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Question I found in "Introduction to Graph Theory" by Douglas B. West:

Let $G$ graph on $n$ vertices with connectivity $\kappa(G)=k \geq 1$. Prove that $n \geq k(\operatorname{diam}(G)-1)+2$ where $\operatorname{diam}(G)$ is the the maximal (edge) distance between a pair of vertices in $G$

Seems easy and I've seen the answer, too, which goes the same as thought: by taking the vertices the end vertices of the path of length $\operatorname{diam}(G)$, and applying Menger's Theorem. I'm having a hard time understanding one part - it seems like all the paths need to be of the same length for the $n \geq k(\operatorname{diam}(G)-1)+2$ to apply. Am I missing some part here?

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Let's write $d$ for the diameter of $G$. Then there is some pair of vertices such that one path between them has exactly $d$ edges, and each path between them has at least $d$ edges. Menger says there are $k$ independent paths. So the number of vertices is at least $k(d-1)+2$.

I'm not sure why you think the paths all have to be the same length. They just have to have length at least $d$ - and, they do.

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    $k$ stands for the connectivity of $G$, which is not the number of components. I don't have $d$ internally disjoint paths, I have $k$ internally disjoint paths, each path having at least $d$ edges.2014-10-13