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Title: sum of consecutive integers Post by Christine on Jul 14th, 2015, 1:04pm Can you find four consecutive numbers whose sum is a perfect square? |
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Title: Re: sum of consecutive integers Post by rloginunix on Jul 14th, 2015, 1:39pm [hide]No[/hide]. |
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Title: Re: sum of consecutive integers Post by rloginunix on Jul 14th, 2015, 5:56pm 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 [hide]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[/hide]. The choices recipes to follow. (The night is falling on the East Coast, I will post my proof tomorrow, hopefully) |
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Title: Re: sum of consecutive integers Post by towr on Jul 14th, 2015, 10:31pm For 4, [hide]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.[/hide] |
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Title: Re: sum of consecutive integers Post by rloginunix on Jul 15th, 2015, 9:47am For N: let x be the first term, (x + N - 1) - the last. We wish that the [hide]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[/hide] |
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