A possible multiplication convention for the octonions is given by the following 7 sets of integers from 1 to 7.
{1,2,3},{1,4,5},{1,6,7},{2,4,6},{2,5,7},{3,4,7},{3,5,6}.
Notice that each of the seven sets of integers above has exactly 3 elements, each element (integer from 1 to 7) is used exactly 3 times, and any two sets share exactly 1 element. These characteristics imply that no element can be repeated in a given set, and that each set must differ from all others.
In an attempt to understand how this relates to the nonassociativity of the octonions I attempted to generalize this. I therefore pose the following question.
Given $s\in\mathbb{I}$ sets, how many elements per set $b\in\mathbb{I}$ and how many total elements $a\in\mathbb{I}$ are necessary such that the sets in every pair of sets share exactly $d\in\mathbb{I}$ elements and each element is used exactly $c\in\mathbb{I}$ times?
Another way of phrasing this is as follows. In the example given, there are 7 sets, each containing 3 elements. Each element is also used 7 times in the entirety of the 7 sets, and each set overlaps with each other set by exactly 1 element. There are exactly 7 unique elements used. I would like to understand the more general case in which there are $s$ sets, each set containing $b$ elements. Each element should be used $c$ times in the entirety of the $s$ sets, and each set should overlap each other set by exactly $b$ elements. The number of unique elements is called $a$.
Attempt at solution:
If $d=0$ (i.e. the sets do not share any elements with one another) then $ac=bs$. This allows only for the trivial solutions: $a=a,b=1,c=1,d=0$ with sets {1},{2},...,{a} and $a=a,b=a,c=1,d=0$ with sets {12...a}.
Relatedly, $c=1$ implies $d=0$.
If $c=2,d=d$ then $$a=d\sum_{i=1}^{b/d(\in\mathbb{I})}i$$
One can see this by considering the case $b=b$, $c=2$, $d=1$. Here consider b=4. To start we can write a group of four elements down, and then maintain the same element in the first position in the second set, changing the other three:
{1,2,3,4},{1,5,6,7}
Since we have used $1$ twice we cannot use it again, so we start the next set with $2$ to ensure it overlaps with the first set, and select the second element to be $5$ to ensure overlap with the second set by one element:
{1,2,3,4},{1,5,6,7},{2,5,8,9}
Continuing this process we finally get
{1,2,3,4},{1,5,6,7},{2,5,8,9},{3,6,8,10},{4,7,9,10}.
For $b=5$
{1,2,3,4,5},{1,6,7,8,9},{2,6,10,11,12},{3,7,10,13,14},{4,8,11,13,15},{5,9,12,14,15}
Using this method one finds that for $b=b$, $c=2$, $b=1$, $b+1-i$ new elements are introduced in the $i$th set. Therefore:
$$a=\sum_{i=1}^{b}i$$
However, one can always take a given solution and multiply $a$, $b$, and $d$ by a common factor to create another valid solution. That is we can parameterize
$$(s,a,b,c,1)\rightarrow(s,ad,bd,c,d)$$ Therefore, we can modify this equation to the original result presented above.
However, I am unsure how to generalize to all $c,d$ in a systematic way. I would ultimately like an algebraic relationship between $a, b, c, d$ and $s$. Any help will be greatly appreciated.
