Date: 10/26/96 at 15:18:8 From: Jae Jang Subject: Math problem Compute the smallest integer n, greater than one, for which the root-mean-square of the first n positive integers is an integer.
Date: 10/29/96 at 7:5:39 From: Doctor Yiu Subject: Re: Math problem Dear Jae, The SMALLEST of such integers is n = 337. There is a simple formula to compute the sum of the squares of the first n positive integers. It is: 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6. Using this, we find the sum of the squares of the first 337 positive integers is (337 * 338 * 775)/6. The average of these 337 squares is: (338*775)/6 = (338/2)*(775/3) = 169 * 225 = (13*15)^2 = 195^2 To check that this is indeed the SMALLEST n, you can write a computer program to verify that for n smaller than 337, the number (n+1)(2n+1)/6 is not a square. The determination of l positive integer n satisfying your requirement amounts to solving the equation (n+1)(2n+1) = 6y^2 for INTEGERS n and y. There is a standard procedure to solve such equations completely. If you want to find ALL possible n so that the root-mean-square of the first n positive integers is a positive integer y, here is an ITERATIVE method. Beginning with (n,y) = (1,1), form: new n = 97(old n) + 168(old y) + 72 new y = 56(old n) + 97(old y) = 42 In this way, the FIRST 5 of such pairs are: (n,y) = (337, 195), (65521, 37829) (12710881, 7338631) (2465845537, 1423656585) (478361323441, 276182038859) You can of course continue this INDEFINITELY. -Doctor Yiu, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum