A geometric problem on number of points and lines

Because of first two condititons we have $$L\times 8 = P\times 8\implies L=P$$

Then from the last condition we have $${8\choose 2}L = {P\choose 2}\implies P= 57$$

Such a configuration really exists, it is a projective plane of order $7$ it has $7^2+7+1$ points and the same number of lines.

There are $\binom P2$ pairs of points, and each pair gives a line. However, each line has $8$ points on it, meaning that the pairs of points are divided into groups of $\binom 82=28$, where any pair in a group gives the same line. So the number of distinct lines is $L= \frac{\binom P2}{28}$.

Similarly, the number of distinct points is $P= \frac{\binom L2}{28}$.

This gives the following set of equations: $$ \cases{P=\frac{L(L-1)}{56}\\L=\frac{P(P-1)}{56}} $$

• 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 $3249$.