I hope that what follows is right: We will work with indices modulo $n$. We also make the following association: $0 \mapsto \textrm{red}, \, 1 \mapsto \textrm{blue}$. $AB = 0$ or $AB = 1$ will mean that $[AB]$ has the corresponding color. (a) Letting $j = i+k$, we see that $P_{i+2m}P_{i+2m+k}= P_{i}P_{i+k}$ and that $P_{i+2m+1}P_{i+2m+1+k}= 1-P_{i}P_{i+k}$. Thus, the color of a segment of "length" (i.e. the absolute value of the diferrence of the indices of its endpoints --- lots of "of", sorry ) determines completely the color of all segments of length $k$. So the only contradiction that could arise is that $P_{i}P_{i+k}= 0$ implies $P_{i}P_{i+k}= 1$ (or the other way around). We have two cases (all computations are $\bmod \, n$): Case 1. $i = i+2m+1, \, i+k = i+2m+1+k \, \, \Longleftrightarrow \, \, 2m+1 = 0 \, \, \Longleftrightarrow \, \, n \equiv 1 \left( \bmod \, 2 \right)$. Case 2. $\left. i = i+2m+1+k, \, i+k = i+2m+1 \, \, \Longleftrightarrow \, \, 2m+1=-k, \, 2m+1=k \, \, \Longleftrightarrow n \equiv 2 \left( \bmod \, 4 \right) \right.$ (Don't forget that $k = j-i \neq 0$). Thus, the only "good" $n$ are those that satisfy $n \equiv 0 \left( \bmod \, 4 \right)$. (b) Make some cyclic permutations and perhaps a "mirror reflection" on the indices of the $P_{i}$'s such that $P_{0}P_{1}= 0$, where $P_{0}$ is the point from where we start our journey. We'll prove that we can reach $2k, \, 2k+1$, $\forall k \in \overline{1, \frac{n}2-1}$ ($P_{1}$ can obviously be reached). Clearly, we have that $P_{2k}P_{2k+1}= 0$. If $P_{0}P_{2k}= 0$, then we can take the route $P_{0}\mapsto P_{2k}\mapsto P_{2k+1}$. Otherwise, $P_{0}P_{2k}= 1$, so $P_{1}P_{2k+1}= 0$. Thus, we can choose: $P_{0}\mapsto P_{1}\mapsto P_{2k+1}\mapsto P_{2k}$. So... is it OK? And also: what's $AMO$?

@Perfect_radio After ten years I think may be you have made a little mistake I think the answer of $(a)$ is just $2|n$ You may ask $n=6$ you choose the longest diagonal will be a contradiction,but the problem didn't say you can "mod", for example,it didn't say $P_{n+1}=P_{1}$ so may be $P_{j+1}$ doesn't exist at all. If you do this problem like this the answers will be different,and the difficulty will increase a lot,but it's a P5 so what do you think?(or you can just ignore this post and consider me as a stupid boy who couldn't figure out this problem after thinking for 10 years) Your sincerely $ CeuAzul $

Who can resolve this proof?

Math1331Math wrote: Who can resolve this proof? If the points aren't taken cyclicly, then part (a) of the problem makes zero sense, since any value of $n$ would work. So I think it's safe to assume that the points are taken cyclicly. I doubt part (b) would be true if the points aren't taken cyclicly since the problem would be far too weak, though I may be wrong. (a): If $n$ is odd, then $j = i + 1$ is enough to form a contradiction by going around in a circle, since a line segment cannot be both red and blue. If $n$ is 2 mod 4, then by considering the main diagonals, the same contradiction occurs. So $n$ must be a multiple of $4$. It's easy to form a construction in this case. (b) Suppose there are $4k$ vertices. Let $u$ be a vertex with degree $2k$ red edges and $v$ be a vertex with degree $2k-1$ red edges. Suppose for the sake of contradiction that there exists no path of length 3. Let $w$ be any point such that $v$ and $w$ are connected with a red edge. Then, there cannot exist an edge from $v$ to $v$, $u$ or any one of the possible $w$'s. This rules out $2k + 2$ vertices, so at most $2k - 2$ vertices can be connected to $v$ with a red edge. Since $v$ is of degree $2k - 1$, this is impossible. So, there must exist a path of length at most 3.