Number Theory In Complex Number Set

Relevant wiki: Diophantine Equations - Solve by Factoring

Notice that by equating the real and imaginary parts we get, $a^4 + b^4 - 6a^2 b^2 = n$ and $4a^3b - 4ab^3 = 4ab(a + b)(a - b) = 2016$ .

So, we need integer pairs $(a, b)$ such that $ab(a+b)(a-b) = 504$ .

Notice that if $(a, b)$ is a solution, then so are $(-a, -b),(b, -a),(-b, a)$ .

So, we’ll assume W.L.O.G that $a > b > 0$ .

Let $d = a-b$ . Then $db(d+b)(d+ 2b) = 504$ .

Notice that the highest power of three dividing $504$ is $2$ . So clearly, exactly one of the four factors must be divisible by $9$ .

Observe that $db(d+b)(d+ 2b) = 504 < 546 = 1 \cdot6\cdot(1 + 6)\cdot(1 + 12)$ . Thus $b \le 5$ .

Similarly, $db(d + b)(d + 2b) = 504 = 7 \cdot1\cdot(7 + 1)\cdot(7 + 2)$ . Thus $d \le 7$ .

It is evident that neither $d$ nor $b$ can be a multiple of $9$ . So, either $d + b = 9$ or $d + 2b = 9$ (because $d + 2b \le 7 + 2 \cdot 5 = 17 < 18$ ).

So, the possible pairs are $(7, 2),(5, 4),(4, 5)$ for $(d, b)$ when $d + b = 9$ . But $5\nmid 504$ and so we reject the last two pairs.

It is seen that for $(d, b) = (7, 2), d - b = 5$ , but clearly $5 \nmid 504$ . So, no solutions for the case $d + b = 9$ .

Moving on, for $d + 2b = 9$ , we have the following possible pairs $(7, 1),(5, 2),(1, 4)$ .

Since $5 \nmid 504$ , we reject the middle pair, while $d + b = 5$ for $(d, b) = (1, 4)$ and since $5 \nmid 504$ , we reject the last pair.

It is clear that the first pair $(7, 1)$ satisfies the equation.

This gives us $(a, b) = (8, 1)$ and so in summary the possible solutions are $(a, b) = (8, 1),(-8, -1),(-1, 8),(1, -8)$ .

Hence the only possible value of $n$ is $8^4 + 1^4 - 6(8\cdot 1)^2 = \boxed{3713}$ .