I want a $K_5$-free $2$-coloring of $K_{42}$ to prove $R(5, 5) > 42$. The only construction I could find is from Exoo, G. (1989). A lower bound for R (5, 5). Journal of graph theory, 13(1), 97-98:
A cyclic coloring is one in which the vertices of $K_n$ are identified with the integers from $0$ to $n - 1$, and in which the color of an edge depends only on the numerical differences between its end-vertices. These differences are often called chord lengths. A cyclic coloring can be specified by giving the chord lengths (from $\mathbf{1}$ to $\mathbf{[n/2]}$) for each color.
Q1: Don't we need the chord lengths from $1$ to $n - 1$? How else will we know what color to assign to the edge $0 \leftrightarrow n - 1$, which has chord length $n - 1$?
The coloring of $K_{42}$ starts with a coloring of $K_{43}$. Assign the chord lengths $1, 2, 7, 10, 12, 13, 14, 16, 18, 20, 21$ to red and $3, 4, 5, 6, 8, 9, 11, 15, 17, 19$ to blue. The resulting $K_{43}$ has $0$ blue $K_5$s and $43$ red $K_5$s, which have vertices of the form $$\{i, i + 1, i + 2, i + 22, i + 23 \} \quad \text{for } i = 0, 1, \dots, 42$$
Q2: Does this use congruence classes mod $43$? When $i = 41$, the vertices would be $\{ 41, 42, 0, 20, 21 \}$?
The next step is to delete vertex $0$, producing a colored $K_{42}$ with only $43 - 5 = 38$ red $K_5$s. Finally, change the following edges from red to blue: $11 \leftrightarrow 32$ and $j \leftrightarrow j + 1$ for $j = 4, 5, 6, 7, 13, 14, 15, 16, 23, 24, 30, 33, 39, 40, 41$. One can verify with a computer that this coloring of $K_{42}$ is $K_5$-free.
I really want to see a picture of this coloring, but I can't find one.