Let $K$ be a field and let $F={P\over Q}\in K(X)$ be a rational fraction, for simplicity we denote also by $F$ the rational function associated to the rational fraction $F$. It is clear that if $P$ and $Q$ are both even or both odd polynomial functions, then $F$ is an even rational function and we have also that if one of the two polynomial functions is even and the other is odd, then $F$ must be odd. Now is the converse true, I mean do we have that if $F$ is an even rational function then necessarely both $P$ and $Q$ are both odd or both even and that if $F$ is odd then necesserely one is odd and the other is even ?
I think it is true, but i don't know how to prove it, suppose that $F$ is even then $${P(x)\over Q(x)}={P(-x)\over Q(-x)}$$ hence $$P(x)Q(-x)=P(-x)Q(x)$$ But how to go from here ?
You may suppose that $P$ and $Q$ have no commun divisors in $K[x]$. Your equality $P(x)Q(-x)=P(-x)Q(x)$ show then that as $P(x)$ divide $P(-x)Q(x)$ and is prime to $Q(x)$, $P(x)$ must divide $P(-x)$. As they have the same degree, there exists $c$ such that $P(-x)=cP(x)$. Replacing $x$ by $-x$, we get $c^2=1$, hence $c=1$ or $c=-1$ and we are done.