to find the distance i am trying to find the power graphs of cycles $C_n$ and then calculation of distances between vertices. for cycles $C_n$ we can find power graphs upto power greatest integer function of n/2. Square of $C_n$ yielding result that distance between any two vertices is same. my question is how to prove that distance between any two vertices is same on squaring cycles.
 A: Thanks for the description above; that clears it up.  (The square of $C_n$ is a special case of a Cayley graph $\mathrm{Cay}(\mathbb{Z}_n,\{\pm 1,\pm 2\})$.  Since the group here is the cyclic group $\mathbb{Z}_n$, this is also known as a circulant graph.)
Here's an example, shaded according to their distance from the top vertex:

It would be possible to find the distance between any two vertices in "$C_n$ squared" by induction.  Firstly, we need to check the small cases, then assume $n \geq 5$.  After which, the following proof will work:


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*Base case: Check $\mathrm{dist}(u,u+1)=0$, $\mathrm{dist}(u,u+1)=1$, $\mathrm{dist}(u,u+2)=1$, $\mathrm{dist}(u,u-1)=1$, $\mathrm{dist}(u,u-2)=1$, and prove that there are no other vertices of distance $\leq 1$.

*Inductive step: for $k \in \{2,3,\ldots,\lfloor n/4 \rfloor-1\}$, prove that $\mathrm{dist}(u,u+2k-1)=k$, $\mathrm{dist}(u,u+2k)=k$, $\mathrm{dist}(u,u-2k+1)=k$, $\mathrm{dist}(u,u-2k)=k$, , and prove that there are no other vertices of distance $\leq k$.  This part will be made easier using the following property:


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*The distance between two distinct vertices $u$ and $v$ satisfies $$\mathrm{dist}(u,v)=1+\min_{v' \in N(v)} \mathrm{dist}(u,v')$$ where $N(v)$ denotes the set of vertices adjacent to $v$.


*End case: In this problem, the number of vertices of distance $\lfloor n/4 \rfloor$ from $u$ can be $1,\ldots,4$, depending on the value of $n$.  This needs to be accounted for separately.  (This is simply a bookkeeping issue that could arise.)
Other comments:


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*"...distance between any two vertices is same..."  If this implies that all pairs of vertices $C_n$ and the square of $C_n$ have the same distance, then this is not true: e.g. when $n=5$, the square of $C_n$ has $4$ vertices of distance $1$.  In fact, squaring any graph with an induced $3$-node path $uvw$ will decrease the distance between $u$ and $w$.

