TITLE: Generalization of finite-projective-plane with more than one intersection point
QUESTION [5 upvotes]: In a finite projective plane, each two points appear together in exactly one line, and each two lines intersect in exactly one point. It is known that, if each line contains $n+1$ points, then the total number of points and of lines is $n^2+n+1$.
What is known about a generalization of this concept, in which each two points appear together in exactly $k$ lines, and each two lines intersect in exactly $k$ points? In particular, for what values of $n,k$ such planes exist, and what is the number of lines and points in that case?

REPLY [4 votes]: Let me adjust notation slightly -- the $k$ in the original post is more usually a $\lambda$ in the literature. Thus the concept you want is this:

Definition. A symmetric $2-(v,k,\lambda)$ design is a pair $(\Omega, \mathcal{B})$ where $\Omega$ is a set of size $v$ and $\mathcal{B}$ is a set of $k$-subsets of $\Omega$ such that:

any 2 points of $\Omega$ lie in $\lambda$ elements of $\mathcal{B}$;
any 2 elements of $\mathcal{B}$ intersect in $\lambda$ elements of $\Omega$.


A simple counting argument asserts that an object has the property that $b=|\mathcal{B}|=v$. If you want to know when these things exist, then the following theorem should be your starting point:

The Bruck-Ryser-Chowla Theorem. If a symmetric $2-(v,k,\lambda)$ design exists, then

if $v$ is even, then $k-\lambda$ is a square;
if $v$ is odd, then the following Diophantine equation has a nontrivial solution:
  $$x^2-(k-\lambda)y^2 - (-1)^{(v-1)/2}\lambda z^2=0.$$


More is known in special cases. For instance there is a famous result of Lam, using a computer, that asserts that a symmetric $2-(111,11,1)$ design does not exist (there is no projective plane of order $10$).