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?
2026-03-25 03:00:53.1774407653
Bounding $x^2+6x$ between consecutive cubes when solving $y^3=x^2+6x$
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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...