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Topic: Squares to zero (Read 397 times) |
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pcbouhid
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Squares to zero
« on: Dec 12th, 2005, 5:02pm » |
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Add only (+) and (-) signs between 1^2, 2^2, 3^2... and 2005^2 and make it finally equal zero.
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| « Last Edit: Jan 4th, 2006, 8:11am by pcbouhid » |
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Eigenray
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Re: SQUARES TO ZERO
« Reply #1 on: Dec 12th, 2005, 10:18pm » |
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Can't be done. Any such sum will always be odd. Unless you had something else in mind?
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pcbouhid
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Re: SQUARES TO ZERO
« Reply #2 on: Dec 13th, 2005, 9:21am » |
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on Dec 12th, 2005, 10:18pm, Eigenray wrote:| Can't be done. Any such sum will always be odd. |
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Right! Now, make it equal to 1.
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| « Last Edit: Dec 13th, 2005, 9:22am by pcbouhid » |
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Eigenray
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Re: SQUARES TO ZERO
« Reply #3 on: Dec 13th, 2005, 11:59am » |
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| hidden: | First note that since -n2+(n+1)2=(2n+1), we can replace a pair n2,(n+1)2 by 2n+1. Furthermore, since
n2 - (n+1)2 - (n+2)2 + (n+3)2 = -(2n+1) + (2n+5) = 4,
we can replace any block of four consecutive squares by a 4. Thus we can reduce to the numbers
1, 5, 9, 13, 17, 4,4,4,...,4 (499 4's),
where each of 5,9,13,17 represents 2 squares of the original, and each 4 is a block of 4 squares. But
1 = 1 + 5 + 9 + 13 + 17 + 4*244 - 4*255,
which corresponds the the sequence of signs
+, -+, -+, -+, -+, [+--+, +--+, ... (244 times)], [-++-, -++-, ... (255 times)]. |
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Barukh
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Re: SQUARES TO ZERO
« Reply #4 on: Dec 14th, 2005, 10:29am » |
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Another arrangment would be:
+-+ (+-) [375] -- (+-) [248] ++ (+-) [375] -+
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