Let \(n > 3\) be a positive integer. Suppose that \(n\) children are arranged in a circle, and \(n\) coins are distributed between them (some children may have no coins). At every step, a child with at least \(2\) coins may give \(1\) coin to each of their immediate neighbors on the right and left. Determine all initial distributions of the coins from which it is possible that, after a finite number of steps, each child has exactly one coin.
Encode the sequence \(c_{i}\) as a polynomial \(p(x)=\sum_{i} a_{i} x_{i}\). The cyclic nature of the problem makes it natural to work modulo \(x^{n}-1\). Child \(i\) performing a step is equivalent to adding \(x^{i}(x-1)^{2}\) to the polynomial, and we want to reach the polynomial \(q(x)=1+x+\ldots+\) \(x^{n-1}\). Since we only add multiples of \((x-1)^{2}\), this is only possible if \(p(x)=q(x)\) modulo the ideal generated by \(x^{n}-1\) and \((x-1)^{2}\), i.e.
\[\left(x^{n}-1,(x-1)^{2}\right)=(x-1)\left(\frac{x^{n}-1}{x-1}, x-1\right)=(x-1) \cdot(n,(x-1))\]
This is equivalent to \(p(1)=q(1)\) (which simply translates to the condition that there are \(n\) coins) and \(p^{\prime}(1)=q^{\prime}(1)(\bmod n)\), which translates to the invariant described in solution \(1\).