Problem 1062

Proof 1 First find the fixed point, $f(x)=3 x+2, f(x)=x, x_{0}=-1, f(x)$
$$\begin{array}{l} +1=3(x+1) . \\ f^{[2]}(x)+1=3(f(x)+1)=3^{2}(x+1), \cdots, f^{[100]}(x)+1 \\ =3^{100}(x+1) . \end{array}$$

The problem to be proved easily becomes: $\exists m \in \mathbf{N}$, such that $1988 / 3^{100}(m+1)-1$, i.e., to prove
$$3^{100}(x+1)-1988 y=1$$

has integer solutions for $x, y$, and $x \in \mathbf{N}$.
Since $\left(3^{100}, 1988\right)=1$, the above equation has integer solutions.
Let $\left(x_{0}, y_{0}\right)$ be one of the integer solutions, then
$$\left\{\begin{array}{l} x=x_{0}+1988 t, \\ y=y_{0}+3^{100} t \end{array}\right.$$

are also integer solutions, where $t$ is any integer. Clearly, we can take $t$ large enough to make $x_{0}+1988 t \in \mathbf{N}$. Let $x=m$ at this time, then $1988 \mid f^{[100]}(m)$.
Note For the indeterminate equation $s x-t y=1$, if $\left(x_{0}, y_{0}\right)$ is a solution of the equation, then
$$\left\{\begin{array}{l} x=x_{0}+t k, \\ y=y_{0}+s k \end{array} \quad(k \in \mathbf{Z})\right.$$

are all solutions of the indeterminate equation.
If $(s, t)=1$, then there must be a solution.
Proof 2 From Proof 1, we know
$$f^{[100]}(n)=3^{100}(x+1)-1,$$

So we only need to prove that there exists $n \in \mathbf{N}^{+}$, such that
$$3^{100}(n+1) \equiv 1(\bmod 4 \times 7 \times 71 .)$$

Take $n+1=3^{m}$, consider whether there exists $m$, such that $3^{100+m} \equiv 1(\bmod 4 \times 7 \times 71)$. By Fermat's Little Theorem $3^{6 k} \equiv 1(\bmod 7) 3^{70 k}=1(\bmod 71)$, and $3^{2 k} \equiv 1(\bmod 4)$. Therefore, take $100+m=[2 k, 6 k, 70 k]=210 k$. Taking $m=210 k-100$ meets the requirement.