Given a finite projective plane that contains a quadrangle OXYZ (i.e. no three 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 and OY, which must exist by the axioms. By the assumption OXYZ is a quadrangle these 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 a line through Z). Conversely every line through Z must intersect $x$ in a unique 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 one of them). By the same reasoning, using $y$ , there are $q+1$ lines through Z so $p=q$ . We also found $q+1$ points (including O) on $y$ and the same number on $x$ . Intersecting the $q+1$ lines through Z with XY (on which Z does not lie, the quadrangle again) reveals at least $q+1$ distinct points there and at 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 them is a distinct line, because there is only one line through a given X$_i$ and Y$_j$ . 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 $q$ lines through 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 line PX and one line PY. Note we did not use the line XY. Conversely for any P not on XY there must be some PX and some PY, so there are exactly $q^2$ points not on 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, Y and Z), by the same games as before that implies for each of them $q+1$ points on 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 point not on that line. Such reasoning covers $q^2$ items on first application and rapidly mops up stragglers on repeated application.

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