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Forum index » Science and Technology » Math » Research
A curious identity
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ymchee@gmail.com
science forum beginner


Joined: 26 Jun 2006
Posts: 1

PostPosted: Mon Jun 26, 2006 6:31 am    Post subject: A curious identity Reply with quote

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(1-3) + abs(2-6) + abs(3-4) +
abs(4-1) + abs(5-2) + abs(6-5) = 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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Oscar Lanzi III
science forum Guru Wannabe


Joined: 30 Apr 2005
Posts: 176

PostPosted: Thu Jul 06, 2006 12:38 am    Post subject: Re: A curious identity Reply with quote

Not if L = 1. If I'm interpreting this right, all terms of S_1 are
|i-i|, 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 |i-j| 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 |i-i| occur specifically in the L = 1 sum. The sum of all the
absolute values is twice a tetrahedral number, specifically
(n-1)*n*(n+1)/3. If this partitions equally among the (n-1) 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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