let me try by induction on .
If then for every it suffices to take and the statement holds.
Suppose that the statement holds for , that is, for every there exists some such that if the interval is -coloured then there will be a monochromatic arithmetic progression of length and the common difference of the arithmetic progression is of the same colour.
We'll show that for , for every it will suffice to take , where denotes van der Waerden number.
Let the interval be -coloured, it must contain a monochromatic arithmetic progression of length , let's denote it and let,WLOG, blue be its colour.
If there's such that is blue, then is a blue arithmetic progression of length with blue common difference , and we're done.
So suppose this is not the case, then is an arithmetic progression of length coloured with colours. By the induction hypothesis, -coloured interval contains a monochromatic arithmetic progression with the common difference of the same colour. This means that is a monochromatic arithmetic progression of length with common difference of the same colour. The induction step is completed.