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Section 49.1 The Theorem

Remark 49.1.2. Importance of No Constant Terms.

Suppose p1(x)=2x and p2(x)=2x+1 and N is 2–coloured such that even numbers are red and odd numbers are blue.

  • Suppose that a is even. Then a is red and for any natural number d, a+p1(d) is red because a+p1(d)=a+2d is even, but a+p2(d) is blue because a+p2(d)=a+2d+1 is odd.

  • Suppose that a is odd. Then a is blue and for any natural number d, a+p1(d) is blue because a+p1(d)=a+2d$ is odd, but a+p2(d) is red because a+p2(d)=a+2d+1 is even.

Contradiction!

Remark 49.1.3. Is the restriction of integer coefficients necessary?

Suppose the polynomials p1,p2,,pm have rational coefficients and no constant term. Then define qi(x)=pi(cx), for 1im where c is the least common multiple of the denominators of the coefficients of all the polynomials. Now the polynomials q1,q2,qm have integer coefficients and no constant term.

For example, let p1(x)=23x2+3x, p2(x)=512x3+4x2+28x, p3(x)=724x4, p4(x)=83x3+92x.

Then c=24.

The coefficient of the term xn is multiplied by cn, so the obtained polynomials are with integral coefficients:

q1(x)=p1(24x)=23(24x)2+3(24x)=384x2+72x,
q2(x)=p2(24x)=512(24x)3+4(24x)2+28(24x)=5760x3+2304+6x,
q3(x)=p3(24x)=724(24x)4=96768x4, and
q4(x)=p4(24x)=83(24x)3+92(24x)=36864x3+108x.

If a,dN are such that a+q1(d),a+q2(d),a+q3(d), and a+q4(d) are of the same colour, then a+p1(24d),a+p2(24d),a+p3(24d), and a+p4(24d) are of the same colour too.

Remark 49.1.4. Why the name ``Polynomial van der Waerden Theorem''.

Suppose that pi(x)=ix for 1il1. Then whenever N is finitely coloured there exists natural numbers a and d such that the set {a,a+p1(d),a+p2(d),,a+pl1(d)}is monochromatic.

However we can also view this as a monochromatic set {a,a+d,a+2d,...,a+(l1)d.

This is van der Waerden's theorem!