# Let K be a field and let F = <mrow class="MJX-TeXAtom-ORD"> P Q </mf

Let $K$ be a field and let $F=\frac{P}{Q}\in K\left(X\right)$ 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
$\frac{P\left(x\right)}{Q\left(x\right)}=\frac{P\left(-x\right)}{Q\left(-x\right)}$
Hence
$P\left(x\right)Q\left(-x\right)=P\left(-x\right)Q\left(x\right)$
But how to go from here?
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Jordin Church
You may suppose that $P$ and $Q$ have no commun divisors in $K\left[x\right]$. Your equality $P\left(x\right)Q\left(-x\right)=P\left(-x\right)Q\left(x\right)$ show then that as $P\left(x\right)$ divide $P\left(-x\right)Q\left(x\right)$ and is prime to $Q\left(x\right)$, $P\left(x\right)$ must divide $P\left(-x\right)$. As they have the same degree, there exists c such that $P\left(-x\right)=cP\left(x\right)$. Replacing $x$ by $-x$, we get ${c}^{2}=1$, hence $c=1$ or $c=-1$ and we are done.
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Grimanijd
Take $Q=x+{x}^{2}$, $P=xQ$. Then $F=x$, and $F$ is odd, but $Q$ isn't odd or even.
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