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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!   
Associated Topics:
High School Basic Algebra

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