15. (1991 Sichuan Province High School Mathematics League Question) Let the positive real number $a>1$, the natural number $n \geqslant 2$, and the equation $[a x]=$ $x$ has exactly $n$ different solutions. Try to find the range of values for $a$.
15. From the given conditions, 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 can be transformed into
$$
x=[a x]=[a] x+[\{a\} x] .
$$

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, $[a x]=x$ holds for only $n-1$ positive integers.
If $\{a\} x<1$ and $x^{\prime}<x$, then obviously
$\{a\} x^{\prime}<1$.
Therefore, the positive integer solutions to $\{a\} x<1$ must be $x=1,2, \cdots, n-1$.
At this time, $\{a\}<\frac{1}{n-1}$.
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 $x \geqslant n$, it follows that $\{a\} n \geqslant 1$, i.e.,
$\{a\} \geqslant \frac{1}{n}$.
Thus, $\frac{1}{n} \leqslant\{a\}<\frac{1}{n-1}$.
Since $[a]=1$, then
$1+\frac{1}{n} \leqslant a<1+\frac{1}{n-1}$.