Roots united and enumerated

Every root of this equation is a root of $x^n - 1 = 0$ for some $1 \le n \le 10$ , and so is a primitive $n$ th root of unity for some $1 \le n \le 10$ . There are $\phi(n)$ primitive $n$ th roots of unity and so there are $\sum_{n=1}^{10}\phi(n) = \boxed{32}$ roots to the original equation.

For positive integer $n$ , $x^n-1=0$ $\implies x^n = e^{2\pi i}$ $\implies x = e^{\frac {2k\pi}ni} = e^{\frac {360^\circ k}ni}$ , where $k=0, 1, 2 ... n-1$ , a total of $n$ complex roots. Then the number of distinct roots of $\prod_{n=1}^{10} (x^n-1)=0$ is the sum of distinct roots of all factors $x^n-1=0$ for $1 \le n \le 10$ .

• $x=1$ add a distinct root: $+1$
• $x^2=1$ , $\implies x = \pm 1$ add $-1$ as an distinct root: $+1$
• $x^3=1$ , add two distinct roots $e^{\pm \frac {2\pi}3i}$ : $+2$
• $x^4=1$ , add two distinct roots $e^{\pm \frac {\pi}2i}$ : $+2$
• $x^5=1$ , add four distinct roots $e^{\pm \frac {2\pi}5i}, e^{\pm \frac {4\pi}5i}$ : $+4$
• $x^6=1$ , add two distinct roots $e^{\pm \frac \pi3i}$ : $+2$
• $x^7=1$ , add six distinct roots $e^{\pm \frac {2\pi}7i}, e^{\pm \frac {4\pi}7i}, e^{\pm \frac {6\pi}7i}$ : $+6$
• $x^8=1$ , add four distinct roots $e^{\pm \frac \pi 4i}, e^{\pm \frac {3\pi}4i}$ : $+4$
• $x^9=1$ , add six distinct roots $e^{\pm \frac {2\pi}9i}, e^{\pm \frac {4\pi}9i}, e^{\pm \frac {8\pi}9i}$ : $+6$
• $x^{10}=1$ , add four distinct roots $e^{\pm \frac \pi 5i}, e^{\pm \frac {3\pi}5i}$ : $+4$

Therefore, the number of distinct roots of $\prod_{n=1}^{10} (x^n-1)=0$ is $=1+1+2+2+4+2+6+4+6+4 = \boxed{32}$ .