Kvant Math Problem 214

Let $f(x)=ax^{2}+bx+c$ and assume the equation $f(x)=x$ has no real roots.

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Problem

Quadratic trinomial $f(x) = ax^{2} + bx + c$ is such that the equation $f(x) = x$ has no real roots. Prove that the equation $f(f(x)) = x$ also has no real roots.

Yu. I. Ionin

All-Union Mathematical Olympiad of School Students (VII, 10th grade)

Exploration

Let $f(x)=ax^{2}+bx+c$ and assume the equation $f(x)=x$ has no real roots. This is equivalent to the quadratic

$g(x)=f(x)-x=ax^{2}+(b-1)x+c$

having no real zeros, hence having constant sign on $\mathbb{R}$. Since $g$ is continuous and nonvanishing, either $g(x)>0$ for all $x$ or $g(x)<0$ for all $x$.

The equation $f(f(x))=x$ can be rewritten using $g$ as

$f(f(x)) - x = \bigl(f(f(x)) - f(x)\bigr) + \bigl(f(x)-x\bigr) = g(f(x)) + g(x).$

The key point is that $g(x)$ keeps a fixed sign for all real arguments, so both $g(x)$ and $g(f(x))$ inherit that same sign. Hence their sum cannot vanish.

This suggests that the composition cannot introduce a sign change, so no fixed point or 2-cycle can appear.

The only subtlety is ensuring that substituting $f(x)$ into $g$ preserves the sign argument correctly without assuming monotonicity of $f$.

Problem Understanding

This is a Type B problem. One must prove that if the quadratic $f(x)$ has no real fixed points, then the second iterate equation $f(f(x))=x$ also has no real solutions.

The core difficulty is that the condition concerns only $f(x)-x$, while the conclusion involves a composition $f(f(x))$, which could in principle create new real solutions unrelated to fixed points of $f$.

The correct expectation is that the sign of $f(x)-x$ is globally fixed, and this rigidity propagates through composition so that $f(f(x))-x$ cannot vanish.

Proof Architecture

First, prove that the quadratic $g(x)=f(x)-x$ has constant sign on $\mathbb{R}$.

Second, rewrite $f(f(x))-x$ as $g(f(x))+g(x)$.

Third, show that $g(f(x))$ also has the same constant sign as $g(x)$ for all real $x$.

Fourth, conclude that $f(f(x))-x$ is strictly positive everywhere or strictly negative everywhere, hence has no real zeros.

The most delicate step is the decomposition $f(f(x))-x=g(f(x))+g(x)$ and the justification that both summands share a fixed sign independent of $x$.

Solution

Let $f(x)=ax^{2}+bx+c$ and define

$g(x)=f(x)-x=ax^{2}+(b-1)x+c.$

The assumption that the equation $f(x)=x$ has no real roots means that the quadratic equation $g(x)=0$ has no real solutions. Since $g$ is a continuous quadratic polynomial and never vanishes on $\mathbb{R}$, it follows that it has constant sign on $\mathbb{R}$, meaning either

$g(x)>0 \quad \text{for all } x\in\mathbb{R},$

or

$g(x)<0 \quad \text{for all } x\in\mathbb{R}.$

Consider the expression $f(f(x))-x$. Using $g(x)=f(x)-x$, we rewrite

$f(f(x)) - x = f(f(x)) - f(x) + f(x) - x = g(f(x)) + g(x).$

Since $g(t)$ has constant sign for every real $t$, the value $g(f(x))$ also has the same sign for every real $x$, because $f(x)$ is real for every real $x$ and thus lies in the domain where $g$ has fixed sign.

If $g(x)>0$ for all real $x$, then both $g(f(x))>0$ and $g(x)>0$ for all real $x$, hence

$f(f(x)) - x = g(f(x)) + g(x) > 0 \quad \text{for all } x\in\mathbb{R}.$

If $g(x)<0$ for all real $x$, then both $g(f(x))<0$ and $g(x)<0$ for all real $x$, hence

$f(f(x)) - x = g(f(x)) + g(x) < 0 \quad \text{for all } x\in\mathbb{R}.$

In both cases the expression $f(f(x)) - x$ never vanishes on $\mathbb{R}$, so the equation $f(f(x))=x$ has no real solutions.

This completes the proof. ∎

Verification of Key Steps

The central decomposition $f(f(x)) - x = g(f(x)) + g(x)$ follows directly from adding and subtracting $f(x)$, and no additional assumptions are used.

The crucial property is that $g(x)$ having no real zeros implies strict positivity or strict negativity everywhere. This follows from continuity of polynomials: if $g$ changed sign, the intermediate value theorem would force a real root.

The second delicate point is that $g(f(x))$ preserves the same sign. This holds because the sign condition is global in the codomain of $f$, and $f(x)\in\mathbb{R}$ for all real $x$, so substitution does not introduce new values where $g$ could behave differently.

Alternative Approaches

One alternative approach uses dynamical systems language: since $f(x)-x$ has constant sign, every iteration of $f$ moves all real points strictly in one direction relative to the line $y=x$, preventing periodic points of any finite period. One then formalizes this monotonic displacement to exclude solutions of $f^{2}(x)=x$.

Another approach is to analyze the geometry of the parabola $y=f(x)$ relative to the line $y=x$, showing that it lies entirely above or below, and then studying intersections of iterates via geometric inequalities, though this requires more careful control than the direct sign decomposition used above.