More fun in 2015, Part 27

We first observe that $a^{2014}\equiv 1 \pmod{2015}$ iff $a^2\equiv {1} \pmod {2015}$ since $\phi(2015)=1440$ and $\gcd(1440, 2014)=2$ .

Next, $a^2\equiv {1} \pmod {2015}$ iff $a^2\equiv {1}$ modulo 5, 13, and 31 since $2015=5*13*31$ . Now each of those congruences has the solutions $a=\pm 1$ , so that the congruence $a^2\equiv {1} \pmod {2015}$ has $2^3=8$ solutions.

This is equivalent to $a^{2014}\equiv 1 \pmod{k}$ for $k=5,13,31$ . This is necessary since $2015=5\times13\times31$ . We can further reduce this to $a^{10} \equiv 1 \pmod{13}$ , $a^{2}\equiv 1 \pmod {5}$ , $a^4 \equiv 1 \pmod{31}$ .

Let $a=g^i$ for some generator $g$ . We get $12|10i, 4|2i, 30|4i$ since the generator can only give one when raised to some power divisible by $p-1$ . ('|' means divides). For each case, we get two values of $i$ less than $p-1$ : 0, and $\frac{p-1}{2}$ . Hence, we get two solutions for $a$ for each case.

Combining these solutions, we get $2^3=8$ total.