, 2012

Let $A_0A_1...A_{p-1}$ be the regular polygon and let $(a_0, a_1, ..., a_{p-1})$ be the integers written at its vertices, $a_k$ at $A_k$ for $k = 0, 1, ..., p-1$. Consider the polynomial with integer coefficients
$$H(x) = a_0 + a_1x + ... + a_{p-1}x^{p-1}.$$
Applying the transformation
$$a_0 \to a_{p-1} + a_1 - a_0, \quad a_1 \to a_0 + a_2 - a_1, \dots, a_{p-1} \to a_{p-2} + a_0 - a_{p-1}$$
on the coefficients of the polynomial $H(x)$ we obtain the polynomial
$$H_1(x) = H(x)(x^{p-1} + x - 1) \pmod{(x^p - 1)}.$$
After $s$ steps we obtain the polynomial
$$\begin{aligned} H_s(x) &= H_{s-1}(x)(x^{p-1} + x - 1) \pmod{(x^p - 1)} \\ &= H(x)(x^{p-1} + x - 1)^s \pmod{(x^p - 1)}. \end{aligned}$$
Finally, after $p$ steps we get
$$H_p(x) = (x^{p-1} + x - 1)^p H(x) \pmod{(x^p - 1)} \quad (1)$$
But
$$(x^{p-1} + x - 1)^p = p \cdot Q(x) + x^{p(p-1)} + x^p - 1$$
$= p \cdot Q(x) + (x^p)^{p-1} - 1 + x^p - 1 + 1 = p \cdot Q(x) + 1 \pmod{(x^p - 1)}$,
where $Q$ is a polynomial with integer coefficients. From (1) we get
$$H_p(x) = p \cdot Q(x)H(x) + H(x) \pmod{(x^{p-1})},$$
Therefore $H_p$ and $H$ have the same coefficients modulo $p$.

The transformation in the problem is described by the following matrix relation
$$\begin{pmatrix} a_0^{(1)} \\ \vdots \\ a_{p-1}^{(1)} \end{pmatrix} = \begin{pmatrix} -1 & 1 & 0 & \cdots & 0 & 1 \\ 1 & -1 & 1 & \cdots & 0 & 0 \\ 0 & 1 & -1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & 0 & 0 & \cdots & 1 & -1 \end{pmatrix} \begin{pmatrix} a_0 \\ \vdots \\ a_{p-1} \end{pmatrix} \quad (1)$$
After $p$ steps we get
$$\begin{pmatrix} a_0^{(p)} \\ \vdots \\ a_{p-1}^{(p)} \end{pmatrix} = A^p \cdot \begin{pmatrix} a_0 \\ \vdots \\ a_{p-1} \end{pmatrix}, \quad (2)$$
where $A$ is the square matrix in relation (1).
We can write $A = -I_p + X + X^{p-1}$, where $X$ is the permutation matrix
$$X = \begin{pmatrix} 0 & 0 & \dots & 0 & 1 \\ 1 & 0 & \dots & 0 & 0 \\ 0 & 1 & \dots & 0 & 0 \\ \dots & \dots & \dots & \dots & \dots \\ 0 & 0 & \dots & 1 & 0 \end{pmatrix} .$$
In $M_p(\mathbb{Z}_p)$ we have
$$\begin{align*} A^p &= (-I_p + X + X^{p-1})^p = (-I_p)^p + X^p + X^{p(p-1)} \\ &= -I_p + I_p + I_p = I_p, \end{align*}$$
since $X^p = I_p$. From (2) it follows that in $M_{p,1}(\mathbb{Z}_p)$ we have
$$\begin{pmatrix} a_0^{(p)} \\ \vdots \\ a_{p-1}^{(p)} \end{pmatrix} = \begin{pmatrix} a_0 \\ \vdots \\ a_{p-1} \end{pmatrix},$$
and we are done.