THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

$a\ge 2^4+2^3>2$ is a prime number, hence it is odd. Then precisely one of the primes $b$ or $c$ equals $2$.

If $b=2$, then $a=16+c^3\le 2017$, hence $c\le 11$, so $c\in \{3,5,7,11\}$.

Considering each of these values, we obtain: for $c=3$, $a=43$, which is a prime; for $c=5$, $a=141$, which is not a prime; for $c=7$, $a=359$, which is a prime; for $c=11$, $a=1347$, which is not a prime.

The solutions in this case are $a=43$, $b=2$, $c=3$, respectively $a=359$, $b=2$, $c=7$.

If $c=2$, then $a=b^4+8\le 2017$, hence $b\le 5$, so $b\in \{3,5\}$. For $b=3$, we have $a=89$, which is a prime, and for $b=5$, we obtain $a=633$, which is not a prime. So the only solution in this case is $a=89$, $b=3$, $c=2$.