Hello I'm hoping to understand where I went wrong in the below proof. This was given on a quiz and I received 5/40 points for it. My professor said I just gave examples and that I didn't actually give a proof. Any thoughts or comments that help me understand my error in thinking would be greatly appreciated.
$$\textbf{Problem}$$
Suppose $G$ is a simple undirected graph with $n$ vertices. If $G$ is self-complementary, prove that either $n = 4t$ or $n=4t+1$ for some $t\in Z^{+}$. To receive credit for this problem you must write complete sentences, show all of your work, explain all of your reasoning, and include all details.
$$\textbf{Proof}$$
Assume BWOC that $G$ is self-complementary and that $n$ is neither of the form $4t$ or $4t+1$ for any $t\in Z^{+}$. Then $n$ can only be one of the following cases:
Case 1: $n=2$. If $n=2$ then the complete graph on $G$ would have $\frac{2(2-1)}{2}=1$ edge and thus $G$ cannot be self-complementary because $G$ must have an even number of edges to be self-complementary.
Case 2: $n=3$. If $n=3$ then the complete graph on $G$ would have $\frac{3(3-1)}{2}=3$ edges and thus $G$ cannot be self-complementary because G must have an even number of edges to be self=complementary.
Case 3: $n=4t+2$ for any $t\in Z^{+}$. If $n=4t+2$ for any $t\in Z^{+}$ then the complete graph on $G$ would have $\frac{(4t+2)(4t+2-1)}{2}=\frac{16t^{2}+12t+2}{2}=8t^{2}+6t+1$ number of edges and since $8t^{2}+6t$ is even the total number of edges on $G$ must be odd and thus $G$ cannot be self=complementary because $G$ must have an even number of edges to be self-complementary.
Case 4: $n=4t+3$ for any $t\in Z^{+}$. If $n=4t+3$ for any $t\in Z^{+}$ then the complete graph on $G$ would have $\frac{(4t+3)(4t+3-1)}{2}=\frac{16t^{2}+20t+6}{2}=8t^{2}+10t+2+1$ number of edges and since $8t^{2}+10t+2$ is even the total number of edges on $G$ must be odd and thus $G$ cannot be self=complementary because $G$ must have an even number of edges to be self-complementary.
Thus we have arrived at a contradition. We assumed that $n$ is neither of the form $4t$ or $4t+1$ for any $t\in Z^{+}$ and yet $G$ is self-complementary and have proved that this is not possible. Thus if a graph $G$ with $n$ vertices is self-complementary n must either be of the form $4t$ or $4t+1$ for some $t\in Z^+{}$.
Best Answer
The core of your proof is correct, but it's very difficult to read. My primary complaint would be the unnecessary use of contradiction. I might proceed directly as follows:
Let $G$ be a self-complementary graph. First note that the number of edges in $G$ must be exactly $\frac{1}{2}\binom{n}{2}$ since there are a total of $\binom{n}{2}$ possible edges on $n$ vertices. It follows that $\binom{n}{2}$ must be even. Observe the following cases:
In all these cases, we find that $\binom{n}{2}$ is odd. This shows that $n$ must be of the form $4t$ or $4t+1$ for $t \in \mathbb{Z}^+$, as desired.