If a rational function satisfies $h(t)=h(1/t)$, then $h(t)=f(t+1/t)$

Question

As shown in the title, I want to show that if a rational function satisfies $h(t)=h(1/t)$, then h(t) can be express in the variable $1+1/t$, that is, $h(t)=f(t+1/t)$ for some rational function $f$. Any help is apreciated.

Answer 1

Let $S$ denote the set of images of $\mathbb{Q}\backslash\{0\}$ under $t\mapsto t+1/t$. For $x\in S$, the equation $t+1/t=x$ is equivalent to $t^2-xt+1=0$, whose roots are of the form $u,\,1/u$ for $u\in\mathbb{Q}\backslash\{0\}$. Define $g(x)$ as the root of minimum modulus, so $g:S\mapsto (\mathbb{Q}\backslash\{0\})\cap[-1,\,1]$ and$$t=g(x)\implies x=t+1/t.$$Finally, define $f(x):=h(g(x))$, so$$t=g(x)\implies h(t)=f(x)=f(t+1/t).$$For each $t\in\mathbb{Q}\backslash\{0\}$, the above calculation works for some $x\in S$.