If two real polynomials $f(x)$ and $g(x)$ of degrees $m \geq 2$ and $n \geq 1$ satisfy $f(x^2+1)=f(x) g(x) ~~\forall~~x \in \mathbb R,$ then :

If two real polynomials $f(x)$ and $g(x)$ of degrees $m \geq 2$ and $n \geq 1$ respectively satisfy $$f(x^2+1)=f(x) g(x) ~~\forall~~x \in \mathbb R,$$ then :

$(A)~ f$ has exactly one real root $x_0$ such that $f~'(x_0) \ne 0$

$(B)~ f$ has exactly one real root $x_0$ such that $f~'(x_0) = 0$

$(C)~ f$ has $m$ distinct real roots

$(D)~ f$ has no real root

Attempt:

Given that $f(x^2+1)=f(x) g(x) ~~\forall~~x \in \mathbb R,$

$\deg f(x^2+1) = 2m$ and $\deg f(x) g(x) = m+n$

$\implies 2m = m+n$

$\implies m=n$.

First, what you wrote is correct, even though I do not believe it is a part of a solution.

If $f(x) = 0$ then $f(x^2+1) = 0$. So this means that given any real root $x_0$ of $f$, the reals $x_1 = x_0^2 + 1$, $x_2 = x_1^2 + 1, \dots$ will also be roots. Since for all $x \in \mathbb R$, $x^2 + 1 > x$, the $x_i$ form a strictly increasing sequence and therefore an infinite number of roots of $f$.

On the other hand, to show that the statement of the problem is not itself contradictory, here is an example of such functions $f$ and $g$: $f(x) = x^2-x+1$, $g(x) = x^2+x+1$, and we check that $f(x^2+1) = f(x) g(x)$.

Let $x_0$ be a real root of $f(x)$.

Then you have $f(x_0^2+1)=f(x_0)g(x_0)=0$

That is, $x_0^2+1$ ia a real root of $f(x)$ as well.

Since $h(x)=x^2+1>x$ for $x \in \mathbb{R}$, it is clear that any $h^k(x_0)$ is a real root of $f(x)$, even with $k>m$.

But $f(x)$ has a maximum of $m$ roots, hence contradiction, $f(x)$ has no real root.

• What you use is not that $x^2+1$ is monotone, but that $x^2 + 1 > x$ for all reals $x$. Commented Apr 27, 2015 at 11:49
• @Circonflexe You are right, thank you, I edit! Commented Apr 27, 2015 at 11:50
• Instead of induction, you could use that $f$ has at most finitely many roots and start with $x_0$ as maximal real root. Then as above $x_0^2+1$ would be an even bigger root - contradiction. - Remarkably, only $m\ge1$ was used and $n\ge1$ was not used at all. Commented Apr 27, 2015 at 11:54
• @Martigan Thank you for the answer! Commented Apr 27, 2015 at 15:42