Any clues on how to solve this? I'm stuck

Here's how i did it.

$5^5 = 3125 \ $ so for positive integer $ \ a \ $ , $ \ 1 \ \le \ a \ \le \ 4 $

We check a maximum of four values of $a$

** **__ $a = 4$ __
$$4^5 + b^2 + c^2 = 2012$$

$$b^2 + c^2 = 988$$

$$\sqrt{988} < 32 $$

$$ \sqrt{ \frac{988}{2}} > 22 $$

So WLOG , we test $ \ 23 \ \le \ b \ \le \ 31 $

Nine values , none of them work.

** **__ $a = 3$ __
$$ 3^5 + b^2 + c^2 = 2012$$

$$b^2 + c^2 = 1769$$

$$ \sqrt{1769} < 43$$

$$ \sqrt{\frac{1769}{2}} > 29 $$

So WLOG , we test $ \ 30 \ \le \ b \ \le \ 42$

Thirteen values , $37^2 + 20^2 = 1769$ works AND $37 - 20 = 17$ is prime

We can stop , no need to test any other values of $a$ since the conditions are satisfied.

THEREFORE

$$a = 3 $$

$$b = 37$$

$$c = 20$$

$$a + b + c = 60$$

Coice E.

Note: $3^5 + 40^2 + 13^2 = 2012$ but $40 - 13 = 27$ is NOT prime so choice D is disqualified.