(a) Observe that $0$ must be blue, due to (1) (otherwise suppose $0$ wasn't blue, take $i$ to be a blue number (which exists) then apply (2) and (3)). Now, assume $1$ were also blue, if $k$ were the smallest orange number then $k+1$ would be orange as well and due to $0$ and $1$ being blue, $k \geq 2$. The same argument works for red, so this means all integers $i \leq 1$ are blue, and after that it can have at most 2 colors (blue, with orange or red but not both). Therefore, $0$ and $1$ must be of different colors. (b) Observe that there must be at least 2 blue integers, since if there were only one, and all three colors are used; there will be at least 1 color with 2 different elements, so using (1) we can construct $2 \times 1$ distinct blue numbers (so there is a blue number that's not 0). Using this observation and problem (a), we can generalize the fact above, in the following way: Lemma. The coloring is periodic, with period $|x| > 1$, where $|x|$ is the smallest positive magnitude such that $x$ has a blue color. ProofNote that by (2) and (3), if $x$ is blue then $-x$ must be blue, since $x + (-x) = 0$ and $0$ is blue. Continue to induct to deduce that all integer multiples of $x$ are blue (take $(kx) + (-x) = (k-1)x$ knowing that $-x$ and $(k-1)x$ are both blue). Next, look at $k|x| + a$, where $k$ is an integer and $1 \leq a \leq |x|$. Since $|x|$ is blue, $(k+1)|x|+a = (k|x|+a) + |x|$ so they have the same color (due to (2) and (3)). From the lemma, it follows that blue colors only exist at integer multiples of $x$ $(\star)$. Suppose $x > 0$ since $-x$ and $x$ are the same color. So consider the coloring in $\pmod{x}$. Now we'll show that $x = 3$. It's clear that $x = 2$ doesn't work since we know all 3 colors are used, and due to periodicity. If otherwise $x > 3$, pick $0 < a < b < x$ of the same color and $0 < c < x$ of a different color (neither color is blue). By (1), $a+c$ and $b+c$ must be both blue, however $0 < (b+c)-(a+c) = b-a < x$ which violates $(\star)$. So all possible colorings are blue for all integers $0 \pmod{3}$, and for $1 \pmod{3}$ and $2 \pmod{3}$ we can color them red, orange respectively or vice versa, and we've shown that these are the only ones that work.