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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!

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    $\begingroup$ Same idea. Suppose $x=3w$ for $w\in\mathbb Z$... $\endgroup$ – Simply Beautiful Art Jan 3 '17 at 15:56
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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 $9$...

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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.

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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.

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