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

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Which provides the solution of all infinite sequences of positive integers as:
 
Which provides the solution of all infinite sequences of positive integers as:
  
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<math>a_{n}=a_{1}+(n-1)m</math>, <math>\forall m\geq 0 \mid m \in \mathbb{Z}</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 $a_{1}\geq 0 \mid a_{1} \in \mathbb{Z}
  
 
~ Tomas Diaz
 
~ Tomas Diaz

Revision as of 14:07, 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=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 $i$ because $g(j)$ needs to be greater than 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: $P(a_{n})=a_{n}^{k}+C_{k-1}a_{n}^{k-1}+...+C_{1}a_{n}+k!m^{k}$

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 0 \mid a_{1} \in \mathbb{Z}

~ Tomas Diaz

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