Show that $K_{4,4}$ contain no subdivision of $K_5$ Show that $K_{4,4}$ contain no subdivision of $K_5$
Here is What I think the proof look like
Assume the contrary that $K_{4,4}$ contain a subdivision $F$ of $K_5$ let $k$ be the number of vertices of degree $2$, then we have $5$ vertices of degree $4$ and $k$ vertices of degree $2$, so 
$V(F) = 5+k$
$m(F)= 10+k$
$m(K_{4,4})=16$
From here can't I jump to conclusion that $m(F)=10+k \leq 16$ so $k \leq 6$? and then $V(F) \leq 11$ but $V(K_{4,4}) =8$ ?
 A: Hint Assume by contradiction that $K_{4,4}$ contains a subdivision of $K_5$.  Mark the vertices of the $K_5$ as being red.
Look at the splitting of the $8$ vertices of $K_{4,4}$ into the two bipartite groups. By the Pigeonhole principle, one of those two bipartite groups must contain at least three red vertices.
Now, the problem can be split into two cases : the case where there are three red vertices in one bipartite group (and two in the other), and the case where one bipartite group contains four red vertices (and the other contains one). In the first case, you have at least 3 connections between the red vertices in the 'loaded' bipartite group; each of these connections must go through a vertex of degree 2 from the other group.  But this is not possible (Why?). In the second case you can make a similar argument.
A: As you say, $V(F)=5+k$ and $V(K_{4,4})=8$, so clearly $k$ is at most 3.

That means that at most 3 edges of the $K_5$ have been subdivided.
Now notice that the 5 vertices of the (original) $K_5$ must be partitioned among the two parts of the $K_{4,4}$, either as 3+2 or as 4+1.
When two vertices of the $K_5$ appear in the same part, the edge between them must be subdivided.
In the 3+2 case, at least 4 edges need to be subdivided (3 on one side and 1 on the other).
In the 4+1 case, at least 6 edges need to be subdivided (all on the side with 4 vertices).
Both cases thus yield the desired contradiction.
