Problem A. 928. (February 2026)
A. 928. Let \(\displaystyle a_0=0<a_1<a_2<\ldots<a_n\) be integers such that the sequence \(\displaystyle b_k=\frac{a_{k+1}-a_k}{2k+1}\) (\(\displaystyle k=0\), 1, \(\displaystyle \ldots\), \(\displaystyle n-1\)) is non-decreasing. Suppose that \(\displaystyle c_1\), \(\displaystyle c_2\), \(\displaystyle \ldots\), \(\displaystyle c_n\) are real numbers such that the polynomial \(\displaystyle 1+\sum_{k=1}^n c_kx^{a_k}\) is divisible by the polynomial \(\displaystyle (x+1)^n\). Show that \(\displaystyle 2>|c_1|>|c_2|>\ldots>|c_n|\).
Proposed by Géza Kós, Budapest
(7 pont)
Deadline expired on March 10, 2026.
Solution. Define the coefficient \(\displaystyle c_0=1\) as well, and let
\(\displaystyle p(x) = \sum_{k=0}^n c_kx^{a_k} \)
the polynomial in the statement.
Claim 1.
\(\displaystyle (1a)\) \(\displaystyle p(-1)=p'(-1)=\ldots=p^{(n-1)}(-1)=0\).
\(\displaystyle (1b)\) For every polynomial \(\displaystyle q(x)\) with \(\displaystyle \deg q<n\), we have \(\displaystyle \displaystyle \sum_{k=0}^n c_k \cdot(-1)^{a_k} q(a_k)=0\).
Proof.
\(\displaystyle (1a)\) We show by induction that for every \(\displaystyle 0\le k\le n-1\), the polynomial \(\displaystyle p^{(k)}(x)\) is divisible by \(\displaystyle (x+1)^{n-k}\). By the condition, that is true for \(\displaystyle k=0\).
If the induction hupothesis holds true for some \(\displaystyle 0\le k<n-1\), so
\(\displaystyle p^{(k)}(x) = (x+1)^{n-k}\cdot h(x) \)
with some polynomial \(\displaystyle h(x)\), then
\(\displaystyle p^{(k+1)}(x) = \Big((x+1)^{n-k}\cdot h(x)\Big)' = (x+1)^{n-k-1} \cdot \Big((n-k)h(x)+(x+1)h'(x)\Big), \)
so the statement is true for \(\displaystyle (k+1)\) as well.
After all, the polynomials \(\displaystyle p(x)\), \(\displaystyle p'(x)\), \(\displaystyle p''(x)\), ..., \(\displaystyle p^{(n-1)}(x)\) are all divisible by \(\displaystyle (x+1)\), so they have a common root at \(\displaystyle -1\).
\(\displaystyle (1b)\) Write \(\displaystyle q(x)\) as
\(\displaystyle q(x) = \sum_{\ell=0}^{n-1} d_\ell \cdot x(x-1)\cdots(x-\ell+1) \)
whith some suitable real numbers \(\displaystyle d_0,d_1,\ldots,d_{n-1}\). Then
