if $x^2=y^3$, then $x=a^3$ and $y=b^2$

Question

We are given that $x^2 = y^3$ and that $x,y$ are positive integers.

We need to show that there are $a,b$ positive integers such that

$x=a^3$, and $y=b^2$.

I tried to do modular arithmetic in order to solve this problem but I got nowhere. I can't see how one would proceed for this question.

Any help is greatly appreciated

Answer 1

As $x^2=y^3$, $y^3$ is a perfect square. Hence all its prime factors have an even multiplicity. At the same time, these multiplicities are multiple of $3$, hence they are multiple of $6$. Both $x^2$ and $y^3$ are perfect sixth powers, from which the claim follows.

Answer 2

If your question is correct, then:

$$ \begin{align} x^2 &= y^3 \\ \left(a^3\right)^2 &= \left(b^3\right)^2 \\ a^6 &= b^6 \\ a &= \pm b \end{align} $$

Because $a, b \in \mathbb{N}$, then $a = b$.

So in conclusion: $x^2 = y^3$ if and only if $a = b$ with $a, b \in \mathbb{N}$.