When I am learning Number Theory, I saw this question:
"If $p$ is a prime number, then $$2x^2+1\equiv 0\mod p$$ has solution iff $p\equiv 1 \pmod8$ or $p\equiv 3\pmod8$."
Where should I start?
---After first edit---
I have considered when $p=5,7,11,13,17,19,23$ and found out the equation actually have solutions when $p=11,17,19$ which fulfill the condition.
I simplified the equation to $$x^2\equiv \frac{p-1}{2} \pmod p,$$ I am familiar with equation like $x^2\equiv \text{constant}\pmod p$ but this time the remainder will change at the time $p$ change.
I have started finding the Legendre symbol $(\frac{p-1}{2}/p)$ but do not know how to continue.
---After second edit---
I have tried letting $p=8k+1$ which satisfy $p\equiv 1\pmod8$ then substitute in Legendre symbol $$(\frac{p-1}{2}/p)=(4k/8k+1)=(4/8k+1)(k/8k+1)=(k/8k+1)$$ After that I still haven't got any further calculation...
We want $\big(\frac{-1 \cdot 2^{-1}}{p}\big) = \big(\frac{-1 \cdot 2}{p}\big) = 1$ which is the case iff $\big(\frac{-1}{p}\big) = \big(\frac{2}{p}\big)$ since the Legendre symbol is multiplicative. Now use the First and Second Nebensatz.
edit: this is only true for odd primes; there is no solution for p=2.