Difference between revisions of "2016 UMO Problems/Problem 5"

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==Problem ==
 
==Problem ==
  
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Let <math>a_0,a_1,a_2,\ldots</math> be a sequence of integers (positive, negative, or zero) such that for all nonnegative integers <math>n</math> and <math>k</math>, <math>a_{n+k}^2-(2k + 1)a_{n}a_{n+k} + (k^2 + k)a_n^2 = k^2-k</math>. Find all possible sequences <math>(a_n)</math>.
  
  
== Solution ==
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== Solution 1 ==
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We can factor the equation as <math>[a_{n+k}-k\cdot a_n][a_{n+k}-(k + 1)\cdot a_n] = k(k-1)</math>.
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Substituting in <math>k = 1</math>, we find that either <math>a_{n+1} = 2a_n</math> or <math>a_{n+1} = a_n</math> for all integers <math>n</math>. If <math>p</math> is a prime such that <math>p \vert a_r</math>, then by repeatedly applying <math>a_{n+1} = 2a_n</math> or <math>a_{n+1} = a_n</math>, we find that <math>p</math> divides <math>a_n</math> for all integers <math>n >= r</math>. Substituting <math>k = 2</math> and <math>n = r</math> into (1) we notice that the LHS is divisible by <math>p^2</math>, and the RHS is equal to <math>2</math>; hence <math>p^2 \vert 2</math>, which is a contradiction. Thus <math>a_n</math> is never divisible by any prime for any <math>n</math>, so <math>a_n = \pm 1</math>. In particular, <math>a_{n+1} = a_n</math> for all integers <math>n</math>, and <math>a_n</math> is constant either 1 or -1. Substituting either of these sequences into (1) shows that both are valid solutions.
  
 
== See Also ==
 
== See Also ==
 
{{UMO box|year=2016|num-b=4|num-a=6}}
 
{{UMO box|year=2016|num-b=4|num-a=6}}
  
[[Category:]]
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[[Category: Intermediate Number Theory Problems]]

Latest revision as of 03:20, 22 January 2019

Problem

Let $a_0,a_1,a_2,\ldots$ be a sequence of integers (positive, negative, or zero) such that for all nonnegative integers $n$ and $k$, $a_{n+k}^2-(2k + 1)a_{n}a_{n+k} + (k^2 + k)a_n^2 = k^2-k$. Find all possible sequences $(a_n)$.


Solution 1

We can factor the equation as $[a_{n+k}-k\cdot a_n][a_{n+k}-(k + 1)\cdot a_n] = k(k-1)$. Substituting in $k = 1$, we find that either $a_{n+1} = 2a_n$ or $a_{n+1} = a_n$ for all integers $n$. If $p$ is a prime such that $p \vert a_r$, then by repeatedly applying $a_{n+1} = 2a_n$ or $a_{n+1} = a_n$, we find that $p$ divides $a_n$ for all integers $n >= r$. Substituting $k = 2$ and $n = r$ into (1) we notice that the LHS is divisible by $p^2$, and the RHS is equal to $2$; hence $p^2 \vert 2$, which is a contradiction. Thus $a_n$ is never divisible by any prime for any $n$, so $a_n = \pm 1$. In particular, $a_{n+1} = a_n$ for all integers $n$, and $a_n$ is constant either 1 or -1. Substituting either of these sequences into (1) shows that both are valid solutions.

See Also

2016 UMO (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6
All UMO Problems and Solutions
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