If we let $x, y$ be two integers which satisfy $x^2 + 6 = y^3$, I want to show that $x$ is odd and not divisible by $3$. I have attempted the first part but can't figure out how to show that $x$ isn't divisible by $3$, my attempt for proving it is odd follows below:
If $x$ is even, then we can say it must be equal to some $2w$ for $w \in Z$; we can rewrite the expression $x$ satisfies as $4w^2 + 6 = y^3$. Now considering two cases:
- $y$ is even $\implies y^3$ is divisible by $2^3$ and thus by $4$ also. But this would imply that $6$ is divisble by $4$ also if $y^3 = 4z$ for some $z \in Z$
- $y$ is odd $\implies y^3$ is also odd but the equation from above implies it must be even as it as an even number summed with $6$
Thus both cases lead to a contradiction and $x$ must be odd, but how can I prove it is not a multiple of 3?