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Let $x,y$ be positive integers, such that $$3x(x^2+2)=y^2$$ since $$3\cdot 1(1^2+2)=3\times 3=9=3^2$$ $$3\cdot 2(2^2+2)=6\cdot 6=36=6^2$$ $$24\cdot 3(24^2+2)=72\cdot 578=204^2$$ so I have found three solutions $(x,y)=(1,3),(2, 6),(24, 204)$

Are there any other solutions?

ADD: In fact $$LHS=(x-1)^3+x^3+(x+1)^3$$

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  • $\begingroup$ There's a vast literature on finding integer points on elliptic curves. It's very likely that this curve has been analyzed, although I don't know a reference offhand. In any case, you should add tags for nt.number-theory and for elliptic-curve. $\endgroup$
    – Joe Silverman
    Commented Mar 11, 2017 at 0:18
  • $\begingroup$ @JoeSilverman True, but most software (Magma, Sage), likes $y^2 = f(x),$ where $f(x)$ is a monic polynomial. If you would tell the OP how to transform his question to that form, I am sure s/he would be delighted. $\endgroup$
    – Igor Rivin
    Commented Mar 11, 2017 at 0:32
  • $\begingroup$ You can multiply both sides by $9$. Then set $X = 3x$ and $Y = 3y$ and get $Y^2 = f(X)$, where $f$ is monic. $\endgroup$
    – Jeremy Rouse
    Commented Mar 11, 2017 at 0:47
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    $\begingroup$ @JeremyRouse You get $X((X/3)^2+2) = Y^2,$ So setting $\mathfrak{Y} = 3 Y,$ we finally get $X^3 + 18 X = \mathfrak{Y}^2.$ $\endgroup$
    – Igor Rivin
    Commented Mar 11, 2017 at 1:02
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    $\begingroup$ Whereup Magma says: > IntegralPoints(E); [ (0 : 0 : 1), (3 : -9 : 1), (6 : 18 : 1), (72 : -612 : 1) ] [ <(0 : 0 : 1), 1>, <(3 : -9 : 1), 1>, <(6 : 18 : 1), 1>, <(72 : -612 : 1), 1> ] $\endgroup$
    – Igor Rivin
    Commented Mar 11, 2017 at 1:04

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See: A Diophantine Equation by J. W. S. Cassels

The only solutions are $x = 0, 1, 2, 24$.

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    $\begingroup$ ...as confirmed by magma, as in my comment. $\endgroup$
    – Igor Rivin
    Commented Mar 11, 2017 at 2:48

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