2016 UMO Problems/Problem 5
Problem
Let be a sequence of integers (positive, negative, or zero) such that for all nonnegative integers and , . Find all possible sequences .
Solution 1
We can factor the equation as . Substituting in , we ﬁnd that either or for all integers . If is a prime such that , then by repeatedly applying or , we ﬁnd that divides for all integers . Substituting and into (1) we notice that the LHS is divisible by , and the RHS is equal to ; hence , which is a contradiction. Thus is never divisible by any prime for any , so . In particular, for all integers , and is constant either 1 or -1. Substituting either of these sequences into (1) shows that both are valid solutions.
See Also
2016 UMO (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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