[Solved]: Team construction in tri-partite graph

What is the minimal $k$ such that for every red/blue-coloring of the edges of $K_{k,k,k}$ (the complete 3-partite graph with $k$ vertices in each partition) there is either a red triangle or a blue $K_{3,3}$?

Erel proved that the lower bound on $k$ is at least 5, and then using a SAT formulation that $k ge 8$. frafl showed that the upper bound on $k$ is at most 15. Aravind sketched a nice argument for a better upper bound. Here is a more detailed form of Aravind’s argument. If a vertex $u$ in partition $A$ is red-connected to 3 vertices $S$ in partition $B$ and 3 vertices $T$ in partition $C$, then there is either a red triangle involving $u$ and one vertex from each of $S$ and $T$, or otherwise $Scup T$ induces a blue $K_{3,3}$. So no vertex can have more than 2 red-connected neighbours in both of its neighbour partitions. Hence every vertex has at least $k-2$ blue-connected neighbours in at least one of its neighbour partitions. Let $S$ be the vertices in $A$ which have at least $k-2$ blue-connected neighbours in $B$, and $T$ be those vertices in $A$ which have at least $k-2$ blue-connected neighbours in $C$; note that $A = Scup T$. If $Scap T$ is non-empty, then switching colours yields a contradiction since $kge 5$. So assume $S$ and $T$ are disjoint. In fact, each vertex in $S$ must be blue-connected to at most 2 vertices in $C$ (so red-connected to at least $k-2$ vertices in $C$), and each vertex in $T$ must be blue-connected to at most 2 vertices in $B$ (and red-connected to at least $k-2$ vertices in $C$). Now $k ge 6$ so without loss of generality suppose that $S$ contains a subset $S’$ with at least 3 vertices. They are each blue-connected to at least $k-2$ vertices in $B$, so these neighbourhoods must have a common intersection $U$ with at least $k-6$ vertices. If $kge 9$, then $U$ contains at least 3 vertices, so $S’cup U$ induces a blue $K_{3,3}$. This shows that $kge 9$ is enough to always meet the conditions, and 9 is therefore an upper bound on the desired quantity. What remains is to either demonstrate a counterexample with $k=8$ (which would show that the desired quantity is 9), or to show that $k=8$ is always enough to guarantee a red triangle or a blue $K_{3,3}$ (which would show it is 8).