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This seems very simple, but I have not been able to come up with an answer to it.

Suppose $a+bi \mid n$ in $\mathbb{Z}[i]$, where $a,b \in \mathbb{Z}$ with $(a,b)=1$. Prove that $a^2 + b^2 \mid n$ in $\mathbb{Z}$.

My attempt was: $a+bi \mid n \implies \overline{a+bi}\mid n$ (otherwise $n$ is not an integer), and if we have that $(a+bi,a-bi)=1$ (or a unit) in $\mathbb{Z}[i]$, then by unique factorization this implies that $(a+bi)(a-bi)\mid n$. However, I found it difficult to prove $(a,b)=1$ in $\mathbb{Z} \implies$ $(a+bi,a-bi)=1$ in $\mathbb{Z}[i]$.

Does anyone have a nice solution to this? Any solution is welcome.

For additional context, this question came up as I was trying to prove that $\mathbb{Z}[i]/(a-ib) \cong \mathbb{Z}/(a^{2}+b^{2})\mathbb{Z}$ when $a,b$ are coprime in the integers. A way of doing this using natural injections is given in Quotient ring of Gaussian integers, where there are many very good answers. One way is to use the first isomorphism theorem with the canonical map $\mathbb Z \to \mathbb Z[i]/(a - i b)$, and I need to prove that the kernel of that map is $\langle a^2+b^2 \rangle$. In every answer I saw on this, the solution skips over that step, making me suspect that there's something very simple I'm missing.

metamorphy
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Albert
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  • Well, taking norms immediately shows that $a^2+b^2\mid n^2$. (I'm not sure about the stronger statement $a^2+b^2\mid n$.) You can also think about using the Euclidean algorithm on $(a+bi,a-bi)$ perhaps. – YiFan Tey Jan 16 '21 at 12:40
  • @YiFan Thanks for the comment, yes, I noticed that as well. But I was uncertain how to proceed from there. But the fact that the isomorphism ("additional context" part) holds is proof that indeed $a^2+b^2 | n$, so I think it holds. I was trying to prove the above in order to prove the isomorphism, so I still don't have a proof. – Albert Jan 16 '21 at 12:43
  • @metamorphy Thank you for the hint. I tried something like that where $d$ was a prime in $\mathbb{Z}[i]$, but I didn't get anywhere, and I tried it in a different approach. Could you be a bit more specific with your hint? (I'm rather very slow with hints....) – Albert Jan 16 '21 at 12:46
  • Hint: rationalize the denominator of $,n/(a+bi).,$ See our answer in the linked dupe for details. – Bill Dubuque Jan 16 '21 at 13:21

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Some $c,\,d\in\Bbb Z$ satisfy $(a+bi)(c+di)=n$. For this to be real, $c+di=k(a-bi)$ for some $k\in\Bbb Q$, whence $k(a^2+b^2)=n$. Now we just need $k\in\Bbb Z$, which follows from $(a,\,b)=1$.

J.G.
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  • Thank you for your answer! Could you help me understand why $(a,b)=1$ implies $k\in \mathbb{Z}$? – Albert Jan 16 '21 at 12:49
  • @Albert $k=c/a=d/b$, so $bc=ad$. Since $a$ is coprime to $b$ yet divides $bc$, $a|c$, so $k\in\Bbb Z$. – J.G. Jan 16 '21 at 13:01