XVIII OBM

We have $p(x)-p(y)=(x^3+14x^2-2x+1)-(y^3+14y^2-2y+1)=(x-y)(x^2+xy+y^2+14x+14y-2)$. Since $101$ is prime, $p(x)\equiv p(y)(\mathrm{mod}101)\iff x\equiv y(\mathrm{mod}101)$ or $x^2+xy+y^2+14x+14y-2\equiv 0(\mathrm{mod}101)$.

Completing squares, we have $$\begin{array}{rl}{x}^2+xy+y^2+14x+14y-2& \equiv 0(\mathrm{mod}101)\\ & \iff (2x+y+14{)}^2+3y^2+28y-2\equiv 0(\mathrm{mod}101)\\ & \iff (2x+y+14{)}^2\equiv -3(y-29{)}^2(\mathrm{mod}101)(\ast )\end{array}$$ By the quadratic reciprocity law, $\left(\frac{-3}{101}\right)\cdot \left(\frac{101}{-3}\right)=(-1{)}^{\frac{101-1}{2}\cdot \frac{-3-1}{2}}=1\iff \left(\frac{-3}{101}\right)=\left(\frac{2}{3}\right)=-1$, so $-3$ is not a quadratic residue modulo $101$. Thus $$(\ast )\iff 2x+y+14\equiv y-29\equiv 0(\mathrm{mod}101)\iff x\equiv y\equiv 29(\mathrm{mod}101)$$ Thus $p(x)\equiv p(y)(\mathrm{mod}101)\iff x\equiv y(\mathrm{mod}101)$, which means that $p(x)$ admits an inverse function.

To finish the problem, fix $x$ and consider $x,p(x),p^{(2)}(x),\dots (\mathrm{mod}101)$. Since there are infinite numbers and $101$ remainders, there are $m$ and $n$ such that $m>n$ and $p^{(m)}(x)=p^{(n)}(x)(\mathrm{mod}101)\iff p^{(m-n)}\equiv x(\mathrm{mod}101)$. So for each $k(\mathrm{mod}101)$ there is a positive integer $n_k$ such that $p^{(n_k)}(k)\equiv k(\mathrm{mod}101)$. Choose $N=\text{lcm}(n_1,n_2,\dots ,n_{101})$ and we are done.