This problem and the following generalisation appeared 1979 in Ars Combinatoria (thanks to Darij who found it): Let $ (G, + )$ be a finite abelian group of order $ n$. Let also $ a_1,a_2,...,a_{n - 1} \in G$ be arbitrary. Then there exist pairwise distinct $ b_1,b_2,...,b_{n - 1} \in G$ and pairwise distinct $ c_1,c_2,...,c_{n - 1} \in G$ such that $ a_k = b_k + c_k$ for $ k = 1,2,...,n - 1$. [Moderator edit: The Ars Combinatoria paper is: F. Salzborn, G. Szekeres, A problem in Combinatorial Group Theory, Ars Combinatoria 7 (1979), pp. 3-5.]

so could someone post a proof of either the problem or its generalization?

you can find the proof in file shortlist 2005 which has been posted by orl The main idea is given two sequence $a_{1}...a_{n}$ and $b_{1}...b_{n}$ s.t $\sum a_{i}\equiv 0(mod n)$ and $\sum b_{i}\equiv 0(mod n)$ and there are exactly two i;j s.t $a_{i}\neq\ b_{i}(modn)$ and $a_{j}\neq\ b_{j}(modn)$.Then if we know two permutation good for the sequence (a_1...a_n) then we can build two permuttionm good for (b_1...b_n) i will come back with detail if you need

Well, I will post the solution from Ars Combinatoria if a re-find that two sheets of paper... It's a bit different from the ISL one.

I think you didnt keep promise,Zetax Please post it here and now!

to spdf: that was also my idea when i first saw the problem but i can't find a good way to contruct those 2 permutations for $ b_j$.you said you can post details.please do so

Sorry for not repsonding (I merely forgot...). But I just saw that problem again: In a slightly different manner (but being equivalent to this one here) it is solved in "The Mathematics of Juggling", called the "Converse of the Average Theorem". Main ideas: You show that this property (being a sum of two permutations) is invariant under the operations $ a_{i,j,d}$ that add $ d$ to $ a_i$ and subtract $ d$ from $ a_j$. For this, you need to do it algorithmically (but describing it is a bit hard without that graphics...).

ZetaX wrote: Sorry for not repsonding (I merely forgot...). But I just saw that problem again: In a slightly different manner (but being equivalent to this one here) it is solved in "The Mathematics of Juggling", called the "Converse of the Average Theorem". Main ideas: You show that this property (being a sum of two permutations) is invariant under the operations $ a_{i,j,d}$ that add $ d$ to $ a_i$ and subtract $ d$ from $ a_j$. For this, you need to do it algorithmically (but describing it is a bit hard without that graphics...). that is what I tried to do, but I can not prove that we can do it for the general case... can someone please help?

Since it's almost twelve years without complete solution, here's the official solution: Suppose there exists permutations $\sigma$ and $\tau$ of $[n]$ for some sequence $\{ a_i\}_{i\in [n]}$ so that $a_i\equiv_n \sigma (i)+\tau (i)$ for all $i\in [n]$. Given a sequence $\{ b_i\}_{i\in [n]}$ with sum divisible by $n$ that differ, in modulo $n$, from $\{ a_i\}_{i\in [n]}$ in only two positions, say $i_1$ and $i_2$. We want to construct permutations $\sigma'$ and $\tau'$ of $[n]$ so that $b_i\equiv_n \sigma' (i) +\tau' (i)$ for all $i\in [n]$. Recall that $b_i\equiv a_i\pmod{n}$ for all $i\in [n]$ that $i\neq i_1,i_2$. Construct a sequence $i_1,i_2,i_3,...$ by, for each integer $k\geq 2$, define $i_{k+1}\in [n]$ to be the unique integer satisfy $\sigma (i_{k-1})+\tau (i_{k+1})\equiv_n b_{i_k}$. Let (clearly exists) $p<q$ are the indices that $i_p=i_q$ with minimal $p$, and then minimal $q$. If $p>2$. This means $i_j\not\in \{ i_1,i_2\} \implies \sigma (i_j) +\tau (i_j) \equiv_n b_{i_j}$ for all $j\in \{ p,p+1,...,q\}$. Summing the equation $\sigma (i_{k-1})+\tau (i_{k+1})\equiv_n b_{i_k}$ for $k\in \{ p,p+1,...,q-1\}$ gives us $$\sum_{j=p-1}^{q-2}{\sigma (i_j) } +\sum_{j=p+1}^{q}{\tau (i_j)} \equiv_n\sum_{j=p}^{q-1}{b_{i_j}} \implies \sigma (i_{p-1}) +\sigma (i_p) +\tau (i_{q-1}) +\tau (i_q) \equiv_nb_{i_p}+b_{i_{q-1}}.$$Plugging $i_p=i_q$ and use $\sigma (i_p) +\tau (i_p)\equiv_n b_{i_p}$ gives us $\sigma (i_{p-1}) +\tau (i_{q-1})\equiv_n b_{i_{q-1}} \equiv_n \sigma (i_{q-1})+\tau (i_{q-1})$. Hence, $\sigma (i_{p-1}) \equiv_n \sigma (i_{q-1})\implies i_{p-1}=i_{q-1}$, contradiction to the definition of $p,q$. So, we've $p\in \{ 1,2\}$. Let $p'=3-p$. Define the desired permutations $\sigma'$ and $\tau'$ as follows: $$\sigma' (i_l)=\begin{cases} \sigma (i_{l-1}), & \text{ if } l\in \{ 2,3,...,q-1\} \\ \sigma (i_{q-1}), & \text{ if } l=1 \end{cases} ,\tau' (i_l)= \begin{cases} \tau (i_{l+1}), & \text{ if } l\in \{ 2,3,...,q-1\} \\ \tau (i_{p'}), & \text{ if } l=1 \end{cases} $$and $\sigma' (i) =\sigma (i),\tau' (i)=\tau (i)$ for the rest $i\in [n]$ that $i\not\in \{ i_1,i_2,...,i_{q-1}\}$. Note that the reason we choose $\tau (i_{p'})$ is just to not use $\tau (i_p)=\tau (i_{(q-1)+1})$ more than one time. This construction gives us $\sigma' (i)+\tau' (i)\equiv_n b_i$ for all $i\in [n]$ except when $i=i_1$. But since both $\sigma'$ and $\tau'$ are permutations of $[n]$, we've $\sum_{i\in [n]}{(\sigma' (i)+\tau' (i))} \equiv_n 2\times \frac{n(n-1)}{2}\equiv_n 0\equiv_n \sum_{i\in [n]}{b_i}$. This guarantee that $\sigma' (i) +\tau' (i)\equiv_n b_i$ when $i=i_1$ too. This prove the validity of permutations we constructed.

The case with $n$ prime is also resolved in this paper

any other solutions ?