66th 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 $$d_1(a_1^3+b_1^3)=2016p.(1)$$ There are exactly two possible cases:

1) $a_1+b_1\ne p$.

2) $a_1+b_1=p$.

1) If $a_1+b_1\ne p$, then $$\begin{array}{rl}p\le a_1+b_1& ⟹p^3\le (a_1+b_1{)}^3\le 4(a_1^3+b_1^3)\le 4\cdot 2016p\\ & ⟹p^2\le 4\cdot 2016\\ & ⟹p\le 2\cdot \sqrt{2016}\le 2\cdot 45=90.\end{array}$$ 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 $a_1^3+b_1^3\ne p$. Since $$a_1^3+b_1^3=(a_1+b_1)(a_1^2-a_1{b}_1+b_1^2)=(a_1+b_1)((a_1+b_1{)}^2-3a_1{b}_1)(2)$$ and $a_1+b_1\nmid p$, we have $(a_1+b_1{)}^2-3a_1{b}_1\ne p$. Hence $$p\le (a_1+b_1{)}^2-3a_1{b}_1.(3)$$ 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 $$p\le (a_1+b_1{)}^2-3a_1{b}_1<(a_1+b_1{)}^2\le 2016^2.$$ Thus, in this case there are also only a finite number of $p$ satisfying the problem condition.