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

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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>
 
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>
  
−
Let <math>P=\prod_{i=1}^{k}\left ( a_{n+i} \right ) = \prod_{i=1}^{k}\left ( a_{n}+g(i)) \right )</math>
+
Let <math>P=\prod_{j=1}^{k}\left ( a_{n+i} \right ) = \prod_{j=1}^{k}\left ( a_{n}+g(j)) \right )</math>
  
−
If we want the coefficients of <math>P(a_{n})</math> to be positive, then <math>g(i)\geq 0</math> for all <math>i</math> which will give the following value for <math>P</math>:
+
If we want the coefficients of <math>P(a_{n})</math> to be positive, then <math>g(j)\geq 0</math> for all <math>j</math> which will give the following value for <math>P</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>
+
<math>P=a_{n}^{k}+C_{k-1}a_{n}^{k-1}+...+C_{1}a_{n}+\prod_{j=1}^{k} g(j) = P(a_{n})</math>
  
−
Thus for every <math>i</math> and <math>n</math> we need the following:
+
Thus for every <math>j</math> and <math>n</math> we need the following:
  
−
<math>a_{n}+g(i)=a_{n+i}=a_{1}+f(n+i)</math>
+
<math>a_{n}+g(i)=a_{n+j}=a_{1}+f(n+j)</math>
 +
 
 +
Solving for <math>g(j)</math> we get:
 +
 
 +
<math>g(j)=a_{1}+f(n+j)-a_{n}=a_{1}+f(n+j)-a_{1}-f(n)</math>
 +
 
 +
<math>g(j)=f(n+j)-f(n)\geq 0</math> for all <math>n</math> and <math>i</math>
 +
 
 +
This means that <math>f(n)</math> needs to be increasing with n or staying constant with <math>f(1)=0</math> because <math>a_{1}=a_{1}+f(1)</math>  In addition, since we need all coefficients to be integer then all <math>f(n)</math> and <math>g(j)</math> must also be integers
 +
 
 +
So, if we set <math>f(n)=(n-1)d</math> with <math>d\geq0 \mid d \in \mathbb{Z}</math> then <math>f(n)</math> is increasing or constant, with <math>f(1)=0</math>
 +
 
 +
Then, <math>g(j)=f(n+j)-f(n)=(n+j-1)d-(n-1)d=jd</math>
 +
 
 +
This gives:
 +
<math>P(a_{n})=a_{n}^{k}+C_{k-1}a_{n}^{k-1}+...+C_{1}a_{n}+k!d^{k}</math>

Revision as of 12:13, 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 $P=\prod_{j=1}^{k}\left ( a_{n+i} \right ) = \prod_{j=1}^{k}\left ( a_{n}+g(j)) \right )$

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:

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

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 $i$

This means that $f(n)$ needs to be increasing with n or staying constant 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

So, if we set $f(n)=(n-1)d$ with $d\geq0 \mid d \in \mathbb{Z}$ then $f(n)$ is increasing or constant, with $f(1)=0$

Then, $g(j)=f(n+j)-f(n)=(n+j-1)d-(n-1)d=jd$

This gives: $P(a_{n})=a_{n}^{k}+C_{k-1}a_{n}^{k-1}+...+C_{1}a_{n}+k!d^{k}$

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