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This question already has an answer here:

I am trying to prove that if $a^3 \mid b^2$ then $a\mid b$, where $a,b \in \mathbb{Z}$. Let $PDC(x)$ be the set of all primes in the prime decomposition of $x$.

So far, I am using the fundamental theorem of arithmetic to try to see what I can do with it. Proving contrapositively, I have $a\nmid \ b \implies \exists p $ a prime, $\exists k \geq 1$ such that $p^k \in PDC(a)$ and $\exists r \in \mathbb{Z}$ such that $p^r$ with $r < k$. So, $p^{3k} \in PDC(a^3)$ and $p^{2r} \in PDC(b)$ where $2r < 3k$ which implies that $a^3\nmid \ b^2$.

Is this valid? Thank you in advance.

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marked as duplicate by Bill Dubuque elementary-number-theory May 26 '15 at 1:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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This answer has the same spirit as yours but I feel that it might be a bit clearer.

Let $\{p_1,\ldots,p_n\}$ collect the prime factors of $a$ and $b$ and write $$ a=\prod_{t=1}^np_t^{\alpha_t}\quad\text{and}\quad b=\prod_{t=1}^np_t^{\beta_t}. $$ Then $a^3\mid b^2$ implies $3\alpha_t\leq 2\beta_t$. But then $3\alpha_t\leq2\beta_t\leq 3\beta_t$ which gives $\alpha_t\leq\beta_t$ and $a\mid b$ follows.

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