$f(x^2 + 1) = f(x)g(x)$ $\forall$ x $\in\mathbb{R}$ $\Rightarrow$ no. of roots of $f(x)=$?

Question

If two real polynomials $f(x)$ and $g(x)$ of degrees $m$ $(\ge2)$ and $n$ $(\ge1)$ respectively, satisfy $f(x^2 + 1) = f(x)g(x)$; for every $x \in \mathbb{R}$ , then what can be said about the root(s), if any, of $f(x)$?

Answer 1

Hint: If $x$ is a root of $f$, then $f(x) = 0$. That means that $f(x^2 +1) = f(x)g(x) = 0$. So that means that $x^2 + 1$ is a solution. But $x < x^2 + 1$. You can see by noting that $x = x^2 +1$ doesn't have any roots and since $1^2 + 1 > 1$ you have $x < x^2 +1$ for all real numbers $x$.

So you must now have two roots, so ...

Answer 2

Suppose $r=r_0$ is a root. Then $r_1=r_0^2+1$ if distinct, is another, and $r_2=r_1^2+1$, if distinct, is another, and so on. Hence, either two situations arise. Which are...?

Answer 3

Clearly, if $x$ is a root then so is $x^2+1$. Thus if $x_0$ is the largest real root, then we need $x_0\ge x_0^2+1$, that is $x_0^2-x_0+1\le0$. But $x_0^2-x_0+1=(x_0-\frac12)^2+\frac34>0$

Answer 4

I guess the answer is that $f(x)$ has no roots.
Firstly we can see that degree of $f(x)$ must be even. Why?. Using the left hand side of the definition $dgr(f(x))=2m$.
Then if we had one root $(r_0)$, we could build the increasing sequence $r_{n+1}=r_n^2+1$ of solutions. Which it's a contradiction with the fact of a finite degree of the polynomial.