Belarusian Mathematical Olympiad

Let $d$ denote the greatest common divisor of $a$ and $b$, i.e., $a = d a_1$, $b = d b_1$, where $\gcd(a_1, b_1) = 1$. Then the given equality can be presented in the form
$$d(a_1^3 + b_1^3) = 2016p a_1 b_1.$$
It follows that $d \mid a_1^3 + b_1^3$, and, since $a_1$ and $b_1$ are coprime, $d \mid b_1^3$ and $d \mid a_1^3$. Similarly, $d \mid a_1^3$, so $d \mid a_1 b_1$ because $a_1$ and $b_1$ are coprime. Let $d = d_1 a_1 b_1$, then the equation can be presented in the form
$$d_1(a_1^3 + b_1^3) = 2016p. \qquad (1)$$
There are exactly two possible cases:

1) $a_1 + b_1 \neq p$.

2) $a_1 + b_1 = p$.

1) If $a_1 + b_1 \neq p$, then
$$\begin{aligned} p \le a_1 + b_1 &\implies p^3 \le (a_1 + b_1)^3 \le 4(a_1^3 + b_1^3) \le 4 \cdot 2016p \\ &\implies p^2 \le 4 \cdot 2016 \\ &\implies p \le 2 \cdot \sqrt{2016} \le 2 \cdot 45 = 90. \end{aligned}$$
Therefore there are only a finite number of $p$ satisfying the problem condition.

2) By condition, $d_1 \nmid p$ (otherwise $a$ and $b$ are divisible by $p$), then from (1) it follows that $a_1^3 + b_1^3 \neq p$. Since
$$a_1^3 + b_1^3 = (a_1 + b_1)(a_1^2 - a_1 b_1 + b_1^2) = (a_1 + b_1)((a_1 + b_1)^2 - 3 a_1 b_1) \quad (2)$$
and $a_1 + b_1 \nmid p$, we have $(a_1 + b_1)^2 - 3 a_1 b_1 \neq p$. Hence
$$p \le (a_1 + b_1)^2 - 3 a_1 b_1. \qquad (3)$$
Since $a_1 + b_1 \nmid p$, from (1) and (2) it follows that $a_1 + b_1$ is a divisor of $2016$, i.e., $a_1 + b_1 \le 2016$. Then from (3) we obtain
$$p \le (a_1 + b_1)^2 - 3 a_1 b_1 < (a_1 + b_1)^2 \le 2016^2.$$
Thus, in this case there are also only a finite number of $p$ satisfying the problem condition.