TITLE: A 4-Regular graph with 7 vertices is non planar
QUESTION [8 upvotes]: A Graph with 7 vertices each having degree 4 cannot be planar. Any hints on the proof?

REPLY [4 votes]: Instead of trying to find $4$-regular graphs on $7$ vertices, find complements of $4$-regular graphs on $7$ vertices. These are $2$-regular graphs, hence a $C_7$ and a $C_3 \cup C_4$. 
In $C_7$ we can take vertices $(1,2,3)$ and $(4,5,6)$ in two partitions. Although $3$ and $4$ are connected, we will have a path between $3$ and $4$ via $7$ in $\overline{C_7}$ hence has a minor isomorphic to $K_{3,3}$. 
In $C_3 \cup C_4$, we will take $(1,2,3)$ [from $C_3$] and $(4,5,6)$ [from $C_4$] in two partitions. So in $\overline{C_3 \cup C_4}$, we have a $K_{3,3}$ present.
Hence there are no  planar $4$-regular graphs on $7$ vertices.