TITLE: Does $a^n \mid b^n$ imply $a\mid b$?
QUESTION [10 upvotes]: Does $a^n \mid b^n$ imply $a\mid b$? I think it does but haven't been able to prove it. 
I don't know much number theory so an elementary answer would be great.

REPLY [7 votes]: If you can assume the fundamental theorem of arithmetic (that each integer has a unique factorization in prime numbers), you can write:
$$
\begin{align*}
a &= p_1^{e_1} p_2^{e_2} \ldots p_r^{e_r} \\
b &= q_1^{d_1} q_2^{d_2} \ldots q_s^{d_s}
\end{align*}
$$
Here the $p_i$, $q_i$ are primes, and $e_i$ and $d_i$ are all greater than 0.
If $a^n \mid b^n$, then $p_i^{n e_i}$ must have a counterpart in a $q_j^{n d_j}$, in that $p_i = q_j$ and $n e_i \le n d_j$, so it must then also be that $e_i \le d_j$; and this means $a \mid b$.

REPLY [4 votes]: Hint: $p$ is a prime factor of $k$ if and only if $p^n$ is a factor of $k^n$. This holds for any prime $p$, integer $k$, and positive integer $n$.