Is there a quick way to solve $3^8 \equiv x \mod 17$? 
Is there a quick way to solve $3^8 \equiv x \mod 17$?

Like the above says really, is there a quick way to solve for $x$? Right now, what I started doing was $3^8 = 6561$, and then I was going to keep subtracting $17$ until I got my answer.
 A: I usually just do it in steps if it's reasonable enough.
$3^3=10$
$3^4=10*3=13$
$3^5=13*3=5$
$3^8=5*3^3=5*10=16$
A: My immediate response was that $3^2=9$, so $3^4=81\equiv-4\pmod{17}$, and $3^8\equiv16\pmod{17}$, though if I had any further computing to do, I’d probably convert that to $-1\pmod{17}$.
A: Strictly speaking, the fastest way would probably be just to use some sort of computer/software.
For example, here is the answer found in Wolfram|Alpha: $3^8 \equiv x \mod 17$
A: When dealing with powers, squaring is a good trick to reduce computations (your computer does this too!) What this means is:
$ \begin{array}{l l l l l}
3 &             &\equiv 3 &\pmod{17}\\
3^2 &\equiv 3^2 & \equiv 9 &\pmod{17}\\
3^4 & \equiv 9^2 & \equiv 81 \equiv 13 & \pmod{17}\\
3^8 & \equiv 13^2 & \equiv 169 \equiv 16 & \pmod{17}\\
\end{array} $

Slightly irrelevant note: By Euler's theorem, we know that $3^{16} \equiv 1 \pmod{17}$. Thus this implies that $3^{8} \equiv \pm 1 \pmod{17}$. If you know more about quadratic reciprocity, read Thomas Andrew's comment below.
A: Here is a simple way of calculating it, which unfortunately only works for these numbers.
Note that $2^4=16=-1 \pmod {17}$. 
Then
$$3^8=-2^43^8=-2^49^4=-18^4=-1^4=-1 \pmod{17}$$
