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Source: Polish MO Finals P1 2023

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191 posts

#1 h Mar 29, 2023, 2:14 PM • 1 Y

Y by rightways

Given a sequence of positive integers $a_1, a_2, a_3, \ldots$ such that for any positive integers $k$ , $l$ we have $k+l ~ | ~ a_k + a_l$ . Prove that for all positive integers $k > l$ , $a_k - a_l$ is divisible by $k-l$ .

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9029 posts

#2 h Mar 29, 2023, 2:47 PM • 3 Y

Y by StarLex1, rightways, Capryon

Solution

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#3 h Mar 29, 2023, 8:14 PM

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This problem was proposed by Burii.

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#4 h Apr 13, 2023, 3:27 PM

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For $n \geq l+1$ , taking $k= n-l$ give $n ~ | ~ a_{n-l} + a_{l}$ . But $n ~ | ~ 2n ~ | ~ a_{n+l} + a_{n-l}$ , so $n ~ | ~ a_{n+l} - a_{l}$ $\implies$ $k-l ~ | ~ a_{k} - a_{l}$ .

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levifb

#5 h Jun 6, 2023, 1:36 PM • 1 Y

Y by Chiaquinha

@above to do it you need $k \geq 2l$ , not guaranteed.

Fixing this idea:

Let $\dfrac{k}{l} = 1 + \epsilon$ and take $n \in \mathbb{N}$ with $n > \dfrac{1}{\epsilon}$ . Hence, $\dfrac{k}{l} > \dfrac{n+1}{n} \Rightarrow nk - (n+1)l >0$ .

Now,

$k-l \mid nk - nl \mid a_l + a_{nk - (n+1)l}$ and

$k-l \mid (n+1)k - (n+1)l \mid a_k + a_{nk - (n+1)l}$ .

Subtracting these two gives the conclusion.

This post has been edited 1 time. Last edited by levifb, Jun 6, 2023, 1:37 PM

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Avron

#6 h Mar 29, 2025, 2:51 PM

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For any $k,l$ take $m$ big enough such that $n=m(k-l)-l$ is positive. Then $k+n=(m+1)(k-l) \mid a_k+a_n$ and $l+n=m(k-l) \mid a_l+a_n$ so:
$k-l\mid (a_k+a_n)-(a_l+a_n)=a_k-a_l$ and we're done.

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