Van Der Waerden’s Theorem

Abstract

Let us show that if [1,325] is colored with 2 colors, then there is a monochromatic AP of length 3. Assume indirect, and divide [1,325] into 65 blocks of length 5, say B1, . . . , B65. Each block can be colored 2 = 32 ways thus there are two blocks of the same color, Bi, Bi+d, such that 1 ≤ i < i + d ≤ 33. Within the first 3 elements of Bi two of them a, a + e are of the same color, say Red. Then all elements of the set {a, a + e, a + d, a + d + e} are Red. Note that {a, a + e, a + 2e} ⊂ Bi and {a + d, a + e + d, a + 2e + d} ⊂ Bi + d thus both a + 2e and a + 2e + d must be Blue. Now, if a + 2e + 2d is Blue, then the AP {a+2e, a+2e+d, a+2e+2d} is Blue, while if it is Red then the AP {a, a+e+d, a+2e+2d} is Red. This contradicts our indirect assumption.