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

Revision as of 11:29, 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 $a_{n}=a_{1}+f(n)$ , then $a_{n+1}=a_{1}+f(n+1)$ , $a_{n+k}=a_{1}+f(n+k)$

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

@@ Line 11: / Line 11: @@
 https://www.youtube.com/watch?v=CmJn5FKxpPY [Video contains another solution to problem 3]
-Let <math>a_{n}=a_{1}+f(n)</math>
+Let <math>a_{n}=a_{1}+f(n)</math>, then <math>a_{n+1}=a_{1}+f(n+1)</math>, <math>a_{n+k}=a_{1}+f(n+k)</math>
-<math>a_{n+1}=a_{1}+f(n+1)</math>
+Let <math>P=\prod_{i=1}^{k}\left ( a_{n+i} \right ) = \prod_{i=1}^{k}\left ( a_{n}+g(i)) \right )</math>
-<math>a_{n+k}=a_{1}+f(n+k)</math>

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

Revision as of 11:29, 3 October 2023

Problem

Solution