Let $G$ contains $2n$ vertices and $n^2+1$ edges.
G contains a triangle by Turan theorem.
Prove, that there is an edge that takes part in two or more triangles
If I assume the inverse, then every triangle has at most $2n-3$ edges going out of its vertices. Let there be $k$ triangles
That means, that we get a triangle-free part of the graph with $2n – 3k$ vertices and at least $n^2 – 2kn$ edges.
if $\left[\frac{(2n-3k)^2}{4}\right] \leq n^2-3kn$ then that means that the triangle-free part contains a triangle which leads to contradiction. But as I plotted the graphs I didn't find anything valuable.
May be the main fact is that the number of edges is exactly $n^2+1$, but I don't know how to use it.
Best Answer
We prove a bit more general
Proposition. Let $n\ge 4$, $G$ be a graph with $n$ vertices and more than $ \lfloor n^2/4\rfloor$ edges. Then $G$ contains no triangles sharing an edge.
Proof. Suppose to the contrary that there are no triangles sharing an edge. Proceed as follows. As far as $G$ contains a triangle we remove its vertices from $G$. Assume we stopped when in $G$ remains $k$ vertices. Since $G$ became triangle-free, by Turán’s theorem, there remains at most $\lfloor k^2/4\rfloor$ edges.
Consider a graph $G$ at some step of the removal. Let $G$ contains $p$ vertices and $T$ be any triangle of $T$. Since $G$ contains no triangles sharing an edge, no vertex of $G\setminus T$ can be adjacent to two vertices of $T$. So when we remove vertices of a triangle $T$ from $G$ we remove at most $3+p-3=p$ edges. That is in total we have removed at most
$$n+n-3+\dots+k+3=\frac{k(n-k)}{3}+3\sum_{i=1}^{(n-k)/3} i=$$ $$\frac{k(n-k)}{3}+3\frac{\frac{n-k}{3}\left(\frac{n-k}{3}+1\right)}{2}=$$ $$\frac{k(n-k)}{3}\left(k+\frac{n-k+3}{2}\right)=$$ $$\frac{(n-k)(n+k+3)}{6}$$ edges of $G$. Thus $$\left\lfloor \frac{k^2}4\right\rfloor+\frac{(n-k)(n+k+3)}{6}>\left\lfloor \frac{n^2}4\right\rfloor.$$
So
$$\frac{k^2}4+\frac{(n-k)(n+k+3)}{6}\ge \frac{n^2-1}{4}+\frac{1}{4}$$
$$3k^2+2(n-k)(n+k+3)\ge 3n^2$$
This inequality transforms to
$$(k-3)^2\ge (n-3)^2.$$
Since $k\le n-3<n$, we have $k-3<0$, which easily implies $k=1$ and $n=4$. But then $G$ is a graph with four vertices and at least five edges, so it contains two triangles sharing an edge, a contradiction.$\square$