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

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!

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Try factoring: $(a+\sqrt{2}b)(a-\sqrt{2} b)\equiv 0 \mod p$ $\implies$ $a \equiv \sqrt{2} b \mod p$. If $a$ is an integer $\mod p$, then $\sqrt{2} b$ is an integer $\mod p$. Thus, can we conclude that $b$ is a square $\mod p$? – zzzzzzzzzzz Jun 7 '12 at 16:27
I don't think one can simply work modulo $p$. Notice that $1^2-2\cdot 3^2\equiv 0 \bmod 17$, but $3$ is not a square mod $17$. – Álvaro Lozano-Robledo Jun 7 '12 at 16:33
Can someone explain to me the fault my logic? $a=1$, $b=3$ doesn't satisfy $p \equiv 1 \mod 8$ as a counterexample, but I still feel unsure about my idea. – zzzzzzzzzzz Jun 7 '12 at 16:55
@Matt Do you want to work with negative primes, too? $$a^2-2b^2=1^2-2\cdot 3^2=-17$$ – DonAntonio Jun 7 '12 at 16:59
@Matt Your logic is unclear. What is "an integer mod $p$"? And why is being "an integer mod $p$" imply being a square mod $p$? Also, when you say $a\equiv \sqrt{2}b\pmod p$, what exactly do you mean? (I am not saying that there is no meaning, but from your comment, it's not clear that you understand how to make sense of it.) – Alex B. Jun 7 '12 at 16:59

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 $p$.
You need Alvaro's original condition $p \equiv 1 \bmod 8$, not just $p \equiv 1 \bmod 4$, to handle the case $q = 2$. – KCd Nov 28 '14 at 16:26