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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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    Is $n$ an integer or an natural? – 2017-01-31
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    $n$ is integer ... updating the question :) missed that – 2017-01-31
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    I'm not sure $y^3=n^2+3$ even has a solution. Look up Mordell Curves to see what I'm talking about. – 2017-01-31
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    @JyrkiLahtonen That is my current answer, but it seems you have to divide the cases. I'm working on a cleaner solution. – 2017-01-31
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    @S.C.B. Sorry about that. Your answer came while I was typing, so I missed it. +1 of course. – 2017-01-31
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    @Xam How this has tag "diophantine-equations" :O :| – 2017-02-01
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    @RezwanArefin look at both solutions of your question. – 2017-02-01

2 Answers 2

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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

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

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    Does this covers the negative numbers also? – 2017-01-31
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    @RezwanArefin Yes, it does. But wait for a moment, I made a mistake in one of my inequalities. – 2017-01-31
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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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    I like your solution. +1 – 2017-01-31
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    This solution doesn't involve polynomials of degree $>3$, so I think it's not overkill. – 2017-10-16