Congruence relation

Question

I did not have any number theory course, so can someone help me solve the following congruence relation for odd prime $p$ $$(-7)^{\frac{p-1}{2}} \equiv 1 \mod{p}$$ I want to find prime $p$ such that this is true. I am not sure what operations are allowed but I thought I could do this: $$(-7)^{\frac{p+1}{2}} \equiv -7 $$ where we can see that $\frac{p+1}{2}$ is an even power since primes are and so this does not have a solution. I am not sure if it is okay.

Answer 1

By Euler's criterion, $$(-1)^{(p-1)/2}\equiv\left(\frac{-7}p\right)\pmod p$$ where $\left(\frac{-7}p\right)$ is the Legendre symbol. By quadratic reciprocity $$\left(\frac{-7}p\right)=\left(\frac p7\right)$$ and so your identity holds iff $p$ is a square modulo $7$ (as long as $p\ne7$).