$$\begin{align*} \sum_{k=0}^n c_k \cdot(-1)^{a_k} q(a_k) &= \sum_{k=0}^n (-1)^{a_k}c_k \cdot \bigg(\sum_{\ell=0}^{n-1} d_\ell \cdot a_k(a_k-1)\cdots(a_k-\ell+1) \bigg) \\ &= \sum_{\ell=0}^{n-1} (-1)^\ell d_\ell \cdot \bigg( \sum_{k=0}^n c_k \cdot a_k(a_k-1)\cdots(a_k-\ell+1) \cdot (-1)^{a_k-\ell} \bigg) \\ &= \sum_{\ell=0}^{n-1} (-1)^\ell d_\ell \cdot p^{(\ell)}(-1) \\ &=0. \end{align*}$$Claim 2. For every integer \(\displaystyle 1\le k\le n\),
\(\displaystyle c_k = (-1)^{a_k+k} \prod_{\substack{1\le\ell\le n\\\ell\ne k}} \dfrac{a_\ell}{|a_k-a_\ell|}. \)
Proof. Apply Claim (1b) to the polynomial
\(\displaystyle q(x)=\prod_{\substack{1\le\ell\le n\\\ell\ne k}}(x-a_\ell). \)
The sum in (1b) contains only two non-zero terms, and we obtain
\(\displaystyle \sum_{\ell=0}^n c_\ell \cdot(-1)^{a_\ell} q(a_\ell) = c_0\cdot q(0)+c_k\cdot(-1)^{a_k}q(a_k) = 0, \)
hence
\(\displaystyle c_k = \dfrac{c_0\cdot(-1)^{a_k+1}q(0)}{q(a_k)} = (-1)^{a_k+k} \prod_{\substack{1\le\ell\le n\\\ell\ne k}} \dfrac{a_\ell}{|a_k-a_\ell|}. \)
Claim 3. For every triple \(\displaystyle u,v,w\) with \(\displaystyle 0\le u<v<w\le n\) we have
| \(\displaystyle \dfrac{a_v-a_u}{v^2-u^2} \le \dfrac{a_w-a_u}{w^2-u^2} \le \dfrac{a_w-a_v}{w^2-v^2}, \) | \(\displaystyle (2) \) |
or equivalently,
\(\displaystyle \dfrac{a_v-a_u}{a_w-a_u}\le\dfrac{v^2-u^2}{w^2-u^2}, \quad \dfrac{a_w-a_u}{a_w-a_v}\le\dfrac{w^2-u^2}{w^2-v^2} \quad\text{és}\quad \dfrac{a_v-a_u}{a_w-a_v}\le\dfrac{v^2-u^2}{w^2-v^2}. \)
Proof. By the condition, the sequence \(\displaystyle \dfrac{a_{k+1}-a_k}{2k+1}=\dfrac{a_{k+1}-a_k}{(k+1)^2-k^2}\) is non-decreasing, so the points
\(\displaystyle (0^2,a_0), (1^2,a_1), (2^2,a_2), \ldots, (n^2,a_n) \)
lie along the graph of a convex function. The relation \(\displaystyle (2)\) is just another form of the same convexity.
Claim 4.
\(\displaystyle |c_1| < 2. \)
Proof. By Claim 2 we have
\(\displaystyle |c_1| = \prod_{\ell=2}^n \dfrac{a_\ell}{a_\ell-a_1}, \)
and by Claim 3, for every \(\displaystyle 2\le\ell\le n\) we have
\(\displaystyle \dfrac{a_\ell}{a_\ell-a_1} = \dfrac{a_\ell-a_0}{a_\ell-a_1} \le\dfrac{\ell^2-0^2}{\ell^2-1} = \dfrac{\ell^2}{(\ell-1)(\ell+1)}. \)
Therefore,
\(\displaystyle |c_1| = \prod_{\ell=2}^n \dfrac{a_\ell}{a_\ell-a_1} \le \prod_{\ell=2}^n \dfrac{\ell^2}{(\ell-1)(\ell+1)} = \prod_{\ell=2}^n \dfrac{\ell}{\ell-1} \cdot \prod_{\ell=2}^n \dfrac{\ell}{\ell+1} = n\cdot\frac2{n+1} < 2. \)
Claim 5. For \(\displaystyle 1\le k\le n-1\) we have
\(\displaystyle |c_k| > |c_{k+1}|. \)
Proof. By Claim 2. we have
