1. Initial Setup and Definitions:
- Let n be a natural number and consider a circle with circumference n.
- On the circle, in a clockwise direction, numbers 0,1,2,…,n−1 are written, each at equal distances.
- Each number is colored either red or blue, with at least one number of each color.
- There exists a set S⊊{0,1,2,…,n−1} with ∣S∣≥2 such that if (x,y) is a circle sector with endpoints of distinct colors and y−x∈S, then y∈S.
2. Claim:
- We claim that there exists a divisor d of n different from 1 and n such that if (x,y) are different points of distinct colors and their distance is divisible by d, then both x and y are divisible by d.
3. Basic Additive Number Theory:
- For any set of integers {s1,s2,…,s∣S∣} such that gcd(s1,s2,…,s∣S∣,n)=d, we can find coefficients c1,c2,…,c∣S∣ such that:
i=1∑∣S∣cisi≡d(modn)
- This implies that for d=1, x and x+d must be the same color for every d=x.
4. Induction Proof:
- Base case: For ∣S∣=1, this is just the Euclidean algorithm.
- Inductive step: Assume it is true for k=∣S∣. This means we can find coefficients c1,…,c∣S∣ such that:
i=1∑∣S∣cisi≡dS(modn)
where dS is the gcd of S={s1,…,s∣S∣}.
- For k=∣S∣+1, let the new set S′={s1,…,s∣S∣}∪{t}. The gcd of the new set is (dS,t)=dS′. By the Euclidean algorithm, we can find coefficients l and m such that:
ldS+mt=dS′
- Setting ci′=l⋅ci and c∣S∣+1′=m, we verify that the claim holds true.
5. Extremal Chaining:
- Consider the remaining numbers (attacking numbers) in the set E={0,1,…,n−1}−S.
- We show that E can be divided into two sets EA and EB which behave similarly to A and B.
6. Chaining Down:
- If e∈EA, then e−b (for b∈B) also belongs to EA, and their sizes are equal.
- This is proven by considering the minimal representation of e and showing that it can be chained down to s1.
7. Set Theory Argument:
- If a∈A and b∈B, then a+b must not be an attacking number, or else it would lead to a contradiction.
- This implies that all members of E must be of unison color.
8. Final Argument:
- Suppose there exists some d∣n with d=n such that S+d=S.
- The minimal d′ which satisfies this equation will work for the problem.
- If not, then all numbers from {0,1,…,n−1}−{0} have the same color, contradicting the problem statement.
9. Conclusion:
- We have shown that there exists a divisor d of n different from 1 and n such that if (x,y) are different points of distinct colors and their distance is divisible by d, then both x and y are divisible by d.
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