Proof of $m \leqslant n \times \frac{\sqrt{4n-3} + 1}{4}$ in a graph with no cycle with length 4. (m: edges, n: vertices)

Asked Jul 02 '20 at 12:15

Active Jul 02 '20 at 12:18

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I have faced a problem and do not know how to prove the following.

In a graph with no cycles with length $4$, prove that $$m \leqslant n \times \frac{\sqrt{4n-3} + 1}{4}$$ where $m$ is the number of edges and n is the number of vertices.

I tried to count number of paths with length $2$ and relate it to $m$, but couldn't consider the graph has no cycle with length $4$ in the equation.

Can anyone please help me?

edited Jul 02 '20 at 12:18

Anand

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asked Jul 02 '20 at 12:15

MMM

$\frac{\sqrt{4n-3} + 1}{4}$ is a root of $4 x^2 - 2 x + 1=n$ if that helps. – Alexey Burdin Jul 02 '20 at 12:50
@AlexeyBurdin You mean $4 x^2 -2x + 1 = 0$ ? – MMM Jul 02 '20 at 13:06
No. – Alexey Burdin Jul 02 '20 at 13:10
Some ideas might be gleaned from the previous Question Maximum edges in a square free graph. – hardmath Jul 02 '20 at 13:33
Your question is already answered here. – Alex Ravsky Jul 03 '20 at 02:05
1

@AlexRavsky Thanks a lot Alex, I hadn't seen that post previously. – MMM Jul 03 '20 at 17:22

Proof of $m \leqslant n \times \frac{\sqrt{4n-3} + 1}{4}$ in a graph with no cycle with length 4. (m: edges, n: vertices)

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