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Oscar Lanzi III science forum Guru Wannabe
Joined: 30 Apr 2005
Posts: 176

Posted: Thu Jul 06, 2006 12:38 am Post subject:
Re: A curious identity



Not if L = 1. If I'm interpreting this right, all terms of S_1 are
ii, hence zero. It follows that you should require n >= 2 (requiring
2n+1 to be prime does not of itself eliminate n = 1), and place L in
G\{1}.
The primality of 2n+1 enforces that ij with i and j in the indicated
order occurs in *one* S_L sum for any (i,j) in GxG, and that all zero
elements ii occur specifically in the L = 1 sum. The sum of all the
absolute values is twice a tetrahedral number, specifically
(n1)*n*(n+1)/3. If this partitions equally among the (n1) values of L
that have the nonzero terms, we will get n*(n+1)/3 for each.
FWIW, my *guess* is that the modified proposition is true.
OL 

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ymchee@gmail.com science forum beginner
Joined: 26 Jun 2006
Posts: 1

Posted: Mon Jun 26, 2006 6:31 am Post subject:
A curious identity



Let n be a positive integer such that 2n+1 is prime. Elements of the
factor group G = Z^*_{2n+1}/H, where H = {1,+1}, may be taken to be
{1,2,...,n}. For every x in Z^*_{2n+1}, let x' \in {1,2,...,n} denote
its image in G. For every L in {1,2,...,n}, let
S_L = \sum_{a=1}^n abs (a(aL)'),
where abs(.) denotes the usual absolute value. For example, when n=6,
and L=3,
S_3 = \sum_{a=1}^6 abs(a(3a)') = abs(13) + abs(26) + abs(34) +
abs(41) + abs(52) + abs(65) = 14.
Is it true that S_L = n(n+1)/3 always, for every L?
Computational evidence supports this.
Thanks!  Y. M. Chee 

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