Planar graph $G$ whose repeated strong products with itself are planar Is there a planar graph $G$ whose repeated strong products with itself are planar? 
 A: Recall that the mothers of all non-planar graphs are $K_5$ and $K_{3,3}$. In what follows we need only that $K_5$ is not planar.

If $G$ has no edges then all repeated strong products of $G$ with itself have no edges either, hence are planar.

So assume $G$ contains at least one edge.
Then $G\boxtimes G$ contains a $K_4$ and $G\boxtimes G\boxtimes G$ a $K_8$, which is not planar. Hence we need only consider $G\boxtimes G$, no higher products.
If $G$ contains a three-cycle (or $K_3$), then $G\boxtimes G$ contains a $K_9$ and is not planar.
Assume $G$ contains a path  $a-b-c-d$ of length $3$ (with $a,b,c,d$ distinct).
Then we find $K_5$ in  $G\boxtimes G$: We have direct edges between most of the five vertices $(a,b)$, $(b,a)$, $(b,b)$, $(b,c)$, $(c,b)$ except $(a,b)-(c,b)$ and $(b,a)-(b,c)$, which are obtainable from the non-intersecting paths $(a,b)-(a,c)-(b,d)-(c,d)-(d,c)-(c,b)$ and $(b,a)-(c,a)-(d,b)-(c,c)-(b,c)$. This gives us a planar embedding of $K_5$, which is absurd.

Now assume $G$ contains a vertex of degree $3$ (i.e., a $K_{1,3}$), say $a$ is neighbour of $b,c,d$. Then in $G\boxtimes G$ most of the five points $(a,a),(a,b),(a,c),(b,a),(c,a)$ are directly connected, except $(a,b)-(a,c)$ and $(b,a)-(c,a)$, which can but be obtained from the non-intersecting paths
$(a,b)-(d,b)-(d,a)-(d,c)-(a,c)$ and $(b,a)-(b,d)-(a,d)-(c,d)-(c,a)$.
Again, this gives us a planar embedding of $K_5$, which is absurd.


The only graphs that remain are those whose connected components are single vertices, or single edges $a-b$ (aka. $K_2$)  or two adjacent single edges $a-b-c$ (aka $K_{1,2}$). The components of $G\boxtimes G$ are then subgraphs of $K_{1,2}\boxtimes K_{1,2}$, which is planar:


Summary. The strong product $G\boxtimes G$ is planar if and only if the connected components of $G$ are subgraphs of $K_{1,2}$. For $n\ge 3$, the $n$-fold strong product of  $G$ with itself is planar if and only if $G$ has no edges.
