Can Odd Primes Solve x^2 ≡ 2 (mod p)?

[ElDavidas]

Homework Statement

Let $p$ be an odd prime. Show that $x^2 \equiv 2 (mod p)$ has a solution if and only if $p \equiv 1 (mod 8)$ or
$p \equiv -1 (mod 8)$

The Attempt at a Solution

Ok, I figured the more of these I try, the better I'll get at them. Assuming that
$x^2 \equiv 2 (mod p)$ has a solution first. I get

$$1 = ( \frac{2}{p} )= -1^{\frac{p^2 - 1}{8}}$$

So $$1 = -1^{\frac{p^2 - 1}{8}}$$.

This implies that $\frac{p^2 - 1}{8}$ must be even. So

$\frac{p^2 - 1}{8} = 2k$ for an integer $k$.

I'm not sure if I'm on the right track though. Any hints would be appreciated.

 

[mighty2000]

Thank you for your post. You are on the right track with your attempt at a solution. Let me guide you through the rest of the proof.

To show that x^2 ≡ 2 (mod p) has a solution, we need to find an integer x that satisfies this congruence. One way to do this is by using the quadratic reciprocity theorem, which states that for an odd prime p,

( \frac{2}{p} )= (-1)^{\frac{p^2 - 1}{8}}

This means that if p ≡ 1 (mod 8) or p ≡ -1 (mod 8), then the Legendre symbol (2/p) will be equal to 1, and therefore there exists an integer x such that x^2 ≡ 2 (mod p).

Now, let's assume that p ≡ 3 (mod 8) or p ≡ 5 (mod 8). In this case, the Legendre symbol (2/p) will be equal to -1, which means that there is no integer x that satisfies x^2 ≡ 2 (mod p). Therefore, p must be either 1 or -1 (mod 8) for the congruence to have a solution.

To summarize, if p ≡ 1 (mod 8) or p ≡ -1 (mod 8), then x^2 ≡ 2 (mod p) has a solution. If p ≡ 3 (mod 8) or p ≡ 5 (mod 8), then the congruence has no solution. I hope this helps. Keep up the good work!