Difference between revisions of "2018 IMO Problems/Problem 2"
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==Solution== | ==Solution== | ||
| − | 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 | + | 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 | ||
| + | <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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| + | We subtract the first equation from the second and get: <cmath>a_2 (a_3 – a_1) = (a_1 – a_3).</cmath> | ||
| + | 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. | ||
Revision as of 01:03, 16 August 2022
Find all numbers 
for which there exists real numbers 
satisfying 
and
![\[a_{i}a_{i+1} + 1 = a_{i+2}\]](http://latex.artofproblemsolving.com/d/b/3/db3ac32b874d4123e08da8e59006ab7b4d64e2b7.png)
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
![\[\begin{cases} a_1 a_2 + 1 = a_3 \\a_2 a_3 + 1 = a_1 \\a_3 a_1 + 1 = a_2 \end{cases}\]](http://latex.artofproblemsolving.com/9/a/5/9a538c38d276127025c05d8703195344c61b35ca.png)
We subtract the first equation from the second and get: ![\[a_2 (a_3 – a_1) = (a_1 – a_3).\]](http://latex.artofproblemsolving.com/e/7/c/e7ca9f0b62d4be598fa6ae274ea2c47717d2676e.png)
So 
Case 1' Let 
The sequence 
is the desired sequence.
