Question: Consider the congruence $x^n =2($mod $13)$. Then for which n does it have a solution?

1. n= 5
2. n=6
3. n=7
4. n=8

Can we take any help from this fact $x^{12 }=1 ($ mod $13),\forall x $?

Please give some hints.

Answer: 
As $\displaystyle2^2\equiv4;2^3\equiv8;2^4=16\equiv3;2^6=64\equiv-1\pmod{13}$

$\displaystyle2$ is itself a primitive root $\pmod{13}$

Taking discrete logarithm wrt $2\pmod{13}$ on $\displaystyle x^n\equiv2\pmod{13}$

$\displaystyle n\cdot\text{ind}\_2x\equiv1\pmod{12}\implies (n,12)=1$

Using [Bézout's Lemma](https://proofwiki.org/wiki/B%C3%A9zout%27s_Lemma), we can always find integer $\text{ind}\_2x$ for $(n,12)=1$
From the fact you were given, you know that $$(x^6)^2=1\pmod{13}$$
So, for example, $x^6\neq 2\pmod{13}$ since $2^2\neq 1\pmod{13}$.

Because $13$ is prime, $x^2-1=0\pmod{13}$ has at most two solutions, the degree of the polynomial. In fact it has exactly two: $1,-1$, found by inspection.

If $x^6=1$, then $x^7=2$ is equivalent to $x^1=2$. $x^8=2$ is equivalent to $x^2=2$. For $x^5=2$, we multiply both sides by $x$ to get $1=2x$. Now we have three simpler modular equations to solve.

If $x^6=-1$, the situation is similar.