Show divisibility by 7

HINT:

If $n\equiv0,\pm1,\pm2,\pm3;n^2\equiv0,1,4,2\pmod7$

So, $a^2,b^2\equiv1,2,4$

Check for $c^2\pmod7$ when $a^2\not\equiv b^2\pmod7$

But my greater concern is how the problem, specifically $\pmod7$ was conceived!

Using Euclid's formula, $a=2mn, b=m^2-n^2$

We have $7\nmid2mn(m^2-n^2)$

Now, $(m^2-n^2)^2-(2mn)^2=m^4+n^4-6m^2n^2\equiv m^4+n^4+m^2n^2\pmod7$

But $(m^2-n^2)(m^4+n^4+m^2n^2)=(m^2)^3-(n^2)^3\equiv1-1\pmod7$ using Fermat's Little Theorem as $(m,7)=(n,7)=1$

$\implies7|(m^4+n^4+m^2n^2)$ as $7\nmid(m^2-n^2)$

Can you take it from here?