section above,
that for an odd number as inradius, there are always at least two primitive Pythagorean triangles.
The number of primitive Pythagorean triangles with inradius from 1 to 100 is as follows, where the odd numbers
are in red and the primes are in blue:
We see all the prime inradii have just 2 primitive triangles and all the odds after 1 have at least 2.
Are they all just 1, 2 or 4? Further investigation shows that
there are 8 with inradius 105 and 165 and the next new value is 16.
Neville Robbins gives the exact formula T(r) for the number of
primitive Pythagorean triangles
with inradius r ≥ 2 as:
T(r) = 2 ^{(number of distinct prime factors of r)}, if r is odd T(r) = 2 ^{(number of distinct prime factors of r) – 1}, if r is even
For even numbers, we merely ignore the factor of 2 and count only the other prime factors.
Example: r = 105
105 = 3 × 5 × 7 and three distinct prime factors tells us there are
2^{3}=8 primitive Pythagorean triangles
with inradius 105.
Example: r = 45
45 = 3 × 3 × 5 so it has
2
prime factors: 3 and 5.
The number
of primitive Pythagorean triangles with inradius 45 is
T(45) = 2^{2} = 4.
They are 91, 4140, 4141; 140, 171, 221;
92, 2115, 2117 and
115, 252, 277.
Example: r = 32
32 = 2 × 2 × 2 × 2 × 2,
so 32 has no odd prime factors and
T(32) = 2^{0 } = 1.
What is the smallest number that is the inradius of 16 primitive Pythagorean triangles?
The only numbers r
where T(r) = 1
(numbers r which are the inradii
of just one primitive Pythagorean triangle) are the numbers r = a power of 2.
Which primitive Pythagorean triangles are they?
The smallest r which occurs in n primitive Pythagorean triangles
is 1 (1 triangle), 3 (2 triangles), 15 (3 triangles) and the complete series starts 1, 3, 15, 105, 1155, ...
A070826.
If we factorize these numbers we have 1, 3, 3×5, 3×5×7, 3×5×7×11, ...
which, after 1, is just the product of the first n – 1 consecutive odd prime numbers.
Whenever we see a count of things which are powers of two, we generally find that the things counted are
sets where each item may independently be in the set or not.
For instance, there are 8 baskets (sets)
of fruit we can make if we can optionally include 3 pieces of fruit: an apple, an orange and a pear say.
2^{3} gives 8 possible sets (baskets) which includes the empty set too:
{}, {apple}, {orange}, {pear}, {apple, pear}, {apple, orange}, {orange, pear}, {apple, orange, pear}
The number of Pythagorean triangles with a given inradius r
depends on the prime numbers that are the factors of r.
On the Number of Primitive Pythagorean Triangles with a Given Inradius
Neville Robbins, Fibonacci Quarterly
(2006) 44, pages 368369.
Excircles
Apart from the incircle and the circumcircle
there are three more triangles defined by any triangle.
These are the circles outside the triangle
that have all three sides, when extended, as tangents, called
excircles.
If these three circle's radii are called r_{a}, r_{b}
and r_{b}
and if the incircle's radius is r then all four are related by the
formula:
1
+
1
+
1
=
1
r_{a}
r_{b}
r_{c}
r_{}
We also have a relationship between these 4 radii and the area A:
r r_{a} r_{b} r_{c} = A^{2}
Also, if the triangle's sides are whole numbers, then so are the excircle radii (the exradii)!
In particular, for a PPT generated by m,n, the three exradii are:
n (m + n), m (m – n), m (m + n)
which proves that they are whole numbers because m and n are.
The incircle and the three excircles touch all three sides but
the circumcircle touches all three vertices. Its
centre is equidistant from all three vertices and, in a rightangled triangle,
is found at the midpoint of the hypotenuse
since that point is equidistant from the two ends of the hypotenuse and,
by symmetry or geometrical arguments,
that point is also the same distance from the other vertex at the rightangle.
The radius of the incircle is called the inradius and denoted r (as we saw above);
the radius of the circumcircle is called the circumradius and denoted R.
So Pythagorean triangles will have whole number circumradii only if the hypotenuse is an even number
Lines from the centre of the incircle to the vertices divide each angle into two.
Lines from the centre of the circumcircle perpendicular to each side divide those sides into two.
This calculator will find Pythagorean triples for you, either primitive or all with
any combination of sides with a fixed value or in a given range of values.
You can find the actual triples or else just count
the number found.
If you give a range of values, the total in that range can be counted (Total count of).
A separate count, with one count for each value in the range, is given if you select
Separately count. Sizes gives a list of those values in the range for which
the requested type of triangle (all or primitive) exist. Values are
repeated where there are different triples of the same size.
For example all (primitive and nonprimitive)
triangles with hypotenuse
= 20 up to 30:
The Total count of all Pythagorean triangles with a hypotenuse in the range 2030 is 6
If we Show all of them they are:
The information shown is : number of solution found, the triple,
if it is primitive or a multiple of a primitive,
the values on m,n if possible,
the perimeter = a+b+h,
the area = ab/2,
the incircle radius,
the altitude of the triangle if h is the base=a b/h,
the sides product a×b×h,
the legs difference (ab), the hypotenuseshortest leg difference,
the hypotenuselongest leg difference, the excess=a+bh=2×inradius
Find each count shows the individual counts for each value in the range.
There is 1 of size 20, 0 of size 21 to 24, 2 of size 25, and 1 each of sizes 26, 29 and 30
so the separate counts are reported as
1, 0, 0, 0, 0, 2, 1, 0, 0, 1, 1: a count for each of the values from 20 to 30.
The sizes of the 6 hypotenuses are 20, 25, 25, 26, 29, 30.
Since there are infinite number of PTs with any given side difference
 see above on Hypotenuse and Longest leg are consecutive
and The two legs are consecutive 
these options are marked with the ∞ sign and
an extra input box will appear for difference searches to limit the search
to a given maximum side length. Sizes reports the sizes (of the sideperimeterareainradius requested) in the
given range, so that if a sideperimeterareainradius is found in
more than one triple, it is reported once for each separate triple. Show all lists all the triples found but if you want just one example
use Show one.
The results
are printed in the Results box, triples being given with their area,
perimeter and inradius. Select and copy from this area to use the output
as text or in other applications.
A General Pythagorean Calculator
Here is a generator of all things Pythagorean which can search for various conditions on the triangles sides and has some
very fast counting algorithms for some cases.
We now turn our attention from the sides to the angles in a Pythagorean triangle.
We can call an angle (not a rightangle) in a Pythagorean triangle a Pythagorean angle.
Since the triangle has integer sides, such angles have sine, cosine and tangent that are
pure fractions ("rational").
We can start with any fraction, say 2/3, as the tangent of an angle α
and use it to generate a Pythagorean triangle which has 2α as an angle
using the formula
tan 2α
=
2 tan α
1 – tan^{2} α
So tan α = ^{2}/_{3} gives tan 2α = ^{5}/_{12}
so the Pythagorean triangle generated has legs 5 and 12 and hypotenuse 13.
If tan α = ^{n}/_{m} then
tan 2α = 2mn/(m^{2}–n^{2}).
tan α = ^{n}/_{m} if and only if
tan 2α = 2mn/(m^{2}–n^{2}).
Also, if α and β
are Pythagorean angles, then so are α + β and
α – β.
If α and β are Pythagorean angles,
then let tan( α ) =
n
and
tan( β ) =
N
m
M
then
tan( α + β ) =
n M + m N
m M – n N
tan( α – β ) =
n M – m N
m M + n N
1809. On Note 1719 A G Walker, The Mathematical Gazette vol 29 (1945), page 26.
59.11 More Properties of Pythagorean Triplets L E Ellis
The Mathematical Gazette vol 59 (1975) pages 186189.
Finding a Pythagorean Triangle approximating a given Angle
Can we find a Pythagorean triangle with a given angle? Sometimes this may not be possible with small numbers
but we will always be able to find some Pythagorean triangles with an angle almost equal to
any angle you require.
Here is a Calculator to find a primitive triangle with better and better approximations to a given angle.
It only generates primitive triangles since all its multiples have identical angles but bigger sides.
You can use Pi in the input box e.g. for the angle ^{π}/_{3} (radians).
If you want Pythagorean triangles
with a specific ratio of sides, e.g. ^{1}/_{3}, then use a function to
find angle with a given sine or cosine or tangent.
These are called the inverse trigonometric functions
arcsine, arccosine and arctangent often abbreviated to asin, acos and atan
and, given the since, cosine or tangent find the associated angle as a small positive number.
Buttons for these inverse function are found on all but the most basic of calculators:
3
5
4
e.g. all of these describe the 3 4 5 triangle but don't forget they all find the angle in radians
not degrees:
Remember that sines and cosines are in the
range 1 to 1
so asin(15/8) gives an error
asin( 3/5 ) is the angle (radians) whose sine is 3/5,
i.e. the ratio of the leg opposite the angle, 3, to the hypotenuse, 5;
acos( 4/5 ) for the ratio of the leg next to the angle, 4, to the hypotenuse, 5;
atan( 3/4 ) for the ratio of the leg opposite, 3, to the leg next to the angle, 4.
