For a given $2k-1$ edge coloring of the complete graph $K_{2k}$, say a Hamiltonian path $P$ is a rainbow path if every color appears exactly once in $P$.
My question is
"For each even integer $2k$, 2k (k \geq 2)$, is there a proper $2k-1$ edge coloring of $K_{2k}$ with no rainbow paths?"
It is easy to see that every proper edge-coloring of $K_4$ or $K_6$ does not contain any rainbow paths.
For $K_8$, I checked that the statement is still true but some proper 7-edge-coloring contains rainbow paths.
When the number of vertices is a power of 2(say $2^k$), , then the statement is true, but I do not know if it is true for every even number (I don't even know for $2k = 10$).
Here's why the statement is true for $2^k$.
Label the vertices of $K_{2^k}$ by the elements of the group $((Z_2)^n, ((Z_2)^k, +)$ and label each edge by the sum(or difference) between two ends. Then the edge-labels give us a proper $2^k-1$ edge coloring, and this coloring does not have any rainbow paths.
