A Little Complex Stuff

Let $F(x)=\prod_{1\le a,b\le 2015} (x-\omega^{ab})$ where $\omega$ is a primitive $2015$th root of unity. We want to find $\log_2\left(-F(-1)\right)$. Then we can write (after checking some symmetry thing) $F(x)=\prod_{d\mid 2015} \Phi_d(x)^{h(d)}$ for some $h$ mapping from the divisors of $2015$ to $\mathbb{Z}_{\ge 0}$ . But $\Phi_{d}(-1)=1$ unless $d=1\ 0$ for $d\mid 2015$ , so it suffices to find $h(1)$ . This is the number of pairs $(a,b)$ with $2015\mid ab$ , which is easily computed to be $(2\cdot 5-1)(2\cdot 13-1)(2\cdot 31-1)=13725$ . Thus our answer is $\log_2\left(-(-2)^{13725}\right)=13725$