Let $k$ be the number of edges of a graph $G$, I’d want to show that $G$ can contain at most $(2k)^\frac{5}{2}$ cycles of length $5$.
I thought about showing this for the Erdös-Renyi graph model $G(2k,k)$ where $2k$ is the amount of nodes.
Now ideally I’d show that $P(\#5 $-cycles $ $ in $G > (2k)^\frac{5}{2})=0$ where G is a realisation of $G(2k,k)$
However, when considering the expression $(2k)^\frac{5}{2}$, the exponent puzzles me. I see no straightforward way to get the factor $\frac{1}{2}$ in the exponent like say by using markov’s inequality or so, whereas $2k$ shows up naturally by the fact that one can choose both a direction and a starting node for a cycle.
Any ideas on how to further this direction or another way to show this?