2009 USAMO Problems/Problem 6
Problem
Let be an infinite, nonconstant sequence of rational numbers, meaning it is not the case that Suppose that is also an infinite, nonconstant sequence of rational numbers with the property that is an integer for all and . Prove that there exists a rational number such that and are integers for all and .
Solution
Suppose the can be represented as for every , and suppose can be represented as . Let's start with only the first two terms in the two sequences, and for sequence and and for sequence . Then by the conditions of the problem, we have is an integer, or is an integer. Now we can set , because the least common denominator of is and of is , and multiplying or dividing appropriately by will always give an integer.
Now suppose we kept adding and until we get to in sequence and in sequence so that is an integer for all with , where is a positive integer. At this point, we will have = , because these are the least common denominators of the two sequences up to . As we keep adding and , will always have value , and we are done.
See Also
2009 USAMO (Problems • Resources) | ||
Preceded by Problem 5 |
Followed by Last question | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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