Three little integers

Step 1: By observation, $17\times 4^b$ when divided by $3$ will always remain $2$ .

The only way to make $c^2$ possible is to make $3^a = 1$ .

$a = 0$

Step 2: That makes $c = 3n$ .

$\because 1 + 17\times 4^b = 9n^2$ $(3n)^2 = 9n^2$

$17\times 4^b \equiv 8 \pmod 9$

Step 3: We will have to make $17\times 2^x$ and $2^y$ differ by 2.

The only solution set is $(32,34)$ , which is $17\times 2^1$ and $2^5$ . $2^1\times 2^5 = 4^3$

$b=3$

Step 4: Conclude out that $c=33$

$a+b+c = 36$