Krdytkyu

I feel like I'm meant to use the uniqueness of prime factorization and show that since $text{lcm}(3,2)=6$ each prime should appear 6 times, but how would I justify this?

Use the unique prime factorization.

First not that if $p$ is a prime factor of $x$ then $p|x^2$ so $p|y^3$ so $p|y$. And if $q$ is a prime factor of $y$ then $q|y^3$ so $q|x^2$ so $q|x$. So $x$ and $y$ have the exact same prime factors.

So if the distinct prime factors of $x$ are $P={p_1, p_2, ..... p_n}$ the $x = prodlimits_{k=1}^n p_k^{a_i}$ for some $a_i ge 1$ and $y = prodlimits_{k=1}^n p_k^{b_i}$ for some $b_i ge 1$.

Now $x^2 = y^3$ so

$(prodlimits_{k=1}^n p_k^{a_i})^2 = (prodlimits_{k=1}^n p_k^{b_i})^3$ so

$prodlimits_{k=1}^n p_k^{2a_i}=prodlimits_{k=1}^n p_k^{3b_i}$ so (because the is a unique prime factorization of the value $x^2$ which is $y^3$) we have:

$2a_i = 3b_i$ which means $2|3b_i$ for each $b_i$ and $3|2a_i$ for each $a_i$. And as $2,3$ are primes that means $2|b_i$ and $3|a_i$ for each $b_i$ and each $a_i$.

So each $b_i = 2c_i$ for some integer $c_i$. So $y = prodlimits_{k=1}^n p_k^{b_i} = prodlimits_{k=1}^n p_k^{2c_i}= (prodlimits_{k=1}^n p_k^{c_i})^2$. So $y$ is a perfect square.

So each $a_i = 3d_i$ for some integer $d_i$. So $x = prodlimits_{k=1}^n p_k^{a_i} = prodlimits_{k=1}^n p_k^{3d_i}= (prodlimits_{k=1}^n p_k^{d_i})^3$. So $x$ is a perfect square.

That's it.

Write $x^2 = prod_{i=1}^k p_i^{l_i}$, for some positive finite integer $k$ where $p_i$ is the $i$-th prime and $l_i$ is a nonegative integer. Then as $prod_{i=1}^k p_i^{l_i}$ is a square [because $x$ is an integer] each $l_i$ must be a multiple of 2.

As $prod^k_{i=1} p_i^{l_i}$ is also a cube [because $y$ is an integer] each $l_i$ must also be a multiple of 3. Thus each $l_i$ must be a multiple of 6.

Then $x = prod_{i=1}^k p_i^{frac{l_i}{2}}$ [because $x^2 = prod_{i=1}^k p_i^{l_i}$] and $y = prod_{i=1}^k p_i^{frac{l_i}{3}}$ [because $y^3 = prod_{i=1}^k p_i^{l_i}$], which as $l_i$ is a multiple of 6, implies that $frac{l_i}{2}$ is a multiple of 3, which implies that $x$ is a cube; namely
$x= left(prod_{i=1}^k p_i^{frac{l_i}{6}}right)^3$ with each $frac{l_i}{6}$ an integer. Likewise $y$ is a square, namely $y= left(prod_{i=1}^k p_i^{frac{l_i}{6}}right)^2$ with each $frac{l_i}{6}$ an integer..

Here's an alternate proof which uses unique factorization in a different way: Given $x^2 = y^3$ with $x,y ne 0$, define $a := frac{x}{y}$. Then $a^2 = frac{x^2}{y^2} = frac{y^3}{y^2} = y$ and $a^3 = frac{x^3}{y^3} = frac{x^3}{x^2} = x$.

It remains to show that $a in mathbb{Z}$. We do know that $a in mathbb{Q}$ and that $a^2 = y in mathbb{Z}$. Now, look at one of the standard proofs that $sqrt{2}$ is irrational using unique factorization; this will generalize in a fairly straightforward manner to show the desired fact that if $a in mathbb{Q}$ and $a^2 in mathbb{Z}$, then $a in mathbb{Z}$.

(This can also be viewed as a corollary of the fact that $mathbb{Z}$ is integrally closed, since $mathbb{Z}$ is a unique factorization domain and any unique factorization domain is integrally closed.)

Here's a proof by infinite descent that $x$ is necessarily a perfect cube. The proof that $y$ is a perfect square is of the same exact spirit.

Without loss of generality assume that $x,y$ are positive. Suppose there were to exist pairs of positive integers $(x,y)$ with $x^2=y^3$ but $x$ not a perfect cube. Choose the pair $(x_0,y_0)$ with $x$ minimal. Clearly $x_0,y_0>1$. Consider some prime $p$ dividing $x_0$ and write $x_0=px_1$. Then $pmid y_0^3implies y_0=py_1$ for some $y_1inmathbb Z_{geq 0}$. So $$p^2x_1^2=p^3y_0^3implies x_1^2=py_1^3implies pmid x_1$$ Write $x_1=px_2$, again with $x_2$ a positive integer. Then $$p^2x_2^2=py_1^3implies px_2^2=y_1^3implies pmid y_1$$ Next let $y_1=py_2$. We have that $$px_2^2=p^3y_2^3implies x_2^2=p^2y_3^3implies pmid x_2$$ Finally, take $x_2=px_3$. Our equation becomes $$p^2x_2^2=p^2y_2^3implies x_3^2=y_2^3$$
So $(x_3,y_2)$ is a pair of positive integers satisfying $x_3^2=y_2^3$, but clearly $x_3<p^3x_3=x_0$. A contradiction $blacksquare$