TITLE: A geometric problem on number of points and lines
QUESTION [5 upvotes]: There are P points and L lines such that

Each line contains 8 points
Each point lies on 8 lines
Any two distinct lines intersect in a unique point
Any two distinct points lie on a unique line.

Lines may be straight or curved. What is $P\times L$ ?

My answer, is that, using last two conditions, if $P$ is no of points then L = $\binom{P}{2}$. Also if no of lines is $L$ then $P = \binom{L}{2}$, from here we need $\binom{\binom{L}{2}}{2} = L$ which gives 3 as L. This is certainly incorrect.
How to approach this geometric problem?

REPLY [2 votes]: Any two distinct lines intersect in a unique point
Any two distinct points lie on a unique line.


These two rules are, essentially, the incidence axioms for two-dimensional projective geometry. As such, we have a standard model for the system described here: the projective plane over the field $\mathbb{F}_7=\mathbb{Z}/7$ with seven elements.
We can use that model to count the points and lines - or we can just do things directly. Suppose we have a projective plane such that each line has $n+1$ points and each point is on $n+1$ lines.
Then, consider the system of all $n+1$ lines through a given point $P$. These lines each contain $n$ points other than $P$. Also, each point other than $P$ is on exactly one of these lines. That's $n(n+1)$ points other than $P$; add $P$ back in, and we have a total of exactly $n^2+n+1$ points.
By symmetry between points and lines, we also have exactly $n^2+n+1$ lines; we could run a similar count using some particular line $\ell$, the $n+1$ points on it, and the $n$ other lines through each of those points.
For our example with $n+1=8$, that's $7^2+7+1=57$ each points and lines, for a product of $3249$.