Problem
For $n=2020$, show that we can 4-colour the elements of the set $ V = \{1, \dots , n\}$ in such a way that any arithmetic progression of ten terms is not monochromatic.
What I have so far
Let $S_{10}$ be the set of all arithmetic progressions of length ten in $V$. Let $E$ be the event that no set in $S_{10}$ is monochromatic. Then $P(E) = 1 - P(\bar{E})$, where $\bar{E}$ is the event that some set in $S_{10}$ is monochromatic . We have
$P(\bar{E}) = \bigcup_{i = 1}^{|S_{10}|} P(E_i)$
where $E_i$ is the event corresponding to $i$th set of $S_{10}$ being monochromatic. We then have
$P(E_i) = \frac{4}{4^{10}} = \frac{1}{4^{9}}$
Now, there are a most $n^2 = 2020^2$ arithmetic progressions by the union bound $P( \bar{E} ) \leq \sum_{i=1}^{2020^2} P(E_i) = \frac{2020^2}{4^9} \approx 15$, which is no help as any probability must lie in $[0,1]$.
This is where I'm stuck, as I'm not sure which bound to use. Also, this is one of the first Exercises in of our graduate module so I don't think we're expected to invote any power theorems or the like.
(There is a similar question about 2-colouring arithmetic progressions of size 18 here however I don't understand the method of counting the APs, at least not enough to extend it to the lenth 10 case)
Any help would be greatly appreciated.
An a.p. of $10$ terms is $(a, a+d, \ldots, a + 9 d)$ where $a \ge 1$, $d \ge 1$ and $a + 9 d \le n$. Thus $9 d = (a + 9 d) - a \le n - 1$, so $d \le \lfloor (n-1)/9 \rfloor$. Given such $d$, there are $n-9d$ possibilities for $a$. Thus if $m = \lfloor (n-1)/9 \rfloor$, the number of a.p.'s of $10$ terms is $$ \sum_{d=1}^m (n - 9 d) = m n - 9 \frac{m (m+1)}{2} $$ In this case $m=224$ and the number of a.p.'s is $225680$.
Now consider a random $4$-colouring of $V$. For each a.p. of $10$ terms the probability of it being monochromatic is $1/4^9 = 1/262144$. The probability that at least one such a.p. is monochromatic is at least $1 - 225680/262144 > 0$. Therefore some $4$-colouring has no monochromatic a.p. of $10$ terms.