Can $n^2+3n+3$ be a perfect cube? Let $n$ be a positive integer. Can $n^2+3n+3$ be a perfect cube?
I think the answer is no. My proof: Let $b^3 = n^2 +3n + 3$. Then $nb^3 + 1 = n^3 + 3n^2 + 3n + 1 = (n+1)^3$.
So I just need to show that
$$ (n+1)^3 = nb^3 + 1$$
has no solutions. This seems to be true, but I have trouble proving it.
Can anyone help me prove or disprove this?
Edit As pointed out by WhatsUp in his answer, if you use elliptic curves you can compute that the only integral solution to the curve $y^2 + 3y = x^3 - 3$ is $n=17$. So with the additional condition that $n > 17$, $n^2 + 3n + 3$ is not a cube.

Yes, it can.
$17^2+3\times17+3=7^3$.
