Let $p$ be a prime number such that $p \equiv 3 \pmod 4$. Show that $x^2 \equiv -1 \pmod p$ has no solutions.

Question

Let $p$ be a prime number such that $p \equiv 3 \pmod 4$. Show that $x^2 \equiv -1 \pmod p$ has no solutions.

I noticed that this is equivalent to proving $x^2\equiv 2(2k+1) \pmod p$. I also know that $x^2 \neq 2(2k+1)$. But I still can't prove it. Any help would be greatly appreciated.

Answer 1

If there is an $x$ such that $x^2=-1\pmod p$ then ${\rm ord}_p(x)=4$. Thus $4$ must divide the order of the multiplicative group modulo $p$ which is $p-1$.

Answer 2

Hint: What happens if you raise both sides to the power $\frac{p-1}2$ in the congruence $x^2\equiv-1$?