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

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==Problem==
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Find all numbers <math>n \ge 3</math> for which there exists real numbers <math>a_1, a_2, ..., a_{n+2}</math> satisfying <math>a_{n+1} = a_1, a_{n+2} = a_2</math> and  
 
Find all numbers <math>n \ge 3</math> for which there exists real numbers <math>a_1, a_2, ..., a_{n+2}</math> satisfying <math>a_{n+1} = a_1, a_{n+2} = a_2</math> and  
 
<cmath>a_{i}a_{i+1} + 1 = a_{i+2}</cmath>
 
<cmath>a_{i}a_{i+1} + 1 = a_{i+2}</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>
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<i><b>Case 1a</b></i>
  
 
Let <math>n = 3k, k={1,2,...}.</math>  
 
Let <math>n = 3k, k={1,2,...}.</math>  
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<cmath>{a1a2a3+a3=a32a2a3a4+a4=a42a3a4a1+a1=a12a4a1a2+a2=a22</cmath>
 
<cmath>{a1a2a3+a3=a32a2a3a4+a4=a42a3a4a1+a1=a12a4a1a2+a2=a22</cmath>
 
We multiply each equation by a number that precedes a pair of product numbers in a given sequence <math>a_1, a_2, a_3, a_4, a_1, a_2.</math> So we multiply the equation with product <math>a_1 a_2</math> by <math>a_4</math>, we multiply the equation with product <math>a_4 a_1</math> by <math>a_3</math> etc. We get:
 
We multiply each equation by a number that precedes a pair of product numbers in a given sequence <math>a_1, a_2, a_3, a_4, a_1, a_2.</math> So we multiply the equation with product <math>a_1 a_2</math> by <math>a_4</math>, we multiply the equation with product <math>a_4 a_1</math> by <math>a_3</math> etc. We get:
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<cmath>{a1a2a4+a4=a3a4a2a3a1+a1=a4a1a3a4a2+a2=a1a2a4a1a3+a3=a2a3</cmath>
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We add all the equations of the first system, and all the equations of the second system. The sum of the left parts are the same! It includes the sum of all the numbers <math>a_i</math> and the sum of the triples of consecutive numbers <math>a_i a_{i + 1} a_{i + 2}.</math> Hence, the sums of the right parts are equal, that is,
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<cmath>a_1^2 + a_2^2 + ... + a_4^2 – a_1 a_2 – a_2 a_3 – a_3 a_4 – a_1 a_4 = 0.</cmath>
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It is known that this expression is doubled
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<cmath>(a_1 – a_2)^2 +  (a_2 – a_3)^2 + ... + (a_4 – a_1)^2 = 0 \implies a_1 = a_2 = a_3 = a_4 .</cmath>
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Substituting into any of the initial equations, we obtain the equation <math>a_1^2 + 1 = a_1,</math> which does not have real roots. Hence, there are no such real numbers.
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<i><b>Case 2a</b></i>
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Let <math>n = 5.</math> We get system of equations
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<cmath>{a1a2+1=a3a2a3+1=a4a3a4+1=a5a4a5+1=a1a5a1+1=a2</cmath>
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We repeat all steps of <i><b>Case 2</b></i> and get: there are no such real numbers.
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<i><b>Case 2b </b></i>
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Let <math>n = 3k \pm  1.</math> We repeat all steps of cases <math>2</math> and <math>2a</math> and get: there are no such real numbers.
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'''vladimir.shelomovskii@gmail.com, vvsss'''
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==See Also==
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{{IMO box|year=2018|num-b=1|num-a=3}}

Latest revision as of 01:44, 19 November 2023

Problem

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 1a

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 $a_{i}a_{i+1} + 1 = a_{i+2}$.

Case 2

Let $n = 4.$ We get system of equations \[\begin{cases} a_1 a_2 + 1 = a_3 \\a_2 a_3 + 1 = a_4 \\a_3 a_4 + 1 = a_1\\a_4 a_1 + 1 = a_2 \end{cases}\] We multiply each equation by the number on the right-hand side and get: \[\begin{cases} a_1 a_2 a_3 + a_3 = a_3^2 \\a_2 a_3 a_4 + a_4 = a_4^2 \\a_3 a_4 a_1 + a_1 = a_1^2 \\a_4 a_1 a_2 + a_2 = a_2^2 \end{cases}\] We multiply each equation by a number that precedes a pair of product numbers in a given sequence $a_1, a_2, a_3, a_4, a_1, a_2.$ So we multiply the equation with product $a_1 a_2$ by $a_4$, we multiply the equation with product $a_4 a_1$ by $a_3$ etc. We get: \[\begin{cases} a_1 a_2 a_4 + a_4 = a_3 a_4 \\a_2 a_3 a_1 + a_1 = a_4 a_1 \\a_3 a_4 a_2 + a_2 = a_1 a_2 \\a_4 a_1 a_3 + a_3 = a_2 a_3 \end{cases}\] We add all the equations of the first system, and all the equations of the second system. The sum of the left parts are the same! It includes the sum of all the numbers $a_i$ and the sum of the triples of consecutive numbers $a_i a_{i + 1} a_{i + 2}.$ Hence, the sums of the right parts are equal, that is, \[a_1^2 + a_2^2 + ... + a_4^2 – a_1 a_2 – a_2 a_3 – a_3 a_4 – a_1 a_4 = 0.\] It is known that this expression is doubled \[(a_1 – a_2)^2 + (a_2 – a_3)^2 + ... + (a_4 – a_1)^2 = 0 \implies a_1 = a_2 = a_3 = a_4 .\] Substituting into any of the initial equations, we obtain the equation $a_1^2 + 1 = a_1,$ which does not have real roots. Hence, there are no such real numbers.

Case 2a

Let $n = 5.$ We get system of equations \[\begin{cases} a_1 a_2 + 1 = a_3 \\a_2 a_3 + 1 = a_4 \\a_3 a_4 + 1 = a_5\\a_4 a_5 + 1 = a_1 \\a_5 a_1 + 1 = a_2\end{cases}\] We repeat all steps of Case 2 and get: there are no such real numbers.

Case 2b

Let $n = 3k \pm 1.$ We repeat all steps of cases $2$ and $2a$ and get: there are no such real numbers.

vladimir.shelomovskii@gmail.com, vvsss

See Also

2018 IMO (Problems) • Resources
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 3
All IMO Problems and Solutions
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