$14 \cdot 10$ Let $a > 1$ be a positive real number, and $n \geqslant 2$ be a natural number, and the equation $[a x]=x$ has exactly $n$ different solutions. Try to find the range of values for $a$.
(China Sichuan Province High School Mathematics Competition, 1992).
[Solution] From the given, we know that $x$ must be an integer, and $x \geqslant 0$.
Let $\{a\}=a-[a]$, then
$$
a x=[a] x+\{a\} x,
$$

thus the original equation transforms into
$$
x=[a x]=[a] x+[\{a \mid x] .
$$

Since $[a] \geqslant 1$, the necessary and sufficient condition for (1) to hold is
$$
[a]=1 \text {, and }\{a\} x<1 \text {. }
$$

Since $x=0$ is clearly a solution to the equation, we have
$$
[a x]=x
$$

which holds for only $n-1$ positive integers.
If $\{a\} x<1, x^{\prime}<x$, then it is obvious that
$$
\{a\} x^{\prime}<1 \text {. }
$$

Therefore, the positive integer solutions to $\{a\} x<1$ must be $x=1,2, \cdots, n-1$.
At this time, we have
$$
\{a\}<\frac{1}{n-1} \text {. }
$$

Furthermore, when $x \geqslant n$, $x$ is not a solution to (1) (otherwise there would be at least $n+1$ solutions), which means $\{a\} x \geqslant 1$ must hold.
From $\{a\} x \geqslant 1$ and $x \geqslant n$, we have $\{a\} n \geqslant 1$, i.e.,
$$
\{a\} \geqslant \frac{1}{n} \text {. }
$$

Thus,
$$
\frac{1}{n} \leqslant\{a\}<\frac{1}{n-1} \text {. }
$$

Also, since $[a]=1$, we have
$$
1+\frac{1}{n} \leqslant a<1+\frac{1}{n-1} .
$$