1986 USAMO Problems/Problem 3
Problem
What is the smallest integer , greater than one, for which the root-mean-square of the first positive integers is an integer?
The root-mean-square of numbers is defined to be
Solution
Let's first obtain an algebraic expression for the root mean square of the first integers, which we denote . By repeatedly using the identity , we can write and We can continue this pattern indefinitely, and thus for any positive integer , Since , we obtain Therefore, Requiring that be an integer, we find that where is an integer. Using the Euclidean algorithm, we see that , and so and share no factors greater than 1. The equation above thus implies that and is each proportional to a perfect square. Since is odd, there are only two possible cases:
Case 1: and , where and are integers.
Case 2: and .
In Case 1, . This means that for some integers and . We proceed by checking whether is a perfect square for . (The solution leads to , and we are asked to find a value of greater than 1.) The smallest positive integer greater than 1 for which is a perfect square is , which results in .
In Case 2, . Note that is an odd square, and hence is congruent to . But for any , so Case 2 has no solutions.
Alternatively, one can proceed by checking whether is a perfect square for . We find that is not a perfect square for , and when . Thus the smallest positive integers and for which result in a value of exceeding the value found in Case 1, which was 337.
In summary, the smallest value of greater than 1 for which is an integer is .
See Also
1986 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
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