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Difference between revisions of "2018 IMO Problems/Problem 2"

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We find at least one series of real numbers for <math>n = 3,</math> for each <math>n = 3k</math> and we prove that if <math>n = 3k \pm 1,</math> then the series does not exist.
 
We find at least one series of real numbers for <math>n = 3,</math> for each <math>n = 3k</math> and we prove that if <math>n = 3k \pm 1,</math> then the series does not exist.
  
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<i><b>Case 1</b></i> Let <math>n = 3.</math> We get system of equations  
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<i><b>Case 1</b></i>
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Let <math>n = 3.</math> We get system of equations  
 
<cmath>\begin{cases} a_1 a_2 + 1 = a_3 \\a_2 a_3 + 1 = a_1 \\a_3 a_1 + 1 = a_2 \end{cases}</cmath>
 
<cmath>\begin{cases} a_1 a_2 + 1 = a_3 \\a_2 a_3 + 1 = a_1 \\a_3 a_1 + 1 = a_2 \end{cases}</cmath>
  
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So <math>a_2 = – 1 \implies  a_1 = 2, a_3 =  – 1.</math>
 
So <math>a_2 = – 1 \implies  a_1 = 2, a_3 =  – 1.</math>
  
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<i><b>Case 1'</b></i> Let <math>n = 3k, k={1,2,...}.</math> The sequence <math>{ 2, – 1, – 1, ..., 2,  – 1, – 1}</math> is the desired sequence.
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<i><b>Case 1'</b></i>
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Let <math>n = 3k, k={1,2,...}.</math>  
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Real numbers <math>a_1 =a_4 =...=2, a_2 = a_3 = a_5=...=-1</math> satisfying <math>a_{n+1} = a_1, a_{n+2} = a_2</math> and <math>a_{i}a_{i+1} + 1 = a_{i+2}</math>.

Revision as of 01:12, 16 August 2022

Find all numbers $n \ge 3$ for which there exists real numbers $a_1, a_2, ..., a_{n+2}$ satisfying $a_{n+1} = a_1, a_{n+2} = a_2$ and \[a_{i}a_{i+1} + 1 = a_{i+2}\] for $i = 1, 2, ..., n.$

Solution

We find at least one series of real numbers for $n = 3,$ for each $n = 3k$ and we prove that if $n = 3k \pm 1,$ then the series does not exist.

Case 1

Let $n = 3.$ We get system of equations \[\begin{cases} a_1 a_2 + 1 = a_3 \\a_2 a_3 + 1 = a_1 \\a_3 a_1 + 1 = a_2 \end{cases}\]

We subtract the first equation from the second and get: \[a_2 (a_3 – a_1) = (a_1 – a_3).\] So $a_2 = – 1 \implies a_1 = 2, a_3 = – 1.$

Case 1'

Let $n = 3k, k={1,2,...}.$ Real numbers $a_1 =a_4 =...=2, a_2 = a_3 = a_5=...=-1$ satisfying $a_{n+1} = a_1, a_{n+2} = a_2$ and $a_{i}a_{i+1} + 1 = a_{i+2}$.

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