Final answer:
To show that a self-complementary graph has 4n or 4n + 1 vertices, we use the property that the sum of the edges in the graph and its complement equals the edges in a complete graph with the same number of vertices. The proof relies on algebraic manipulation and the properties of numbers modulo 4.
Step-by-step explanation:
The question revolves around proving why every self-complementary graph has 4n or 4n + 1 vertices, for some integer n ≥ 0. By definition, a graph is self-complementary if it is isomorphic to its complement. Let's consider a self-complementary graph G with v vertices and e edges. A complete graph Kv with v vertices has ec = v(v - 1) / 2 edges. Since G is self-complementary, the number of edges in G plus the number of edges in its complement G' equals ec. Therefore, 2e = v(v - 1) / 2, which simplifies to 4e = v(v - 1). Now, v must be of the form 4n or 4n + 1 because any number squared is either 0 or 1 mod 4, and the only way v(v - 1) is divisible by 4 is if v is in one of these two forms. This ensures that the number of vertices v in a self-complementary graph must satisfy the condition of being 4n or 4n + 1.