1987 IMO Problems/Problem 3
Problem
Let be real numbers satisfying . Prove that for every integer there are integers , not all 0, such that for all and
.
Solution
We first note that by the Power Mean Inequality, . Therefore all sums of the form , where the is a non-negative integer less than , fall in the interval . We may partition this interval into subintervals of length . But since there are such sums, by the pigeonhole principle, two must fall into the same subinterval. It is easy to see that their difference will form a sum with the desired properties.
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