# to check a property of square of a graph

I am trying to work out a problem. Given a self-centered graph, is the square of the graph also a self-centered graph? I tried numerically on few graphs given in http://www.matf.bg.ac.rs/~zstanic/indexdiam.html and I got the result that, yes, they are still self-centered. But I am not getting any idea to prove it theoretically.

A self-centered graph is a graph whose diameter equals its radius. Or where the eccentricity of every vertex is the same.

The $k$th power of a graph $G$ is a graph with the same set of vertices as $G$ and an edge between two vertices iff there is a path of length at most $k$ between them.

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Since the square of the graph will add edges between every pair of vertices a distance 2 apart, you should be able to see that if the distance between two vertices in $G$ is $2k$ for some integer $k$ then the distance in $G^2$ will be $k$. What happens for vertices an odd distance apart?
Added extra detail: Suppose $u$ and $w$ are a distance $d$ apart in $G$. Then in $G^2$, they will be a distance $\lceil \frac{d}{2} \rceil$ apart. If the path was $u v_1 v_2 v_3 v_4 \ldots w$ in $G$ then there will exist a path in $G^2$ which skips every other vertex: $u v_2 v_4 \ldots w$. If a path of length less than $\frac{d}{2}$ exists in $G^2$ between $u$ and $w$ then we can conclude that in $G$ there was a path between them of length less than $d$, a contradiction.