Prove: If $a^5|b^3$, then $a|b$ for integers $a, b$.

Question

I know that if $a^5|b^3$, then there exists some integer t such that $ta^5=b^3$. I also know that if I can show there exists some integer s such that $sa=b$, then $a|b$. I believe I should use prime factorization in some way. I can write $a^5 = (p_1^{r_1}p_2^{r_2}...p_n^{r_n})^5$ and $b^3=(p_1^{s_1}p_2^{s_2}...p_m^{s_m})^3$ if I allow some of the exponents to possibly b zero. Logically, I would think that some of the $p_i$ in the factorization of $a$ would be in the list of prime factors of b. However, I am not sure which direction I should go with this.

Answer 1

Hint: If $a^5|b^3$, $a^3|b^3$.