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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$.

Could someone please tell me how to proceed ahead?

Thank you for your help.

MathMan
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2 Answers2

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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 $f(x^2+1) = f(x) g(x)$.

Circonflexe
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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.

Martigan
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