I came across this homework problem and honestly have no idea where to begin.
Prove that there is no rational function in $F(x)$ such that its square is $x$.
First of all, I am confused about the notation. I'm accustomed to seeing $F(x)$ denote the field $F$ adjoined with some element $x$, but $x$ is not defined here.
My only thought was to take $f(x)=\frac{g(x)}{h(x)}$, then have $\left(\frac{g(x)}{h(x)}\right)^2=\frac{g(x)^2}{h(x)^2}$, which doesn't get very far.
A hint on how to start would be greatly appreciated.
Here's a hint: if $g(x)$ has degree $p$ and $h(x)$ has degree $q$, then what are the degrees of $(g(x))^2$ and the degree of $x(h(x))^2$?