Example 4. Let $G$ be a simple graph of order 10 and does not contain a cycle of length 4, $C_{4}$. Then the maximum number of edges in $G$ is 16.
Proof: Let $f(n)$ be the maximum number of edges in an $n$-order simple graph without $C_{4}$. Clearly, $f(n)=4$. Now we prove that $f(5)=6$. First, if $G$ is two triangles sharing exactly one common vertex, then $G$ is a 5-order graph with 6 edges and no $C_{4}$. Second, if a 5-order graph $G$ has 7 edges, then there must be a vertex with a degree no greater than 2. Removing this vertex and the edges associated with it, we get a 4-order graph $G^{\prime}$. $G^{\prime}$ still has at least 5 edges, which must contain a quadrilateral. Now, we will prove that $f(10)=16$.

Let $G$ be a 10-order graph without $C_{4}$, and let $A$ be the vertex with the highest degree in $G$. Let $S=\{x \mid x$ is adjacent to $A\}$, $T=\{x \mid x$ is not adjacent to $A\}$, and assume $|S|=k$, then $|T|=9-k$, where $1 \leqslant k \leqslant 9$. For any point $x$ in $S \cup T$, it can be adjacent to at most one point in $S$, otherwise, adding $A$ would form a $C_{4}$. Thus, $S$ can have at most $\left[\frac{k}{2}\right]$ edges. Also, $S$ and $T$ can have at most $9-k$ edges between them, and $T$ can have at most $f(9-k)$ edges. Therefore,
$$
\begin{array}{l}
f(10) \leqslant k+\left[\frac{k}{2}\right]+(9-k)+f(9-k) . \\
\text { Let } g(k)=9+\left[\frac{k}{2}\right]+f(9-k) . \text { Then } \\
f(10) \leqslant g(k) .
\end{array}
$$

The relevant data is calculated as follows:
\begin{tabular}{c|c|c|c|c|c|c}
\hline$k$ & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline$f(9-k)$ & 6 & 4 & 3 & 1 & 0 & 0 \\
\hline$g(k)$ & 17 & 15 & 15 & 13 & 13 & 13 \\
\hline
\end{tabular}

From the table, we can see that when $k \geqslant 5$, $g(k) \leqslant 15$, so $f(10) \leqslant 15$; also, when $k \leqslant 3$, $f(10) \leqslant \frac{3 \times 10}{2}=15$.

Thus, we only need to prove that when $k=4$, $f(10) \leqslant 16$. Since $T$ does not contain $C_{4}$ and has 5 vertices, from the initial discussion, $f(5)=6$. If there is a vertex with degree 1, removing this vertex would leave a 4-order graph with 5 edges, which must contain a $C_{4}$. Therefore, each vertex must have a degree of at least 2. If there is a vertex $A$ with degree 3, let it be adjacent to $B, C, D$ but not to $E$. Since $E$ has a degree of at least 2, $E$ must be adjacent to two of $B, C, D$, which would form a $C_{4}$. Therefore, each vertex must have degrees of 4, 2, 2, 2, 2, meaning $T$ forms two triangles sharing exactly one common vertex. Since $|S|=4$, if $S$ and $T$ are connected by at least 5 edges, then there must be a vertex in $S$ that is connected to two vertices in $T$, forming a $C_{4}$. Thus,
$$
\begin{array}{l}
\quad f(10) \leqslant 4+\left[\frac{4}{2}\right]+4 \\
+f(5)=16 .
\end{array}
$$

Finally, the graph on the right is a 10-order graph without $C_{4}$ and with 16 edges, so $f(10)=16$.
For $n \leqslant 21$, the values of $f(n)$ are as follows:
\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
\hline$n$ & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline$f(n)$ & 4 & 6 & 7 & 9 & 11 & 13 & 16 & 18 & 21 \\
\hline
\end{tabular}
\begin{tabular}{r|l|l|l|l|l|l|l|l}
\hline 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 \\
\hline 24 & 27 & 30 & 33 & 36 & 39 & 42 & 46 & 50 \\
\hline
\end{tabular}

Where $f(8)=11$ is the 5th problem of the 1992 CMO.
For a general $n \in \mathbb{N}$, what is the expression for $f(n)$? This remains an open problem, and people have only found an upper bound for $f(n)$, which is given below.