$-1$ isn't a quadratic residue $\pmod{p}, p=4k+3$.

Question

I remember coming across this fact a while ago in a pdf somewhere, but I haven't been able to find it again. Can someone show me how to prove it? I would appreciate easier proofs.

EDIT: I'm very sorry, I forgot to actually state it.

Prove that $-1$ is not a quadratic residue modulo primes of the form $4k+3$.

Thanks!

Answer 1

$x^2=-1$ holds if and only if $x$ has order $4$. An element of order $4$ exists in $(\mathbb Z/p\mathbb Z)^* \cong C_{p-1}$ if and only if $4|p-1$.

Answer 2

Hint $\ \overbrace{{\rm mod}\ P=4K\!+\!3\!:}^{\large\ \ \ \ P\ -\ 1\,\ =\,\ 2(\color{#c00}{2K+1})}\,\ \ X^{\large 2}\equiv -1\!\!\!\!\!\!\overset{\ \ \ \ \ \ \ (\ \ )^{\Large\color{#c00}{2K+1}}}\Longrightarrow\! X^{\large P\,-\,1}\equiv -1\ $ contra little Fermat