Let M be a finte projective plane so that all lines in M have the same number of points lying on them, call this number N+1
Prove:the toall number of poins in M is N^2+N+1 and total number of lines
in M is N^2+N+1.
Projective stuff... Great
From what I can recall, a line passes through every two points.
For the total number of lines, define a_ij to indicate the line passing through points i and j. All lines have N+1 points; So the cardinality of (a_ij) is (N+1)^2. But also, the diagonal (a_ii) does not define a line; So exclude N lines from this list, to get a total of (N+1)^2-N=N^2+N+1 lines.
For the number of points, I think there should be more than just N^2+N+1... Check again plz.
Choose a point P and line L with P not lying on L (if you cannot do this it is not a plane but a line).
The lines through P are in 1-1 correspondence with the points on L. Any line through P meets in in just one point, and any point on L defines a line through P. So there are N+1 lines, each of which has N points other than P. So there are N(N+1)+1 points in the plane.
Now there are (N+1)(N^2+N+1) pairs of the form (P,T) with P a point and T a line through P. Since each line T has N+1 points P, each line occurs in this list (N+1) times. So the total number of lines is (N^2+N+1).
There's another way of seeing the latter. Use duality to observe that every property of the plane remains true if you interchange the words point and plane. Since there are N+1 points on each line, and we showed N+1 lines through each point, there are the same number of points as lines by duality.