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Can both $n+3\; \text{and}\; n^2+3$ both be cubic number at same time? Where $n$ is an integer number. Not necessarily positive.

I tried writing $x^3 = n+3$ and expressing $n^2+3$ in terms of $x$. I found $x^6 -6x^3+12$ but this doesn't help. How do I prove this?

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  • $\begingroup$ Is $n$ an integer or an natural? $\endgroup$
    – S.C.B.
    Jan 31, 2017 at 6:30
  • $\begingroup$ $n$ is integer ... updating the question :) missed that $\endgroup$
    – Rezwan Arefin
    Jan 31, 2017 at 6:34
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    $\begingroup$ I'm not sure $y^3=n^2+3$ even has a solution. Look up Mordell Curves to see what I'm talking about. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Jan 31, 2017 at 6:38
  • $\begingroup$ @JyrkiLahtonen That is my current answer, but it seems you have to divide the cases. I'm working on a cleaner solution. $\endgroup$
    – S.C.B.
    Jan 31, 2017 at 6:41
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    $\begingroup$ @S.C.B. Sorry about that. Your answer came while I was typing, so I missed it. +1 of course. $\endgroup$
    – Jyrki Lahtonen
    Jan 31, 2017 at 6:42

2 Answers 2

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Slightly overkill, but if $n + 3$ and $n^2 + 3$ are both cubes, then so is their product, and so

$$ (n + 3)(n^2 + 3) = n^3 + 3n^2 + 3n + 9 = (n + 1)^3 + 2^3 $$

would be a cube. But as is well known, the only possible solutions to this can occur when one of the cubes is $0$, and so we have that either $n = -3$ or $n = -1$, and we can verify that neither of these yield solutions.

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    $\begingroup$ I like your solution. +1 $\endgroup$
    – Xam
    Jan 31, 2017 at 15:53
  • $\begingroup$ This solution doesn't involve polynomials of degree $>3$, so I think it's not overkill. $\endgroup$
    – Rosie F
    Oct 16, 2017 at 7:57
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Your question is asking if there is $x,a \in \mathbb{Z}$ such that $$x^6-6x^3+12=a^3$$ Note that if we have $x \ge 2, x \le -2$, then we have that $$(x^2-1)^3 = x^6-3x^4+3x^2-1 < x^6-6x^3+12=a^3$$ And also $$(x^2+1)^3 =x^6+3x^4+3x^2+1 > x^6-6x^3+12=a^3$$ This give us $x^2-1<a<x^2+1$, thus forcing $a$ to be $x^2$. However, as $x^3 \neq 2$ for any integer $x$, this is a contradiction.

The only cases left are $x=-1, 0,1 $, which can be manually checked to be never be cubes.

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  • $\begingroup$ Does this covers the negative numbers also? $\endgroup$
    – Rezwan Arefin
    Jan 31, 2017 at 6:48
  • $\begingroup$ @RezwanArefin Yes, it does. But wait for a moment, I made a mistake in one of my inequalities. $\endgroup$
    – S.C.B.
    Jan 31, 2017 at 6:49

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