Divisibility of unknown numbers #2

Assume that there is an integer $a$ such that $2016n+3=a^3$ .

$\Leftrightarrow 2016n=a^3-3$ $(1)$

Let $a=7m+r$ , where $m$ is an integer and $r \in \{0,\pm 1, \pm 2, \pm 3 \}$

Testing all the cases, we can conclude that $a^3-3$ is not divisible by 7.

However, $2016n$ is divisible by 7. According to $(1)$ , this is clearly a contradiction.

Therefore, there is no such $a$ that $2016n+3=a^3$ . In other words, $2016n+3$ isn't a perfect cube.

Hence, the answer is $\boxed{no}$ .