Show that $x^{2} \equiv 1 \pmod{15}$ has four incongruent solutions mod 15. Show that $x^{2} \equiv 1 \pmod{15}$ has four incongruent solutions mod 15.
My instinct was to find the primitive root and then use a theorem to directly show the number of incongruent solutions, which follows from knowing the primitive root. But, apparently there are no primitive solutions to $mod 15.$ So what's another option?

$x^2\equiv 1 \pmod 3$ and $x^2\equiv 1 \pmod 5$. Now, $x\equiv 1 \text{ or } 2 \pmod 3$. There are $2$ solutions. Also $x\equiv 1 \text{ or } 4 \pmod 5$. There are $2$ solutions. By Chinese remainder theorem: $x^2\equiv 1 \pmod {15}$ has $2\cdot 2=4$ incongruent solutions.
