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Christine
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sum of consecutive integers
« on: Jul 14th, 2015, 1:04pm » |
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Can you find four consecutive numbers whose sum is a perfect square?
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rloginunix
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Re: sum of consecutive integers
« Reply #1 on: Jul 14th, 2015, 1:39pm » |
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No.
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rloginunix
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Re: sum of consecutive integers
« Reply #2 on: Jul 14th, 2015, 5:56pm » |
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Further. We generalize this quickly into a beautiful theorem:
When may the sum of N consecutive integers be represented as a perfect square?
If my arithmetic holds then the sum does not exist if the non-zero power of 2 of the prime factorization of N is even and, conversely, the sum does exist if the power of 2 of the prime factorization of N is either odd or zero. The choices recipes to follow.
(The night is falling on the East Coast, I will post my proof tomorrow, hopefully)
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towr
wu::riddles Moderator
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Some people are average, some are just mean.
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Re: sum of consecutive integers
« Reply #3 on: Jul 14th, 2015, 10:31pm » |
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For 4, the sum of 4 consecutive integers is always 4x+6, where x is the first number. Squares are always 0 or 1 modulo 4, but 4x+6 is 2 modulo 4, so it's never square.
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Wikipedia, Google, Mathworld, Integer sequence DB
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rloginunix
Uberpuzzler
Posts: 1029
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Re: sum of consecutive integers
« Reply #4 on: Jul 15th, 2015, 9:47am » |
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For N: let x be the first term, (x + N - 1) - the last. We wish that the sum of N such terms be a perfect square:
N(x + x + N - 1)/2 = N(2x + N - 1)/2 = s2
Solve for x:
x = s2/N + 1/2 - N/2 (1)
N is either 1) odd or 2) even.
1). N = 2q + 1. From (1) it follows that its last two terns add up to be whole and even. For x then to be whole (1)'s first term must be whole - choose s = N and put it into (1):
x = (N + 1)/2 = (2q + 2)/2 = q + 1
Example: q = 1, N = 3, x = 2, sum = 2 + 3 + 4 = 9 = 32
2). N = 2q. Put it into (1):
x = s2/(2q) + 1/2 - q (2)
From the prime factorization theorem it follows that any integer, q including, may be represented as 2k*d where d is odd and k is positive or zero. For k then we have two choices - odd or even.
2.1). Let k = 2z. Then q = 22z*d. From (2) it follows then that for x to be whole s2/q must be odd (why?). The choice for s then is obvious: s = 2z*d. Put it back into (2):
x = d/2 + 1/2 - 22z*d
It means that it works and N has an odd number of 2s in its factorization.
Example: d = 3, z = 1, q = 12, N = 24 = 23*3, x = -10, sum = -10 + ... + 13 = 36 = 62
2.1). Let k = 2z - 1. Then no matter how many 2s s has s2/q is never odd and whole (what is it?):
s = 2r*t
s2/q = 22r - 2z + 1*t/d
for whole r and z the power of 2 above is never zero (why?).
qed
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