TITLE: Show that $m+3$ and $m^2 + 3m +3$ cannot both be perfect cubes.
QUESTION [6 upvotes]: Show that $m+3$ and $m^2 + 3m +3$ cannot both be perfect cubes.
I've done so much algebra on this, but no luck. Tried multiplying, factoring, etc.

REPLY [3 votes]: The first observation is that $m=-2$ is a counterexample. Presumably $m$ should be non-negative.
Assume that $m+3=n^3$ is a perfect cube. Then $n>1$ and
$$
m^2+3m+3=n^6-3n^3+3.
$$
This is strictly less than $n^6=(n^2)^3$. The previous cube below this is
$$
(n^2-1)^3=n^6-3n^4+3n^2-1,
$$
which is $<n^6-3n^3+3$ as $n\ge2$. Therefore $m^2+3m+3$ is strictly between two consecutive perfect cubes, and cannot be a perfect cube itself.