Partition of the set $\mathbb{Z}_p^*=\{1,2,3,...,p-1\}$ in pairs $(a,b)$ where $b-a=$ constant $\pmod{p}$, with $p$ an odd prime. I would like to know if someone already studied the subject where one need to partition the set $\{1,2,3,...,p-1\}$, $p$ odd prime,  in sets of pairs $(a,a+k)$ where $a$ and $a+k$ are considered $\bmod p$, and $k$ is a fixed number in each set of pairs.
I give some examples:  in $\mathbb{Z}_{13}$ I may partition the set $\{1,2,3,...,p-1\}$  in pairs
$$(1,2), (3,4),(5,6),(7,8),(9,10),(11,12)$$ where the second element equals the first $+1$  (here $k=1$).
But the pairs $$(1,10),(3,12),(6,2),(8,4),(9,5),(11,7)$$ are also a solution; in this case $k=9$.
I think that I have understood the situation when the constant $k$ is the same for all pairs.
But what if we partition the set $\{1,2,3,...,p-1\}$  in two sets of pairs: the first has pairs $(a,a+k) \pmod{p}$ for some $k$, and the second has pairs $(b,b+k_1)\pmod{p}$ for some constant $k_1$ distinct from $k$?
Example: in $\mathbb{Z}_{13}$ I have the pairs $(1,12), (11,9), (10,8)$ and their difference $\bmod 13$ is $k =11$. The remaining elements may belong to the pairs $(7,4), (6,3), (5,2)$ and their difference is $k\equiv 10\pmod{13} $ $(2-5=3-6=4-7=-3 \equiv 10 \pmod{13}$.
I hope that the explanation is clear. My question is if it is possible that, for a given $p$, to know something about the possible $k$ such that partitions like this make sense and the number of solutions (i.e. partitions) that exist. The simplest case occurs when there is only one constant; but for a big enough prime $p$ partitions may occur which involve many constants.
 A: For every $p$ an odd prime and $k$ which without loss of generality assume to be in $\{1,...,p-1\}$ there is a unique way to partition the set $\{1,2,...,p-1\}$ into pairs $(a,a+k)$. Note $k$ is either in the bracket $(0,k)$ or $(k,2k \mod p)$, since $0 \notin \{1,...,p-1\}$ it is in $(k,2k \mod p)$. If $3k \neq 0 \mod p$ then it is either in $(2k \mod p, 3k \mod p) $ or $(3k \mod p, 4k \mod p)$, $2k \mod p$ is already taken thus it is in the bracket $(3k\mod p,4k \mod p)$...we can continue this chain to note that the only viable partition is
$$(k,2k\mod p), (3k\mod p, 4k \mod p),...((p-2)k\mod p, (p-1)k \mod p)$$.
A: Let $a=\frac{p+1}{2} $
$a-1=\frac{p+1}{2}-1=\frac{p-1}{2} $
$a-1=(a+1)-2=(a+2)-3=\dots $
So $1$ can be paired with $a$, $2$ with $a+1$, etc. up to $\frac{p-1}{2} $ with $p-1$ and their differences will all be the same, i.e. $\frac{p-1}{2} $, hence they will all be the same $\bmod p$
A: I don't really understand the question, but every nonzero class $k$ modulo$~p$ is a generator of the additive group $\def\Z{\Bbb Z}\Z/p\Z$. The means that the relation (between classes $x,y$) defined by the condition $x-y\equiv k\pmod p$ determines an oriented graph on $\Z/p\Z$ that abstractly is a single oriented cycle. After removing the class of $0$ and the edges involving it, what is left is a linear graph on $p-1$ elements. Clearly removing every other edge is now the unique way of reducing this graph to $\frac{p-1}2$ components consisting of two vertices linked by one (oriented) edge.