All of the above will find the 3, 4, 5 triple as will asin(4/5), acos(3/5), atan(4/3).
Calculations are slightly more accurate if radian measure is used.
It turns out that the only angles that are a rational number of turns and
that are angles in a Pythagorean triangle (that is, their sines and cosines
are rational too) are the ones met at school:
30°=^{1}/_{10} turns and 60°=^{1}/_{5} turns
and their sines and cosines are sin(30°)=cos(60°)=1/2 as well as 0° 90° and 180°.
So all rational number of turns will have to be approximated as an angle in a Pythagorean triangle.
The Calculator below will show those approximations.
Find Pythagorean triangles with a given Angle Calculator
The Shapes and Sizes of Pythagorean Triangles P Shiu, Mathematical Gazette
vol. 67 (1983) pages 3338. This describes the algorithm behind the anglefinder calculator above. To find a
Pythagorean triangle with angles close to θ let
u = tan(θ)+ sec(θ) and find its
continued fraction. If the
successive convergents to it are m_{k} / n_{k} then
a suitable Pythagorean triangle is
x = 2 m_{k} n_{k} ,
y = m_{k}^{2} – n_{k}^{2},
z = m_{k}^{2} + n_{k}^{2}.
Using Pythagorean Triangles to Approximate Angles W S Anglin
The American Mathematical Monthly vol 95 (1988) pages 540541.
Further Triple Patterns
The Pythagorean Triangle Angle Calculator above finds some interesting number patterns too.
For instance, there is
no Pythagorean triangle with an angle of 45° (type in 45 into the "angle of..." box,
make sure degrees is on and then click on the Find button)
and if we ask the Calculator to find approximations
it finds the sequence with legs differing by one that we
found above.
If we try it on a series of angles such as 0.1, 0.01 and 0.001 radians,
we discover two series
of easilyremembered and visually striking patterns
as in An easy method of writing down a
series of triples above:
399
40
401
180
19
181
39999
400
40001
19800
199
19801
3999999
4000
4000001
1998000
1999
1998001
399999999
40000
400000001
199980000
19999
199980001
Try it with 0.2, 0.02, 0.002 radians and you'll find at least two more
patterns like this!
40
9
41
99
20
101
4900
99
4901
9999
200
10001
499000
999
499001
999999
2000
1000001
49990000
9999
49990001
99999999
20000
100000001
There are more complex patterns here too:
88501
17940
90301

899850001
17999400
900030001
446930400
8939801
447019801
8999985000001
17999994000
9000003000001
4496993004000
8993998001
4497001998001
With 0.3, 0.03 and 0.003 there are patterns in the first four approximations:
12
5
13
24
7
25
2112
65
2113
2244
67
2245
221112
665
221113
222444
667
222445
22211112
6665
22211113
22224444
6667
22224445
532
165
557
391
120
409
55432
1665
55457
39991
1200
40009
5554432
16665
5554457
3999991
12000
4000009
555544432
166665
555544457
399999991
120000
400000009
The first triple in the final series is the one suggested by the rest of the pattern. It is indeed
a Pythagorean triple  check it with the Test a Triangle  is it Pythagorean?
calculator above.
The series of angles 0.4, 0.04, 0.004 and 0.0004 radians generates:
12
5
13
24
10
26
1200
49
1201
2499
100
2501
124500
499
124501
249999
1000
250001
12495000
4999
12495001
24999999
10000
25000001
The 0.5 radians sequence gives:

15
8
17
760
39
761
1599
80
1601
79600
399
79601
159999
800
160001
7996000
3999
7996001
15999999
8000
16000001
and with 0.6 radians and its tenths:
4
3
5
91
60
109
544
33
545
9991
600
10009
55444
333
55445
999991
6000
1000009
5554444
3333
5554445
99999991
60000
100000009
380
261
461
5436800
326601
5446601
55443668000
332666001
55444666001
555444366680000
333266660001
555444466660001
0.7 radians seems to give only one simple pattern:
351
280
449
39951
2800
40049
3999951
28000
4000049
399999951
280000
400000049
but 0.8 gives two:

624
50
626
31000
249
31001
62499
500
62501
3122500
2499
3122501
6249999
5000
6250001
312475000
24999
312475001
624999999
50000
625000001
and 0.9 gives several:
4
3
5
3
4
5
220
21
221
264
23
265
483
44
485
24420
221
24421
24864
223
24865
49283
444
49285
2466420
2221
2466421
2470864
2223
2470865
4937283
4444
4937285
246886420
22221
246886421
246930864
22223
246930865
493817283
44444
493817285
20
21
29
48
55
73
319
360
481
2240
201
2249
6148
555
6173
39919
3600
40081
222440
2001
222449
617148
5555
617173
3999919
36000
4000081
22224440
20001
22224449
61727148
55555
61727173
399999919
360000
400000081
2222244440
200001
2222244449
6172827148
555555
6172827173
39999999919
3600000
40000000081
0.010 will again give the first example above for 0.1.
However, do use the Calculator above and repeat the experiment. This time you will probably notice at least
two more pattern series to add to your collection!
How about 0.11, 0.011, 0.0011, ...
and then 0.12, 0.012, 0.0012, ...
and so on?
Also try 2/3, 2/30, 2/300, etc. (the Calculator handles expressions as input)
or you can input it as 0.6666, 0.06666, 0.006666, etc.
which has a one very simple pattern in particular.
You can also do this with any other (small) sequence of numbers making a decimal.
Remember though that the biggest angle is 90° which is 1.57079632679489 radians.
The reason the patterns are so "obvious" above is that our numbers are written in base 10 and we are taking
angles onetenth as large each time.
An interesting mathematical Project is to find formulae for each of these series.
It will then be easy to verify that
all the triples in the series are Pythagorean by
summing squares of the two legs and checking it equals the square on the hypotenuse.
You might even be able to find a formula that encompasses several of the series above.
What else can you find?
Email me at the address at the foot of this page
and I'll add any interesting series of triples that you find, with your name.
Curious Connections
This section explores some of the curious connections between Pythagorean triangles and
other mathematical topics. The places where they turn up is sometimes very surprising!
The 3 4 5 triple in an unusual guise
Here is a circle and a ring all with the same circle centres. The disc has radius 3 and the ring is between circles of radius 4 and 5.
Which is bigger (in area, that is, which would take more paint to colour it)  the disc or the ring?
A little algebra on the area of a circle (π r^{2} for a circle of radius r) and knowledge of the
3 4 5 triple will give you the perhaps surprising result that they are the same!
Try this variation:
which is bigger this time  the disc or the ring?
Special Digit Triples
Side Reversals
The Pythagorean triple 88209, 90288, 126225 has two legs which are integers in reverse order.
There seems to be only one more with a hypotenuse less than 1,000,000: 125928, 829521, 839025
although you could also include 20691, 196020, 197109 and 82863, 368280, 377487.
What is the next? Can you find any more like this?
An Interesting Pythagorean TriangleAmerican Math Monthly 66 (1959) page 65.
You Do The Maths...
Find another triple with each leg being the reverse of the other leg ("legs" are the sides forming the rightangle).
[The legs and the hypotenuse have 7 digits.]
8053155 5513508 9759717
33 56 65 and 3333 5656 6565 have a hypotenuse which when reversed
becomes a side.
By multiplying the first triple by various values, write down several more.
If we insert a "1" before the numbers in the triple 5 12 13 we get
15 112 113 which is also Pythagorean. The problem was posed and solved
for a triangle with all sides less than 100
in ...
E472 V Thebault, W E Buker
The American Mathematical Monthly vol 49 (1942) page 196.
Both 5 12 13 and
15 112 113 are primitive too! Is this the only case?
Update 16 August 2013:
John McMahon has discovered that this prefix is part of a more general pattern for 5, 12, 13:
5
12
13
15
112
113
25
312
313
35
612
613
45
1012
1013
55
1512
1513
The general pattern is ( n×10 + 5 )^{2} + ( n(n+1)×50 + 12 )^{2} = ( n(n+1)×50 + 13 )^{2}.
The question is now, what other Prefix Patterns exist for other PTs?
There are other nonprimitive solutions:
We can multiply these two triples by 10 or 100 or any power of 10 and still get a valid solution:
1 before 50,120,130 = 10 ×5,12,13 → 150,1120,1130 = 10 ×15,112,113,
1 before 500,1200,1300 = 100 ×5,12,13 → 1500,11200,11300 = 100 ×15,112,113,
...
