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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?

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    $\begingroup$ The fact that such a prime $p$ must satisfy $$p\equiv 1\pmod{4}$$ may be helpful. $\endgroup$ Commented Jun 10, 2016 at 0:44
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    $\begingroup$ @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? $\endgroup$ Commented Jun 10, 2016 at 0:47
  • $\begingroup$ look... $25$ appears in two Pythagorean triple $(15,20,25)$ and $(7,24,25)$ but $29$ appear in just one triple... $\endgroup$ Commented Jun 10, 2016 at 0:50
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    $\begingroup$ 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. $\endgroup$ Commented Jun 10, 2016 at 0:54
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    $\begingroup$ 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 $s\pm ti$. $\endgroup$ Commented Jun 10, 2016 at 1:18

3 Answers 3

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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.

enter image description here

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  • $\begingroup$ This jstor.org/stable/40391253 ? $\endgroup$
    – lhf
    Commented Jun 10, 2016 at 1:00
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    $\begingroup$ Here is a better scan, from JSTOR: i.sstatic.net/DDymG.gif $\endgroup$
    – lhf
    Commented Jun 10, 2016 at 1:02
  • $\begingroup$ @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. $\endgroup$
    – Will Jagy
    Commented Jun 10, 2016 at 1:04
  • $\begingroup$ 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 $p$... $\endgroup$ Commented Jun 10, 2016 at 1:06
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    $\begingroup$ @WillJagy oh... yes... if it's prime... it's primitive... ;) ☺ $\endgroup$ Commented Jun 10, 2016 at 1:09
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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 $\mathbb Z[i]$.

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  • $\begingroup$ very interesting! $\endgroup$
    – Vincent
    Commented Jun 14, 2016 at 18:40
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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.

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