Primes of the form $p=a^2-2b^2$.

Question

I've stumbled upon this and I was wondering if anyone here could come up with a simple proof:

Let $p$ be a prime such that $p\equiv 1 \bmod 8$, and let $a,b\geq 1$ such that $$p=a^2-2b^2.$$ Question: Is $b$ necessarily a square modulo $p$?

I have plenty of numerical data to support an affirmative answer, but the proof eludes me so far. For instance: \begin{align*} 17 & = 5^2 - 2\cdot 2^2\\ &= 7^2 - 2\cdot 4^2\\ & = 23^2 - 2\cdot 16^2\\ & = 37^2 - 2\cdot 26^2\\ & = 133^2 - 2\cdot 94^2\\ \end{align*} and $2\equiv 36$, $4$, $16$, $26\equiv 9$, $94\equiv 9 \bmod 17$ are squares.

Thanks!

Answer 1

Since $p\equiv 1\pmod 8$, $2$ is a square modulo $p$. It will therefore be enough to show that any odd prime divisor of $b$ is also a square modulo $p$. Then any prime divisor of $b$ will be a square modulo $p$, therefore $b$ itself will be.

Let $q$ be an odd prime divisor of $b$, and consider your equation modulo $q$. You find that $p\equiv a^2 \pmod{q}$, so that $p$ is a square modulo $q$. By quadratic reciprocity (using that $p\equiv 1\pmod 4$), $q$ is a square modulo $p$.