This applies to the following too:
1 before
500,12495,12505
=
5×[100,2499,2501]
→
1500,112495,112505
=
5×[300,22499,22501]
1 before
7500,21875,23125
=
625×[12,35,37]
→
17500,121875,123125
=
625×[28,195,197]
2 before
600,1045,1205
=
15×[120,209,241]
→
2600,21045,21205
=
5×[520,4209,4241]
2 before
600,11242,11258
=
2×[300,5621,5629]
→
2600,211242,212258
=
2×[1300,105621,105629]
3 before
7500,11375,13625
=
125×[60,91,109]
→
37500,311375,313625
=
125×[300,2491,2509]
3 before
9000,15675,18075
=
75×[120,209,241]
→
39000,315675,318075
=
30×[1300,105621,105629]
3 before
900,16863,16887
=
3×[300,5621,5629]
→
3900,316863,316887
=
3×[1300,105621,105629]
Are there any Pythagorean triples formed by prefixing a number larger than 3 at the front of all sides?
Pythagorean Triples and Prime Numbers
What about the prime numbers as sides of a Pythagorean triangle?
Clearly they only occur in the primitive Pythagorean triangles.
All the primitive triangles are generated by the m,n formula
in which one side is 2 m n.
So our prime number sides must be the odd side of a primitive Pythagorean triple (since there is no Pythagorean triangle with a side of 2)
and/or the hypotenuse.
Here are some with the prime numbers shown like this:
Prim PT
m,n
3
4
5
2,1
5
12
13
3,2
8
15
17
4,1
7
24
25
4,3
20
21
29
5,2
12
35
37
6,1
9
40
41
5,4
28
45
53
7,2
11
60
61
6,5
48
55
73
8,3
13
84
85
7,6
39
80
89
8,5
65
72
97
9,4
Prim PT
m,n
20
99
101
10,1
60
91
109
10,3
15
112
113
8,7
88
105
137
11,4
17
144
145
9,8
51
140
149
10,7
85
132
157
11,6
52
165
173
13,2
19
180
181
10,9
95
168
193
12,7
28
195
197
14,1
In The new book of prime number records 3rd edition (1995) SpringerVerlag, P Ribenboim
conjectures there are an infinite number of Pythagorean triangles with two prime number sides.
Pythagorean Triangles and Egyptian Fractions
Egyptian Fractions
are a different way of writing fractions as
used by the ancient Egyptians who built the Pyramids and before them the ancient Babylonians.
They did not use
the ratio of two whole numbers as we do, e.g. "fourfifths" or
^{4}/_{5} which is
the ratio of 4 to 5 and
also the result of 4 divided by 5, of 4 things divided into 5 equal parts.
Instead they wrote fractions as a sum of unit fractions.
For example ^{3}/_{4}
would be ^{1}/_{2} + ^{1}/_{4}, a
sum of two different fractions each with a
numerator (top number) of 1.
Fractions which are the reciprocal of an integer (i.e. have a numerator of 1)
are called unit fractions.
Every fraction ^{a}/_{b}
can indeed be written as a sum of distinct unit fractions and in many ways
and these are called Egyptian Fractions.
The simplest kinds of fractions are those that can be written as a sum of two unit fractions.
For n=3
the number of ways is one since
^{2}/_{3} = ^{1}/_{2} + ^{1}/_{6}
is the only way to write ^{2}/_{3} as the sum of two unit fractions.
For n=8
we have ^{2}/_{8} which is, of course,
^{1}/_{4}
and we have two pairs of different unit fractions with this sum:
^{1}/_{4} = ^{1}/_{12} + ^{1}/_{6} = ^{1}/_{20} + ^{1}/_{5}.
Surprisingly this number of ways to write 2/n as a sum of two different unit fractions
is exactly the same as the number of Pythagorean triangles having n as a leg:
For n=3 we have just one Pythagorean triangle having 3 as a leg: 3, 4, 5
For n=8 there are two Pythagorean triangles having 8 as a leg: 6, 8, 10 and
8, 15, 17
The number of ways we can write 2/n as a sum of two different unit fractions gives the series
The Fibonacci Numbers
are a simple series of numbers that appear in lots of places in nature:
1, 2, 3, 5, 8, ... where each number is the sum of the previous two in the series.
Often mathematicians start this series with 0 and 1 and we get the series:
0, 1, 1, 2, 3, 5, 8, 13, 21, ... A000045.
Fibonacci Numbers as sides of Pythagorean triangles
It is easy to see that
there is no triangle with all 3 sides being consecutive Fibonacci numbers
If the sides are a<b<c
then c is at least a + b by the Fibonacci rule.
However,
in any triangle the two shorter sides must add to more than the longest side or else the sides will not meet:
this is called the Triangle Inequality: a+b>h.
We know of two Pythagorean triangles with 2 Fibonacci numbers as sides:
3 4 5 5 12 13
It is thought that there are no more but this remains an open question.
Pythagorean Triples Containing Fibonacci Numbers: Solutions for
F_{n}^{2} ± F_{k}^{2} = K^{2} by M BicknellJohnson,
Fibonacci Quarterly17 (1979) pages 112, (Addenda: page 293).
Although no Pythagorean triangles contain 2 Fibonacci numbers as separate sides,
there is a series of Pythagorean triangles with many Fibonacci relationships
where the m,n generators are successive Fibonacci numbers and
each has a hypotenuse that is a Fibonacci number.
The perimeter of each is equal to the length of the longest leg in the next triangle.
Problem B725, Russell Jay Hendel and Herta Freitag, the Fibonacci Quarterly 32 (1994) page 97
Not all Fibonacci numbers can be a hypotenuse: there are no Pythagorean triangles with a hypotenuse of
2, 3, 8, 21, 144, 987, ... .
Areas and Fibonacci Numbers
In terms of the areas of Pythagorean triangles, Mohanty and Mohanty, mentioned earlier,
use the term Pythagorean number for a whole number that is the area of a Pythagorean triangle. They
showed that
F(2n) F(2n+1) F(2n+2) is always a Pythagorean number (the area of a Pythagorean triangle)
and, if the middle term, F(2n + 1),
is even, then the product is a primitive Pythagorean number (the area of a primitive Pythagorean triangle) (Corollary 2.4)
F(2n) F(2n+2) F(2n+4) is a Pythagorean number and
if the middle term, F(2n+2) is even, then the product is a primitive Pythagorean number (Corollary 2.4).
The product of 4 consecutive Fibonacci numbers is a Pythagorean number and it is primitive if, in the centre,
one Fibonacci number is even and the other is odd (Corollary 2.5).
The last result is true for four consecutive Lucas numbers too but
They also conjecture that no Fibonacci number is Pythagorean
i.e. no single Fibonacci number can be the area of a Pythagorean triangle.
Pythagorean Triples and Pi
Use the buttons in this section to change the graph
Earlier we saw that
of the number of primitive Pythagorean triangles with a hypotenuse less than N was virtually a
straight line when plotted against N:
Since the "straight line" graph goes through the origin,
we can also
of (the number of primitive Pythagorean triangles with hypotenuse less than N) / N.
This ratio seems to be settling down to a particular value as N gets larger: what is this value?
Discovered by D N Lehmer in 1900 it is
1
=
1
=
1
= 0.1591549..
2π
2×3.1415926..
6.283185..
The number of primitive Pythagorean triangles with hypotenuse less that N is approximately
N
= 0.1591549 N
2 π
For example, for N = 100, this approximation formula gives
0.1591549 × 100 = 15·92 primitive Pythagorean triangles
with hypotenuse less than 100 whereas
the exact value is 16.
But this is not the only relationship between Pythagorean triangles and π!
of the number of primitive Pythagorean triangles with a perimeter less than N
is also a "straight line".
What is the ratio this time? Have a look at
shown in the graph area above.
Again it was D N Lehmer who proved that the limit of this ratio also involved π:
The number of primitive Pythagorean triangles with perimeter less than N is
approximately
ln(2) N
= 0.07023 N
π^{2}
ln(2) means log_{e}(2), the natural log of 2,
i.e. the number which, when e is raised to that power, gives 2.
Since e^{ 0·693147...} = 2 then
ln(2) = 0·693147...
So
ln(2)
=
0·693147
=
0·693147
= 0·07023
π^{2}
3·141592^{2}
9·869604^{ }
For example, Lehmer's formula for N = 1000
gives the value of 0·07023 × 1000 = 70.23
as the approximate number of primitive Pythagorean triangles with a perimeter less than 1000 whereas
the exact value is 70: so it is not bad as an estimate!
It seems remarkable that π should appear in this context, but it does
have an amazing tendency to appear in many formulae for approximations in different areas of mathematics.
Estimate the number of PPTs with hypotenuse or perimeter<N Calculator
C A L C U L A T O R
the number of primitive Pythagorean triangles with
less than
R E S U L T S
we generalised it to find many others of the same form.
But Tim Schumacher of Christchurch, New Zealand sent me an email in Nov 2014 with a
marvellous connection between the Fibonacci method above and such formulas for π.
If we take any starting values for the Fibonacci method, we found that
a, b, a+b, a+2b generated the PT 2b(a+b), a(a+2b), b^{2} + (a+b)^{2}.
