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        riddles >> easy >> Sum of squares of 2007 consecutive integers
        
(Message started by: ecoist on Dec 1^st, 2007, 6:14pm)

Title: Sum of squares of 2007 consecutive integers
Post by ecoist on Dec 1^st, 2007, 6:14pm

Show that the sum of the squares of 2007 consecutive integers cannot be the n-th power of an integer, for any integer n>1.

Title: Re: Sum of squares of 2007 consecutive integers
Post by Grimbal on Dec 2^nd, 2007, 12:35am

[hideb]2007 = 3·3·223
If you take the sum of the squares mod 9, it doesn't matter where you start. In fact, you sum several time over the same numbers.
sum_{[i=n...n+2006]} i²
= sum_[i=0...2006] i² (mod 9)
= 223·sum_[i=0...8] i² (mod 9)

sum_[i=0...8] i² = 0+1+4+0+7+7+0+4+1 = 24 = 6 (mod 9).

So, sum_[i=0...8] i² is a multiple of 3 but not of 9.
and 223 is not a multiple of 3
so the original sum is a multiple of 3 but not of 9. It has only one prime factor 3, so it can not be a power >1 of an integer.[/hideb]

Title: Re: Sum of squares of 2007 consecutive integers
Post by Hippo on Dec 2^nd, 2007, 1:02pm

Nice reasoning ;)

Title: Re: Sum of squares of 2007 consecutive integers
Post by ecoist on Dec 2^nd, 2007, 3:46pm

And self-contained! Avoids using the formula for the sum of the first n squares. The clarity is refreshing as well!