International Mathematics Competition
for University Students
2026

IMC2026: Day 1, Problem 5

Problem 5. Prove that there exists a constant \(\displaystyle C>0\) such that for every pair \(\displaystyle A,B\) of positive integers, there is a real polynomial \(\displaystyle p(x)\) with

\(\displaystyle p(0)^2 > \sum_{i=1}^A p(-i)^2 + \sum_{i=1}^B p(i)^2 \quad\text{and}\quad \deg p < C\sqrt{AB}. \)

Géza Kós, Loránd Eötvös University, Budapest

Solution. Without loss of generality we can assume \(\displaystyle A\ge B\).

Let \(\displaystyle k\) be an odd integer with \(\displaystyle 7\sqrt{AB}\le k\le7\sqrt{AB}+2\), and consider the Chebishev polynomial \(\displaystyle T_k(x)\). Let \(\displaystyle \omega=\arccos\dfrac{A-B}{A+B}\), and let \(\displaystyle u_0=\cos\omega_0\) be the greatest root of \(\displaystyle T_k(x)\) in the interval \(\displaystyle \left[-1,\dfrac{A-B}{A+B}\right]\). Since \(\displaystyle k\) is odd, we have \(\displaystyle T_k(0)=0\), so \(\displaystyle u_0\ge0\). Moreover, \(\displaystyle \omega\le\omega_0<\omega+\dfrac{\pi}{k}\).

The requested polynomial will be constructed as

\(\displaystyle p(x) = \frac{T_k\bigg(u_0+\displaystyle\frac{1+u_0}{A}x\bigg)}{x}. \)

Notice that for all \(\displaystyle x\in[-A,B]\) we have

\(\displaystyle u_0+\dfrac{1+u_0}{A}x \ge u_0+\dfrac{1+u_0}{A}(-A) = -1 \quad\text{and}\quad u_0+\dfrac{1+u_0}{A}x \le \dfrac{A-B}{A+B}+\dfrac{1+\frac{A-B}{A+B}}{A}B = 1. \)

Hence, for \(\displaystyle x\in[-A,B]\) we have \(\displaystyle \left|T_k\bigg(u_0+\displaystyle\frac{1+u_0}{A}x\bigg)\right|\le1\) and \(\displaystyle |p(x)|\le\dfrac1{|x|}\), and therefore

\(\displaystyle \sum_{i=1}^A|p(-i)|^2 + \sum_{i=1}^B|p(i)|^2 < 2\sum_{i=1}^\infty\frac1{i^2} = \frac{\pi^2}{3}<4. \)

In order to estimate \(\displaystyle p(0)\), notice that

\(\displaystyle |p(0)| = \frac{1+u_0}A|T_k'(u_0)| \ge \frac{1}A|T_k'(u_0)|. \)

From \(\displaystyle \cos(kt)=T_k(\cos t)\) we get

\(\displaystyle -k\sin(kt)=T_k'(\cos t)\cdot(-\sin t) \)

\(\displaystyle T'(u_0) = T'(\cos\omega_0) = k\dfrac{\sin(k\omega_0)}{\sin\omega_0} = \pm\frac{k}{\sin\omega_0}. \)

Since \(\displaystyle \omega_0\le\min\left(\omega+\frac\pi{k},\frac\pi2\right)\),

\(\displaystyle \sin\omega_0 < \sin\omega + \frac\pi{k} \le \sqrt{1-\cos^2\omega} +\frac\pi{7\sqrt{AB}} \)

\(\displaystyle =\sqrt{1-\bigg(\frac{A-B}{A+B}\bigg)^2} + \frac{\pi/7}{\sqrt{AB}} = \frac{2\sqrt{AB}}{A+B}+\frac{\pi/7}{\sqrt{AB}} < 3\sqrt{\frac{B}{A}}. \)

Hence,

\(\displaystyle |p(0)| \ge \frac1A |T_k'(u_0)| = \frac1A\cdot\frac{k}{\sin\omega_0} > \frac{7\sqrt{AB}}{A\cdot3\sqrt{\frac{B}{A}}} > 2, \)

so indeed

\(\displaystyle |p(0)|^2 > 4 > \sum_{i=1}^A|p(-i)|^2 + \sum_{i=1}^B|p(i)|^2. \)

The degree of \(\displaystyle p\) is

\(\displaystyle \deg p = k-1 < 7\sqrt{AB}+1 \le 8\sqrt{AB}. \)

So, \(\displaystyle C=8\) is suitable.