If $\frac{p(\pi)}{q(\pi)}=\frac{p(\pi+n)}{q(\pi+n)}$ for all $n\in\mathbb{Z}$, is $\frac{p(\pi)}{q(\pi)}$ rational?

Question

Let $p(x)$ and $q(x)\neq 0$ be polynomials with coefficients in $\mathbb{Q}$. If $\frac{p(\pi)}{q(\pi)}=\frac{p(\pi+n)}{q(\pi+n)}$ for all $n\in\mathbb{Z}$, then can we conclude that $\frac{p(\pi)}{q(\pi)}$ must be a rational?

Answer 1

$p(\pi)q(x+\pi)-q(\pi)p(x+\pi)$ vanishes at all integer points. Since it is a polynomial it must vanish identically. Put $x =-\pi$ to get $p(\pi)q(0)=q(\pi)p(0)$. Hence $\frac {p(\pi)} {q(\pi)}=\frac {p(0)} {q(0)}$ which is rational.

Answer 2

The rational function $$ \frac{p(x)}{q(x)}-\frac{p(\pi)}{q(\pi)} $$ (a priori with real coefficients, not necessarily rational) has infinitely many zeroes. That means it must be constantly $0$. So $\frac{p(x)}{q(x)}$ must also be a constant, which clearly has to be a rational number.