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I'm asking for verification of the following proof:

We start with the following equality, working entirely in $\mathbb{Z}$:

$$a(m^2+n^2)=b(p^2+q^2)$$

With $x=m^2+n^2,y=p^2+q^2; a,b$ coprime and $x,y$ coprime.

Claim: Under these constraints, both $a$ and $b$ must be sums of two squares.

Proof: Temporarily assume the claim is false, and assume that $a$ is not the sum of two squares. Therefore $a$ has some non-squared prime factor of the form $4k-1$, so that the prime factorization of $ax$ looks like this:

$$ax=(4k_1+1)^{e_1}(4k_2+1)^{e_2}(4k_2+1)^{e_2}\cdots (4k_n-1)$$

There may be other factors of the form $4k_i-1$, but they are extraneous to the proof; further, we can represent any factor of the form $(4k_i-1)^{2n+1}$ as $(4m+1)^n(4k_i-1)$ and move the first portion to the first portion of the product (though it is no longer prime), leaving us with a bare $4k_n-1$.

Because $ax=by$, the prime factorization of $ax$ is the same as that of $by$. Assuming there are sufficient factors to rearrange, we can rearrange these factors into $b$ and $y$. Since $y=p^2+q^2$ is the sum of two squares, the factor $4k_n-1$ cannot be in the factorization of $y$, and must be in the factorization of $b$. But if this is the case, then $a$ and $b$ share a factor and are not coprime, contradicting the initial constraints. Therefore, our assumption was incorrect, and $a$ must be the sum of two squares. By symmetry, $b$ is also the sum of two squares. $\blacksquare$

Have I made any mistakes, major or minor?

Bill Dubuque
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Eric Snyder
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  • Follows immediately by applying unique fractionization to $,\dfrac{a}b = \dfrac{y}x,,$ (assuming $a$ or $b > 0)\ \ $ – Bill Dubuque Mar 22 '23 at 02:28
  • For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be a proof checking machine. – Bill Dubuque Mar 22 '23 at 02:29
  • Thanks for the pointer to the duplicate, @BillDubuque. Seeing that makes me think I've actually completely misdevised the actual thing to be proven--the assumed coprimality may actually be wrong. – Eric Snyder Mar 22 '23 at 23:27
  • That thought crossed my mind too. What is the "actual thing to be proven"? Are you familiar with the Gaussian integers (being Euclidean so a UFD)? – Bill Dubuque Mar 22 '23 at 23:51
  • I am semi-familiar with them and how they relate to the sum of two squares. What I've been noodling around with is the magic square of squares problem, along with trying to help answer a different question along similar lines (can't find it right now, essentially about parametrizing the form as part of $a(m^2+n^2)=b(p^2+q^2)=2x^2$). I have a hunch that $a$ and $b$ must also be SOS, but there's a kink in the math where $m,n,p,q$ aren't necessarily integers in the parametrization. – Eric Snyder Mar 23 '23 at 05:55

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