$-1$ is not a square in $\mathbb F_p$ if $p$ is congruent to $3$ modulo $4$

Question

Consider the finite field $\mathbb F_p:=\mathbb Z/p\mathbb Z$, where the prime integer $p$ is assumed congruent to $3$ modulo $4$. Below there is an argument to prove that $-1$ cannot be a square in $\mathbb F_p$: I know that proof verifies are not appreciated on MSE, but this is quite short and I'm new to number theory. I have doubts on this proof because the notes of my course prove this result very concisely in a few lines, using only the case $n=1$ below. But if you think that my proof works too I won't bother with the argument of the notes for the moment. Thanks in advance.

Let $u$ be a generator of the cyclic group $\mathbb F_p^\times$, so that if $-1$ is a square then $u^{2m}=-1$ for some $1\leq m\leq p-1$. Then $u^{4m}=1$, meaning that there is $1\leq n\leq 4$ such that $4m=n(p-1)$.

• $n=1$: so $4m=p-1$ and $p$ is congruent to $1$ modulo $4$, a contradiction.
• $n=2$: so $4m=2p-2$ and $2m=p-1$; hence $1=-1$ and $p=2$, a contradiction.
• $n=3$: so $4m=3p-3$, meaning that $3p$ is congruent to $3$ modulo $4$; but if $p$ is congruent to $3$ modulo $4$, then $3p$ is congruent to $1$ modulo $4$, a contradiction.
• $n=4$, so $m=p-1$ and $u=1$; hence $-1=u^2=1$ and $p=2$, a contradiction.

Answer 1

A shorter proof can sometimes be trickier to interpret, but more enlightening. If $p = 4k+3$ then $p-1 = 4k+2$. Squaring preserves parity so we only need to check the even cases. If we square an even number do we ever get $4k+2$?

Answer 2

Suppose $-1$ is a square, then $-1\equiv_p x^2$. Raising both sides, and noticing that $p=3+4k$ yields: $$-1\equiv_p (-1)^{1+2k}\equiv_p (-1)^\frac{p-1}{2}\equiv_p x^{p-1}$$ This contradicts Fermat’s Little Theorem.