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.
Here's a result : 

> 
> If $o(a)=p-1$ in the group $U(p)$ then the system $$x^n\equiv a (\text{mod } p)$$ has a solution if and only if $\text{gcd}(n,p-1)=1$ 
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> 
> 

Verify that $o(2)=12$ in $U(13)$. Let's prove the result. 

Suppose $x$ is a solution. Then $o(x^n)=p-1$. Which means $o(x^n)=o(x)=p-1$ (every element must have order $\leq p-1$). So $\text{gcd}(n, p-1)=1$.

Conversely, let $\text{gcd}(n, p-1)=1$. Then $o(a^n)=o(a)=p-1$. So $a=a^{nk}$ for some $k$. Take $x=a^k$.