Difference between revisions of "1986 USAMO Problems/Problem 3"
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(Added that Case 2 has no solutions by analyzing mod 4. Also changed variable name "p" to "k", as p commonly denotes a prime.) |
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Requiring that <math>I_n</math> be an integer, we find that | Requiring that <math>I_n</math> be an integer, we find that | ||
<cmath> | <cmath> | ||
− | (2n+1 ) (n+1) = | + | (2n+1 ) (n+1) = 6k^2, |
− | </cmath> where <math> | + | </cmath> where <math>k</math> is an integer. Using the Euclidean algorithm, we see that |
<math>\gcd(2n+1, n+1) = \gcd(n+1,n) = 1</math>, and so <math>2n+1</math> and <math>n+1</math> share no | <math>\gcd(2n+1, n+1) = \gcd(n+1,n) = 1</math>, and so <math>2n+1</math> and <math>n+1</math> share no | ||
factors greater than 1. The equation above thus implies that <math>2n+1</math> | factors greater than 1. The equation above thus implies that <math>2n+1</math> | ||
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which <math>(4b^2-1)/3</math> is a perfect square is <math>b=13</math>, which results in <math>n=337</math>. | which <math>(4b^2-1)/3</math> is a perfect square is <math>b=13</math>, which results in <math>n=337</math>. | ||
− | In Case 2, <math>2n+1 = 12b^2 - 1 = a^2</math>. | + | In Case 2, <math>2n+1 = 12b^2 - 1 = a^2</math>. Note that <math>a^2 = 2n+1</math> is an odd square, and hence is congruent to <math>1 \pmod 4</math>. But <math>12b^2 -1 \equiv 3 \pmod 4</math> for any <math>b</math>, so Case 2 has no solutions. |
+ | |||
+ | Alternatively, one can proceed by checking whether <math>12b^2 -1</math> is a perfect square for <math>b=1, 2 ,3 ,\dots</math>. We find that <math>12b^2 -1</math> is not a perfect square for <math>b = 1,2, 3, ..., 7, 8</math>, and <math>n= 383</math> when <math>b=8</math>. Thus the smallest positive integers <math>a </math> and <math>b</math> for which <math>12b^2- 1 = a^2</math> result in a value of <math>n</math> exceeding the value found in Case 1, which was 337. | ||
In summary, the smallest value of <math>n</math> greater than 1 for which <math>I_n</math> is an integer is <math>\boxed{337}</math>. | In summary, the smallest value of <math>n</math> greater than 1 for which <math>I_n</math> is an integer is <math>\boxed{337}</math>. |
Latest revision as of 12:19, 19 July 2023
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.