Let $K$ be a finite field with $p$ elements.
show that for any $k\in K$, $x^n=k$ has a solution in $K$ where $\gcd(n,p-1)=1 $
If $f(x)\in K[x]$ has no roots in $K$, then $f$ and $x^p-x$ are relatively prime.
For the first one, I thought since $K$ is finite multiplicative group of $K\setminus\{0\}$ is cyclic. so for any $k\in K$, $k=k_1^n$ for some $k_1\in K$ and for some $n\in\{1,2,3,...p\}$.
But here ai have to show this property for any $n$ s.t $\gcd(n,p-1)=1 $. I think here $p-1$ part has to do something with $|K\setminus\{0\}|=p-1$ . But I am really stuck.
And for the second part I have not any lead.
If $k=0$ it is trivial, just take $x=0$.
Now suppose $k\ne 0$. Then $k^{p-1}=1$, because the multiplicative group has order $p-1$. Now, $n$ being relatively prime to $p-1$ means that $n$ has an inverse mod $p-1$. That means there is some $t$ such that $nt\equiv 1$(mod $p-1$). Let $u=tp$, then $nu\equiv p$(mod $p-1$). And then we get $k^{nu}=k^p=k^{p-1}k=k$. So $x=k^u$ is a solution.
Now the second part. Note that $x^p-x=x(x^{p-1}-1)$, and hence every element of $K$ is a root of this polynomial. (because again, if $0\ne k\in K$ then $k^{p-1}=1$). Since a polynomial of degree $p$ over a field can't have more than $p$ roots we conclude that these are all the roots of the polynomial. So now, suppose $f\in K[x]$ has no roots. Assume $f$ and $x^p-x$ are both divisible by a non constant polynomial $g$. This means that $f$ and $x^p-x$ have a common root in the algebraic closure of $K$. ($g$ is non constant, so it has a root in the algebraic closure). But that is a contradiction, because all the roots of $x^p-x$ are in $K$, and $f$ has no roots in $K$. So they can't have common roots.