But each and every one of these leads to a formula for π too using the ratios
of the middle two terms and the outer two terms as follows:
π
= arctan
a
+ arctan
b
4
a + 2 b
a + b
It is not difficult to use the trig formula for the tangents of a sum of two angles to prove this is true but
Tim with his student Dingcheng Luo found a wonderfully simple diagram generalising
Ko Hayashi's diagram to prove it. It
almost needs no words at all:
tan(α) =
a
a + 2 b
tan(β) =
√2 b
=
b
√2 (a + b)
a + b
tan(α + β) =
a + b
= 1
a + b
Note that if all edges are the same length, say a = b = 1
then we get Euler's formula of 1738 seen above:
π
= arctan
1
+ arctan
1
4
2
3
so this is a generalisation of his result.
Pythagorean Triples and Partitions
Jack Garfunkel proposed that there is a relationship between partitions of a number
(the number of ways we can write that number as a sum of positive integers) and Pythagorean triples:
P_{3}(a) +
P_{3}(b) =
P_{3}(h) for any Pythagorean triple a, b, h
P_{3}(n) is the number of ordered lists of 3 whole numbers
whose sum is n i.e.
the number of ways of writing n as n = i + j + k
for whole numbers i, j and k where
0 < i ≤ j ≤ k
For instance, we can only write 1 + 1 + 1 to make 3, so this list is the only one: P_{3}(3) = 1.
For 4 we again have only one list: 1 + 1 + 2 = 4 so P_{3}(4) = 1.
There are two solutions for 5 since 5 = 1 + 1 + 3 = 1 + 2 + 2 so P_{3}(5) = 2.
Find all the sumsof3 that total
6, 7, 8, ... and check your results with this table:
n
3
4
5
6
7
8
9
10
11
12
...
P_{3}(n)
1
1
2
3
4
5
7
8
10
12
...
So Garfunkel's result is that since 3, 4, 5 is a Pythagorean triple then P_{3}(3) + P_{3}(4) = P_{3}(5)
and, indeed, from our table, P_{3}(3) = 1, P_{3}(4) = 1 and
2 = P_{3}(5) = P_{3}(3) + P_{3}(4) = 1 + 1 = 2.
Problem 512 Jack Garfunkel, Pi Mu Epsilon Journal (1981) pg 31.
Pythagorean or Babylonian?
A tiny block of clay, about the size of a postcard (5 inches x 3.5 inches or 12cm x 9cm) with 15 rows
of 4 columns of "numbers" is dated to about 1800 BC and so is probably the world's oldest surviving mathematical artefact.
Plimpton 322 is one of 600 such tablets
donated to Columbia University's Rare Book and Manuscript
Library by George Plimpton and was item 322 in his catalogue, hence its name.
(Have a look at their other treasures too.)
Bill Casselman's page on
The Babylonian tablet Plimpton 322 from University of British Columbia has the best image of the tablet and
an excellent explanation of how to read Babylonian numbers and what the tablet contains.
And what does it contain? A list of Pythagorean Triangles arranged in order of triangles which are
approximately 1 degree apart! They are written in the Babylonian scale of base 60 and involve base 60 "fractions".
Such tables were probably used in surveying.
Mansfield and Wildberger of UNSW Australia have found exact connections between the
figures and how they can be used practically. See the references below for a short Youtube video and a link to the fuller
explanation in their journal article.
However, some of this is a little "hyped" and Evelyn Lamb's article
Don't Fall for Babylonian Trigonometry Hype in the Scientific American of August 2017
is a good review of
these Youtube vides and their claims.
Words and Pictures: New Light on Plimpton 322
by Eleanor Robson in American Mathematical Monthly vol. 109 (2002), pages 105120
explores three theories for the
meaning of the numbers on Plimpton 322, one of which is that it is a trigonometric table.
A more detailed explanation is in their paper on this...
The Exact Sciences in Antiquity
by Otto Neugebauer, Dover, (1969) 240 pages argues that the Plimpton 322 tablet contains Pythagorean triples for
triangles for each degree from 30 to 45 and so detects several simple errors in the tablet's table.
Pythagorean Jigsaws
Pythagorean Triangles in a Square
Can we divide a square into Pythagorean triangles?
Charles Jepsen and Roc Yang showed that we could, with a minimum of 5 such triangles.
The smallest square with just 5 Pythagorean triangles that they found is shown here.
They prove also that there is no dissection of a square into just 4 Pythagorean triangles.
So they then ask the question
Is this the smallest square that can be dissected into five Pythagorean triangles?
It seems this question is still "open". Can you find the answer?
Amazingly, Penny Drastik, a primary student at The Illawarra Grammar School in Australia, aged 10,
found 12 smaller solutions for squares of side less than 9000 including
one which she thinks is the smallest square
with this pattern of triangles [April 2008].
Penny also found one square (of side less than 9000) that can be dissected in two different
ways with this pattern of triangles.
I leave you with the challenge of finding the sides of the triangles and the square.
Can you find a smaller square, perhaps with a different arrangement of 5 triangles?
Making Squares from Pythagorean Triangles C Jepsen, R Yang The College Mathematics Journal vol 29 (1998), pages 284288.
Squares with more than 5 Pythagorean Triangles?
Jepsen and Yang's articles (above) gives a beautifully simple argument that it is possible to dissect a square into
any number of Pythagorean triangles, from 5 upwards.
We have the solution for 5 triangles (which they prove is the smallest number of Pythagorean triangles
that we need to fill a square).
So take any one the Pythagorean triangles in any square's dissection
and find the altitude from the right angle to the hypotenuse:
We have thus divided that Pythagorean triangle into two right angled triangles but these will only be Pythagorean if
the new side's lengths h and w are integers. What are these two lengths in term of
a, b, c?
Since the area of the triangle is ab/2 and also ch/2
then h = ab/c.
A small amount of algebra (left to the reader) shows that w = a^{2}/c.
By symmetry, we also have c – w = b^{2}/c
So h and w might not be integers unless c divides exactly into
both a^{2}, ab andb^{2}. But we can make them integers by expanding
the whole square dissection containing the triangle by factor c!
Thus we obtain another square, c times larger, with an
one Pythagorean triangle replaced by two.
Interestingly, if we try this on the 5triangle dissection above, we find the top triangle neatly divides into two with no expansion needed:
Clearly we can do this as often as we like to get squares with an ever increasing number of Pythagorean triangles in them.
There is always a square that can be dissected into n Pythagorean triangles for every n from 5 upwards
However, the triangle that is divided into two makes two smaller Pythagorean triangles both of exactly the same shape
as the original, that is, the angles in the two new triangles are equal to those in the triangle that was split.
So we are guaranteed to have two or more Pythagorean triangles of the same shape in our jigsaw
if we use this method.
Triangles with the same angles in each are called similar triangles; they need not be the same size but they do have the same shape.
Triangles with matching sides and angles are identical in both size and shape and are called congruent triangles.
Is it always possible to dissect a square into any number (≥ 5) of different (i.e. not similar) Pythagorean triangles?
Another proof of the Pythagoras Theorem
Using the diagram above with the altitude (height) h marked in triangle a b c,
we can find another proof of the Pythagoras Theorem a^{2} + b^{2} = c^{2}:
Show that the triangles ABC, CDB and ACD are all similar (that is, the same three angles occur in each)
Identify the other two angles equal to the angle at A and the other two angles equal to the angle at B
Find the sine of angle B in triangle BCD and the equivalent angle in ABC and its sine.
Connect them with an equation involving
b, c and c–w, b
Find the cosine of angle B in triangle ABC and the equivalent angle in the third triangle, ADC, and its cosine.
Connect them with an equation involving
w, a, and a, c
Rewrite your two equations so that there are no fractions by crossmultiplying if necessary
Add the two equations so that one side is a^{2} + b^{2}
Can you show that the other side of the equation is now c^{2}?
Pythagorean Triangles in a Circle
We can always draw a circle through any set of 3 points. There is a very nice animation of how to do this at
Math Open Reference site. The method is easy and was given
by Euclid's Elements Book 3, Proposition 1.
However, something special happens for Pythagorean triangles because of the rightangle.
The circle through the three points of a rightangled triangle has its centre at the midpoint of
the hypotenuse
In other words, pick a point on a circle and connect it to two ends of any diameter and the angle you have just made is always a right
angle.
So if we can find several Pythagorean triangles with the same hypotenuse we can place them on the same diameter of a circle
so that their rightangles will lie on the circle.
By orienting the triangles so that the smallest angles are all at one end of the diameter, the rightangles will lie on a quarter circle.
By duplicating each triangle with each nonrightangle at each end of the diagonal, the rightangles will all lie on a semicircle.
By including the triangles twice on each side of the diagonal, the rightangles will lie on a circle.
And, because they are all Pythagorean triangles, the points on the circle will all have integer coordinates.
The simplest example is 7 24 25 and 15 20 25 which have a common hypotenuse of 25.
Clearly we can double all the lengths and treble them and get many more examples.
Here is a set of 4 with a common hypotenuse of 65: 16 63 65, 25 60 65, 33 56 65, 39 52 65
Can you find another set of 4 having a hypotenuse of 85?
There is a set of 3 all having a hypotenuse of 125  which are they?
