0
$\begingroup$

First, I want to declare that this problem have been solved by myself, I just want to share my work to everyone.

It has known that a prime has the form $4k+3$ cannot be sum of two squares, and also a composite number of the form $n=N^2m$, where $m$ is a square-free integer, admits a expression of sum of two squares when $m$ didn't have a prime factor of the form $4k+3$.

But then, I found that if two distinct prime $p,q$ both have the form $4k+3$, namely $p=4m+3,q=4n+3$, then their product $pq$ will be of the form $4k+1$, I believe it will have the expression of sum of two squares, but my textbook seems like disagree with it, even if it didn't actually wrote it down.

Therefore, I will prove that if $p,q$ are distinct primes have the form $4k+3$, then their product cannot be expressed as sum of two squares.

$\endgroup$
  • $\begingroup$ The correct criterion is : A positive integer can be expressed as a sum of at most two squares, if and only if every exponent in the prime decomposition belonging to a prime of the form $4k+3$ is even. $\endgroup$ – Peter Apr 28 '18 at 7:55
0
$\begingroup$

Let $p=4m+3,q=4n+3,pq=16mn+12m+12n+9=4k+1$, and assume that $pq=a^2+b^2$ for some positive integers $a,b$, so

$$a^2+b^2\equiv 0 \pmod {pq}$$

If $p|a$ will leads to $p|b$, by letting $a=pr,b=ps$ will leads to $p^2(r^2+s^2)=pq$ which says $p|q$, contradiction. Thus, $p\nmid a$, same as $b$.

but this indicates that

$$\begin{cases}a^2+b^2\equiv0\pmod p\quad(1)\\\\a^2+b^2\equiv0 \pmod q\quad(2)\end{cases}$$

For equation $(1)$, because $\gcd(a,p)=1$, so there exist an integer $a_1$ which $$a_1a\equiv1\pmod p$$ By multiple $a_1^2$ on the equation $(1)$, we get $$(ba_1)^2\equiv-1 \pmod p$$ which can't occur because $p\equiv3 \pmod4$. Same as the equation $(2)$.

Therefore, $pq$ can't be sum of two squares.

This completes the proof.

$\endgroup$
  • 1
    $\begingroup$ Your idea is basically correct, but exactly what justifies the assertion "which can't occur because $p\equiv3$ (mod $4$)" if all you know is that such a $p$ cannot be written as the sum of two squares? $\endgroup$ – Barry Cipra Apr 27 '18 at 13:05
  • $\begingroup$ @BarryCipra I thought about it, but $a^2+b^2$ is equal to $pq$, not $p$, so I think the assertion is required because it didn't mean that $a^2+b^2=p$. $\endgroup$ – kelvin hong 方 Apr 27 '18 at 13:15
  • 1
    $\begingroup$ Yes, the assertion is required. What I'm asking you is what justifies it. That is, how do you know that $-1$ is not a quadratic residue mod $p$ if $p\equiv3$ mod $4$? $\endgroup$ – Barry Cipra Apr 27 '18 at 13:24
  • $\begingroup$ Should I need to add details to it? Because at this stage, one must know that $-1$ is not a quadratic residue of prime of the form $4k+3$. $\endgroup$ – kelvin hong 方 Apr 28 '18 at 2:52

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.