[gangsta316]

How would I find all solutions to an equation like
x^2 = y^3
for integers x and y.

This is what I did:
x = y^(3/2)
So y needs to be a perfect square for x to be an integer. So solution should be
(x, y) = (t^3, t^2) for any integer t.
Is that all there is? Is that the proper way of solving such an equation? I've been told that there's a proper method where you use prime factorization, but I can't see it. I know that all the powers of the primes on the LHS will be multiples of 2, and all the powers of the primes on the RHS will be multiples of 3. But what can we do with that?

[Unbounded]

I suppose something like the following might be a little more rigorous; by the fundamental theorem of arithmetic, we can write: , where are distinct prime divisors of x and y respectively, and are positive integers. So,



Therefore from the definition of a prime, for some j.

At the same time,

Since pairs of in each prime decomposition are coprime, clearly, for some (and each) i and j. And m = n.