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: 
If $x$ is of the form $2^m$, then we have $$2^{mn}=2\mod 13\\ \Rightarrow 2^{mn-1}=1\mod 13$$ So, we need to have $12|mn-1$i.e. we need to solve the linear Diphontine equation $mn-12k=1$ clearly then, $n$ cannot be $6$ or $8$. For $n=5,$ a solution to $m,k$ is $m=5,k=2$. For $n=7$, a solution is $m=7, k=4$. So, for $n=5,7$ the congruence equation has solutions for $x$ and such solutions are $2^5,2^7$ respectively.
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.