Can you find the surprising 7 having a hypotenuse of 325?
There are 13 with a hypotenuse of 1105
and 22 for 5525!
are the hypotenuses of Pythagorean triangles. So all integers are in the sequence A004144 or this one.
This series can be further split into the following classes.....
are the only possible counts of the number of Pythagorean triangles with the same hypotenuse for hypotenuses up to 200,000.
There are many numbers that are the hypotenuses of exactly 4 triangles, the next most frequent count being 2,
then 13 and so on, with 160225 being the hypotenuse of 67 Pythagorean triangles. Up to a hypotenuse of 200,000 we cannot find a number
that is the hypotenuse of exactly 8, 9 or 11 triangles. If we do not limit the size of triangles,
then all numbers are possible as counts of Pythagorean triangles with the same hypotenuse according to
A097756
these are the smallest hypotenuses of exactly 1, 2, 3, ... Pythagorean triangles in order of number of triangles,
so 5 is the smallest
hypotenuse of a single Pythagorean triangle, and 25 the smallest of exactly two triangles, etc.
are the counts of the recordbreakers of the previous sequence (the number of triangles of which each is the hypotenuse)
Pythagorean Triangles in a Triangle
Earlier we saw how to split any Pythagorean triangle
into two Pythagorean triangles which are similar to each other and also to the original
(that is, the 3 angles in each triangle are the same: the triangles
are called similar).
Here is a rightangled triangle split into 3 similar triangles. All have the 3 4 5 shape.
You can check this by dividing the sides by each of the factors shown in red in each rightangled triangle.
We can make a diagram just like this containing three similar triangles derived from any Pythagorean triangle.
To show this, first in this diagram;
Label the sides of one triangle a, b and h;
Keeping the sides in the same proportions, and using the angles identified as being equal, label the other sides;
If you have any divisions, multiply the whole diagram by the divisors to make every side a product of a, b and h
You Do The Maths...
The investigations here would make an excellent topic for a Science Fair or Maths Project
Can you make a rightangled triangle jigsaw with more than one shape of Pythagorean triangle?
What about a jigsaw for a nonrightangled triangle?
What is the largest number of rightangled triangle pieces you can fit into a triangular jigsaw?
By looking at the earlier method we used to split one rightangled triangle into two, and comparing it with the
diagram with three here, we can split a rightangled triangle into 4 rightangled pieces, all similar as shown here.
Can you extend it again to five triangles? to six? and to as many as we like?
Make you own patterns using one Pythagorean triangle in a range of sizes to make a nice tiling pattern as in the previous investigation question.
Send some to me at the email address at the foot of this page and I will include them here.
Pythagorean Triangles in a Kite
A kite is a quadrilateral with two pairs of touching sides equal in length.
If we have rightangles where the unequal sides meet, we can
make any Pythagorean triangle into a kite using three differentlysized copies of it
as shown here.
What is the formula for the sides of the kite in terms of the Pythagorean triangle a b h?
What is special about the hypotenuse of the largest triangle (the vertical strut of the kite)?
The kite with 3 Pythagorean triangles can easily be transformed into a rectangle:
First reflect the red and yellow triangles in a horizontal mirror
then move them across to fit on the two legs of the green triangle
Pythagorean Triangles in a Rectangle
So now how about rectangles dissected into Pythagorean triangles?
The simplest method is to put two identical triangles together along their hypotenuses to make a rectangle:
So a rectangle can be dissected into two Pythagorean triangles if its sides are the two legs of a Pythagorean triangle.
But both the triangles in this dissection are identical (congruent).
Here are two rectangles each containing three rightangled triangles. Show that in each diagram
all the triangles are similar.
In each of these two 4triangle rectangle dissections find two similar triangles.
Can you find two more rectangular dissections into 4 rightangled triangles each with
4 similar triangles?
Can you find a rectangle that dissects into different (nonsimilar) Pythagorean triangles?
Pythagorean Triangles round a point
This section investigates putting 4 Pythagorean triangles round a point with their rightangles meeting at the point
and other variations.
Four triangles with their rightangles meeting
Can we fit four Pythagorean triangles together round a point at which all the rightangles meet?
You should find it easy to answer the question if you use these two diagrams on the left as guides.
However, what if the four triangles are all different as shown here on the right?
These shapes are quadrilaterals since none of their sides are equal in general.
The smallest perimeter is 176 and has two pairs of similar Pythagorean triangles
if you want to find it for yourself and check your answer with this button:
The smallest solution (smallest perimeter) with 4 different Pythagorean triangles (that is, no two triangles are similar)
has a perimeter of 950:
Three triangles with their rightangles meeting
If we try just three triangles meeting at their rightangles, we have to go into 3 dimensions.
The three triangles will meet as if in the corner of a room as shown here.
Again this is possible and the smallest perimeter is 636 so it is harder to find than the 4 different triangles meeting at a point of the
previous section.
There is another interesting connection here:
The square of the area of the triangle formed from the 3 hypotenuses = the sum of the squares of areas of the three Pythagorean triangles
This theorem applies even if the 3 rightangled triangles were not Pythagorean.
More than 4 triangles with their rightangles meeting
We can fit more than 4 Pythagorean triangles together with their rightangles meeting at a point if we allow three to be flat
"on the floor" and the other two to be perpendicular as if on two "walls".
Imagine looking at the corner of a building from outside and concentrate on the point where the two vertical
walls meet the ground. The diagram shows a blue and green vertical wall at the corner on the red ground.
Three triangles meet and lie flat on the ground; a fourth triangle is on the green vertical wall and the fifth is on the
blue vertical wall, all 5 rightangles touching at the corner point where the three surfaces meet.
Can you imagine 6 meeting at a "corner"? What about 7 or 8? Are more possible?
Find the only two Pythagorean triangles with an area equal to their perimeter.
Which are the only 3 numbers that cannot be the shortest side of any Pythagorean triangle?
[Check your answer with A009005.]
Find three consecutive numbers which can be the hypotenuses of
Pythagorean triangles.
Can you find four consecutive numbers which are hypotenuses?
What about five?
and how about a set of nine?
[Check your answers with A099799.]
Find a few Pythagorean triangles whose shortest side is a square number e.g.
9=3^{2}, 12, 15, and 25=5^{2}, 312, 313.
Of the primitive ones in your list, what is special about their m
and n values?
A remarkable property of the m,n formula for triples:
( m^{2} – n^{2} )^{2} + (2 m n)^{2} = ( m^{2} + n^{2} )^{2}
is illustrated in this puzzle:
Find triples with a hypotenuse that is a square numberH^{2}.
There are two triangles with a hypotenuse of 5^{2}=25 for instance: 15 20 25 and
7 24 25.
List the values H that are squared to make these hypotenuses:
where have you seen this series before on this page?
It seems there are two triples for each and every hypotenuse that is a square numberH^{2}.
One is easily explained as it has a simple relationship with the triple with hypotenuse H:
what is that relationship?
For the second, look at its generators to find a
proof that it always exists.
Can you guess how many triples there will be with a hypotenuse that is a fourth power: H^{4}?
Following on from the previous Puzzle, what can you find out about triples with a hypotenuse of the form
H^{3}?
What about Pythagorean triples having a smallest side which is a cube? e.g.
27=3^{3}, 36, 45. What is special about their m
and n values?
How many Pythagorean triangles have a side of length 48?
Find a number that can be the side of even more Pythagorean triangles.
(Hint: there are 5 answers less than 100)
What is the highest number of triples you can find with the same side in each?
Which number less than 250 occurs in 32 triples?
What is the smallest number that is the hypotenuse of more than one triple?
What is the greatest number of triples you can find with the same hypotenuse?
Is there one that is not a multiple of 5?
Find some numbers which are the odd sides of more than one primitive Pythagorean triangle.
The first two are
15: which is a side in both 8 15 17 and 15 112 113 and
21: which is a side of both 20 21 29 and 21 220 221.
Can you find a property to describe the factorizations into primes of each number in this series?
[Check your answer with A061346.]
33,44,55 is a triple where all the numbers are palindromes, that is,
they are the same when written backwards.
We can find a whole series of Pythagorean triples where all the numbers are palindromes:
3,4,5
33,44,55
333,444,555
...
Also we have the triple 303,404,505.
What is the series of factors that has been used to generate these from 3 4 5?
Another is 66,88,110 if we include an initial 0 in front of the hypotenuse:
66,88,0110.
What about the Pythagorean triples 606,808,01010 and 666,888,01110?
Can you find any more infinite series of palindromic Pythagorean triples?
[Hint: You might find A057148 useful here.]
4/3/05 is a date and also a Pythagorean triple.
There was another one that year ('05)  when was it?
Assuming that the years are in this century and are just two digits long,
when is the next Pythagorean Triple Date?
How many other such days are there in this century?
If the date is any set of 3 numbers that are a Pythagorean triple (that is, the numbers need not be in order),
how many dates are there in one century? [Why not organise a 'Pythagoras Day' at your school/college/maths department on the next such date?]
