## Pappus and Desargues solution

So far, I have put up the following posts related to finite projective planes:

The objective is to give the apparently relevant background to the long-standing conjecture that there are no projective planes of order $$n$$ unless $$n$$ is a prime power.

The material still to come is:
(1) Pappus implies Desargues (this post);
(2) Wedderburn’s theorem
(3) Desargues implies the plane is over a division ring
(4) Pappus implies the plane is over a field
(5) Stinson’s proof that there are no orthogonal Latin squares of order 6
(6) the construction of non-Desarguesian planes
(7) Lee’s proof that there are two orthogonal Latin squares for any order $$n\gt 6$$

Showing that Pappus implies Desargues was set as a problem here.

To recap, the Pappus configuration is:

0 1 2 = $$L_0$$
0 3 6 = $$L_1$$
0 4 7 = $$L_2$$
1 3 8 = $$L_3$$
2 5 6 = $$L_4$$
4 6 8 = $$L_5$$
1 4 5 = $$L_6$$
2 7 8 = $$L_7$$
3 5 7 = $$L_8$$

If we remove any one of these lines from the list, then its existence can be recovered from the other 8. The Desargues configuration is:

0 1 2 = $$L_7$$
0 3 4 = $$L_8$$
0 5 6 = $$L_9$$
7 8 9 = $$L_0$$
3 5 7 = $$L_2$$
4 6 7 = $$L_1$$
1 5 8 = $$L_4$$
2 6 8 = $$L_3$$
1 3 9 = $$L_6$$
2 4 9 = $$L_5$$

Here it is important that the point $$i$$ does not lie on any of the lines apart from those listed, except possibly $$L_i$$. Again if the line $$L_i$$ is removed, then it can be recovered from the other 9.

So suppose that $$L_0$$ is removed from the Desargues list. That leaves us with:

0 1 2 = $$L_7$$
0 3 4 = $$L_8$$
0 5 6 = $$L_9$$
3 5 7 = $$L_2$$
4 6 7 = $$L_1$$
1 5 8 = $$L_4$$
2 6 8 = $$L_3$$
1 3 9 = $$L_6$$
2 4 9 = $$L_5$$

Take: point 10 to be the intersection of $$L_3,L_6$$; line $$L_{10}$$ to be the line through points 3,6; point 11 to be the intersection of $$L_7,L_{10}$$; line $$L_{11}$$ to be the line through points 0,10; point 12 to be the intersection of $$L_2,L_{11}$$; and point 13 to be the intersection of $$L_1,L_{11}$$. That takes us to:

$$L_7$$ = 0 1 2 11
$$L_8$$ = 0 3 4
$$L_9$$ = 0 5 6
$$L_2$$ = 3 5 7 12
$$L_1$$ = 4 6 7 13
$$L_4$$ = 1 5 8
$$L_3$$ = 2 6 8 10
$$L_6$$ = 1 3 9 10
$$L_5$$ = 2 4 9
$$L_{10}$$ = 3 6 11
$$L_{11}$$ = 0 10 12 13

These points and lines are a Pappus configuration:

$$L_7$$ = 0 1 11
$$L_9$$ = 0 5 6
$$L_{11}$$ = 0 10 12
$$L_6$$ = 1 3 10
$$L_4$$ = 1 5 8
$$L_2$$ = 3 5 12
$$L_{10}$$ = 3 6 11
$$L_3$$ = 6 8 10

from which we deduce the existence of the line

$$L_{12}$$ 8 11 12

We also have this Pappus configuration:

$$L_7$$ = 0 2 11
$$L_8$$ = 0 3 4
$$L_{11}$$ = 0 10 13
$$L_5$$ = 2 4 9
$$L_3$$ = 2 6 10
$$L_{10}$$ = 3 6 11
$$L_6$$ = 3 9 10
$$L_1$$ = 4 6 13

from which we deduce the existence of the line

$$L_{13}$$ = 9 11 13

Using these two lines we have finally this Pappus configuration:

$$L_{10}$$ = 3 6 11
$$L_2$$ = 3 7 12
$$L_6$$ = 3 9 10
$$L_1$$ = 6 7 13
$$L_3$$ = 6 8 10
$$L_{11}$$ = 10 12 13
$$L_{12}$$ = 8 11 12
$$L_{13}$$ = 9 11 13

from which we deduce the existence of the line

$$L_0$$ = 7 8 9

as required.