finite projective planes have $q^2+q+1$ points and $q^2+q+1$ lines

Given a finite projective plane that contains a quadrangle OXYZ (i.e. nothree of these four points are on one line). To prove: the plane has$q^2+q+1$ points and $q^2+q+1$ lines for some integer $q$, and there are $q+1$points on each line and $q+1$ lines through each point.

Let $x$ and $y$ be the lines OX andOY, which must exist by the axioms. By the assumption OXYZ is a quadranglethese lines are distinct and Z is not on them. Let there be $p$ points X${}_i$ on$x$ other than O, for each of them one line ZX${}_i$ exists, and is distinct(one lines cannot pass through two X${}_i$ unless it is $x$ but that’s not aline through Z). Conversely every line through Z must intersect $x$ in aunique point (two lines intersecting in Z cannot intersect at another point,and Z is not a point on $x$). So there are $p+1$ lines through Z (OZ is oneof them). By the same reasoning, using $y$, there are $q+1$ lines through Zso $p=q$. We also found $q+1$ points (including O) on $y$ and the same numberon $x$. Intersecting the $q+1$ lines through Z with XY (on which Z does notlie, the quadrangle again) reveals at least $q+1$ distinct points there andat most $q+1$ because for each point there there is a line through it and Z.

The lines not through O intersect $x$ in one of the $q$ points X${}_i$ and $y$in one of the $q$ points Y${}_j$. There are $q^2$ possibilities and each of themis a distinct line, because there is only one line through a given X${}_i$ andY${}_j$. The lines that do pass through O intersect XY in one of the $q+1$points there, again one line for each such point and vice versa. That’s $q+1$lines through O and $q^2$ not through O, $q^2+q+1$ in all.

There are $q+1$ lines through X (to each of the points of $y$) and $q+1$lines through Y (to each of the points of $x$). Intersect the $q$ linesthrough X other than XY with the $q$ lines through Y other than XY, these$q^2$ intersections are all distinct because for any P there’s only one linePX and one line PY. Note we did not use the line XY. Conversely for any P noton XY there must be some PX and some PY, so there are exactly $q^2$ points noton XY. Add the $q+1$ points on XY for a total of $q^2+q+1$.

The constructions above already showed $q+1$ lines through some points (X, Yand Z), by the same games as before that implies for each of them $q+1$ pointson every line not through that point. We also saw $q+1$ points on some lines($x$, $y$, XY) which implies for each of them $q+1$ lines through every pointnot on that line. Such reasoning covers $q^2$ items on first application andrapidly mops up stragglers on repeated application.

Some form of this proof is standard math lore; this version was half remembered and half reconstructed.