Perfect before 2016 ends

Working in modulo $5$ ,

Perfect squares can only be $0(mod 5)$ or $\pm 1(mod 5)$

$a^4 \equiv 0, 1 (mod 5)$

$5bc(a^2 + b) \equiv 0(mod 5)$

$2 \equiv 2(mod 5)$

The expression can be either $2 (mod 5)$ or $3 (mod 5)$ but as I said earlier there are no perfect squares that are either $2(mod 5)$ or $3(mod 5)$

Therefore, there are no possible perfect square that would satisfy the expression.