Brazilian Mathematical Olympiad

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, $p(x) \equiv p(y) \pmod{101} \Leftrightarrow x \equiv y \pmod{101}$ or $x^2 + xy + y^2 + 14x + 14y - 2 \equiv 0 \pmod{101}$.

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

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