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I need a source where I can find a proof of Reiman's theorem:

If the graph G is quadrilateral($C_4$)-free, then

$$|E(G)| \leq \frac{|V(G)|}{4}(1 + \sqrt{4\cdot|V(G)| - 3})$$

Here is the idea of the proof:

Consider the set of ordered triplets $$F= \{(u, v, w) : uv \in E(G), uw \in E(G)\}$$ $$|F| = \sum_{u \in V(G)}(d_G(u)(d_G(u) - 1))$$ On the other hand, as $G$ doesn't have $C_4$, then (can someone explain this inequality?) $$|F| \leq n(n - 1)$$ where $n = |V(G)|.$

This part is already clear to me:
Each triplet is defined by its $2$ ends $u$ and $w$, you can choose these $2$ vertices in $\binom{n}{2}$ ways. These triplets are ordered, so we are getting the maximum capacitance of $F$ is $2\binom{n}{2}$. And, therefore, the above inequality holds.
Then, using Cauchy–Schwarz Inequality

we're getting full proof of the theorem.

Thanks in advance.

RobPratt
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Heghine
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    There are a few links to papers in the answers to this question that might be helpful: http://math.stackexchange.com/questions/571823/maximum-edges-in-a-square-free-graph – Casteels Jun 04 '14 at 17:56
  • Béla Bollobás Extremal Graph Theory (1978), section 6.2, pp. 309-313 gives a proof. If this isn't easily available to you, I can try to summarize it when I get a free moment. – Perry Elliott-Iverson Jun 04 '14 at 18:41
  • Thank you, @PerryIverson. But I couldn't find there this theorem exactly. – Heghine Jun 05 '14 at 09:52
  • @Heghine At the bottom of page 313, Bollobás notes that Theorem 2.3 has this consequence. I believe it's a consequence of Theorem 2.3 together with Theorem 2.6(i). – Perry Elliott-Iverson Jun 05 '14 at 12:01
  • I'm interested in this proof but don't have access to this book. Is it easy to give a quick intuition about how to view the graph ? Or just the starting point ? – Manuel Lafond Jun 05 '14 at 18:10
  • @manuellafond, (http://books.google.am/books?id=q2MAF1vYuWQC&printsec=frontcover#v=onepage&q&f=false) here is the book. – Heghine Jun 06 '14 at 06:54
  • @manuellafond, I've also added the idea of the proof. – Heghine Jun 06 '14 at 07:20

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The key is to determine, for each ordered pair of vertices $v$, $w \in V(G)$, how many vertices $u \in V(G)$ are such that $(u, v, w) \in F$.

Our assumption that $G$ is $C_4$-free means that for every ordered pair $v$, $w \in V(G)$, there exists at most one vertex $u$ such that $uv$, $uw \in E(G)$. This is because if there existed distinct $v$, $w$, $u$, $u' \in V(G)$ such that $uv$, $uw$, $u'v$, $u'w \in E(G)$, then $u$, $u'$, $v$, and $w$ would form a copy of $C_4$, a contradiction.

It follows that $|F|$ is at most the number of ordered pairs of vertices of $G$. That is, $$|F| \leq 2\binom{n}{2} = n(n - 1),$$ as we wanted to show.

Andrew Uzzell
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