Let $x, y, z$ be a primitive pythagorean triple. Show that exactly two of $x^2, y^2, z^2$ are congruent mod 7

Question

So, I know that if x,y, and z are a pythagorean triple, then

$x = m^2-n^2$, $y = 2mn$, $z = m^2+n^2$

How can I show that exactly two of $x^2, y^2, z^2$ are congruent mod 7?

Answer 1

We have $x^2+y^2=z^2$.

What are the square elements modulo $7$? They are $0,1,4,2$.

So, there are $\binom4 2+4\ =10$ possibilities of (nonnecessarily disctint) pairs $x^2,y^2$ modulo $7$, check at each one, that their sum is a square element or not, and verify the claim.

(E.g. $z^2$ cannot be $0$, as neither $6,3,5$ are squares mod $7$, so, if, say $x\equiv 0\pmod7$, then $y^2\equiv z^2$. And so on..)