Nonexistence of Pythagorean triple with largest side prime. 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.
 A: We use "Euclid's" formula for generating the Pythagorean triples. The job can be done much more elegantly, with much less machinery. Please see Bill Dubuque's answer. 
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.
A: 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 $\rm\,a<p$
A: If $k,x,y\in\Bbb Z^+$ and $A=k(x^2-y^2), B=2kxy, C=k(x^2+y^2)$  and $A$ is prime, then, since $A=k(x-y)(x+y)$ is prime with $x+y>1$, we must have  $x-y=1.$ So $$\frac B A=\frac {2xy}{x^2-y^2}=\frac {2(y+1)y}{(y+1)^2-y^2}=$$ $$=\frac {2y^2+2y}{2y+1}>\frac {2y^2+y}{2y+1}=y\ge 1.$$
