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I've been researching cubes and I'm trying to solve this Diophantine equation over the integers.

$$ax^3 + bx^2 + cx + d = y^3$$where a, b, c, d are parameters for a given $n$. For example, for $n = 5$, you have$$5x^3 + 30x^2 + 90x + 100 = y^3$$

Is there any way to solve this Diophantine equation? Or does anyone know any references or links to point me to? (Either for general $a,b,c,d$ or for the specific one above). Thanks!

Like I said above, $a, b, c, d$ are functions of n: $$a = n$$ $$b = (3/2)n(n-1)$$ $$c = (1/2)n(2n-1)(n-1)$$ $$d = (1/4)n^2(n-1)^2$$

It's theorized that there are no solutions (to this specific one). But I want to prove it. Surely, I can't check for every single x, I'm mainly looking for a bound on x as a function of n. Essentially, "After some finite x, it will never be a cube" ...

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    $\begingroup$ Because it's cubic, barring various degeneracies it'll be an elliptic curve, and many things are known about rational points (Mordell-Weil), while integral points are subtler. Searching on "elliptic curve" should give you a broader context. $\endgroup$
    – paul garrett
    Aug 14, 2014 at 0:46
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    $\begingroup$ math.stackexchange.com/questions/896661/… $\endgroup$
    – Will Jagy
    Aug 14, 2014 at 1:33
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    $\begingroup$ You posted on m.se and MO without linking each to the other. That's rude. $\endgroup$
    – Gerry Myerson
    Aug 14, 2014 at 2:45
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    $\begingroup$ @Pietro: I thinks he means the (textspeak) shorthand for "laughing out loud", it's not a Mathematical symbol! $\endgroup$
    – Geoff Robinson
    Aug 14, 2014 at 7:59
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    $\begingroup$ You probably know already that $P_n(x) := ax^3+bx^2+cx+d$ is identically a cube if $n=0$ or $\pm1$. Otherwise $P_n(x)$ has no repeated factors so $y^3=P_n(x)$ is an elliptic curve and thus has only finitely many integer solutions, but it can be hard to provably list them all and we don't expect to be abls to do it uniformly in $n$. I suppose you know already that $x=(1-n)/2$ always works if $n$ is odd, while both $x=-n/2$ and $x=1-(n/2)$ work if $n$ is even. There are also occasional sporadic solutions like $x=-34,-22,3,15$ for $n=20$. $\endgroup$
    – Noam D. Elkies
    Aug 14, 2014 at 12:58

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