Given a graph $G$, $G^2$ is defined as the graph with the same vertex set and $E(G^2) =E(G) \cup {xy: \exists z s.t. xy,zy \in E(G)}$ or simply we add edges between vertices of distance 2. I need to prove that this is planar or non-planar for each n.
As I can see, $C_3^2 = C_3$ is planar and for all $n \geq 4$ when n is odd the graph is non-planar and when n is even the graph is planar.
However, I am not sure how to prove $C_{2n}^2$ is planar in a rigorous way. Can I just say, once we enumerate all the vertices we can connect k == 1 (mod 2) among each other from outside and k==0 (mod 2) from inside? It is obvious but I am not sure if I should instead try to prove there is no $K_{3,3},K_{5}$ minors in $C_{2n}^2$ . Thanks
Once we enumerate all the vertices we can connect k == 1 (mod 2) among each other from outside and k==0 (mod 2) from inside. This proves it is planar