$\lg_{2} \left( \prod\limits_{a=1}^{2015} \prod\limits_{b=1}^{2015} (1 + e^{\frac{2\pi iab}{2015}}) \right)$?

Question

How can this problem be solved?

$$ \lg_{2} \left( \prod\limits_{a=1}^{2015} \prod\limits_{b=1}^{2015} (1 + e^{\frac{2\pi iab}{2015}}) \right) $$

Answer 1

You can do it as follows: $$ \begin{align*}\prod_{a=1}^{2015}\prod_{b=1}^{2015}\left(1+e^{\frac{2\pi i a b}{2015}}\right)=&\prod_{d\mid 2015}\prod_{\substack{1≤a≤2015\\ \space (2015,a)=d}}\prod_{b=1}^{2015}\left(1+e^{\frac{2\pi i a b}{2015}}\right)\\=& \prod_{d\mid 2015}\prod_{\substack{1≤a≤2015\\ \space (2015,a)=d}}2^d\\=&\prod_{d\mid 2015}2^{d\varphi\left(\frac{2015}{d}\right)} \end{align*}$$ Thus, the logarithm evaluates at: $$ \lg_2\left(\prod_{d\mid 2015}2^{d\varphi\left(\frac{2015}{d}\right)}\right)=\sum_{d\mid 2015}d\varphi\left(\frac{2015}{d}\right)=13725 $$ This can be calculated by hand. Furthermore, as r9m pointed out, this sum is multiplicative. By observing that $$ \sum_{d\mid p^k}d\varphi\left(\frac{p^k}{d}\right)=p^{k-1}((k+1)p-k) $$ we obtain $$ \sum_{d\mid 2015}d\varphi\left(\frac{2015}{d}\right)=(2\cdot5-1)(2\cdot13-1)(2\cdot 31 -1)=13725 $$

Explanation:

If $(2015,a)=d$, we can see that $$ \prod_{b=1}^{2015}\left(1+e^{\frac{2\pi i a b}{2015}}\right)=\left[\prod_{b=1}^{\frac{2015}{d}}\left(1+\exp\left(\frac{2\pi i b}{\frac{2015}{d}}\right)\right)\right]^d $$ By factoring the polynomial $$x^n-1=\prod_{b=1}^{n}\left(x-e^{\frac{2\pi i b}{n}}\right)$$we can see that $$(-1)^n-1=\prod_{b=1}^{n}\left(-1-e^{\frac{2\pi i b}{n}}\right)=(-1)^n\prod_{b=1}^{n}\left(1+e^{\frac{2\pi i b}{n}}\right)$$and thus $$\prod_{b=1}^{n}\left(1+e^{\frac{2\pi i b}{n}}\right)=1-(-1)^n$$Since every divisor of $2015$ is odd, we obtain $$ \left[\prod_{b=1}^{\frac{2015}{d}}\left(1+\exp\left(\frac{2\pi i b}{\frac{2015}{d}}\right)\right)\right]^d=2^d $$ Leading to the result described above.