prove there doesn't exist a nonconstant rational function f with real coefficients so that $f(x^2/(x+1))$ is a real polynomial

Question

A rational function is a quotient of two polynomials. Prove there doesn't exist a nonconstant rational function f with real coefficients so that $f(x^2/(x+1))$ is a polynomial with real coefficients.

Assume for the sake of contradiction that such a function f exists. Let $f(x) = a(x)/b(x)$ for some (coprime) polynomials $a$ and $b$. Write $a(x) = A\prod_{i=1}^{n_a} (x-r_i)$ and $b(x) = B \prod_{i=1}^{n_b}(x-s_i)$ where $r_i, s_i \in \mathbb{C}$ for all $i$ and $A,B\in\mathbb{C}$. One can probably make some simplifying assumptions about these values using the fact that f has real coefficients, but I'm not sure if I can assume any of the above values are real. We know that $p(x) := f(x^2/(x+1)) = A/B \prod_{i=1}^{n_a} (x^2-r_i (x+1))/\prod_{i=1}^{n_b} (x^2 - s_i (x+1)) (x+1)^{n_b-n_a}.$ In order for the above rational function to be a polynomial, the coefficient of $x^i$ must be zero for any negative i. We know that by coprimeness, the sets $\{r_i\}_{i=1}^{n_a}$ and $\{s_i\}_{i=1}^{n_b}$ are disjoint. Using the fact that $p(x)$ is a polynomial, I think one can determine whether $n_b \ge n_a$ or some other rough comparison between the two. I'm not sure how to proceed from here.

Answer 1

Suppose $f = \dfrac{p}{q}$ such that

$ f (\dfrac{x^2}{1+x}) = \dfrac{p(\dfrac{x^2}{1+x})}{q(\dfrac{x^2}{1+x})} = g(x)$

Let $n$ and $m$ be degree of $p$ and $q$ respectively. Suppose $n < m$. Then you can multiply by $(1 + x)^m$ to numerator and denominator of $p/q$ and you will get in numerator a polynomial of degree $m + n$ whereas in denominator a polynomial of degree $2m$ but since $m + n < 2m$ their ratio can't equal a polynomial hence $n \geq m$. Now let $n > m$ and now multiply numerator and denominator by $(1+x)^n$.

If $p(t) = a_nt^2 + \ldots + a_0$ then in the numerator you get

$a_nx^{2n} + a_{n-1}x^{2(n-1)}(1+x) + \ldots + a_0(1+x)^n$

whereas in denominator you get $(1+x)^{n -m}$ $×$ a polynomial

You can see the numerator takes nonzero value at $x = -1$ but denominator is $0$ at same hence function is undefined at $x = -1$ which should not happen if it was polynomial. Hence we must have $ n = m$ but then eventually we get $g(x)$ as ratio of two polynomials of same degree which cannot be a non constant polynomial.