Let $p$ be an odd prime with $(\frac{2}{p})=1$. Then, there exists $c$ with $$0<c<\frac{p}{2}$$ and $$c^2\equiv 2\mod p$$
We can conclude that for every $b$ with $0<b<\frac{p}{2}$ there exists $a$ uniquely with $0<a<\frac{p}{2}$ and $$a^2\equiv 2b^2\mod p$$
Can I use the pigeon-hole principle to show that at least one of the pairs $(a/b)$ satisfies $$|a^2-2b^2|<2p$$ which would imply $a^2-2b^2=\pm p\ $ (which would be sufficient to show that $p$ can be represented as $a^2-2b^2$) ?
If not, can this approach be used to prove that $a^2-2b^2=p$ with positive integers $a,b$ has a solution in another way ?
No, you cannot because the squares are much larger than twice the size of the argument.