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. | ||
− | <i><b>Case 1</b></i> Let <math>n = 3.</math> We get system of equations | + | <i><b>Case 1</b></i> |
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+ | Let <math>n = 3.</math> We get system of equations | ||
<cmath> | <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> | ||
− | <i><b>Case 1'</b></i> Let <math>n = 3k, k={1,2,...}.</math> | + | <i><b>Case 1'</b></i> |
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+ | 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>. |
Revision as of 01:12, 16 August 2022
Find all numbers for which there exists real numbers satisfying and for
Solution
We find at least one series of real numbers for for each and we prove that if then the series does not exist.
Case 1
Let We get system of equations
We subtract the first equation from the second and get: So
Case 1'
Let Real numbers satisfying and .