Problem: Find the maximum number $E$ such that the following holds:
there is an edge-colored graph with $60$ vertices and $E$ edges,
with each edge colored either red or blue, such that in that coloring,
there is no monochromatic cycles of length $3$
and no monochromatic cycles of length $5$.

Solution: The answer is $E = 30^2 + 2 \cdot 15^2 = 6 \cdot 15^2 = 1350$.

First, we prove $E \le 1350$.  Observe that:
\begin{claim*}
  $G$ contains no $K_5$.
\end{claim*}
\begin{proof}
  It's a standard fact that the only triangle-free two-coloring
  of the edges of $K_5$ is the union of two monochromatic $C_5$'s.
\end{proof}
Hence by Tur\'{a}n theorem we have $E \le \binom42 \cdot 15^2 = 1350$.

To show this is achievable, take a red $K_{30,30}$,
and on each side draw a blue $K_{15,15}$.
This graph has no monochromatic odd cycles at all as desired.