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Prove for an integer $n = a^2+b^2$ with prime factorization $n=2^{e}p_{1}p_{2}...p_rM^2$ (where $p_1, p_2, ...p_r$ are distinct odd primes and e = 0 or 1), then $p_i=1 mod 4$ for all i. My first thought is to use quadratic reciprocity but other than that i cannot think of anything else. Any hints?

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  • $\begingroup$ This might be helpful. en.wikipedia.org/wiki/… $\endgroup$
    – dust05
    Commented Jun 9, 2020 at 5:58
  • $\begingroup$ @dust05 thx! But here n is not a prime though. Any further hints? $\endgroup$
    – popp321
    Commented Jun 9, 2020 at 6:00
  • $\begingroup$ Maybe here math.uga.edu/~pete/4400twosquares.pdf. Pag. 4-5 $\endgroup$ Commented Jun 9, 2020 at 6:07
  • $\begingroup$ The main point is that $p_is$ can be written as a sum of two squares. Moreover stuff like $(a^2+b^2)(c^2+d^2)$ can also be written as sum of two squares $\endgroup$ Commented Jun 9, 2020 at 6:10
  • $\begingroup$ Remember of course that for $e=1$, then $2=1^2+1^2$ $\endgroup$ Commented Jun 9, 2020 at 6:27

2 Answers 2

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You have $n\equiv 0\mod p_i$. So, $a^2+b^2\equiv 0\mod p_i$.

If $b$ is invertible mod $p_i$: This implies that $-1=(ab^{-1})^2\mod p_i$, i.e. $-1$ is quadratic residue mod $p_i$. This only happens when $p_i\equiv 1\mod 4$.

If $b$ is divisible by $p_i$: Then $a$ is also divisible by $p_i$. So, $(a')^2+(b')^2=n'$, where $a'=a/p_i$, $b'=b/p_i$ and $n'=n/p_i^2$. Note that $n'\equiv 0\mod p_i$, so you can repeat the same argument.

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If you know some modern algebra, one can see this via:

Lemma. If $p \mid a^2 + b^2$ and $p \equiv 3 \pmod4$ then $p \mid a, b$ in $\mathbb{Z}[i]$.

Proof. Since $p \mid a^2 + b^2 = (a+bi)(a-bi)$ and $p$ is prime in $\mathbb{Z}[i]$. (If $p$ is not prime then $p = \epsilon (A+Bi)(A-Bi)$ form for $A, B \in \mathbb{Z}$ and $\epsilon = \pm1, \pm i$, which leads us to $p = A^2 + B^2$ ) So $p \mid a+bi$ or $p \mid (a-bi)$, and either cases give the conclusion.

With the lemma one have $ p^2 \mid a^2 + b^2$ for $p$ prime factor of sum of two squares and of the form $4k+3$, which was what we wanted.

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