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In 2013 I formulated the following conjecture in additive combinatorics.

Conjecture. Let $G$ be an additive abelian group of odd order, and let $A$ be a subset of $G$ with $|A|=n>2$. Then, there is a circular permutation $(a_1,\ldots,a_n)$ of all the elements of $A$ such that all the adjacent sums $ a_1+a_2,\ldots,a_{n-1}+a_n,a_n+a_1$ are pairwise distinct.

Recently, Mr. Yu-Xuan Ji, a student at Nanjing Univ., verified this conjecture for $|G|<30$.

I'm even unable to show the conjecture for $G=\mathbb Z/p\mathbb Z$ with $p$ an odd prime.

Any ideas towards the solution of this conjecture? Your comments are welcome!

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    $\begingroup$ Perhaps what was meant is "there is an enumeration" rather than "there is a circular permutation"? $\endgroup$
    – Abdelmalek Abdesselam
    Commented Jun 22, 2020 at 15:23
  • $\begingroup$ I use the word "circular" since we consider $a_n$ and $a_1$ adjacent. $\endgroup$
    – Zhi-Wei Sun
    Commented Jun 22, 2020 at 15:25
  • $\begingroup$ Given a choice of ordering of the elements of a set $A$ as $a_1,\ldots,a_n$, the standard understanding of "circular permutation" of that ordering is one of the form $a_i,\ldots,a_n,a_1\ldots,a_{i-1}$. If one understands the conjecture in this way, it does not make sense. $\endgroup$
    – Abdelmalek Abdesselam
    Commented Jun 22, 2020 at 15:28
  • $\begingroup$ The conjecture was contained in my paper available from maths.nju.edu.cn/~zwsun/196a.pdf $\endgroup$
    – Zhi-Wei Sun
    Commented Jun 22, 2020 at 15:28
  • $\begingroup$ Perhaps a better word is a cyclic permutation (in cycle notation) "$(a_1\cdots a_n)$". $\endgroup$
    – Abdelmalek Abdesselam
    Commented Jun 22, 2020 at 15:34

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