"I haven't seen that technique in 20 years!"

To show that, for an undirected graph G with n vertices, where G is self-complimentary, n = 4k or 4k + 1 for some positive integer k, first note that 4k is necessarily even and a multiple of 4, and also that there are n(n-1)/4 edges on graph G (earlier shown). As such, in either case for n, in the numerator of this formula there is a multiple of 4, in that either n or n - 1 is equal to 4k. Now assume the contrary, that there exists some n not of the form 4k or 4k + 1 for which a self-complementary graph can be drawn. In this case, then this graph has a non-integer number of edges (the numerator in the formula contains no multiples of 4), which is impossible, so we've reached a contradiction, and n must be of the form 4k or 4k+1.

Boo-yah!