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Difference between revisions of "2023 IMO Problems/Problem 3"

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<math>P=a_{n}^{k}+C_{k-1}a_{n}^{k-1}+...+C_{1}a_{n}+\prod_{i=1}^{k} g(i) = P(a_{n})</math>
 
<math>P=a_{n}^{k}+C_{k-1}a_{n}^{k-1}+...+C_{1}a_{n}+\prod_{i=1}^{k} g(i) = P(a_{n})</math>
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Thus for every <math>i</math> and <math>n</math> we need the following:
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<math>a_{n}+g(i)=a_{n+i}=a_{1}+f(n+i)</math>

Revision as of 11:45, 3 October 2023

Problem

For each integer $k \geqslant 2$, determine all infinite sequences of positive integers $a_1, a_2, \ldots$ for which there exists a polynomial $P$ of the form $P(x)=x^k+c_{k-1} x^{k-1}+\cdots+c_1 x+c_0$, where $c_0, c_1, \ldots, c_{k-1}$ are non-negative integers, such that \[P\left(a_n\right)=a_{n+1} a_{n+2} \cdots a_{n+k}\] for every integer $n \geqslant 1$.

Solution

https://www.youtube.com/watch?v=JhThDz0H7cI [Video contains solutions to all day 1 problems]

https://www.youtube.com/watch?v=SP-7LgQh0uY [Video contains solution to problem 3]

https://www.youtube.com/watch?v=CmJn5FKxpPY [Video contains another solution to problem 3]

Let $f(n)$ and $g(i)$ be functions of positive integers n and i respectively.

Let $a_{n}=a_{1}+f(n)$, then $a_{n+1}=a_{1}+f(n+1)$, $a_{n+k}=a_{1}+f(n+k)$

Let $P=\prod_{i=1}^{k}\left ( a_{n+i} \right ) = \prod_{i=1}^{k}\left ( a_{n}+g(i)) \right )$

If we want the coefficients of $P(a_{n})$ to be positive, then $g(i)\geq 0$ for all $i$ which will give the following value for $P$:

$P=a_{n}^{k}+C_{k-1}a_{n}^{k-1}+...+C_{1}a_{n}+\prod_{i=1}^{k} g(i) = P(a_{n})$

Thus for every $i$ and $n$ we need the following:

$a_{n}+g(i)=a_{n+i}=a_{1}+f(n+i)$

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