Mathematical Reflections 2016, Issue 6 - Problem S393

Problem:
If $n$ is an integer such that $n^2+11$ is a prime, prove that $n+4$ is not a perfect cube.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Assume that $n+4$ is a perfect cube. Then $n+4=m^3$, where $m \in \mathbb{Z}$, i.e. $n=m^3-4$. It follows that
$$\begin{array}{lll} n^2+11&=&(m^3-4)^2+11\\&=&m^6-8m^3+27\\&=&m^6+m^3+27-9m^3\\&=&(m^2)^3+(m)^3+(3)^3-9m^3\\&=&(m^2+m+3)(m^4-m^3-2m^2-3m+9). \end{array}$$ As $m^2+m+3 \geq 3$ and $m^4-m^3-2m^2-3m+9 \geq 3$ for all $m \in \mathbb{Z}$, then  $n^2+11$ is not prime, contradiction.