If $a$ is a perfect $n$'th power $\bmod 4a^2$ then $a$ is a perfect $n$'th power.

Question

This problem has me very confused because I think that it can be trivially generalized to the following.

Let $m$ be an integer such that $v_p(m)>v_p(a)$ for all $p|a$, then if $a$ is a perfect $n$'th power $\bmod m$ then $a$ is a perfect $n$'th power.

Proof: Notice that we can obtain $v_p(a)$ just by knowing $a\bmod m$, and if $a$ is a perfect $n$'th power $\bmod m$ we must have that $v_p(a)$ is a multiple of $n$ for all $p|m$. So the result follows.

Am I doing something wrong? This is supposed to be problem $4$ of a contest and should thus be non-trivial.

Here is the link to the contest problem.