# 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)

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 $$p \mid 2$$.

• The primes that are $\not\equiv 3\bmod 4$ are exactly the ones that are the sum of two squares. It sounds like you have this backwards. – Arthur Aug 20 at 10:43
• The title is not correct. Those prime numbers ARE the sum of two perfect squares. – Peter Aug 20 at 10:44
• When $p\equiv1\pmod 4$ your argument gives $1\equiv1\pmod p$. – Angina Seng Aug 20 at 10:44
• The more difficult part is to show that primes of the form $\ 4k+1\$ are the sum of two perfect squares. That it is impossible for $\ 4k+3\$ , is almost trivial. And the case $\ p=2\$ is immediate ($\ 2=1^2+1^2\$) – Peter Aug 20 at 10:47
• "is a" is an implication which is only one-directional. Compared with your draft, you edited the title towards the wrong direction. You are trying to prove that any prime $p\equiv3\pmod4$ can't be the sum of two squares. – Wolfgang Kais Aug 20 at 11:19

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 $$1 \equiv -1 \pmod{p}$$.
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.
• A more advanced approach is possible with the theory of quadratic residues : The main idea is to show that $-1$ is not a quadratic residue modulo a prime of the form $4k+3$. Hence , $a^2+b^2 =p$ cannot work for coprime integers $a,b$ , but neither if they are not coprime since then the sum of the two perfect squares cannot be squarefree, in particular not prime. – Peter Aug 20 at 10:55
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 $$p\equiv 3\mod 4$$