How about a special Pythagorean Triple Time in hours:minutes:seconds? How many are there in a
whole day if we use a 24hour clock with hours from 0 to 23?
In the Easy method of writing down a series of Triples section, we found a formula for the
pattern given there and used it with n = 10, 100, 1000, ....
What pattern do you get with n = 20, 200, 2000, ...?
... and with n = 30, 300, 3000, ...?
Find your own Pythagorean Triple Pattern not already mentioned in Further Triple Patterns
above.
Here is another way to do this.
Think of a series of numbers that are like those in the lists above, e.g.
from the 399, 40, 401 pattern we might think of hypotenuses that are in the series
901, 90001, 900001, ... . Plug these numbers into the Triples generator and see if any patterns
emerge. The hypotenuse searches on 901 and 90001 give:
[from Ken Sullins, Empire State College]
Can you find a formula for each of the sets of triples in the columns below?
Hint: In each column, the first legs form an arithmetic sequence (that is, they increase by the same amount each time).
But, looking at the other two sides, what else do you notice is common to all the triples in a column?
A
B
C
D
E
F
G
H
I
3, 4, 5
4, 3, 5
12, 5, 13
15, 8, 17
35, 12, 37
40, 9, 41
60, 11, 61
24, 7, 25
63, 16, 65
5, 12, 13
8, 15, 17
20, 21, 29
21, 20, 29
45, 28, 53
56, 33, 65
80, 39, 89
36, 27, 45
77, 36, 85
7, 24, 25
12, 35, 37
28, 45, 53
27, 36, 45
55, 48, 73
72, 65, 97
100, 75, 125
48, 55, 73
91, 60, 109
9, 40, 41
16, 63, 65
36, 77, 85
33, 56, 65
65, 72, 97
88, 105, 137
120, 119, 169
60, 91, 109
105, 88, 137
11, 60, 61
20, 99, 101
44, 117, 125
39, 80, 89
75, 100, 125
104, 153, 185
140, 171, 221
72, 135, 153
119, 120, 169
...
...
...
...
...
...
...
...
...
2n+1 2n^{2}+2n 2n^{2}+2n+1
4n 4n^{2}–1 4n^{2}+1
8n+4 4n^{2}+4n–3 4n^{2}+4n+5
6n+9 2n^{2}+6n 2n^{2}+6n+9
10n+25 2n^{2}+10n 2n^{2}+10n+25
16n+24 4n^{2}+12n–7 4n^{2}+12n+25
20n+40 4n^{2}+16n–9 4n^{2}+16n+41
12n+12 4n^{2}+8n–5 4n^{2}+8n+13
14n+49 2n^{2}+14n 2n^{2}+14n+49
A farmer uses fencing panels all of the same size
to cut off a rightangled corner of a field to form a triangular enclosure.
Each side used a whole number of panels.
When she dismantled the fence she found she could reuse it all to surround a square area exactly.
What was the shortest length of the fence for which this is possible?
If in addition to (a) she could also reuse all the panels to surround another cutoff corner
but which had a different shape to the original,
what is the smallest number of panels she would need now?
Find the smallest ten numbers in the series of perimeters which are all answers to puzzle (a).
A farmer had a rightangled triangular field that he sowed with grass seed.
The next year he sowed a differently shaped rightangled triangular field with exactly the same amount of seed.
What shapes could the two fields have been and what was the area?
Is it possible to find yet a different shape for a third year with the same amount of seed?
If not, find another 3 field shapes with the same area.
In the Pythagorean Triangles in a Rectangle section
above we found a dissection of a rectangle into
3 similar triangles (they all contained the same 3 angles) and one with 4 similar triangles.
Find a dissection of rectangle into 5 similar triangles.
What about 6?
Is it possible to extend this to any number of similar triangles in a rectangle?
Sums of more than two squares
Let's generalize Pythagorean triples to sums of 2 or more squares whose sum is a square.
We saw earlier that not all numbers can be the hypotenuse of a PT
but what if we summed more than two squares? Is it possible to write all numbers as a sum of squares?
If so, what is the minimum number of square we need? If not, what numbers are missing?
1^{2} + 2^{2} + 2^{2} = 3^{2} if we allow squares to be repeated
2^{2} + 3^{2} + 6^{2} = 7^{2} with distinct squares
10^{2} + 11^{2} + 12^{2} = 13^{2} + 14^{2} a nice pattern which generalises
62 = 1^{2} + 5^{2} + 16^{2} = 2^{2} + 3^{2} + 7^{2}
is the smallest number which is the sum of two sets of 3 squares
But what about numbers other than squares? Can all numbers be the sum of some squares?
In fact, not all numbers are the sum of three squares, for instance 7 is not.
What about four squares? This is a famous problem studied by many great mathematicians of the past,
including Gauss and Euler.
Every number is the sum of up to four squares.
These are the questions we ask in this section. First, here is a Calculator to help in your investigations.
A Sums of Squares Calculator
set
If you want unique squares in the sum, no number repeated and all numbers to be squared are in order,
choose a set;
list
If repetitions are allowed and numbers to be squared are in order then choose list.
Each repeated number in a list is shown once with the number of repetitions as a subscript,
for example
the list 1 1 2 4 4 4 is shown as
sequence
If you want to count every possibility of numbers to be squared, negative, zero and positive
and every ordering
of the numbers counted as a different solution, choose a sequence.
Each sequence of numbers is reported once giving the numbers in order together
with the number of permutations of those numbers.
For example: 0 ±2 ±2 ±5 has 2×2×2 = 8 ways of including the signs.
The four numbers including the 0 and the two 2s can be permuted in 12 ways to make 12 different sequences of unsigned numbers:
So there are a total
of 8×12=96 signed number sequences of 4 numbers.
Here is a Calculator to help you investigate further:
Leave the numbers input box (the set/list length) empty for sets/lists of all lengths.
Sequences need a length which is the number of nonzero numbers.
When giving a list of numbers to square and add, use ~ to indicate a range, for example 2~5 means 2,3,4,5.
Some nice patterns in these sums of squares are the following:
3^{2} + 4^{2}
=
5^{2}
10^{2} + 11^{2} + 12^{2}
=
13^{2} + 14^{2}
21^{2}+ 22^{2}+ 23^{2}+ 24^{2}
=
25^{2}+ 26^{2}+ 27^{2}
36^{2}+ 37^{2}+ 38^{2}+ 39^{2}+ 40^{2}
=
41^{2}+ 42^{2}+ 43^{2}+ 44^{2}
...
On each line the total on each side is 25, 365, 2030, 7230, ... ,
n(n+1)(2n+1)(12n^{2}+12n+1)
6
, ... A059255
The left hand sides start with the squares of 3, 10, 21, 36, 55, ..., n(2n + 1), ... A014105
and end with 4, 12, 24, 40, 60, ..., 2n(n+1), ... A046092.
The n squares on the right hand sides start with the squares of 5, 13, 25, 41, 61, ..., 2n(n+1) + 1, ... A001844
and end with the squares of 5, 14, 27, 44, 65, ..., n(2n+3), ... A014106.
All numbers are a sum of up to 4 squares
Lagrange in 1770 proved that,
if we allow 0 as a square then every number is the sum of exactly four squares. This is the same as
saying every number has a representation as a sum of up to 4 nonzero squares.
There are many such forms for a given n usually:
1, 2 and 3 all have just 1 way of writing then as a sum of up to 4 squares, 4 has 2 ways, ... so the
counts start 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 2... A002635.
The first number with 3 such representations is
The smallest numbers with 1, 2, 3, 4, 5, 6, ... representations are 4, 18, 34, 50, 66, 82, ...A124978.
The numbers which only have a single representation as a sum of up to 4 squares:
1, 2, 3, 5, 6, 7, 8, 11, 14, 15, 23, ... A006431.
You Do The Maths...
Some special sums of squares
What is special about the numbers 14, 29, 50, 77, 110 if we look at them as a sum of squares?
Find a formula for these numbers.
14 = 1^{2} + 2^{2} + 3^{2};
29 = 2^{2} + 3^{2} + 4^{2};
50 = 3^{2} + 4^{2} + 5^{2};
They are all the sum of 3 consecutive squares.
(n1)^{2} + n^{2} + (n+1)^{2} = 2 + 3 n^{2} is one formula.
What about the series 30, 54, 86, 126, 174, 230, ...?
Find a formula and thus prove they must all be even.
30 = 1^{2} + 2^{2} + 3^{2} + 4^{2};
54 = 2^{2} + 3^{2} + 4^{2} + 5^{2};
86 = 3^{2} + 4^{2} + 5^{2} + 6^{2};
They are the sums of four consecutive squares.
(n1)^{2} + n^{2} + (n+1)^{2} + (n+2)^{2} = 4 n^{2} + 4 n + 6 = 2(2 n^{2} + 2 n + 3)
which is always even.
There is an interesting surprise if you find a recurrence relation for each of the above,
that is find a formula relating each number in a series to the three before it.
(optional) Can you generalise your result and prove it?
