More fun in 2015, Part 25

We claim that $x^{60}\equiv{1}\pmod{2015}$ iff $gcd(x,2015)=1$ . It follows that the congruency has $\phi(2015)=\boxed{1440}$ solutions.

If $gcd(x,2015)=1$ , then $x^{60}\equiv 1$ modulo 5, 13, and 31, by Fermat's little theorem. Thus $x^{60}\equiv 1 \pmod{2015}$ since $2015=5*13*31$ . (Alternatively, use the Carmichael Lambda $\lambda(2015)=60.)$

If $x^{60}\equiv{1}\pmod{2015}$ , then $x^{60}+2015m=1$ for some $m$ , so that $gcd(x,2015)=1$ .