# On Pythagorean Triplets

The Problem: In the Pythagorean triplets (a,b,c) when a < b then b can't be a prime number.

The Background: While searching the properties of Pythagorean triplets in web I saw quite a few listed, but didn't see the above one which I thought was true, because I had developed a proof.

The Request: As discussed many a times in this site I would request some alternate proofs (or counterexamples) before I share mine for a review.

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$5^2 + 12^2 = 13^2$ –  Will Jagy Aug 7 '12 at 3:31
@Will, $12$ isn't prime, right? Or did Grothendieck say otherwise? –  Ｊ. Ｍ. Aug 7 '12 at 3:36
@J.M., good point, I misread. Let me think. –  Will Jagy Aug 7 '12 at 3:40

Hint $\rm\,\ a^2\! + p^2 = c^2\:\Rightarrow\: p^2 = (c\!-\!a)(c\!+\!a).\:$ Unique factorization $\:\Rightarrow \begin{eqnarray}\rm\:c\!-\!a &=&1\\ \rm c\!+\!a &=&\rm p^2\end{eqnarray}\:$ contra $\rm\,a<p$

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My answer was something along your line -only the end logic was different. Since we know that $a<b<c=a+1$ It shows that $a<b<a+1$ implying $b$,an integer, should lie between two consecutive integers which is a fallacy. –  Barun Dasgupta Aug 7 '12 at 17:29
@Barun Yes, that's the the proof I hinted at (I wanted to leave something for the reader). –  Bill Dubuque Aug 7 '12 at 18:00
The Pythagorean triples all have shape $k(x^2-y^2)$, $k(2xy)$, $k(x^2+y^2)$. Here $x$ and $y$ are positive integers such that $x \gt y$ (and $x$ and $y$ are relatively prime, and of opposite parity, though these side facts do not matter for the proof). If a leg is to be prime we need $k=1$.
Certainly $2xy$ cannot be prime. And $x^2-y^2=(x-y)(x+y)$ cannot be prime unless $x=y+1$. So from now on we may assume that $x=y+1$.
So $x^2-y^2=(y+1)^2-y^2=2y+1$. But $2xy=2(y-1)(y)=2y^2-2$. We cannot have $2y+1\gt 2y^2-2$. So the leg $2y+1$ must be the smallest. In particular, the larger of the two legs cannot be prime.