Today I came up with this problem:
Find all natural numbers $n$ such that there exist a family of $3$-element subsets of the set $\{1,2,...,n\}$ like $\mathbf F$ such that:
a) For all distinct $1\le a,b \le n$ there exists a unique element of $\mathbf F$ like $\mathbf A$ such that it contains both $a$ and $b$.
b) If $1\le a,b,c,x,y,z \le n$ are distinct and $\{a,b,x\},\{b,c,y\},\{c,a,z\}\in \mathbf F$, then $\{x,y,z\}\in \mathbf F$.
It's easy to prove that $n$ should be of forms $6k+1$ or $6k+3$. For $n=3$ it's obvious and for $n=7$ I found this example:
$\{1,2,4\},\{2,3,5\},\{3,1,6\},\{4,5,6\},\{7,1,5\},\{7,2,6\},\{7,3,4\}$.
But I couldn't go any further. Would anyone please tell me, how can I construct such examples for larger $n$'s? Or prove there is no such examples? Thanks