Does $a^3 = b^2$ imply $a$ is a perfect square? Suppose I have $a, b \in \mathbb Z^+$ such that $a^3 = b^2$. How does this imply that $a$ is a perfect square, i.e. $\exists e \in \mathbb Z^+$ such that $a = e^2$ ? I know I have to use prime factorization somewhere in my argument, but I'm stuck. 
 A: Hint. So, start writing down $a$'s and $b$'s prime decomposition, say 
$$ a = \prod_p p^{\nu_p(a)}, \quad b = \prod_p p^{\nu_p(b)} $$
Now $a^3 = b^2$ reads 
$$ \prod_p p^{3\nu_p(a)} = \prod_p p^{2\nu_p(b)} $$
Uniqueness gives you 
$$ \forall p: 3\nu_p(a) = 2\nu_p(b) $$
What does this tell you about the parity of $\nu_p(a)$?
A: The prime factorization of $a=p_1^{n_1}p_2^{n_2}......p_n^{n_n}$
$\implies a^3=p_1^{3n_1}p_2^{3n_2}......p_n^{3n_n}$
Since, $a^3=b^2 \implies a$ and $b$ will have same primes
$\implies b =p_1^{m_1}p_2^{m_2}......p_n^{m_n}$
$\implies b^2 =p_1^{2m_1}p_2^{2m_2}......p_n^{2m_n}$
As, $a^3 =b^2 \implies p_i^{3n_i}= p_i^{2m_i} $, for $i \in \{1,2,....,n\}$
$\implies 3n_i=2m_i \implies n_i=2t_i$
$\implies a=p_1^{2t_1}p_2^{2t_2}......p_n^{2t_m}$
So, $a$ is a perfect square
A: Divide both sides by $a^2$, you get $$a=\frac{b^2}{a^2}=\left (\frac ba \right)^2$$ you say it's integer ,it's rational , it can only be integer if $a$ divides $b$ .  But if $a$ divides $b$ then so do all prime powers dividing $a$ . Further, it means $b=ac$ . Rewriting the equation gives:$$a^3=a^2c^2$$ or $$a=c^2$$
