If $\gcd(7,abc)=1$ and $a^2+b^2=c^2$, prove that $7$ divides $a^2-b^2$ The only information I have on this problem is that for $a^2+b^2=c^2$ that
$$
a = st, b = \frac{s^2-t^2}{2}, c = \frac{s^2+t^2}{2}
$$
and that $\gcd(7,abc)=1$ gives $7x + abcy = 1$
I have no idea how to proceed, so any help welcomed

HINT:
$$x\equiv\pm1,\pm2,\pm3\implies x^2\equiv1,4,2\pmod7$$
So, if $a^2\equiv1,b^2$ must be $\equiv1\pmod7$ and so on
