# Prove that in all graphs with $14$ vertices and $29$ edges there are $4$ vertices $A,B,C,D$ such that they creates the cycle $AB,BC,CD,DA$

I tried to make the degrees of vertices small since the bigger the degrees get, the more possible this cycle occurs. Hence, I first drew $$14$$-sided polygon so now we have a graph with $$14$$ vertices and $$14$$ edges. Then I linked every vertices with itself so we have only $$1$$ edge left. However, if I either put this last edge between two adjacent vertices or connect one vertex with itself again, it seems that this cycle does not occur. Does it? Where is the mistake? Is there another way to approach this question or is my reasoning true?

• You are being asked about a simple graph which disallows edges back to the same vertex and also duplicate edges between the same pair of vertices. – Ross Millikan May 19 at 14:46
• Ohh, okay. Then by the same discussion if we draw every edges that connects two vertices that have one vertex between them, we still have 1 edge left. Can I say that wherever I put this last edge I would have this cycle? – ovunc May 19 at 15:02
• Use @username so that other users can be notified. – Invisible May 19 at 15:19
• You have not justified that you have a $14$ cycle to start with. I think if you start with a $14$ cycle, then draw all the edges that skip a vertex, you don't have any $4$ cycles but any other edge will make one. This does not imply that I have an answer to the question. – Ross Millikan May 19 at 15:44
• How much machinery do you have available to you? It is a theorem of Reiman from the late '50s that a $C_4$-free graph of order $n$ has size at most $\frac{n}{4}$$(1 + \sqrt{4n + 3})$, which immediately solves your problem. The proof isn't too long or difficult, but might be a little overkill depending on your background. An elementary proof of your question doesn't come to mind immediately. – Paralyzed_by_Time May 20 at 0:05