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Prove that if $m$ and $n\in\mathbb{N}$ are both the sums of two squares, and $n|m$, then $\frac{m}{n}$ is also the sum of squares.

I tried to consider prime divisors of m and n and distinct between $p\equiv 1 \pmod 4$ and $p\equiv3\pmod4$ but didn't get to any conclusion

Suzu Hirose
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Daniel Katzan
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  • See https://proofwiki.org/wiki/Brahmagupta-Fibonacci_Identity – lab bhattacharjee Nov 22 '16 at 15:08
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    @labbhattacharjee: that is the converse: if $m$ and $n$ are sums of two squares then $mn$ is also. I don't see an immediate way to use it to prove the OP's question. – TonyK Nov 22 '16 at 15:31
  • Every positive integer $n$ is a sum of squares. Trivially $n$ is the sum of $n$ squares since $1$ is a square. Less trivially, $n$ is the sum of $4$ squares. – bof Jul 12 '22 at 03:49
  • “m and n are both sum sum of squares…” sum of how many squares ? Because every number can be written as a sum of four squares. Do you mean sum of TWO squares? If so, I suggest you edit your question to make this explicitly stated. – insipidintegrator Jul 12 '22 at 03:54

2 Answers2

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Consider this: a positive integer $n$ is a sum of squares (meaning: of two squares of natural numbers) if and only if in its prime-power factorization, primes congruent to $3$ modulo $4$ appear with an even exponent.

Proof

I assume the hard direction is from left to right. So let $p \equiv 3 \pmod{4}$ be a prime, and suppose $p \mid a^{2} + b^{2} = (a + i b) ( a - i b)$. Since $p$ is still a prime in the Gaussian integers, $p$ divides one of the factors. But if $p^{e}$ is the highest power of $p$ that divides $a + i b$, then conjugating we see that $p^{e}$ also divides $a - i b$. It follows that the highest power of $p$ that divides $a + i b$ and $a - i b$ is the same, say $p^{e}$, so the highest power of $p$ that divides $a^{2} + b^{2}$ is $p^{2 e}$.

Andreas Caranti
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  • I have considered and tried to prove this claim, but got stuck, I can only conclude that the sum of all exponents of such primes is even – Daniel Katzan Nov 22 '16 at 15:09
  • I will edit the answer including the proof. – Andreas Caranti Nov 22 '16 at 15:10
  • is there a proof not involving complex numbers? – Daniel Katzan Nov 22 '16 at 15:54
  • Actually I consider right to left more difficult. Left to right can also be seen as a consequence of the fact that $-1$ is not a square modulo $p=4k+3$. @DanielKatzan yes, google "Fermat's Christmas Theorem". ProofWiki gives one. – Bart Michels Nov 22 '16 at 15:56
  • how can you can conclude that from the fact -1 is not a squarmod $p=4k+3$? I can only conclude what i said in my previous comment: sum of all exponents of such primes is even – Daniel Katzan Nov 23 '16 at 05:05
  • Suppose the prime $p \equiv 3 \pmod{4}$ divides $a^{2} + b^{2}$. If $p$ does not divide $a$, then it does not divide $b$, then $a$, say, has an inverse $c$ modulo $p$, so that $a^{2} + b^{2} \equiv 0 \pmod{p}$ implies $1 + (bc)^{2} \equiv 0 \pmod{p}$, and $bc$ is a square root of $-1$ modulo $p$, a contradiction. Then $p$ divides $a$ and $b$, so that $p^{2}$ divides $a^{2}$ and $b^{2}$. Divide $a^{2}$ and $b^{2}$ by $p^{2}$ and repeat. – Andreas Caranti Nov 23 '16 at 15:08
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Another proof: We know that in the prime factorization of a number which is a sum of two squares that $\prod p_i^k$ $k$ can only be odd if $p_i\equiv 1\pmod{4}$.

We prove by considering the contrapositive: Suppose that $n/m$ is not a sum of two squares. Then it has at least one prime factor $q$ with an odd exponent, but that means that either $n$ or $m$ would also have a prime factor $q$ with an odd exponent so either $n$ or $m$ cannot be a sum of two primes.

Donald Hosek
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