Finding integer cubes that are $2$ greater than a square, $x^3 = y^2 + 2$ [duplicate]

Solution 1:

This diophantine equation you refer to is a special case of a more general Mordell's equation. The key is to factorize the left side in $\mathbb{Z}[\sqrt{-2}]$. Since nobody solved it up until now, I am jumping in to skim through the standard solution to these kind of problems

$$x^3 = y^2 + 2 = (y + \sqrt{-2})(y - \sqrt{-2})\tag{1}$$

Note that both of the factors are relatively prime in $\mathbb{Z}[\sqrt{-2}]$, which follows from the fact that any common factor must divide

$$y + \sqrt{-2} - y + \sqrt{-2} = 2 \sqrt{-2}$$

and thus it must have it's norm dividing $\bf N$$ [2 \sqrt{-2}] = 8$ and $y^2 + 2$ both. Assume that $y$ is odd, so that the norm of the common factor forced to be $1$, implying that any such common factor is unit in the ring, i.e., $\pm 1$.

Since $y \pm \sqrt{-2}$ are relatively prime, they are cubes times units, implying

$$y + \sqrt{-2} = (m_1 + n_1\sqrt{-2})^3 \tag{2}$$ $$y - \sqrt{-2} = (m_2 + n_2\sqrt{-2})^3 \tag{3}$$

for some $m_i, n_i \in \mathbb Z$. Computing the $\sqrt{-2}$ and $\sqrt{-2}$-free parts of both sides in $(2)$ yields

$$y = m_1^3 - 6m_1n_1^2\tag{4}$$ $$1 = 3m_1^2n_1 - 2n_1^3\tag{5}$$

Since the left side of $(5)$ is divisible by $n_1$, one needs to check for $n_1 = 1$ and $n_1 = -1$ respectively for the solutions, the first of which yields in the solution you refer to, i.e., $(3, 5)$ and the second gives you none at all.

So, done and done. Wait... not yet, no. I have assumed that $y$ is odd in a previous result. What if it's even? Well, that's your thing to do now :

Excercise: Prove that $y$ is not even

As a hint, this can be done elementarily by parity arguments through modular arithmetic.