They all have the same recurrence relation:
if a,b,c and d are four consecutive numbers in any one of these series then
d = 3 c  3 b + a.
It applies to the sum of K consecutive squares for all K.
To prove this, find the formula for the sum of the K squares starting at n, let's call this S(n).
Then show that S(n) = 3 S(n1)  3S (n2) + S(n3).
We have not used the value of K anywhere in the proof so the recurrence relation
applies no matter value K has.
Numbers needing four squares
1 = 1^{2}; 2 = 1^{2} + 1^{2}; 3 = 1^{2} + 1^{2} + 1^{2};
4 = 2^{2}; 5 = 1^{2} + 2^{2}; 6 = 1^{2} + 1^{2} + 2^{2}
but we cannot write 7 as a sum of three or fewer squares  we need 4: 7=1^{2}+1^{2}+1^{2}+2^{2}).
15 is the next number not a sum of up to three squares.
Looking at the numbers up to 100, how does the series continue?
Since we know that all numbers are the sum of up to 4 squares then this series is all those numbers that need
four squares.
(Harder) Can you find the pattern/formula in the series? (Both Gauss and Legendre found the answer and proved it.)
Look at the remainders when the odd numbers are divided by 8.
There is a related pattern the evens in the list.
The numbers have a remainder of 7 when divided by 8 or are 4 times a number in the list.
As a formula the set is all numbers 4^{m}(8n + 7) where m and n are positive whole numbers or zero.
See A004125
Numbers with are a sum of four squares in just one way
Since every number is a sum of up to 4 nonzero squares, let's have a look at those with only one such sum:
1 = 1^{2}; 2 = 1^{2} + 1^{2}; 3 = 1^{2} + 1^{2} + 1^{2}
However 4 has two representations: 4 = 2^{2} = 1^{2} + 1^{2} + 1^{2} + 1^{2}.
The next with more than 1 sum of up to 4 squares are 9 and 10 but then 11 can has just one such form.
So our list of those numbers with just one way to write them as a sum of up to 4 squares
begins 1, 2, 3, 5, 6, 7, 8, 11.
How does it continue if we don't go beyond 100?
(Harder) Which are the only odd numbers in this list?
Find three separate patterns that cover all the evens.
The evens are all powers of 4 times another factor. What are those factors?
The only odd numbers are 1, 3, 5, 7, 11, 15 and 23.
The evens are 2 or 6 or 14 or a power of four times one of these.
In our sums of squares we have allowed any square to be repeated.
If we insist that each square in a sum appears only once, or in other words, that all the squares are distinct in any sum then
2 and 3 are impossible as a sum of unique squares as are 6, 7, 8, 11, 12 and 15, ... .
It might surprise you to know that this list is finite.
This means we can always write every number beyond a certain limit
as a sum of distinct squares.
What is this limit, the largest number in the list?
128.
See A001422 for the complete list of the 31 numbers and a reference to a proof.
Sums of consecutive squares
Find a formula for the sum of the n squares starting at 1.
If your formula was a product of terms in n, write it as a polynomial in n.
Why must the coefficients of the polynomial add to 1? Why is there no constant coefficient in the polynomial?
Use your answers to find the sum of the squares a^{2} up to b^{2} inclusive.
Expand your answer if necessary so that it only has terms of the form a^{p} or b^{q}.
1^{2} + 2^{2} + ... + n^{2} =
n (n + 1) ( 2n + 1 )
6
1^{2} + 2^{2} + ... + n^{2} =
n
+
n^{2}
+
n^{3}
6
2
3
If n=1 the sum is 1^{2} and the polynomial reduces to the sum of the coefficients which must therefore be 1.
If n=0 the sum is 0 and the polynomial is reduced to just its constant term which must therefore be 0.
a^{2} + ... + b^{2} =
b
+
b^{2}
+
b^{3}
−
a
+
a^{2}
−
a^{3}
6
2
3
6
2
3
Which numbers are not the sum of the squares of a set of 5 numbers? Note that
a set means no number is repeated in a set and sets can have just a single number.
In 1948 Sprague published a proof that there are only 31 such numbers:
2, 3, 6, 7, 8, 11, 12, 15, 18, 19, 22, 23, 24, 27, 28, 31, 32, 33, 43, 44, 47, 48, 60, 67, 72, 76, 92, 96, 108, 112, 128
See A001422
A jigsaw with square pieces
(n + 1)^{2} = n^{2} + 2n + 1 so we can fill a square of side n+1 with one square of side n and (2n + 1) squares of side 1.
Show this in a diagram.
We expand on this in the next section.
Towards a proof of the Four Squares theorem?
If we wanted to prove that every whole number is the sum of four squares (where 0 as a square is allowed), then
we could first prove Euler's Identity of 1749:
(a^{2} + b^{2} + c^{2} + d^{2})
(A^{2} + B^{2} + C^{2} + D^{2})
= (aA + bB + cC + dD)^{2} +
(aB − bA + cD − dC)^{2} + (aC − bD − cA + dB)^{2} +
(aD + bC − cB − dA)^{2}
Verify this identity.
How might it help in a proof that every integer is a sum of four squares?
See the Number Theory reference by Shanks below for more on this on pages 209, exercises 31S and 32S.
Rectangles dissected into square pieces
n
⁝
...
1
Geometry and jigsaw puzzles provide an illustration of algebraic identities.
For instance, in terms of sums of squares we have
(n + 1)^{2} = n^{2} + 1^{2} + ... + 1^{2}
but the diagram shows that the whole square has sides (n+1) and therefore area (n+1)^{2} but is made up of
a square of side n
n squares of side 1 on its right side
n squares of side 1 underneath
with one more square of side 1 in the bottom right
So (n + 1)^{2} = n^{2} + 2n + 1
As a list of squares for (n+1)^{2} we have n once and 1 repeated 2n+1 times.
Rectangles as sums of Squares  Continued Fractions
On the Introduction to Continued Fractions page we showed how to turn
any fraction p/q into a rectangle with sides p and q split into
squares. Here is the diagram for 16/45.
We can always arrange the squares in such diagrams to have one edge along the outside of the rectangle.
So we have seen that ALL rectangles can be split into squares since all fractions have a Continued Fraction form.
But the fractions are in their lowest forms so the Continued Fraction diagram method would be of no use for a square
because the ratio of its sides is just 1.
The closest we get to rectangle split into distinct squares is a Fibonacci rectangle of sides F(n) and F(n+1)
where F(n) is the nth Fibonacci number but then each has a repeated unit square!
The diagram on the right shows a 21x13 rectangle split into squares with sides 13,8,5,3,2,1 and 1.
Rectangles as sums of distinct Squares
In Martin Gardner's More Mathematical Puzzles and Diversions
chapter 17 on Squaring the Squre, contributed by William T Tutte, is all about the hunt for a square or rectangle
which can be dissected into different squares  called a perfect rectangle.
The search led to using electrical network ideas to find
such rectangles. The smallest number of distinct squares we can use is 9 and there are two rectangles with different arrangements
if we forget about their reflections and rotations.
One has an area of 1056 and the other 4209. There are six with 10 squares and 22 with 11.
The series and more information can be found in OEIS on A002839
where it is known up to rectangles with 24 squares.
x
B
y
A
99
78
21
57
77
43
16
41
34
9
25
Here is a rough sketch of how a rectangle might be divided into squares.
We start by labelling two squares x and y, the length of their sides.
We can then deduce that the top of side A is x + y and, since all pieces are supposed to be squares,
so is its height.
Therefore square B is a square of side (side of A) + y = x + 2y.
Continue in this way to deduce the expressions for the sides of each square piece and thus of the whole.
You will end up finding two expressions involving x and y for the same side.
Check that it is 9 x = 16 y.
Now we can let y = 9 and x = 16 to keep all sides as whole numbers
to find a simple solution for the jigsaw.
If we put these numbers back into the diagram we have the almostsquare rectangle 176 by177 shown here.
Is there a square that can be dissected into different squares? Yes, but the smallest seems to have
38 squares in it. TheYou Do The Maths... below look at rectangles with squaresonly dissections.
You Do The Maths...
(Easy) 165 = 6^{2} + 6^{2} + 5^{2} + 5^{2} + 4^{2} + 4^{2} + 3^{2} + 1^{2} + 1^{2}.
These nine squares can be made into a rectangular jigsaw. The area of the rectangle is 165.
Cut out the nine squares and and solve the jigsaw puzzle. What are the dimensions of the rectangle?
A rectangular jigsaw puzzle has exactly nine square pieces, each of a different size, and an area of 1056.
If the square pieces have sides 1, 4, 7, 8, 9, 10, 14, 15 and 18, what is the width and height of the rectangle
and how do the nine pieces fit into it?
According to Beiler's Recreations in the Theory of Numbers
(see Link and References below) this is the smallest rectangular jigsaw
where all the pieces are square and of different sizes.
Check your answer at Doug Williams' Australian Mathematics Task Centre
solution
or on this NRICH page.
