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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 $1,2,\ldots,n$. For every $x \in Z^\*_{2n+1}$, let $x' \in \{1,2,\ldots,n\}$ denote its image in $G$. For every $\lambda \in\{2,\ldots,n\}$, let

$S_\lambda = \sum_{a=1}^n {\rm abs} (a-(a\lambda)')$,

where ${\rm abs}(\cdot)$ denotes the usual absolute value. For example, when $n=6$, and $\lambda=3$, $S_3 = \sum_{a=1}^6 {\rm abs}(a-(3a)') = {\rm abs}(1-3) + {\rm abs}(2-6) + {\rm abs}(3-4) + {\rm abs}(4-1) + {\rm abs}(5-2) + {\rm abs}(6-5) = 14$.

Is it true that $S_\lambda = n(n+1)/3$, for every $\lambda \geq 2$?

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Where does this problem come from? It was mentioned in mathlinks.ro/viewtopic.php?t=125645 , and hipsishopsis has posted a proof. Haven't checked it, however. –  darij grinberg Mar 3 '10 at 15:01
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