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

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<math>P(a_{n})=a_{n}^{k}+C_{k-1}a_{n}^{k-1}+...+C_{1}a_{n}+k!m^{k}</math>
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<math>\prod_{j=1}^{k}\left ( a_{n}+jm \right )=P(a_{n})=a_{n}^{k}+C_{k-1}a_{n}^{k-1}+...+C_{1}a_{n}+k!m^{k}</math>
  
 
with <math>C_{0}=k!m^{k}</math> and coefficients of polynomial <math>\geq 0</math>
 
with <math>C_{0}=k!m^{k}</math> and coefficients of polynomial <math>\geq 0</math>
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<math>a_{n}=a_{1}+(n-1)m</math>, <math>\forall m\geq 0 \mid m \in \mathbb{Z}</math> and <math>a_{1} \geq 1 \mid a_{1} \in \mathbb{Z}</math>
 
<math>a_{n}=a_{1}+(n-1)m</math>, <math>\forall m\geq 0 \mid m \in \mathbb{Z}</math> and <math>a_{1} \geq 1 \mid a_{1} \in \mathbb{Z}</math>
  
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~ Tomas Diaz
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~ Tomas Diaz. orders@tomasdiaz.com
  
 
{{alternate solutions}}
 
{{alternate solutions}}
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==See Also==
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{{IMO box|year=2023|num-b=2|num-a=4}}

Latest revision as of 00:58, 19 November 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=CmJn5FKxpPY [Video contains another solution to problem 3]

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

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

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

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

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

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

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

Solving for $g(j)$ we get:

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

$g(j)=f(n+j)-f(n)\geq 0$ for all $n$ and $j$ because $g(j)$ needs to be greater than or equal to zero for all coefficients to be non-negative.

This means that $f(n)$ needs to be increasing with $n$, or staying constant, and also with $f(1)=0$ because $a_{1}=a_{1}+f(1)$.

In addition, since we need all coefficients to be integer, then all $f(n)$ and $g(j)$ must also be integers. We also need $g(j)$ to not be dependent of $n$, so in the expression $f(n+j)-f(n)$, the $n$ needs to cancel. This mean that the rate of change for $f(n)$ with respect to $n$ needs to be constant. This can only be achieved with $f(n)$ be the equation of a line with slope being either zero or positive integer.

So, we set $f(n)$ to be the equation of a line as $f(n)=mn+b$ with $m$ being the slope with a non-negative value and with $b$ the intercept at $n=0$. We know that $f(1)=0$ so $f(1)=m+b=0$ which means that $b=-m$ and our function becomes $f(n)=mn-m=(n-1)m$. Since $f(n)$ needs to be non-negative integer then $m\geq 0 \mid m \in \mathbb{Z}$ then $f(n)$ is increasing or constant, with $f(1)=0$

Then, $g(j)=f(n+j)-f(n)=(n+j-1)m-(n-1)m=jm$

This gives:

$\prod_{j=1}^{k}\left ( a_{n}+jm \right )=P(a_{n})=a_{n}^{k}+C_{k-1}a_{n}^{k-1}+...+C_{1}a_{n}+k!m^{k}$

with $C_{0}=k!m^{k}$ and coefficients of polynomial $\geq 0$

Then, $a_{n}=a_{1}+f(n)$

Which provides the solution of all infinite sequences of positive integers as:

$a_{n}=a_{1}+(n-1)m$, $\forall m\geq 0 \mid m \in \mathbb{Z}$ and $a_{1} \geq 1 \mid a_{1} \in \mathbb{Z}$

~ Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

2023 IMO (Problems) • Resources
Preceded by
Problem 2 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 4
All IMO Problems and Solutions
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