Decide with justification if there exist positive integers $a < b$ with the following property: Whenever $m, n$ are positive integers with $m^a|n^b$ then $m|n$.
Here's my attempt at a proof:
Assume for the sake of contradiction there are such integers $a < b$. If $m^a = \prod_{i = 1}^r p_i^{ax_i}$ and $n^b = \prod_{i = 1}^r p_i^{by_i}$, we know that $ax_i \le by_i$ for all $1 \le i \le r$ because $m^a \mid n^b$. But, it could be the case that $x_k > y_k$ for some $1 \le k \le r$ which implies that $m \nmid n$, which is a contradiction. Therefore, no positive $a < b$ exist.
Any assistance would be greatly appreciated!
short version. Assuming $a<b$ work, define $$ m = 2^b \, , \; \; n = 2^a \, .$$
Then $m^a = n^b$ but $m$ does not divide $n$