IMC 2026 · Day 1 · P5

Without loss of generality we can assume $A \geq B$.

Let $k$ be an odd integer with $7 \sqrt{A B} \leq k \leq 7 \sqrt{A B} + 2$, and consider the Chebishev polynomial $T_{k}(x)$.

Let $\omega = \arccos \frac{A - B}{A + B}$, and let $u_{0} = \cos \omega_{0}$ be the greatest root of $T_{k}(x)$ in the interval $\left[ -1, \frac{A - B}{A + B} \right]$. Since $k$ is odd, we have $T_{k}(0) = 0$, so $u_{0} \geq 0$. Moreover, $\omega \leq \omega_{0} < \omega + \frac{\pi}{k}$.

The requested polynomial will be constructed as $$p(x) = \frac{T_{k} \left( u_{0} + \frac{1 + u_{0}}{A} x \right)}{x}.$$ Notice that for all $x \in [-A, B]$ we have $$u_{0} + \frac{1 + u_{0}}{A} x \geq u_{0} + \frac{1 + u_{0}}{A} (-A) = -1 \quad \text{and} \quad u_{0} + \frac{1 + u_{0}}{A} x \leq \frac{A - B}{A + B} + \frac{1 + \frac{A - B}{A + B}}{A} B = 1.$$ Hence, for $x \in [-A, B]$ we have $\left| T_{k} \left( u_{0} + \frac{1 + u_{0}}{A} x \right) \right| \leq 1$ and $|p(x)| \leq \frac{1}{|x|}$, and therefore $$\sum_{i=1}^{A} |p(-i)|^{2} + \sum_{i=1}^{B} |p(i)|^{2} < 2 \sum_{i=1}^{\infty} \frac{1}{i^{2}} = \frac{\pi^{2}}{3} < 4.$$

In order to estimate $p(0)$, notice that $$|p(0)| = \frac{1 + u_{0}}{A} \left| T_{k}^{\prime} (u_{0}) \right| \geq \frac{1}{A} \left| T_{k}^{\prime} (u_{0}) \right|.$$ From $\cos (k t) = T_{k} (\cos t)$ we get $$\begin{gathered} -k \sin (k t) = T_{k}^{\prime} (\cos t) \cdot (- \sin t) \\ T^{\prime} (u_{0}) = T^{\prime} (\cos \omega_{0}) = k \frac{\sin (k \omega_{0})}{\sin \omega_{0}} = \pm \frac{k}{\sin \omega_{0}}. \end{gathered}$$

Since $\omega_{0} \leq \min \left( \omega + \frac{\pi}{k}, \frac{\pi}{2} \right)$, $$\begin{gathered} \sin \omega_{0} < \sin \omega + \frac{\pi}{k} \leq \sqrt{1 - \cos^{2} \omega} + \frac{\pi}{7 \sqrt{A B}} \\ = \sqrt{1 - \left( \frac{A - B}{A + B} \right)^{2}} + \frac{\pi / 7}{\sqrt{A B}} = \frac{2 \sqrt{A B}}{A + B} + \frac{\pi / 7}{\sqrt{A B}} < 3 \sqrt{\frac{B}{A}}. \end{gathered}$$ Hence, $$|p(0)| \geq \frac{1}{A} \left| T_{k}^{\prime} (u_{0}) \right| = \frac{1}{A} \cdot \frac{k}{\sin \omega_{0}} > \frac{7 \sqrt{A B}}{A \cdot 3 \sqrt{\frac{B}{A}}} > 2,$$ so indeed $$|p(0)|^{2} > 4 > \sum_{i=1}^{A} |p(-i)|^{2} + \sum_{i=1}^{B} |p(i)|^{2}.$$

The degree of $p$ is $$\operatorname{deg} p = k - 1 < 7 \sqrt{A B} + 1 \leq 8 \sqrt{A B}.$$ So, $C = 8$ is suitable.