A finite projective plane of order $n$ has $n^2+n+1$ points and the same number of lines

Choose a line $\mathrm{\ell }\in \mathcal{L}$. Since no line contains three of the four noncollinear points from [L2], some point $p\in P$ lies outside $\mathrm{\ell }$.

For each point $q\in \mathrm{\ell }$, there is a unique line through $p$ and $q$, and distinct points of $\mathrm{\ell }$ give distinct lines through $p$. Conversely, any line through $p$ meets $\mathrm{\ell }$ in exactly one point by [L2]. Therefore exactly $n+1$ lines pass through $p$.

Each of the $n+1$ lines through $p$ contains exactly $n$ points besides $p$, and the sets of those other points are pairwise disjoint because two distinct lines through $p$ meet only at $p$. Every point distinct from $p$ lies on exactly one of these lines, namely its unique joining line with $p$. Therefore $\mathrm{\mid }P\mathrm{\mid }=1+n(n+1)=n^2+n+1$.

Count incident pairs $(x,{\mathrm{\ell }}^{\mathrm{\prime }})$ with $x\in {\mathrm{\ell }}^{\mathrm{\prime }}$. By [L1], each line contributes $n+1$ such pairs. By step 2.1 and the argument there applied to an arbitrary point, each point also lies on exactly $n+1$ lines. Therefore $\mathrm{\mid }\mathcal{L}\mathrm{\mid }(n+1)=\mathrm{\mid }P\mathrm{\mid }(n+1)$, and step 3.1 gives $\mathrm{\mid }\mathcal{L}\mathrm{\mid }=\mathrm{\mid }P\mathrm{\mid }=n^2+n+1$.