# 2018 IMO Problems/Problem 2

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 1a**

Let Real numbers satisfying and .

**Case 2**

Let We get system of equations We multiply each equation by the number on the right-hand side and get: We multiply each equation by a number that precedes a pair of product numbers in a given sequence So we multiply the equation with product by , we multiply the equation with product by etc. We get: 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 and the sum of the triples of consecutive numbers Hence, the sums of the right parts are equal, that is, It is known that this expression is doubled Substituting into any of the initial equations, we obtain the equation which does not have real roots. Hence, there are no such real numbers.

**Case 2a**

Let We get system of equations
We repeat all steps of * Case 2* and get: there are no such real numbers.

**Case 2b **

Let We repeat all steps of cases and and get: there are no such real numbers.

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