# Why can no prime number appear as the length of a hypotenuse in more than one Pythagorean triangle?

Why is it that no prime number can appear as the length of a hypotenuse in more than one Pythagorean triangle? In other words, could any of you give me a algebraic proof for the following?

Given prime number $p$, and Pythagorean triples $(a,b,p)$ and $(c,d,p)$ where $a<b<p$ and $c<d<p$, then $b=d$.

Please also have a look at the deeper question: Is there any formula to calculate the number of different Pythagorean triangle with a hypotenuse length $n$, using its prime decomposition?

• The fact that such a prime $p$ must satisfy $$p\equiv 1\pmod{4}$$ may be helpful. Commented Jun 10, 2016 at 0:44
• @ZubinMukerjee no... since we already know that... a prime must be of the form of $p=4k+1$ to appear as a hypotenuse... but the question is why this number appear just in one Pythagorean triple? Commented Jun 10, 2016 at 0:47
• look... $25$ appears in two Pythagorean triple $(15,20,25)$ and $(7,24,25)$ but $29$ appear in just one triple... Commented Jun 10, 2016 at 0:50
• An elementary detailed proof is rather lengthy. Please see the uniqueness part of this proof. The proof is more pleasant and more informative if we can use properties of Gaussian integers. Commented Jun 10, 2016 at 0:54
• We give a brief version of the Gaussian integer approach. By the usual representation theorem, the odd prime $p$ is a hypotenuse iff $p=s^2+t^2=(s+ti)(s-ti)$. The two factors are Gaussian primes, so by unique factorization $(u+vi)(u-vi)=p$ only if $u+iv$ is a unit times $s\pm ti$. Commented Jun 10, 2016 at 1:18

This goes back to Euler, who showed that if there are two ways of writing an odd integer $N$ as the sum of two squares, then $N$ is composite. There is a 2009 article on this by Brillhart. Let me try to find a link.

http://www.maa.org/press/periodicals/american-mathematical-monthly/american-mathematical-monthly-december-2009

And if one note that in a primitive triple the hypotenuse is of the form $(u^2+v^2)$, and the legs are of the form $(u^2-v^2)$ and $(2uv)$. So by euler if the hypotenuse is prime it couldn't be written in different ways.

• – lhf
Commented Jun 10, 2016 at 1:00
• Here is a better scan, from JSTOR: i.sstatic.net/DDymG.gif
– lhf
Commented Jun 10, 2016 at 1:02
• @lhf good. I made a jpeg of the relevant page and pasted that in. He also has a 2016 article that allows indefinite quadratic forms, although in that case more care is needed about what is meant by distinct representations. Commented Jun 10, 2016 at 1:04
• when a number is hypotenuse... its square is going to written as sum of squares... so it is of course composite then... it is $p^2$ not $p$... Commented Jun 10, 2016 at 1:06
• @WillJagy oh... yes... if it's prime... it's primitive... ;) ☺ Commented Jun 10, 2016 at 1:09

As noted in the comments and the accepted answer, this comes down to the fact that if a prime $p$ can be written as a sum of two squares, then the representation is unique up to switching and or negating the factors. A fancier explanation for this is the fact that $\mathbb Z[i]$ is a principal ideal domain and its unit group is $\{\pm1,\pm i\}$. (Of course, proving that $\mathbb Z[i]$ is a PID requires some sort of argument like that in the scanned note, but this is a more modern way to think about it.) Once one knows it's a PID, then suppose that $p=u^2+v^2$. Then $p=(u+iv)(u-iv)$, and the fact that $u+iv$ and $u-iv$ have norm $p$ shows that they cannot factor further in $\mathbb Z[i]$. Hence they are irreducible (i.e., they generate prime ideals). So the unique factorization of the ideal $p\mathbb Z[i]$ is as the product of the prime ideals $(u+iv)\mathbb Z[i]$ and $(u-iv)\mathbb Z[i]$. So $u$ and $v$ are unique, up to switching them or replacing them by their negatives, which corresponds to multiplying $u+iv$ by each of the four units in $\mathbb Z[i]$.

• very interesting! Commented Jun 14, 2016 at 18:40

If a Pythagorean triple is primitive, $$B+C$$ is a perfect square as shown by

$$2mn+(m^2+n^2)\quad=\quad m^2+2mn+nn^2\quad=\quad(m+n)^2$$

If $$C$$ is prime, then only one smaller value $$(B)$$ can add to it to make a perfect square. Given $$C\&B$$, there can be only a one $$A$$ to make a Pythagorean triple.