If $x, y$ are two integers which satisfy $x^2 + 6 = y^3$, how can I show that $x$ is odd and not divisible by $3$ 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?
Thanks!
 A: Well,If $3|x$,then $LEFTHAND \equiv 6$(mod 9).
However.let $y$ traverse the residue class of 9,we can find it impossible:
$y^3 \equiv 6$(mod 9).
Please do not think too much,it is easy.
A: Same idea.  Suppose $x=3w$ for $w\in\mathbb Z$.
$$9w^2+6=y^3$$
So $y^3$ must be divisible by $3$, so there is a factor of $9$ on the right.  But $6$ is not divisible by $9$...
A: We first introduce  a
lemma
Lemma :if $$y=mk , m,k \in \mathbb N \rightarrow y^n=m^nk$$,and$$y^n=mk \rightarrow y=mq$$
Or to put it in words ,$y^n$ will be a multiple of $m^n$. 
$$\mathrm{Proof By Contradiction}:$$
Putting  $x=3k$gives:
$$9k^2+6=y^3$$
 Or,$$3(3k^2+2)=y^3$$ By Lemma 1 This would mean that y is a multiple of 3,and if we divide by 3 on both sides it will give:$$3k^2+2={y^3 \over 3}$$
But this will lead to a contradiction,since by lemma 1,$y^3$ is a multiple of 27,or it will give a remainder of zero when divided by 3 ,3 times, but our last equation says second division gives us a remainder of 2,which is a contradiction,hence x is not a multiple of 3.
