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

How can this problem be solved?

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

## closed as off-topic by Servaes, Michael Albanese, user91500, Najib Idrissi, 6005Dec 6 '15 at 17:22

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• By solved you mean evaluated? Let me guess... this is another Putnam problem? – Masacroso Dec 6 '15 at 9:51

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 $$\sum_{d\mid 2015}d\varphi\left(\frac{2015}{d}\right)=(2\cdot5-1)(2\cdot13-1)(2\cdot 31 -1)=13725$$
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.
• @r9m Well, nothing intelligent, but as $2015=5*13*31$ has only $8$ divisors, it takes not to long to calculate it using $\varphi(n)=n\cdot\prod_{p\mid n}\left(1-\frac{1}{p}\right)$. And thanks for the edit! :) – Redundant Aunt Dec 6 '15 at 11:53
• $\sum\limits_{d | n} d\varphi (n/d)$ is convolution of two multiplicative functions, hence it's multiplicative as well. :) – r9m Dec 6 '15 at 12:07