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> Find all primes $p$ and $q$ such that $p^2-2q^2=1.$
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My idea so far was to observe that since $2q^2$ is even, then $q^2$ must be odd or even. If $q^2$ is even, then $q$ is even and the only even prime is $2.$ Thus one pair of primes $(p,q)=(3,2).$

But, if $q^2$ is odd, then $q$ is also odd and $q=2d+1$ for some integer $d.$ Then,

$p^2-2q^2=1$ turns into $p^2=4(2d^2+2d)+3.$

From here I do not know how to proceed.

Any ideas or suggestions would be greatly appreciated.