17. [10] How many ways are there to color every integer either red or blue such that $n$ and $n+7$ are the same color for all integers $n$, and there does not exist an integer $k$ such that $k, k+1$, and $2 k$ are all the same color?
Answer: $\square$
Solution: It suffices to color the integers from 0 through 6 and do all arithmetic mod 7. WLOG, say that 0 is red (we'll multiply by 2 in the end). Then 1 must be blue because $(0,0,1)$ can't be monochromatic. 2 must be red because $(1,2,2)$ can't be monochromatic. Then we have two cases for what 3 is:

Case 1: 3 is red. Then 4 is blue because $(2,3,4)$ can't be monochromatic. This makes 5 red because $(4,5,1)$ can't be monochromatic. Finally, 6 must be blue because $(6,0,5)$ can't be monochromatic. This gives a single consistent coloring for this case.
Case 2: 3 is blue. 4 can't also be blue because this would imply that 5 is red (because of $(4,5,1)$ ) and 6 is red (because of $(3,4,6)$ ), which would make $(6,0,5)$ all red. So 4 must be red. Then we have two possibilities: either 5 is red and 6 is blue, or 5 is blue and 6 is red ( 5 and 6 can't both be red because of $(6,0,5)$, and they can't both be blue because of $(5,6,3))$. These give two consistent colorings for this case.
Overall, we have three consistent colorings: RBRRBRB, RBRBRRB, and RBRBRBR. Multiply this by 2 because 0 could have been blue, and our answer is 6 .