if $x^2=y^3$, then $x=a^3$ and $y=b^2$ 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
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.
If your question is correct, then:
$$
\begin{align}
x^2 &amp;= y^3 \\
\left(a^3\right)^2 &amp;= \left(b^3\right)^2 \\
a^6 &amp;= b^6 \\
a &amp;= \pm b
\end{align}
$$
Because $a, b \in \mathbb{N}$, then $a = b$.
So in conclusion: $x^2 = y^3$ if and only if $a = b$ with $a, b \in \mathbb{N}$.