I need to prove that, assuming $n \in \mathbb{N^+}$

$$\sqrt[3] {3n^2+3n+1} \notin \mathbb{N^+}$$

I'm really stuck with this problem, because so far I haven't managed a clever way to solve it.

Also, I think it really doesn't help the fact that $3n^2+3n+1 = (n+1)^3-n^3$, also because, for example, both $\sqrt {2n+1}$ and $\sqrt[3] {2n+1}$ can be in $\mathbb{N^+}$ for certain values of $n$.

Anyway I'm sure that the relation I wrote is true: first, because it is a consequence of some proved theorems; second, because I tried some billions value of $n$ with the computer, with no results.

Anyone who can enlighten me with some tremendous intuitions? :)

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    $\begingroup$ It's a consequence of Fermat's conjecture (proved by Wiles, special case needed here proved much earlier, by at the latest Euler, iirc), if $\sqrt[3]{3n^2+3n+1} = a \in \mathbb{N}^+$, you'd have $n^3 + a^3 = (n+1)^3$. $\endgroup$ – Daniel Fischer Aug 13 '13 at 9:38
  • $\begingroup$ Sorry it doesn't, I got mistaken for something else. $\endgroup$ – user88595 Aug 13 '13 at 9:42
  • $\begingroup$ @DanielFischer Yeah, you're right :) But you're using Fermat's Last Theorem to prove that, while the theorem I mentioned in my question was indeed that one; I'd like, in fact, not to use it in this case, but to make an answer indipendently $\endgroup$ – Riccardo Del Monte Aug 13 '13 at 9:44
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    $\begingroup$ As mentioned, the special case of exponent $3$ is much easier. You can prove that using the fact that $\mathbb{Z}[e^{2\pi i/3}]$ is a UFD. It's still not trivial, but doable. Maybe the special form here makes it easier. $\endgroup$ – Daniel Fischer Aug 13 '13 at 10:20
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    $\begingroup$ Here is Euler's elementary proof of FLT for n=3. en.wikipedia.org/wiki/… $\endgroup$ – fretty Aug 13 '13 at 16:40

Proof :

Suppose not, then assume that $\sqrt[3]{3n^2 + 3n + 1} = k$ for some positive integer $k$. Then, $k^3 = 3n^2 + 3n + 1$. So, $k^3 = (n + 1)^3 - n^3$. But this implies $k^3 + n^3 = (n + 1)^3$, which contradicts Fermat's Last Theorem. QED

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