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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.

mmm
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    Click this... This seems similar. Didn't know how to use the link code. :( http://math.stackexchange.com/questions/1236862/prove-or-disprove-a3-mid-b2-rightarrow-a-mid-b?rq=1 – randomgirl Jan 19 '17 at 02:58
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    @randomgirl [text you want link to appear as](http:// ~~linkurl~~) – JMoravitz Jan 19 '17 at 03:20
  • The only thing I don't understand is why once I know that $p^r|b$ then why does a|b – mmm Jan 19 '17 at 03:42

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

Carl Schildkraut
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