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 .
The answer is $E=30^{2}+2 \cdot 15^{2}=6 \cdot 15^{2}=1350$.
First, we prove $E \leq 1350$. Observe that:
Claim $-G$ contains no $K_{5}$.

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.

Hence by Turán theorem we have $E \leq\binom{ 4}{2} \cdot 15^{2}=1350$.
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.