Saudi Arabia Mathematical Competitions 2012

Let $A_0{A}_1...A_{p-1}$ 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 $$H(x)=a_0+a_{1}x+...+a_{p-1}{x}^{p-1}.$$ Applying the transformation $$a_0\to a_{p-1}+a_1-a_0,a_1\to a_0+a_2-a_1,\dots ,a_{p-1}\to a_{p-2}+a_0-a_{p-1}$$ on the coefficients of the polynomial $H(x)$ we obtain the polynomial $$H_1(x)=H(x)(x^{p-1}+x-1)(\mathrm{mod}(x^p-1)).$$ After $s$ steps we obtain the polynomial $$\begin{array}{rl}{H}_s(x)& =H_{s-1}(x)(x^{p-1}+x-1)(\mathrm{mod}(x^p-1))\\ & =H(x)(x^{p-1}+x-1{)}^s(\mathrm{mod}(x^p-1)).\end{array}$$ Finally, after $p$ steps we get $$H_p(x)=(x^{p-1}+x-1{)}^{p}H(x)(\mathrm{mod}(x^p-1))(1)$$ But $$(x^{p-1}+x-1{)}^p=p\cdot Q(x)+x^{p(p-1)}+x^p-1$$ $=p\cdot Q(x)+(x^p{)}^{p-1}-1+x^p-1+1=p\cdot Q(x)+1(\mathrm{mod}(x^p-1))$, where $Q$ is a polynomial with integer coefficients. From (1) we get $$H_p(x)=p\cdot Q(x)H(x)+H(x)(\mathrm{mod}(x^{p-1})),$$ Therefore $H_p$ and $H$ have the same coefficients modulo $p$.

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Alternative solution.

The transformation in the problem is described by the following matrix relation $$\left(\begin{array}{c}{a}_0^{(1)}\\ ⋮\\ a_{p-1}^{(1)}\end{array}\right)=\left(\begin{array}{cccccc}-1& 1& 0& \cdots & 0& 1\\ 1& -1& 1& \cdots & 0& 0\\ 0& 1& -1& \cdots & 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 1& 0& 0& \cdots & 1& -1\end{array}\right)\left(\begin{array}{c}{a}_0\\ ⋮\\ a_{p-1}\end{array}\right)(1)$$ After $p$ steps we get $$\left(\begin{array}{c}{a}_0^{(p)}\\ ⋮\\ a_{p-1}^{(p)}\end{array}\right)=A^p\cdot \left(\begin{array}{c}{a}_0\\ ⋮\\ a_{p-1}\end{array}\right),(2)$$ 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 $$X=\left(\begin{array}{ccccc}0& 0& \dots & 0& 1\\ 1& 0& \dots & 0& 0\\ 0& 1& \dots & 0& 0\\ \dots & \dots & \dots & \dots & \dots \\ 0& 0& \dots & 1& 0\end{array}\right).$$ In $M_p({\mathbb{Z}}_p)$ we have $$\begin{array}{rl}{A}^p& =(-I_p+X+X^{p-1}{)}^p=(-I_p{)}^p+X^p+X^{p(p-1)}\\ & =-I_p+I_p+I_p=I_p,\end{array}$$ since $X^p=I_p$. From (2) it follows that in $M_{p,1}({\mathbb{Z}}_p)$ we have $$\left(\begin{array}{c}{a}_0^{(p)}\\ ⋮\\ a_{p-1}^{(p)}\end{array}\right)=\left(\begin{array}{c}{a}_0\\ ⋮\\ a_{p-1}\end{array}\right),$$ and we are done.