More fun in 2015, Part 45

the formula is $\Phi_n(1)=e^{\Lambda(n)}$ where $\Lambda(n)$ is the von mangoldt function .

consider the basic $x^n-1=\prod_{d\mid n} \Phi_d(x)$ divide by $\Phi_1(x)=x-1$ . $x^{n-1}+x^{n-2}+...+x+1=\prod_{d\mid n,d>1} \Phi_d(x)$ put x=1 and take log $\ln(n)=\sum_{d|n,d>1}\ln(\Phi_d(1))$ let $f(n)=\begin{cases} 0&&,n≤1\\\ln(\Phi_d(1))&&,n≥2\end{cases}$ .and $p(n)=\ln(n)$ .then in dirichlet convolution form: $p=f*u$ . convolute both sides with $\mu$ to get: $p*\mu=f*I=f$ . in other words: $\ln(\Phi_n(1))=\sum_{d|n} \ln(d)\mu\left(\dfrac{n}{d}\right)$ it is well known $\ln(n)=\sum_{d|n} \Lambda(n)$ this is easy to prove, try it yourself. in dirichlet convolution form this is: $p=\Lambda*u$ . convolute bothsides with $\mu$ to get: $p*\mu=\Lambda*I=\Lambda$ . in other words $\sum_{d|n} \ln(d)\mu\left(\dfrac{n}{d}\right)=\Lambda(n)$ . we get that $\ln(\Phi_n(1))=\Lambda(n)$ or $\Phi_n(1)=e^{\Lambda(n)}$ . we are done $_\square$ .

Aareyan has written a fine solution. For the sake of variety, let me show another approach.

Note that $\Phi_p(1)=p$ for prime $p$ . Now we show by complete induction on $n$ that $\Phi_n(1)=1$ for square-free $n=p_1*...*p_m$ with $m>1$ . We write $x^n-1=\prod_{d|n}\Phi_d(x)$ , divide by $x-1$ , and evaluate at $x=1$ to see that $n=p_1*...*p_m*\Phi_n(1)$ so $\Phi_n(1)=1$ and $P=\Phi_{2015}(1)=\boxed{1}$ .

This approach easily generalizes to arbitrary $n$ .

$P = \Phi_{2015}(1)$

Thank you Mobius for your wonderful formula.

We have $\begin{aligned} \Phi_{2015}(X) & = \prod_{d | 2015}(X^{\frac{2015}{d}} - 1)^{\mu(d)} \\ & = \frac{(X^{2015} - 1)(X^{5} - 1)(X^{13} -1)(X^{31} - 1)}{(X^{65} - 1)(X^{155} - 1)(X^{403} - 1)(X-1)} \\ & = \frac{\sum_{k = 0}^{4}{X^{403k}} \cdot \sum_{k = 0}^{4}{X^k}}{\sum_{k = 0}^{4}{X^{13k}} \cdot \sum_{k = 0}{X^{31k}}} \\ \Phi_{2015}(1) & = \frac{5 \cdot 5}{5 \cdot 5} = \boxed{1} \end{aligned}$

A short solution could be:

$\Phi_{2015}(x)= \frac{\Phi_{403}(x^5)}{\Phi_{403}(x)}$

then: $\Phi_{2015}(1)= \frac{\Phi_{403}(1^5)}{\Phi_{403}(1)} = 1$