Problem A. 931. (March 2026)

A. 931. Prove that for any integers $\displaystyle k>10$ and $\displaystyle n>k^4$, among the integers $\displaystyle ({n+1^2})$, $\displaystyle ({n+2^2})$, $\displaystyle ({n+3^2})$, $\displaystyle \ldots$, $\displaystyle ({n+k^2})$, there is at least one that has a prime divisor greater than $\displaystyle k$.

Proposed by Géza Kós, Budapest

(7 pont)

Deadline expired on April 10, 2026.

Solution. For every pair $\displaystyle n,k$ of positive integers, let

$\displaystyle L(n,k) = \mathrm{lcm}\big(n+1^2,n+2^2,\ldots,n+k^2\big). $

We have to show that $\displaystyle L(n,k)$ has a prime divisor greater than $\displaystyle k$.

Lemma.

$\displaystyle L(n,k) \ge \frac{(n+1^2)(n+2^2)\cdots(n+k^2)}{(2k)!}. $

$\displaystyle (1) $

Proof. By partial fraction decomposition we have

$\displaystyle \dfrac{1}{\prod\limits_{j=1}^k(x+j^2)} = \sum_{j=1}^k \dfrac{C_j}{x+j^2} \tag{2} $

with some suitable constants $\displaystyle C_1,\ldots,C_k$. A possible way to calculate $\displaystyle C_i$ is multiplying $\displaystyle (2)$ by $\displaystyle (x+i^2)$ then substituting $\displaystyle x=-i^2$:

$$\begin{gather*} \dfrac{1}{\prod\limits_{j\ne i}(x+j^2)} = C_i + \sum\limits_{j\ne i} C_j\dfrac{x+i^2}{x+j^2} \\ C_i = \dfrac{1}{\prod\limits_{j\ne i}(-i^2+j^2)} = \dfrac{1}{\prod\limits_{j\ne i}(j-i)(j+i)} = \dfrac{(-1)^{i-1}\cdot 2i^2}{(k+i)!(k-i)!}. \end{gather*}$$

By multiplying $\displaystyle (2)$ by $\displaystyle (2k)!$ and putting $\displaystyle x=n$,

$\displaystyle \sum_{j=1}^k \dfrac{(-1)^{i-1}2i^2\binom{2k}{k+i}}{n+j^2} = \dfrac{(2k)!}{\prod\limits_{j\ne i}(n+j^2)}. $

A common denominator on the left-hand side is $\displaystyle L(n,k)$, that proves $\displaystyle (1)$. $\displaystyle \Box$

Now let's turn to the problem statment. For the sake of contradiction, suppose that every prime divisor of $\displaystyle L(n,k)$ is at most $\displaystyle k$. Since $\displaystyle k\ge11$, there at at most such $\displaystyle \dfrac{k-1}{2}$ primes. Each prime power divisor of $\displaystyle L(n,k)$ is at most $\displaystyle n+k^2$. Hence,

$\displaystyle L(n,k) \le (n+k^2)^{(k-1)/2}. $

By the Lemma

$\displaystyle L(n,k) \ge \frac{(n+1^2)(n+2^2)\cdots(n+k^2)}{(2k)!} > \dfrac{n^k}{(2k)!}, $

$$\begin{gather*} \dfrac{n^k}{(2k)!} < L(n,k) < (n+k^2)^{(k-1)/2} = n^{(k-1)/2}\cdot \bigg(1+\dfrac{k^2}{n}\bigg)^{(k-1)/2}, \\ n^{(k+1)/2} < (2k)!\cdot \bigg(1+\dfrac{k^2}{n}\bigg)^{(k-1)/2}. \end{gather*}$$

By the AM–GM inequality

$\displaystyle (2k)! = \Big(2\cdot3\cdot\ldots(2k-2)\Big)\cdot (2k-1)\cdot2k < k^{2k-3}\cdot(2k-1)\cdot 2k < 4k^{2k-1}. $

By applying $\displaystyle n>k^4$ and $\displaystyle (1+x)<e^x$,

$\displaystyle \bigg(1+\dfrac{k^2}{n}\bigg)^{(k-1)/2} < \bigg(1+\dfrac{1}{k^2}\bigg)^{(k-1)/2} < \Big(e^{1/k^2}\Big)^{(k-1)/2} < e^{1/2k} < 2, $

and hence we arrive at the contradiction

$$\begin{gather*} k^{2(k+1)} < n^{(k+1)/2} < (2k)!\cdot\bigg(1+\dfrac{k^2}{n}\bigg)^{(k-1)/2} < 4k^{2k-1} \cdot 2 \\ k^3 < 8. \end{gather*}$$

Statistics:

8 students sent a solution.

7 points: Aravin Peter, Bodor Mátyás, Diaconescu Tashi, Sami El-Hajjar (4 students).

4 points: 1 student.

0 points: 2 students.

Not shown because of missing birth date or parental permission: 1 script.

Problems in Mathematics of KöMaL, March 2026

8 students sent a solution.
7 points:	Aravin Peter, Bodor Mátyás, Diaconescu Tashi, Sami El-Hajjar	(4 students).
4 points:	1 student.
0 points:	2 students.
Not shown because of missing birth date or parental permission:	1 script.

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