More fun in 2016, Part 8

We factorize $2016$ into irreducible/prime elements over the Gaussian integers as follows: $2016 \; = \; (1 + i)^5 \times (1 - i)^5 \times 3^2 \times 7 \;.$ Since $1+i$ and $1-i$ are associate, the distinct factors of $2016$ are of the form $a + bi \; = \; z \times (1 + i)^u \times 3^v \times 7^w$ where $z \in \{1,i,-1,-i\}$ is a unit, $0 \le u \le 10$ , $0 \le v \le 2$ and $0 \le w \le 1$ . It is only possible for both $a$ and $b$ to be nonzero when $u$ is odd (since $(1+i)^{2j}$ is either real or purely imaginary). For each odd value of $u$ , and any choice of $v$ and $w$ , there is a single choice of unit $z$ which results in both $a$ and $b$ being positive. Thus there are $5 \times (2+1) \times (1+1) = 30$ factors of $2016$ of the desired form.