Definition: A function $f(x)$ is said to be algebraic if it satisfies the following equation in $y$:
$y^m+R_1(x)y^{m-1}+R_2(x)y^{m-2}+\cdots+R_{m-1}y+R_m(x)=0$, where $R_i(x)'s$ are rational functions i.e. of the form $\frac {p(x)}{q(x)}$, where $p(x),q(x)\ne 0 \forall x\in \mathbb R$ are polynomial functions. $\tag 1$
Let $g:\mathbb R\to \mathbb R$ be a function (non -constant) that has period $T$ i.e., $g(x)=g(x+T)$ for all $x\in \mathbb R$.
How do I prove that $y=g(x)$ can't be a root of the equation of the form $(1)$?
I'm trying to prove by contradiction: WLOG Let $y=g(x)$ be a root of equation $(1)$. Let $g(1)=c$, then
$(g(x))^m+R_1(x)(g(x))^{m-1}+R_2(x)(g(x))^{m-2}+\cdots+R_{m-1}g(x)+R_m(x)=0$ will have infinitely many roots? $x=1,1+T,1+2T,\cdots $etc. How do I take it from here?
Thanks.
Define $F(x,y)=g(x)^m+R_1(y)g(x)^{m-1}+\ldots+R_{m-1}(y)g(x)+R_m(y)$. We know that for each $x$, $F(x,x)=0$. As $g$ is $T$-periodic, for each real number $x$ and each integer $n$, $F(x,x+nT)=F(x+nT,x+nT)=0$. Now, for any given $x$, $F(x,\cdot)$ is a rational function, and has every $x+nT$ as its roots. So it has infinitely many roots so is zero. Thus $F=0$. In other words, let $y$ be any real number, then every element $z \in g(\mathbb{R})$ is a root of the polynomial $Z^m+\sum_{i=1}^m{R_i(y)Z^{m-i}}$. It follows that $g(\mathbb{R})$ is finite -- in particular, if $g$ is continuous, then $g$ is constant.
Note that not much more can be said. For instance, we can write $z(z-1)(z-y)=z^3+R_1(y)z^2+R_2(y)z+R_3(y)$ for some $R_1,R_2,R_3$, and it's not too hard to see that if $g(x)=1(x \in \mathbb{Q})$ (or any other periodic function with values in $\{0,1\}$), $g(x)^3+R_1(y)g(x)^2+R_2(y)g(x)+R_3(y)$ is always zero, that $g$ is periodic nonconstant.