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Topic: Sum of 999 perfect squares (Read 339 times) |
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NickH
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Sum of 999 perfect squares
« on: Jan 27th, 2004, 3:47pm » |
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Can the sum of 999 consecutive squares of integers ever be a perfect power of an integer?
Example: are there integers m, r > 1, such that 10052 + ... + 20032 = mr?
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Eigenray
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Re: Sum of 999 perfect squares
« Reply #1 on: Jan 27th, 2004, 11:38pm » |
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::Nope::
Let f(n) = [sum]k=nn+998 k2.
f(n+1)-f(n) = (n+999)2-n2 = 999(2n+999) == 0 mod 33.
Therefore f(n) == f(1) = 999(1000)(1999)/6 == 18 mod 27.
Now, if mr = f(n) = 27k + 18 = 32(3k+2), then r <= 2.
But if r = 2, then we have (m/3)2 = 3k+2, which is impossible.
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NickH
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Re: Sum of 999 perfect squares
« Reply #2 on: Jan 28th, 2004, 12:48pm » |
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Nice solution!
I had something similar:
If mr = (n - 499)2 + ... + (n + 499)2 = 999n2 + 2(12 + ... + 4992) = 999n2 + 499*500*999/3 = 333(3n2 + 499*500), then 3|mr, but not 32|mr. Contradiction.
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Eigenray
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Re: Sum of 999 perfect squares
« Reply #3 on: Jan 28th, 2004, 4:49pm » |
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on Jan 28th, 2004, 12:48pm, NickH wrote:| If mr = [...] = 333(3n2 + 499*500), then 3|mr, but not 32|mr. |
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But 32 does divide mr.
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NickH
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Re: Sum of 999 perfect squares
« Reply #4 on: Jan 29th, 2004, 1:15am » |
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Oops! I see I carelessly generalised from the problem I'd seen.
The proof I gave is valid for 99 or 9999 or 102n - 1 consecutive squares, but not for 102n-1 - 1 consecutive squares.
Thanks Eigenray.
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