1
$\begingroup$

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 $p \mid 2$.

$\endgroup$
  • 3
    $\begingroup$ 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. $\endgroup$ – Arthur Aug 20 at 10:43
  • 2
    $\begingroup$ The title is not correct. Those prime numbers ARE the sum of two perfect squares. $\endgroup$ – Peter Aug 20 at 10:44
  • $\begingroup$ When $p\equiv1\pmod 4$ your argument gives $1\equiv1\pmod p$. $\endgroup$ – Angina Seng Aug 20 at 10:44
  • $\begingroup$ 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\ $) $\endgroup$ – Peter Aug 20 at 10:47
  • 1
    $\begingroup$ "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. $\endgroup$ – Wolfgang Kais Aug 20 at 11:19
1
$\begingroup$

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}$.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Peter Aug 20 at 10:55
2
$\begingroup$

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 $p\equiv 3\mod 4$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.