TITLE: Diophantine equation, $x^2 = y^3$.
QUESTION [6 upvotes]: I have the following diophantine equation to solve: 
$x^2 = y^3$
I got that if we introduce $ z=a^6 $ then all numbers $ x=a^3$ and $y =a^2$ satisfy the equation. However, I am not sure whether this is an accurate proof or if these are the only solutions to the equation.

REPLY [6 votes]: To show that these are the only solutions, consider the prime factorizations of $x$ and $y$ (where we can write both factorizations over the same set of primes $p_1,\dots,p_n$ by setting some exponents to 0 if necessary):
\begin{align*}
x&=\prod_{i=1}^n p_i^{j_i}\\
y&=\prod_{i=1}^n p_i^{k_i}
\end{align*}
The equation $x^2=y^3$ becomes
$$\prod_{i=1}^n p_i^{2j_i} = \prod_{i=1}^n p_i^{3k_i}$$
The Fundamental Theorem of Arithmetic says that we must have $2j_i=3k_i$ for all $i$. Thus $2\mid k_i$, so we may write $k_i=2l_i$, and then substituting and solving for $j_i$ we get $j_i=3l_i$. Now if we set $a=\prod_{i=1}^n p_i^{l_i}$, then we get $x=a^3$ and $y=a^2$, as desired.