Philippine Mathematical Olympiad

Solution:

It suffices to show that for $n \geq 18$, the total number of colorings (without restriction) exceeds those that make some $n$-term arithmetic sequence monochromatic.

There are $2^{2017}$ colorings of a set with 2017 elements. The number of colorings that make a fixed $n$-term sequence monochromatic is $2 \cdot 2^{2017-n} = 2^{2018-n}$, since the terms not in the sequence can be colored without restriction, while those in the sequence can be colored either all red or all white.

We now find the number of $n$-term arithmetic sequences that can be obtained from $A$. Such a sequence $a, a+r, \ldots, a+(n-1) r$ is completely determined by the first term $a$ and common ratio $r$, subject to the constraint $a+(n-1) r \leq 2017$. For each value of $a$, there are $\left\lfloor\frac{2017-a}{n-1}\right\rfloor$ sequences that start with $a$. This means that the number of arithmetic sequences does not exceed
$$\sum_{a=1}^{2017} \frac{2017-a}{n-1} = \frac{2016 \cdot 2017}{2(n-1)}$$
Therefore, the total number of colorings that make at least one arithmetic sequence monochromatic does not exceed
$$2^{2018-n} \cdot \frac{2016 \cdot 2017}{2(n-1)}$$
But for $n \geq 18$,
$$\begin{aligned} 2^{2018-n} \cdot \frac{2016 \cdot 2017}{2(n-1)} & \leq 2^{2018-n} \cdot \frac{2048 \cdot 2048}{2(n-1)} \\ & = \frac{2^{2039-n}}{n-1} \leq \frac{2^{2021}}{17} < 2^{2017} \end{aligned}$$