Q:

Find all the solutions of the congruence $12^x \equiv 17 \bmod 25$

I need to find all the solutions of the congruence $12^x \equiv 17 \bmod 25$.

I don't really have an idea how to approach this..
I tried to write it as: 
$12^x \equiv 17 \mod 25$ $\Leftrightarrow$ $(3*2*2)^x \equiv -8 \mod 25 $ $\Leftrightarrow$ $ 3^x 2^x 2^x + 2^3 \equiv 0 \mod 25 $ $\Leftrightarrow$
$ 2^x(3^x 2^x 2^{3-x}) \equiv 0 \mod 25  $
Is that the way to solve it? and how should I continue from here? thnk you very much

A:

As said in the comments, you can ultimately reduce to check a finite number of cases. I'll illustrate a standard method which often lessens the number of cases to check. The idea is: since a congruence $\pmod{p^m}$ is also a congruence $\pmod{p^k}$ for each $k<m$, you need not consider the cases for which it fails $\pmod {p^k}$ for $k<m$.
By Euler theorem, since $\operatorname{gcd}(12,\,25)=1$, solutions for $x$ are determined $\pmod{20}$.
Moreover, if $12^x\equiv17\pmod{25}$, then $$12^x\equiv17\pmod5$$
So, $x$ solves $$2^x=2\pmod5$$ as well.
Which is true if and only if $x\equiv 1\pmod{4}$.
So you only need check integers in the form $x=4k+1$ such that $0\le x\le 19$ (hence, $k=0,1,2,3,4$) and such that $12^x\equiv 17\pmod{25}$.
\begin{align}12^{4k+1}&\equiv17\pmod{25}\\12\cdot 12^{4k}&\equiv17\pmod{25}\\12\cdot(-6)^{2k}&\equiv17\pmod{25}\\12\cdot11^k&\equiv17\pmod{25}\\-2\cdot12\cdot11^k&\equiv-34\pmod{25}\\11^k&\equiv16\pmod{25}\end{align}
Now, \begin{array}{c|cc}k&11^k&\mod{25}\\\hline0&1\\1&11\\2&-4\\3&6\\\color{red}4&\color{red}{16}\end{array}
So, $k\equiv4\pmod 5$, hence $x\equiv17\pmod{20}$

