# Coloring numbers from $1$ to $1000$

I mostly just need someone to explain to me this problem:

Prove that it is possible to $2$-color the integers from $1$ to $1000$ so that no monochromatic arithmetic progression of length $17$ is formed.

For some reason when I think about this problem I just think of starting with coloring $1$ red then alternating until $1000$... There is obviously more to this problem, but I can't figure out what it is.

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The problem asks you to find a coloring such that $a_1, a_2, \dots a_{17}$ never share a color, where $a_i = a_1 + (i-1)*d$. If you color even numbers blue and odd numbers red, then take $d=2$ to find a sequence of single color. – Karolis Juodelė Mar 4 '13 at 6:42
What's the reasoning behind the formula? – MITjanitor Mar 4 '13 at 6:49
by an arithmetic progression of 17, do you mean one with 17 terms or with common difference of 17? – Vincent Tjeng Mar 4 '13 at 6:58
What I put above is all that I know for the question. I took it to be 17 consecutive terms. But I may not know what arithmetic progression means. – MITjanitor Mar 4 '13 at 7:02
The formula is what is meant by arithmetic progression. The problem is definitely talking about a progression of 17 terms, with any difference. Even if you've never heard of the length of a sequence, a hint to this is just how trivial the other interpretation would be. – Karolis Juodelė Mar 4 '13 at 7:13

As has been clarified in the comments, by an arithmetic progression of length $17$ they mean an arithmetic progression with $17$ terms, i.e. a progression of $17$ integers given by $a_i=a_1+(i-1)d$ for some integers $a_1$ and $d$, for instance $4,9,14,\dotsc,74,79,84$ for $a_1=4$ and $d=5$.
The claim can be proved using the probabilistic method by showing that the expected number of monochromatic arithmetic progressions of length $17$ upon independently colouring each integer with either colour with equal probability is less than $1$.