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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.

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  • $\begingroup$ What do you mean by "every triangle has at most $2n -3$ edges going out of its vertices"? Do you mean every vertex has degree at most $2n-3$? $\endgroup$ – vxnture Apr 3 at 12:32
  • $\begingroup$ @vxnture I mean that if $a,b,c$ are the degrees of the vertices making a triangle, then $(a-1)+(b-1)+(c-1) \leq 2n -3$. If the inequality doesn't hold then by Dirichlet (pigeonhole) principle there is a vertex that is conected with two vertices of the triangle. $\endgroup$ – Lada Dudnikova Apr 3 at 14:11
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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.$\square$

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