Cyclotomic explosion

The $n^{th}$ cyclotomic polynomial $\Phi_{n} (x)$ has the primitive $n^{th}$ roots of unity as its roots.

Thus, for each $n^{th}$ cyclotomic polynomial, there are $\phi(n)$ distinct roots, where the function $\phi (x)$ is the Euler's totient function . (Note that $\Phi_n (x) \neq \phi ( x)$ ! )

We are looking for the number of roots of the polynomial whose roots are all the primitive $k^{th}$ roots of unity such that $1 \leq k \leq n$ . Since the primitive $k^{th}$ roots of unity are relatively coprime for all integers $k$ such that $1 \leq k \leq n$ , we see that

$\prod\limits_{k=1}^{n} \Phi_{k} (x)$

will have all its roots distinct from one another. Thus, we can find its number of roots as

$\sum\limits_{k=1}^n \phi(k)$

Being the lazy person that I am, I generated a code for that.

However, mathematically and manually, the summation above can be simplified by the summatory function

$= \sum\limits_{m=1}^n m \sum\limits_{d|m} \frac{ \mu (d) }{d}$

where $\mu(d)$ is the Möbius function .

Using such equation above, we can evaluate it to $n=2017$ and arrive at the answer. Either way, that is $\boxed {1237456}$ .