More Mathematical Puzzles and Diversions M Gardner, (1966)
All of Martin Gardner's books are tremendous fun, well written by an "amateur" who is maybe the
best in this field at making maths entertaining to the general reader. His writings have started many people
on the road to becoming professional mathematicians
so beware !
Martin Gardner's Mathematical Games
is a CD of the text of all 15 of Martin Gardner's books based on his Mathematical Games sections in Scientific American:
Hexaflexagons and Other Mathematical Diversions
The Second Scientific American Book of Mathematical Puzzles and Diversions
New Mathematical Diversions
The Unexpected Hanging and Other Mathematical Diversions
Martin Gardners 6th Book of Mathematical Diversions from Scientific American
Mathematical Carnival
Mathematical Magic Show
Mathematical Circus
The Magic Numbers of Dr. Matrix
Wheels, Life, and Other Mathematical Amusements
Knotted Doughnuts and Other Mathematical Entertainers
Time Travel and Other Mathematical Bewilderments
Penrose Tiles to Trapdoor Ciphers
Fractal Music, Hypercards, and more Mathematical Recreations from Scientific American
The Last Recreations: Hydras, Eggs, and Other Mathematical Mystifications.
On the Partition of Numbers into Squares, D H Lehmer,
The American Mathematical Monthly vol 55, (1948), pages 476481 (JSTOR PDF)
Solved and Unsolved Problems in Number Theory
D Shanks (4th edition 2002)
A very readable and accessible book which develops numerous themes in Number Theory from scratch so it is suitable
for any student who wants to start exploring Number Theory.
How many ways can N be a sum of k squares?
There are some interesting formulas for the number of ways a number N can be written as a sum of k nonzero squares. It is
often denoted r_{k}(N).
Sums of two squares
If the sum of two squares is a square, then the three numbers form the sides of the right angled triangles we have looked at earlier on this page.
So now we ask which numbers, square or not, are the sum of two squares?
Fermat stated that any prime greater than 2 is the sum of two squares if and only if it has remainder of 1 when divided by 4
or, using the mod function, if and only iff a prime is congruent to (≡) 1 (mod 4).
We also have 2 = 1^{2} + 1^{2}.
The theorem was proved by Euler in 1749
The primes above 2 are all odd so their remainders when divided by 4 must be either 1 or 3.
primes = a^{2} + b^{2}
prime
3
5
7
11
13
17
19
23
29
31
37
41
43
47
mod 4
1
3
1
1
3
3
1
1
3
1
3
3
1
1
a

1


2
1


2

1
4


b

2


3
4


5

6
5


We can use the identity:
(a + b)^{2} × (A + B)^{2}
= (aA − bB)^{2} + (aB + bA)^{2}
= (aA + bB)^{2} + (aB − bA)^{2}
to show that there is a sumoftwosquares form for
all composite numbers whose prime factors are congruent to 1 mod 4 or, if congruent to 3 mod 4 then have a prime exponent that is even (it is a square).
These numbers are 2, 5, 8, 10, 13, 17, 18, 20, 25, 26, 29, 32, ... A000404.
If we allow 0 as one of the squares, or, what is the same thing, if we want numbers that are the sum
of up to two squares, then we can include 1 itself and 4 and 16, ....
These numbers are 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, 29, 32, ... A001481.
Sums of 3 squares
3, 6, 9, 11, 12, 14, 17, 18, 19, 21, 22, 24, 26, 27, 29, 30, 33, ... A000408
can all be written as a^{2} + b^{2} + c^{2} where a,b and c are not zero.
Sums of 4 squares
There are still some numbers that are neither squares themselves nor can be written as a sum of two or three nonzero squares:
7, 15, 23, 28, 31, 39, 47,... A004215
These need four squares and have the form 4^{A}(8B+7) so they all are either congruent to 7 mod 8 or have a factor which
is a power of 4 or both.
All numbers can be written as a sum of up to four squares!
Links and References
a book
an article, usually in an academic periodical
a link to a web page
Recreations in
the Theory of Numbers  The Queen of Mathematics Entertains
by A H Beiler, Dover, 1964,
was the first book that opened my eyes to the wonderful fun and facts about simple numbers.
There is a whole chapter on Pythagorean triangles: The Eternal Triangle. This book has been in print now for
many years and is a real classic, being both readable and full of interesting facts and tables and certainly accessible
to anyone with an interest in "recreational" mathematics and numbers. The subtitle of the book is
The Queen of Mathematics Entertains
which comes from a quote of Karl Frederich Gauss:
Mathematics is the queen of the sciences and arithmetic the queen of mathematics.
Highly recommended!
Mathematical Recreations
(second revised edition) by Maurice Kraitchik, Dover, 1953,
is another very enjoyable book that will appeal to
anyone who likes "playing with numbers". Apart from a chapter on
the classic numerical pastimes and number puzzles, it has others on Magic Squares, Chess board problems,
Permutations, Geometrical recreations and puzzles and a chapter on the Calendar. But I list it here because of
chapter 4 devoted completely to the Pythagorean Triangles called ArithmeticoGeometrical Questions. It has
the details of the algorithms used in the Calculators on this page.
This continues to be one of my favourite recreational mathematics books, of interest to anyone with
just a basic mathematical knowledge and a love of numbers.
It is available secondhand from as little as less than two US dollars at Amazon.com!
Mathematical Recreations and
Essays W W Rouse Ball, H S M Coxeter, Dover (13th edition 1987), paperback, 428 pages.
This is another
of the few great classics on mathematical recreations many of a geometrical nature. There is a fascinating chapter on
people with the most amazing ability to do arithmetic calculations in their heads. For us here, there is only a short section on
Pythagorean triangles, but, if you have found this web page of interest, I am sure that you
will find much to stimulate your own investigations in this book. It rarely uses mathematics
taught beyond age 16. It really is a book packed full of so many interesting and tantalising titbits of mathematics
that it makes you want to get out a pencil and paper and play with the numbers for yourself.
The Book of Numbers
by John Horton Conway and Richard K. Guy, Copernicus Books (1996), 311 pages, hardback.
This is nothing to do with a book of the Old Testament
as they quip in their introduction, but a collection of interesting mathematics
looking at our attempts to get to grips with the idea of number:
number words in many languages and the many different ways to write numbers as well as
numbers in mathematics. Much of it is at schoolmaths level but some of it goes beyond that
(imaginary, transcendental and infinite numbers). However, don't let that put you off as the book is full of diagrams
and pictures and explanations making it all readily accessible. The chapter on Further Fruitfulness of Fractions
shows how Pythagorean triangles were used by the Babylonians of 1500BC, well before the time of Pythagoras around 600BC
as discovered in a little clay tablet called Plimpton Tablet 322 (now in the Columbia University Library).
Number Theory and Its History
Oystein Ore, Dover (1988), 380 pages, paperback
is a great book if you want to look more seriously at the mathematics of primes and factors, congruences (the arithmetic
of remainders on division)  a topic called Number Theory  all in the context of their history by an excellent writer.
There is a section on the Plimpton Tablet 332 ,
a Babylonian list of Pythagorean triples and how it might have been used by the Babylonians.
The Penguin Dictionary of
Curious and Interesting Numbers David Wells, Penguin (Revised edition 1998), will help
answer some of the puzzle questions above but is a curious and interesting book in its own right! Take a number such as
3.14159.. . No doubt you will recognise it but what about 1634
or 364.2422? (Let your mouse rest on the numbers for the answers!)
And how many number facts do you know about
28? This book is full of wonderful facts about your favourite numbers.
A new algorithm for generating Pythagorean triples
R. H. Dye and R. W. D. Nickalls, The Mathematical Gazette (1998), volume 82, pages 8691.
Introduction to the Theory of numbers, G H Hardy and E M Wright,
Oxford University Press, (6th edition, paperback, 2008)
The m n formula and a proof are given as Theorem 225.
This is a classic book, revised and updated in the sixth edition, that is well worth studying but it
does tend to be at university undergraduate level some of the time.
Angling for Pythagorean Triples Dan Kalman
The College Mathematics Journal 17 (1986), pages 167168.
Height and Excess of Pythagorean Triples
(PDF file)
Darryl McCullough,Mathematics Magazine, (2005), vol 78, pages 2644.
This downloadable PDF file deals with some other aspects of Pythagorean triples apart from those in its title,
and in particular with the
Barning tree which generates all primitive Pythagorean triangles uniquely in a "tree".
Solution to Problem 1447 H Chi, R Killgrove in Crux Math vol 16, September 1990
This is the solution to the problem of finding PTs with an area = n × perimeter that
we examined in the The ratio of Area to Perimeter section on this page.
Online Encyclopedia of Integer Sequences
is Neil Sloane's excellent resource for both checking series of integers and also finding out more about each one. It is the
worldwide resource for such information and Neil welcomes any additional new series as well as more information on the
individual sequences.
All the primitive triads
for hypotenuse up to 10000. Michael Samos has produced this text file table together with
the perimeter and area of all the triangles (and other information on the angles too) if you want a complete list to print.