$p$ a prime and $p\equiv3\mod 4$, if $a^4=1_F$ where $F$ a field with characteristic $p$, then $x=a+a^{-1}\in F$ $p$ a prime and $p\equiv3\pmod 4$, if $a^4=1_F$ where $F$ a field with characteristic $p$, then $x=a+a^{-1}\in F$


I am not quite sure if the statement is true or not. Here is my approach

Clearly we have $x$ lies in extension field $F(a)$. And we have $x^p=(a+a^{-1})^p=a^p+a^{-p}=a^{4k+3}+a^{-4k-3}=a^3+a^{-3}=a^4a^{-1}+a^{-4}a=a^4(a^{-1}+a)=a^{-1}+a=x$. Notice that $a^2=-1_F\in F$, for any $y\in F(a)$, the only cases that $y\notin F$ is that $y$ generated by $1=1_F$, $a$ and $a^3$. Suppose $y=e_1+e_2a+e_3a^3$ with $e_1, e_2, e_3\in F$, then we have $y^p=(e_1+e_2a+e_3a^3)^p=(e_1+e_2a)^p+e_3^pa^{3p}=e_1^p+e_2^pa^p+e_3^pa^{3p}=e_1+e_2a^p+e_3a^{3p}$. However $a^p=a^3=a^{-1}\neq a$ and $a^{3p}=a\neq -a=aa^{2}=a^3$. Hence $y\neq y^p$. Hence $x\in F$. 

Edit: I think I make a assumption that $a^2=-1$. If this is the case then above still hold. If $a^2=1$, then $a^3=a$, but still we have $a^p=a^{-1}\neq a$, so above still hold.
$a \in F$ and $a^{-1} \in F$ then by Closure property $x \in F$
$a \in F$ and $a^{-1} \in F$ then by Closure property $x \in F$