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 2007-04-26, 09:47 #1 Kees     Dec 2005 22×72 Posts odd numbers A nice simple problem $x_{i}$ odd integers, $n>2$ To prove: $4|(\prod_{i=1}^{n}x_{i }x_{i+1})-n$, where $x_{n+1}=x_1$
 2007-04-26, 12:28 #2 Wacky     Jun 2003 The Texas Hill Country 32·112 Posts Kees, I don't think that this is true unless n = 1 mod 4.
 2007-04-26, 12:33 #3 Kees     Dec 2005 C416 Posts Perhaps I made a typo, but could you please give me a counterexample
 2007-04-26, 12:36 #4 Kees     Dec 2005 22×72 Posts Ah, I see the typo, sorry it should be $4|(\sum_{i=1}^{n}x_{i }x_{i+1})-n,\ \mathrm{where}\qquad x_{n+1}=x_1$
 2007-04-26, 15:50 #5 Wacky     Jun 2003 The Texas Hill Country 32·112 Posts That appears to be the same expression as before. Rather than … -n, don't you want … -1 ?
 2007-04-26, 18:53 #6 davar55     May 2004 New York City 2×32×5×47 Posts The PI for multiplication was changed to a SIGMA for addition. And I believe it's true for n > 1, not just n > 2.
 2007-04-26, 19:24 #7 davar55     May 2004 New York City 2×32×5×47 Posts Every odd number is congruent to either 1 or 3 (mod 4). A product xixi+1 is congruent to 1 (mod 4) if both xi and xi+1 have the same 1 or 3 parity and is congruent to 3 (mod 4) if they have different 1 or 3 parity. But the given sum equals (x1x2 - 1) + (x2x3 - 1) + ... + (xnxn+1 - 1). So the terms of this sum are congruent to 0 or 2 (mod 4). If there are an even number 2k of such terms congruent to 2 (mod 4), then the sum will be congruent to 0 + 2*2k = 4k (mod 4), i.e 0 (mod 4). But this must be true, since we just need to consider the number of changes of 1 or 3 parity of the xi as i progresses from 1 to n+1. This must be even since the last xn+1 = x1.
2007-04-27, 11:18   #8
davieddy

"Lucan"
Dec 2006
England

2·3·13·83 Posts

Quote:
 Originally Posted by davar55 The PI for multiplication was changed to a SIGMA for addition.
Or "Crooked E" as Jasong decribed it

2007-04-28, 07:40   #9
davieddy

"Lucan"
Dec 2006
England

145128 Posts

Quote:
 Originally Posted by davar55 Every odd number is congruent to either 1 or 3 (mod 4). A product xixi+1 is congruent to 1 (mod 4) if both xi and xi+1 have the same 1 or 3 parity and is congruent to 3 (mod 4) if they have different 1 or 3 parity.
Neat proof davar.
For completeness:
(4a+1)(4b+3)=16ab+12a+4b+3 (=3 mod 4)
(4a+1)(4b+1)=16ab+4a+4b+1 (=1 mod 4)
(4a+3)(4b+3)=16ab+12a+12b+9 (=1 mod 4)

The result is true for n>0 surely.

David

Last fiddled with by davieddy on 2007-04-28 at 07:53

 2007-04-28, 08:55 #10 davieddy     "Lucan" Dec 2006 England 194A16 Posts Induction? Is the proof by induction more or less instructive? (Said proof left to interested readers ) Last fiddled with by davieddy on 2007-04-28 at 08:56
 2007-05-03, 11:04 #11 davieddy     "Lucan" Dec 2006 England 2·3·13·83 Posts If this proposition is false there is a minimum n for which it fails. Let N be this value. Take N odd numbers. If it fails for this then we can prove that it also fails for N-1 (Excersize for reader). Etc David

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