Problem: Let $a\_1\dots a\_k$ be integers in $Z\_{n}$ such that $n=k(k-1)+1$ and that the list of differences $a\_i-a\_j \bmod n$ is unique to $i,j$ (for $i\neq j$). Such a set exists for $k=1\dots6,8$. No such set exists for $k=7$? (Verified by exhaustive search, but I can't figure out why that case would be different. Is it possible there's an error in my code?)


Does anyone have a quick idea or a short proof of why $k=7, n=43$ would be different? Alternatively, some larger-context work that subsumes this problem and will let me figure out why $n=7$ doesn't work?


Some properties of the $a\_i$: The order doesn't matter, so you might as well list them in increasing order. The set of differences has to equal every number between 1 and $n-1$ exactly once, so there will be an equivalent of any valid set that has a $0$ and a $1$. The set of differences is equivalent under rotation of the $a\_i$ by addition $\bmod n$ (by design), so pick the one that starts $0,1$ as the class representative. The $a\_i$ can also be represented by the $b\_i$ where $b\_i=a\_{i+1}-a\_i$. In this case the $b\_i$ form a partition of $n$ with all partial sums of sequences in order of the $b\_i$ being unique. The $b\_i$ are equivalent under rotation of order or flipping the order, but not in otherwise reordering. (IE, the difference list $[1,2,3,4]$ is equivalent to $[1,4,3,2]$ but not $[1,4,2,3]$.)


Background:
I was doing an analysis of a card game involving restricted pairs, and found some interesting properties for sets of $n=k(k-1)+1$ symbols, taken $k$ at a time, where each two sets of $k$ symbols have exactly one pair in common.


One thing I found was that I could create a maximal set using a partition slicing strategy. But I wanted to create a pattern $(a\_1 \dots a\_k)$ that I could simply increment around the modulus $n$ to create every possible $k-$set. What I found was that for $k=1\dots 8, k\neq 7$, I could have at least two such patterns (removing multiples via various equivalencies). However, I have been unable to find one for $k=7$, even with an exhaustive code-based search of the $[43]^7$ vector space.



Perfect Difference Set of order 7^2-7+1=43
Ok, thanks to @Seva giving me the right name for these, I found out that:


* The sets I'm describing are Perfect Difference sets.
* The one-dimensional version I'm working with are equivalent to projective planes.
* <https://en.wikipedia.org/wiki/Projective_plane> says that there is no plane for $j=6$ $(j^2+j+1)$, which is equivalent to my case where $k=7$ $(k^2-k+1)$.


So, not an error in my code, proven impossible prior to 1938 (as a different problem) and related to projective planes in 1938.


