Mathematical Reflections 2016, Issue 1 - Problem S361

Problem:
Find all integers $n$ for which there are integers $a$ and $b$ such that $(a+bi)^4=n+2016i$.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
We have $(a+bi)^4=(a^4-6a^2b^2+b^4)+4ab(a^2-b^2)i$, so it must be
$$\begin{array}{rcl} a^4-6a^2b^2+b^4&=&n \\ ab(a^2-b^2)&=&504. \end{array}$$
If $(a,b)$ is a solution to this system of equations, then also $(-a,-b)$, $(b,-a)$ and $(-b,a)$ are solutions, so we can assume $a^2-b^2>0$ and so $ab>0$. If $a^2-b^2 \equiv 0 \pmod{4}$, then both $a$ and $b$ are even and $ab(a^2-b^2)$ is divisible by $16$, contradiction.  Moreover, since $(a^2-b^2) \ | \ 504$ and $a^2-b^2 \not \equiv 2 \pmod{4}$ for all integers $a,b$, then $(a^2-b^2) \in \{1,3,7,9,21,63\}$. If $a^2-b^2=1$, then $a=\pm 1$ and $b=0$, which gives $ab=0$, contradiction. If $a^2-b^2=3$, then $a=\pm 2$ and $b=\pm 1$, which gives $ab=2$, contradiction. If $a^2-b^2=7$, then $a=\pm 4$ and $b=\pm 3$, which gives $ab=12$, contradiction. If $a^2-b^2=9$, then $a=\pm 5$ and $b=\pm 4$, which gives $ab=20$, contradiction. If $a^2-b^2=21$, then $a=\pm 11$ and $b=\pm 10$ or $a=\pm 5$ and $b=\pm 2$, which gives $ab \in \{110,10\}$, contradiction. If $a^2-b^2=63$, then $a=\pm 32$ and $b=\pm 31$ or $a=\pm 12$ and $b=\pm 9$ or $a=\pm 8$ and $b=\pm 1$, which gives $ab \in \{992,108,8\}$. We conclude that $a^2-b^2=63$ and $ab=8$, so $$a^4-6a^2b^2+b^4=(a^2-b^2)^2-4(ab)^2=3713,$$ which yields $n=3713$.