Fermat's theorem on sums of two squares (every prime $p$ s.t. $p \not\equiv 3 \pmod 4$ is a sum of two squares)

Question

I'm reflecting the following proof (see below). My question is where it uses the given fact ($p \not\equiv 3 \pmod 4$)? I'm not sure it uses this fact, and it kind of makes me think that something is wrong. Would appriciate your help.

Draft of a possible partial proof. Let $p = 3 \pmod 4$ be a prime number. Assume that $p = a^2 + b^2$. Then $a^2 + b^2 = 0 \pmod p$, implying that $a^2 = -b^2 \pmod p$. By raising both sides in $(p-1)/2$, then using Fermat's little theorem we saw in problem set 6, we conclude that $p \mid 2$.

Answer 1

I assume there is a typo in the question. If $p \equiv 1 \pmod{4}$, $(p-1)/2$ is a even number so you would get $1 \equiv 1 \pmod{p}$ which is not a contradiction. Only when $(p-1)/2$ is odd, you would get $ 1 \equiv -1 \pmod{p}$.

Answer 2

Hint : Every perfect square is congruent to $\ 0\ $ or $\ 1\ $ modulo $\ 4\ $. This can easily be shown by cases. And from this it easily follows that a prime of the form $\ 4k+3\ $ cannot be the sum of two perfect squares.

Answer 3

The question is where are you using the fact that $p\equiv 3\mod 4$. Answer: you are using the fact that $(p-1)/2$ is odd and so

$$(-b^2)^{\frac{p-1}{2}}=-1\mod p.$$

That is only true if $p\equiv 3\mod 4$