# Show that $n + 2$ and $n^2 + n + 1$ cannot both be perfect cubes

Question: If $n$ is a nonnegative integer, prove that $n + 2$ and $n^2 + n + 1$ cannot both be perfect cubes.

Possible solution: Suppose $n+2$ and $n^2 + n + 1$ are perfect cubes, their product $(n+2)(n^2 + n + 1)$ must also be a perfect cube.

However, note that $(n+2)(n^2 + n + 1)=n^3 + 3n^2 + 3n + 2 = (n + 1)^3 + 1^3$

By Fermat's Last Theorem, $a^n + b^n \neq c^n$ if $a,b,c,n$ are positive integers and $n>2$, therefore $a^3 + b^3 \neq c^3$ and $(n + 1)^3 + 1^3$ cannot be a perfect cube (can't be expressed in the form $c^3$ where $c$ is a positive integer)

I'm looking for alternative methods of solution, and some verification that the above proof is correct.

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Fermat seems like overkill. Isn't it enough to notice that the distance between two perfect cubes is always more that one? –  Hans Lundmark Oct 10 '11 at 11:19
Well, except for the distance between 0 and 1 (and the distance between -1 and 0). Hans, why not repost your comment as an answer? –  Gerry Myerson Oct 10 '11 at 11:27
the right side should be $(n+1)^3+1$ –  robjohn Oct 10 '11 at 11:28
@Gerry: OK. Done. –  Hans Lundmark Oct 10 '11 at 12:12
Usually for the homework one should us only methods that were given at the lectures and he can prove by himself after that. The same for the exercises in textbooks. Is it a homework? –  Ilya Oct 10 '11 at 12:13

## 1 Answer

Except for $-1$, $0$ and $1$, the distance between consecutive perfect cubes is always greater than one. This is enough to conclude that $(n+1)^3+1$ is not a perfect cube when $n$ is nonnegative. (No need to invoke Fermat.)

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