Number Theory: Describe all $x\in \mathbb{Z}$ so that $x^m \equiv 0\pmod{n}$

Question

I've been working on this homework problem but I'm not sure how to come up with the answer:

Let the prime factorization of $n\in \mathbb{N}$ with $n>1$ be $n=p_1^{a(1)}p_2^{a(2)}\cdots p_k^{a(k)}$. Describe all $x\in \mathbb{Z}$ so that for some $m>0$, $x^m\equiv 0\pmod{n}$.

What I've come up with so far is:

$x^m\equiv 0\pmod{n}\implies n\mid x^m\implies p_1^{a(1)}p_2^{a(2)}\cdots p_k^{a(k)}\mid x^m$,

But I'm unsure where to go from here.

Thanks!

Answer 1

If there exists an $m$ such that $p_1^{a(1)}p_2^{a(2)}\cdots p_k^{a(k)}\mid x^m$, then that means that $p_1p_2\cdots p_k\mid x$. The reason is this: If $p_i^{a(i)}\mid x^m$, then $p_i \mid x^m$, which in turn implies $p_i \mid x$, by Euclid's lemma. Since this is true for all the primes $p_i$, we have that $\operatorname{lcm}(p_1, p_2,\ldots, p_k)$ divides $x$, and we're done.