$$\begin{align*} |c_k| &= \prod_{\substack{1\le\ell\le n\\\ell\ne k}} \dfrac{a_\ell}{|a_k-a_\ell|} = \bigg(\prod_{\ell=1}^{k-1} \dfrac{a_\ell}{a_k-a_\ell}\bigg) \cdot \dfrac{a_{k+1}}{a_{k+1}-a_k} \cdot \bigg(\prod_{\ell=k+2}^{n} \dfrac{a_\ell}{a_\ell-a_k}\bigg) \quad\text{and} \\ |c_{k+1}| &= \prod_{\substack{1\le\ell\le n\\\ell\ne k+1}} \dfrac{a_\ell}{|a_{k+1}-a_\ell|} = \bigg(\prod_{\ell=1}^{k-1} \dfrac{a_\ell}{a_{k+1}-a_\ell}\bigg) \cdot \dfrac{a_k}{a_{k+1}-a_k} \cdot \bigg(\prod_{\ell=k+2}^{n} \dfrac{a_\ell}{a_\ell-a_{k+1}}\bigg), \quad\text{so} \\ &\frac{|c_{k+1}|}{|c_k|} = \dfrac{a_k}{a_{k+1}} \cdot \prod_{\ell=1}^{k-1} \dfrac{a_k-a_\ell}{a_{k+1}-a_\ell} \cdot \prod_{\ell=k+2}^{n} \dfrac{a_\ell-a_k}{a_\ell-a_{k+1}}. \end{align*}$$By Claim 3,
$$\begin{gather*} \dfrac{a_k}{a_{k+1}} = \dfrac{a_k-a_0}{a_{k+1}-a_0} \le \dfrac{k^2}{(k+1)^2}, \\ \dfrac{a_k-a_\ell}{a_{k+1}-a_\ell} \le\dfrac{k^2-\ell^2}{(k+1)^2-\ell^2} \qquad\text{for \(\displaystyle 1\le\ell\le k-1\),\qquadand} \\ \dfrac{a_\ell-a_k}{a_\ell-a_{k+1}} \le\dfrac{\ell^2-k^2}{\ell^2-(k+1)^2} \qquad\text{for \(\displaystyle k+2\le\ell\le n\).} \end{gather*}$$Hence,
$$\begin{align*} \frac{|c_{k+1}|}{|c_k|} &= \dfrac{a_k}{a_{k+1}} \cdot \prod_{\ell=1}^{k-1} \dfrac{a_k-a_\ell}{a_{k+1}-a_\ell} \cdot \prod_{\ell=k+2}^{n} \dfrac{a_\ell-a_k}{a_\ell-a_{k+1}} \\ &\le \dfrac{k^2}{(k+1)^2} \cdot \prod_{\ell=1}^{k-1} \dfrac{k^2-\ell^2}{(k+1)^2-\ell^2} \cdot \prod_{\ell=k+2}^{n} \dfrac{\ell^2-k^2}{\ell^2-(k+1)^2} \\ &= \dfrac{k^2}{(k+1)^2} \cdot\bigg( \prod_{\ell=1}^{k-1} \dfrac{k-\ell}{k+1-\ell} \cdot \prod_{\ell=1}^{k-1} \dfrac{k+\ell}{k+1+\ell} \bigg)\cdot\bigg( \prod_{\ell=k+2}^{n} \dfrac{\ell-k}{\ell-k-1} \cdot \prod_{\ell=k+2}^{n} \dfrac{\ell+k}{\ell+k+1} \bigg) \\ &= \dfrac{k^2}{(k+1)^2} \cdot \bigg(\dfrac{1}{k} \cdot\dfrac{k+1}{2k} \bigg) \cdot \bigg(\dfrac{n-k}{1} \cdot\dfrac{2k+2}{n+k+1} \bigg) = \dfrac{n-k}{n+k+1} < 1. \end{align*}$$(The calculation above is correct for the boundary cases \(\displaystyle k=1\) and \(\displaystyle k=n-1\) also.)
Claims 4 and 5 prove the problem statement.
Statistics:
9 students sent a solution. 7 points: Bodor Mátyás, Diaconescu Tashi, Kis Ágoston, Rajtik Sándor Barnabás, Xiaoyi Mo (5 students). 5 points: 1 student. 1 point: 2 students. 0 points: 1 student.
Problems in Mathematics of KöMaL, February 2026