Problem: Find the number of solutions to the congruence equation $x^3 \equiv 1 \pmod{280}$.
Attempt: Since $280 = 2^3 \cdot 5 \cdot 7$, we know $x^3 \equiv 1 \pmod{280} \iff x^3 \equiv 1 \pmod{2^3}, x^3 \equiv 1 \pmod{5}$ and $x^3 \equiv 1 \pmod{7}$. I think we can say something about the number of solutions to each of the congruence equations using the fact that there exist primitive roots for $2^3, 5$ and $7$ respectively. However, I am not sure how the argument would go exactly... any help is appreciated!
As you observed, $$x^3\equiv 1\pmod{280}\implies x^3\equiv 1\pmod k, k=5,7,8$$
We find by trial that
$x^3\equiv 1\pmod 8$ has $4$ solutions
$x^3\equiv 1\pmod 5$ has $1$ solution
$x^3\equiv 1\pmod 7$ has $3$ solutions
By the Chinese remainder theorem, each combination of these solutions gives rise to a unique solution modulo 280, so there are $4\cdot1\cdot3=12$ solutions.