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.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
1987 IMO (Problems) • Resources | ||
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