6 Let $n$ be a given positive integer $(n \geqslant 5)$, find the minimum value of the positive integer $m$ such that any $n$-vertex simple graph $G$ with $m$ edges contains two triangles that share exactly one vertex.

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First, prove that when $m=\left[\frac{n^{2}}{4}\right]+1$, there exists a graph $G$ in which no two triangles share exactly one vertex.

Consider a bipartite graph: the vertex set is $M_{1}$ and $M_{2}$, with $M_{1}$ having $\left[\frac{n+1}{2}\right]$ vertices and $M_{2}$ having $\left[\frac{n}{2}\right]$ vertices. Connect each vertex in $M_{1}$ to each vertex in $M_{2}$, and connect two vertices in $M_{1}$, with no other edges between the vertices. In this graph, there are $\left[\frac{n}{2}\right]\left[\frac{n+1}{2}\right]+1=\left[\frac{n^{2}}{4}\right]+1$ edges, but there are no two triangles that share exactly one vertex.

Next, prove that when $m=\left[\frac{n^{2}}{4}\right]+2$, any simple graph $G_{n}$ with $m$ edges and $n$ vertices contains two triangles that share exactly one vertex.

When $n=5$, the number of edges in $G_{5}$ is 8, which is the complete graph $K_{5}$ minus 2 edges. Since each vertex in $K_{5}$ has a degree of 4, removing two edges still leaves one vertex with a degree of 4. Let's assume $d(a)=4$. Thus, the two edges are removed from the remaining $K_{4}$. Among the 4 points $b, c, d, e$, there must be a pair of opposite edges (let's say $b c$ and $d e$) that still exist, so $a, b, c, d, e$ form the required 5 points.
Assume the proposition holds for $n-1 (n \geqslant 6)$, and consider the case for $n$.
Case 1: If $G_{n}$ has a vertex with a degree $\leqslant\left[\frac{n}{2}\right]$, then removing this vertex and its edges leaves at least
$$
\begin{aligned}
{\left[\frac{n^{2}}{4}\right]+2-\left[\frac{n}{2}\right] } & =2+\left[\frac{n}{2}\right]\left(\left[\frac{n+1}{2}\right]-1\right) \\
& =2+\left[\frac{n}{2}\right]\left[\frac{n-1}{2}\right]=2+\left[\frac{(n-1)^{2}}{4}\right],
\end{aligned}
$$
edges among the remaining $n-1$ vertices. By the induction hypothesis, the proposition holds.
Case 2: If every vertex in $G_{n}$ has a degree $\geqslant\left[\frac{n}{2}\right]+1$. Then, when $n=2 k$,
$$
e \geqslant k(k+1)>k^{2}+2=\left[\frac{n^{2}}{4}\right]+2,
$$
and when $n=2 k+1$,
$$
\begin{aligned}
e & \geqslant \frac{1}{2}(2 k+1)(k+1) \\
& =k(k+1)+\frac{k+1}{2} \geqslant\left[\frac{n^{2}}{4}\right]+2,
\end{aligned}
$$
with equality only when $k=3$, i.e., $n=7$.
If $e>\left[\frac{n^{2}}{4}\right]+2$, remove one edge from $G_{n}$ and repeat the discussion. Except for the case $n=7$, all other cases can be reduced to Case 1.
Therefore, we only need to consider the case $n=7$ where each vertex has a degree of 4.
In this case, consider the 4 vertices $b, c, d, e$ adjacent to $a$. In the subgraph formed by $b, c, d, e$, each vertex has a degree of at least 1. Assume $b$ and $c$ are connected. If $d$ and $e$ are also connected, the proposition is proven, so $d$ and $e$ are not connected.

If $d$ and $e$ are connected to $b$ and $c$ (or $c$ and $b$), the proposition is also proven, so $d$ and $e$ are both connected to $b$ (or $c$). Assume $d$ and $e$ are both connected to $b$, and $c, d, e$ are not connected to each other. Consider the other two points $f$ and $g$. Since the degrees of $c, d, e$ are all 4, $c, d, e$ are all connected to $f$ and $g$. Since the degrees of $f$ and $g$ are also 4, $f$ and $g$ are connected (because $f$ and $g$ are not connected to $a$ and $b$). Thus, we get the following graph, where $e a$, $e b$, $e f$, $e g$, $a b$, and $f g$ satisfy the requirements of the problem.
In conclusion, $m=\left[\frac{n^{2}}{4}\right]+2$.