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I am familiar with the method of bounding a polynomial between consecutive squares to prove it is not a square. For example, this method can prove $y^2=x^2+x+1$ has no solutions since $x^2<x^2+x+1<(x+1)^2$. The question is how can this method of bounding be applied to a diophantine equation such as $y^3=x^2+6x$? What would be the method for finding the bounds for this type of equation?

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    $\begingroup$ Do you know if these bounds exist? The cubic equation you wrote down is an elliptic curve ; these have incredibly rich structure, and at first glance it is not clear to me if it has integer solutions or not. $\endgroup$ – Patrick Da Silva Apr 9 '15 at 14:57
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    $\begingroup$ Setting $x+3=a$ gives $y^3=a^2-9$, which is called a Mordell curve. $\endgroup$ – punctured dusk Apr 9 '15 at 14:59
  • $\begingroup$ $y^3 = x^2 + 6 x$ does have positive integer solutions: $ [3, 3], [12, 6], [250, 40]$. $\endgroup$ – Robert Israel Apr 9 '15 at 16:58
  • $\begingroup$ This question concerns the same equation: math.stackexchange.com/questions/1199967 $\endgroup$ – punctured dusk Apr 10 '15 at 9:19
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I don't think it can help for this one: $x$ is near $y^{3/2}$, but that won't generally be an integer. If it was $y^3 = x^3 + 6 x$, on the other hand, you'd be able to do something...

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