More fun in 2015, Part 42

We can decompose $2015$ into non-associate irreducible elements of $\mathbb{Z}[i]$ as follows: $2015 \; = \; (2 + i)(2 - i)(3 + 2i)(3 - 2i)31$ recalling that prime/irreducible elements of $\mathbb{Z}[i]$ are Gaussian integers of the form $a+ib$ where $a^2 + b^2 = 2$ or $a^2 + b^2 = p$ is prime with $p \equiv 1$ modulo $4$ , or else are prime integers $p$ which are congruent to $3$ modulo $4$ . Since the Gaussian integers contain the $4$ units $1,i,-1,-i$ , and since $2015$ is a product of $5$ non-associate prime factors, we deduce that $2015$ has a total of $4 \times 2^5 = 128$ factors in $\mathbb{Z}[i]$ .