Van der Waerden's theorem

Van der Waerden's theorem states that for any given positive integers C and N, there is some number V(C, N) such that if the integers {1, 2, ..., V(C, N)} are colored, each with one of C different colors, then there are at least N integers in arithmetic progression all of the same color.

For example, when C=2, you have two colors of paint, say red and blue. V(2, 3) is bigger than 8, because you can color the integers from {1,...,8} like this:

        1 2 3 4 5 6 7 8
        B R R B B R R B 

and no three integers of the same color form an arithmetic progression. But you can't add a ninth integer to the end without creating such a progression. If you color the number 9 red, you get

        1 2 3 4 5 6 7 8 9
        B R R B B R R B R

and if you color it blue you get

        1 2 3 4 5 6 7 8 9
        B R R B B R R B B

Of course, this doesn't prove that there is no way to color the integers {1,...,9} so that there is no single-colored arithmetic progression.

It is an open problem to determine the values of V(C, N) for various C and N. The proof of the theorem provides only an upper bound. For the case of C=2 and N=3, for example, van der Waerden's theorem says it is sufficient to color the integers {1,...,325} with two colors to guarantee there will be a single-colored arithmetic progression of length 3. But in fact, the bound of 325 is very loose; the minimum required number of integers is only 9. Any coloring of the integers {1,...,9} will have three evenly spaced integers of one color.

For C=3 and N=3, the bound given by the theorem is 7(2·3⁷+1)(2·3^7·(2·37+1)+1), or approximately 4.22·10¹⁴⁶¹⁶. But actually, you don't need that many integers to guarantee a single-colored progression of length 3; you only need 27. (It is possible to color {1,...,26} with three colors so that there is no single-colored arithmetic progression of length 3; for example, RRYYRRYBYBBRBRRYRYYBRBBYBY.)

Anyone who can reduce the general upper bound to any 'reasonable' function can win a large cash prize.

Proof of van der Waerden's Theorem

We will prove the special case mentioned above, that V(2, 3) ≤ 325. Let c(n) be a coloring of the integers {0,...,324}. We will find three elements of {0,...,324} in arithmetic progression that are the same color.

Divide {0,...,324} into the 65 blocks {0,...,4}, {5,...,9}, ... {320,...,324}. Since each integer is colored either red or blue, each block is colored in one of 32 different ways. By the Pigeonhole principle, there are two blocks among the first 33 that are colored identically. That is, there are two integers b₁ and b₂, both in {0,...,32}, such that c(b₁·5 + k) = c(b₂·5 + k) for all k in {0,...,4}.

Among the three integers b₁·5 + 0, b₁·5 + 1, and b₁·5 + 2, there must be at least two that are the same color. (The pigeonhole principle again.) Call these b₁·5 + a₁ and b₁·5 + a₂, where the a_i are in {0,1,2} and a₁ < a₂. Suppose (without loss of generality) that these two integers are both red. (If they are both blue, just exchange 'red' and 'blue' in what follows.)

Let a₃ = 2·a₂ - a₁. If a₃ is red, then we have found our arithmetic progression: b₁·5 + a_i are all red.

Otherwise, b₁·5 + a₃ is blue. Since a₃ ≤ 4, b₁·5 + a₃ is in the b₁ block, and since the b₂ block is colored identically, b₂·5 + a₃ is also blue.

Now let b₃ = 2·b₂ - b₁. b₃ ≤ 64. Consider the integer b₃·5 + a₃, which must be ≤ 324. What color is it?

If it is red, then b₁·5 + a₁, b₂·5 + a₂, and b₃·5 + a₃ form a red arithmetic progression. But if it is blue, then b₁·5 + a₃, b₂·5 + a₃, and b₃·5 + a₃ form a blue arithmetic progression. Either way, we are done.

A similar argument can be advanced to show that V(3, 3) ≤ 7(2·3⁷+1)(2·3^7·(2·37+1)+1). One begins by dividing the integers into 2·3^7·(2·37+1)+1 groups of 7(2·3⁷+1) integers each; of the first 3^7·(2·37+1)+1 groups, two must be colored identically. Divide each of these two groups into 2·3⁷+1 subgroups of 7 integers each; of the first 3⁷+1 subgroups in each group, two of the subgroups must be colored identically. Within each of these identical subgroups, two of the first four integers must be the same color, say red; this implies either a red progression or an element of a different color, say blue, in the same subgroup. Since we have two identically-colored subgroups, there is a third subgroup, still in the same group that contains an element which, if either red or blue, would complete a red or blue progression, by a construction analogous to the one for V(2, 3). Suppose that this element is yellow. Since there is a group that is colored identically, it must contain copies of the red, blue, and yellow elements we have identified; we can now find a pair of red elements, a pair of blue elements, and a pair of yellow elements that 'focus' on the same integer, so that whatever color it is, it must complete a progression.

It should be noted that the proof for V(2, 3) depends essentially on proving that V(32, 2) ≤ 65. We divide the integers {0,...,324} into 65 'blocks', each of which can be colored in 32 different ways, and then show that two blocks must be the same color. Similarly, the proof for V(3, 3) depends on proving that V(3^7·(2·37+1), 2) ≤ 2·3^7·(2·37+1)+1. By a double induction on the number of colors and the length of the progression, the theorem is proved in general.