Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 18: Line 18:
 
Let <math>n = 3k, k={1,2,...}.</math>  
 
Let <math>n = 3k, k={1,2,...}.</math>  
 
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>.
 
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>.
 +
 +
<i><b>Case 2</b></i>
 +
 +
Let <math>n = 4.</math> We get system of equations
 +
<cmath>{a1a2+1=a3a2a3+1=a4a3a4+1=a1a4a1+1=a2</cmath>
 +
We multiply each equation by the number on the right-hand side and get:
 +
<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:

Revision as of 02:21, 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 $a_{i}a_{i+1} + 1 = a_{i+2}$.

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:

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_IMO_Problems/Problem_2&oldid=176986"