I'm understanding transcendental extension and I've found this question: Find $\mathbb{Q}(\pi^2) \cap \mathbb{Q}(\pi^2-\pi)$.
I guess the answer is $\mathbb{Q}$, but I still didn't get the proof. If $\alpha \in \mathbb{Q}(\pi^2) \cap \mathbb{Q}(\pi^2-\pi)$ then there exists $p,q,r,s \in \mathbb{Q}[x]$ s.t
$$\alpha = \frac {p(\pi^2)}{q(\pi^2)} = \frac{r(\pi^2-\pi)}{s(\pi^2-\pi)}$$.
It could help: If we call $\alpha = \phi(\pi)$, since $\alpha \in \mathbb{Q}(\pi^2)$ we have $\phi(\pi) = \phi(-\pi)$ and $\alpha \in \mathbb{Q}(\pi^2- \pi)$ we have $\phi(\pi) = \phi(1-\pi)$.
Do you have any hint about that??
Your helpful observation is on point. First though, for psychological reasons, lets think of rational functions of $x$ rather than of $\pi$. There is no harm here as $\mathbb{Q}(x)\cong \mathbb{Q}(\pi)$. Hence, we have a rational function $\phi(x)$ with the property that $\phi(-x) = \phi(x)$ and $\phi(1-x) = \phi(x)$. Consequently, $\phi(x) = \phi(1-x) = \phi(x-1)$, and $\phi$ is a periodic function. However, the only rational periodic functions are the constant functions: Suppose $\phi = p(x)/q(x)$ is nonconstant, $p(x)$, $q(x)$ relatively prime polynomials. Without loss of generality, we may suppose the degree of $p$ is positive (otherwise replace $\phi$ with $1/\phi$). Then $\phi$ has a positive finite number of zeros, and is not periodic.