I wish to find all b such that $9x^8 \equiv b \pmod{17}$ has a solution using index arithmetic. I have figured out that 3 is a primitive root, giving:
$$\text{ind}_39+8\text{ind}_3x \equiv 2+8\text{ind}_3x \equiv \text{ind}_3b \pmod{16}$$
But, where can I go from here? A hint would be appreciated.
Clearly, if $17|b,17\mid x$
otherwise,
$$\text{ind}_3b-2\equiv8\text{ind}_3x \pmod{16}\begin{cases}0 &\mbox{if } \text{ind}_3x \text{ is even}\iff b\equiv3^2\pmod{17} \\8& \mbox{if }\text{ind}_3x \text{ is odd}\iff b\equiv3^{8+2}\end{cases}$$