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My attempt : Define a map $f$ from $\Bbb{Z}$ to $\Bbb{Z}[i]/\langle a+ib\rangle$ by $f(n) = n+\langle a+ib\rangle$.

Then I have shown that $f$ is ring homomorphism and kernel is $\langle a^2+b^2\rangle$.

For subjectivity let p+iq $\in \mathbb{Z}[i]$. Now a+ib $\in \langle a+ib \rangle$ implies ib+$\langle a+ib \rangle$ = -a+$\langle a+ib \rangle$. Similarly i(a+ib) $\in \langle a+ib \rangle$ implies ia+$\langle a+ib \rangle$ = b+ $\langle a+ib \rangle$.

Now a and b are relatively prime implies there exist integers u, v such that au+bv=1.

Now p+iq +$\langle a+ib \rangle$ = p+iq(au+bv) +$\langle a+ib \rangle$ =p+iqau+iqbv+$\langle a+ib \rangle$ =p+bqu-aqv+$\langle a+ib \rangle$ =$f(p+bqu-aqv) $. Hence f is surjective.

Is proof of subjectivity correct?

Pradip
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    Can you show that if $0\le m<n<a^2+b^2$ then $f(m)\ne f(n)$? – Gerry Myerson Jun 09 '19 at 06:30
  • One way of proving surjectivity is to verify that that quotient ring of $\Bbb{Z}[i]$ has exactly $a^2+b^2$ elements. This has been studied on our site also. See here or here. – Jyrki Lahtonen Jun 09 '19 at 07:13
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    +1 for showing your work. I needed to vote to close as a duplicate as this is a recurring exercise (because it is a nice exercise! See the visual answer in one of the linked threads.). I have a related dupehammer privilege so my vote took effect immediately. – Jyrki Lahtonen Jun 09 '19 at 07:14

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