We know that $a^2 \equiv \frac{1}{2} \pmod{p}$. Now use Thue's lemma (see lemma 1 here for a really simple short proof: http://www.math.uga.edu/~pete/thuelemma.pdf), we find $x$, $y$ such that $0 < |x|, |y| < \sqrt{p}$ and $ax \equiv y \pmod{p}$. This gives us $2y^2 - x^2 \equiv 0 \pmod{p}$, $-p < 2y^2 - x^2 < 2p$ and $x,y$ are nonzero. So it must be $ p=2x^2-y^2 $ as requested.

hajimbrak wrote: Let $p$ be a prime such that $p\mid (2a^2-1)$ for some integer $a$.Show that there exist integers $b,c$ such that $p=2b^2-c^2$. Here.

djmati11 wrote: We know that $a^2 \equiv \frac{1}{2} \pmod{p}$. Can we divide when we are dealing in mod??

well I have a proof using unique factorization theorem on $\mathbb{Z}[\sqrt 2]$.