59th Ukrainian National Mathematical Olympiad

Let $p$ be some prime divisor of the number $abc+1$. Let's prove that $k=p-2$ satisfies the statement of the problem. Since $p$ doesn't divide any of the numbers $a$, $b$, $c$, by Fermat's little theorem: $a^{p-1}\equiv b^{p-1}\equiv c^{p-1}\equiv 1(\mathrm{mod}p)$. Hence $p$ divides each number $a^{p-1}+abc$, $b^{p-1}+abc$ and $c^{p-1}+abc$. Since $p$ doesn't divide any of the numbers $a$, $b$, $c$ and $p$ is prime, then prime number $p$ divides each number $a^k+bc$, $b^k+ca$ and $c^k+ab$. The latter is the consequence of the following equality: $a^k+bc=a^{p-2}+bc=\frac{1}{a}(a^{p-1}+abc)$. The statement is proved.