The Year with so Many Things.

We first claim that if $n$ is odd, then $\prod_{b=1}^{n} (1+e^{2\pi i ab/n}) = 2^{\gcd(a,n)}$ . To see this, write $d = \gcd(a,n)$ and $a = da_1$ , $n=dn_1$ with $\gcd(a_1,n_1) = 1$ .

Then $a_1, 2a_1,\dots,n_1 a_1$ modulo $n_1$ is a permutation of $1,2,\dots,n_1$ modulo $n_1$ , and so $\omega^{a_1},\omega^{2a_1},\dots,\omega^{n_1 a_1}$ is a permutation of $\omega,\omega^2,\ldots,\omega^{n_1}$ ;

it follows that for $\omega = e^{2\pi i/n_1}$ , we have:

$\prod_{b=1}^{n_1} (1+e^{2\pi i a b/n}) = \prod_{b=1}^{n_1} (1+e^{2\pi i a_1 b/n_1}) = \prod_{b=1}^{n_1} (1+\omega^b).$

Now since the roots of $z^{n_1}-1$ are $\omega,\omega^2,\ldots,\omega^{n_1}$ ,

it follows that: $z^{n_1}-1 = \prod_{b=1}^{n_1} (z-\omega^b)$ . Setting $z=-1$ and using the fact that $n_1$ is odd gives $\prod_{b=1}^{n_1} (1+\omega^b) = 2$ .

Finally, $\prod_{b=1}^{n} (1+e^{2\pi i ab/n}) = (\prod_{b=1}^{n_1} (1+e^{2\pi i ab/n}))^d = 2^d$ , and we have proven the claim.

From the claim we find that:

$\begin{aligned} &\log_2 \left( \prod_{a=1}^{2019} \prod_{b=1}^{2019} (1+e^{2\pi i a b/2019}) \right) \\ &= \sum_{a=1}^{2019} \log_2 \left(\prod_{b=1}^{2019} (1+e^{2\pi i a b/2019}) \right) \\ &= \sum_{a=1}^{2019} \gcd(a,2019). \end{aligned}$

Now for each divisor $d$ of $2019$ , there are $\phi(2019/d)$ integers between $1$ and $2019$ inclusive whose $\gcd$ with $2019$ is $d$ . Thus $\sum_{a=1}^{2019} \gcd(a,2019) = \sum_{d|2019} d\cdot \phi(2019/d).$ We factor $2019 = pq$ with $p=3$ and $q= 673$ , and calculate:

\(\begin{align*} &\sum_{d|pq} d\cdot \phi(pq/d) \\ &= 1 \cdot (p-1)(q-1) + p \cdot (q-1) + q\cdot (p-1) + pq \cdot \\

&\quad = (2p-1)(2q-1)= (2\times 3-1)(2\times 673 -1)=6725. \end{align*}\)