Show that if a self-complementary graph contains a pendant vertex, then it must have at least another pendant vertex.
Let $G$ be a graph of order $n$, so it has $n(n-1)/4$ edges, just like its complement. Asume that there is only one pendant vertex $v$, so $d(v)=1$, it means that $G^c$ has a vertex $w$ with $d(w)=n-2$. Then $G$ must also have a vertex of degree $n-2$ and $G^c$ one of degree $1$.
I guess I can reach somehow a contradiction with the fact that $n=4k$ or $n=4k+1$, but I got stuck there.
Can you give me a hint please? Like "this fact may be useful..."