Let $G$ be a self-complementary graph of order $n$, where $n ≡ 1 \pmod{4}$.
Prove that $G$ contains an odd number of vertices of degree $\frac{(n − 1)}{2}$.
My approach
Let $G$ be a self-complementary graph of order n, where $n=4k+1$ i.e. $n$ is odd.
Now size of $G$ is $\frac{n(n-1)}{4}$.
Now if $V(G)=\{v_i:1\leq i \leq n\}$.
Now $\sum d(v_i)=\frac{n(n-1)}{2}$=even.
Then what can i do?
Note that if a vertex $v$ in the graph $G$ doesn't have degree $\frac{(n-1)}2$, then the vertex in $G^c$, the complement, must have a different degree than in $G$, since the total number of potential connections is $(n-1)$.
Hence, if $G$ is self complementary, there has to be a node that will correspond to $v$ that isn't $v$ in $G^c$ to ensure the property. Hence, we can group together each such $v$ with a $v'$ from $G^c$ that holds its properties (i.e., degree, etc).
Note that all of $v'$ are in $G$, and all $v'$ don't have degree $\frac{(n-1)}2$, so since we were able to pair $v$ with $v'$, and all $v'$ also are $v$, the number of $v$ nodes is even. So, the nodes that aren't $v$ must be odd.