Kvant Math Problem 74

Consider small-degree polynomials to detect a pattern.

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Problem

The polynomial $p$ has the following property: for some number $a$, $$p(x)=p(a-x).$$

Prove that $p(x)$ can be represented as a polynomial in $$\left(x-\frac{a}{2}\right)^2.$$.

For example, if $p(x)=x^5+(1-x)^5$, then obviously $p(x)=p(1-x)$ and, as is easy to verify, $$p(x)=5y^2+\frac{5}{2}y+\frac{1}{16},$$ where $$y=\left(x-\frac{1}{2}\right)^2.$$.

Exploration

Consider small-degree polynomials to detect a pattern. For degree zero, a constant function $p(x)=c$ trivially satisfies $p(x)=p(a-x)$ and is already a function of $(x-a/2)^2$. For degree one, write $p(x)=\alpha x + \beta$. Then $p(a-x)=\alpha(a-x)+\beta = -\alpha x + \alpha a + \beta$. Equality $p(x)=p(a-x)$ implies $\alpha = -\alpha$, hence $\alpha=0$. Therefore only constant linear polynomials satisfy the condition, which agrees with the conjecture.

For degree two, let $p(x)=\gamma x^2 + \delta x + \epsilon$. Then $p(a-x)=\gamma(a-x)^2 + \delta(a-x)+\epsilon = \gamma x^2 - 2\gamma a x + \gamma a^2 + \delta a - \delta x + \epsilon = \gamma x^2 + (-2\gamma a - \delta)x + (\gamma a^2 + \delta a + \epsilon)$. Equality with $p(x)$ implies $-2\gamma a - \delta = \delta$, so $\delta=-\gamma a$, and the constant term is then $\gamma a^2 - \gamma a^2 + \epsilon = \epsilon$, which is consistent. Rewriting $x^2 - a x$ in terms of $y=(x-a/2)^2$ gives $x^2 - a x = (x-a/2)^2 - a^2/4$, confirming that the quadratic can be expressed in terms of $y=(x-a/2)^2$.

The crucial step appears to be a change of variable $y=x-a/2$, which centralizes the polynomial at $x=a/2$ and exposes symmetry. The challenge is proving rigorously that every degree polynomial satisfying $p(x)=p(a-x)$ can be written in terms of $(x-a/2)^2$, not just low degrees.

Problem Understanding

The problem asks to prove that if a polynomial $p$ satisfies $p(x)=p(a-x)$ for some $a$, then it is a polynomial in the single variable $y=(x-a/2)^2$. This is a Type B problem, a pure proof. The core difficulty is translating the symmetry condition $p(x)=p(a-x)$ into a structural statement about the polynomial. Intuitively, the polynomial is symmetric with respect to the line $x=a/2$, and any function symmetric about a point can be expressed as a function of the square of the distance from that point.

Proof Architecture

Lemma 1: If $p(x)$ is a polynomial satisfying $p(x)=p(a-x)$, then all odd powers of $x-a/2$ vanish after expansion. This follows because the change of variable $y=x-a/2$ transforms the symmetry condition into $p(a/2+y)=p(a/2-y)$, forcing all coefficients of odd powers of $y$ to zero.

Lemma 2: A polynomial with only even powers of $y$ can be expressed as a polynomial in $y^2$. This follows directly from the definition of a polynomial: $p(y)=\sum_{k=0}^n c_{2k} y^{2k}$ implies $p(y)=\sum_{k=0}^n c_{2k} (y^2)^k$.

The hardest step is Lemma 1, where careless algebra might miss a term in the expansion. Verifying the odd powers vanish for arbitrary degree requires careful coefficient comparison.

Solution

Let $y = x - a/2$. Then $x = a/2 + y$, and the symmetry condition $p(x) = p(a-x)$ becomes $p(a/2 + y) = p(a/2 - y)$. Expand $p(a/2 + y)$ as a polynomial in $y$:

$$p(a/2 + y) = c_0 + c_1 y + c_2 y^2 + \dots + c_n y^n.$$

Then

$$p(a/2 - y) = c_0 - c_1 y + c_2 y^2 - c_3 y^3 + \dots + (-1)^n c_n y^n.$$

Equality $p(a/2+y)=p(a/2-y)$ requires that the coefficients of $y, y^3, y^5, \dots$ vanish. Explicitly, comparing terms gives $c_1 = -c_1$, $c_3=-c_3$, …, hence $c_{2k+1}=0$ for all $k$. Therefore

$$p(a/2 + y) = c_0 + c_2 y^2 + c_4 y^4 + \dots + c_{2m} y^{2m}.$$

Substitute back $y = x - a/2$, obtaining

$$p(x) = c_0 + c_2 (x-a/2)^2 + c_4 (x-a/2)^4 + \dots + c_{2m} (x-a/2)^{2m}.$$

This expresses $p$ as a polynomial in $(x-a/2)^2$, as required.

This completes the proof.

∎

Verification of Key Steps

The key step is verifying that all odd coefficients vanish. Consider $p(x) = x^5 + (1-x)^5$ with $a=1$. Substitute $y = x - 1/2$, then $p(1/2 + y) = (1/2+y)^5 + (1/2-y)^5$. Expand both terms using the binomial theorem: $(1/2+y)^5 = (1/2)^5 + 5 (1/2)^4 y + 10 (1/2)^3 y^2 + \dots$, $(1/2-y)^5 = (1/2)^5 - 5(1/2)^4 y + 10 (1/2)^3 y^2 - \dots$. Adding gives only even powers, confirming the general reasoning. Checking a second-degree example $x^2 + (a-x)^2$ yields the same pattern.

Alternative Approaches

A different approach would use the fact that any polynomial is determined by its values at sufficiently many points. Evaluate $p$ at $x = a/2 + t$ and $x = a/2 - t$ for arbitrary $t$, obtaining a functional equation $p(a/2+t) = p(a/2-t)$. Treating $t$ as a variable shows $p(a/2+t)$ is an even function, hence a polynomial in $t^2$. This method is conceptually shorter but relies implicitly on the even-function characterization. The main approach is preferable because it rigorously handles coefficient-by-coefficient verification and generalizes naturally to any degree.