The diagonal Ramsey number is the least positive integer such that every red-blue edge colouring of the complete graph contains a monochromatic copy of .
For the existence and bound, write for the least forcing either a red or a blue . Given a vertex of , the pigeonhole principle says that at least incident edges are red or at least are blue. In the first case those red neighbours contain a red , which joins to to make a red , or a blue ; the second case is symmetric. Hence
Starting from and using Pascal's identity gives the binomial upper bound for a Ramsey number
Therefore
which proves both existence and the required estimate.
To determine , color a cycle graph red and its complement graph, also a , yellow. This colors without a monochromatic triangle, so . At any vertex of a red-yellow , at least three of the five incident edges have one color. If an edge among their three other endpoints has that color, it completes a monochromatic triangle with the chosen vertex; if none does, those three endpoints themselves form a triangle of the other color. Thus , and
The same local argument proves uniqueness on . No vertex can have three incident edges of one color: their other endpoints would either contain an edge of that color or form a triangle of the other color. Every vertex therefore has vertex degree at most two in each color, and because the two degrees sum to four, each is exactly two. Each color class is consequently a regular graph of degree two on five vertices, hence a single . Up to relabelling, the coloring is exactly the cycle-and-complement coloring above.
Finally, the monochromatic-triangle counting formula gives, for a coloring of with red degrees ,
The bound is attained by coloring all edges of red and the two complementary copies of yellow: there are no red triangles and exactly two yellow triangles. The least requested number is therefore