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Topic: Sum of squares of 2007 consecutive integers (Read 459 times) |
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ecoist
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Sum of squares of 2007 consecutive integers
« on: Dec 1st, 2007, 6:14pm » |
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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.
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Grimbal
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Re: Sum of squares of 2007 consecutive integers
« Reply #1 on: Dec 2nd, 2007, 12:35am » |
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| hidden: | 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. |
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Hippo
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Re: Sum of squares of 2007 consecutive integers
« Reply #2 on: Dec 2nd, 2007, 1:02pm » |
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Nice reasoning 
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ecoist
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Re: Sum of squares of 2007 consecutive integers
« Reply #3 on: Dec 2nd, 2007, 3:46pm » |
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And self-contained! Avoids using the formula for the sum of the first n squares. The clarity is refreshing as well!
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