20070426, 09:47  #1 
Dec 2005
2^{2}×7^{2} Posts 
odd numbers
A nice simple problem
odd integers, To prove: , where 
20070426, 12:28  #2 
Jun 2003
The Texas Hill Country
3^{2}·11^{2} Posts 
Kees,
I don't think that this is true unless n = 1 mod 4. 
20070426, 12:33  #3 
Dec 2005
C4_{16} Posts 
Perhaps I made a typo, but could you please give me a counterexample

20070426, 12:36  #4 
Dec 2005
2^{2}×7^{2} Posts 
Ah, I see the typo, sorry
it should be 
20070426, 15:50  #5 
Jun 2003
The Texas Hill Country
3^{2}·11^{2} Posts 
That appears to be the same expression as before.
Rather than … n, don't you want … 1 ? 
20070426, 18:53  #6 
May 2004
New York City
2×3^{2}×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. 
20070426, 19:24  #7 
May 2004
New York City
2×3^{2}×5×47 Posts 
Every odd number is congruent to either 1 or 3 (mod 4).
A product x_{i}x_{i+1} is congruent to 1 (mod 4) if both x_{i} and x_{i+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 (x_{1}x_{2}  1) + (x_{2}x_{3}  1) + ... + (x_{n}x_{n+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 x_{i} as i progresses from 1 to n+1. This must be even since the last x_{n+1} = x_{1}. 
20070427, 11:18  #8 
"Lucan"
Dec 2006
England
2·3·13·83 Posts 

20070428, 07:40  #9  
"Lucan"
Dec 2006
England
14512_{8} Posts 
Quote:
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 20070428 at 07:53 

20070428, 08:55  #10 
"Lucan"
Dec 2006
England
194A_{16} Posts 
Induction?
Is the proof by induction more or less instructive?
(Said proof left to interested readers ) Last fiddled with by davieddy on 20070428 at 08:56 
20070503, 11:04  #11 
"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 N1 (Excersize for reader). Etc David 
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