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.
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.
