Triplets of Integers

There would be $3$ options for $a$ , for if $a>3$ , then $a^6>2018$ .

Also note that a square, mod $3$ , can either be $0$ or $1$ .

If $a=1$ , then $b^2+c^3=2017 \implies b^2+c^3 \equiv 1 \ (mod \ 3)$ . therefore, we cannot have $c=3k+2$ , for $k$ being a non-negative integer. So, we need to check $8$ possibilities for $0<c< (2017)^{\frac{1}{3}}$ . There is only one solution here $(1,17,12)$ .

For $a=2$ , we have $b^2+c^3=1954$ and we need to check odd $c$ in the range $0<c<(1954)^{\frac{1}{3}}$ . there would be $6$ possibilities and, at the end, the would be one solution $(2,35,9)$ .

For $a=3$ , $b^2+c^3=1289$ . There are $10$ possibilities for $0<c<(1289)^{\frac{1}{3}}$ . Two solution emerges eventually $(3,17,10)$ and $(3,35,4)$ .