17.1.11 ** Let $a>1$ be a real number, and $n \geqslant 2$ be a natural number, such that the equation $[a x]=x$ has exactly $n$ distinct solutions. Find the range of values for $a$.


From the problem, $x$ must be an integer. Also, when $x0 . a \geqslant 2$, $[a x] \geqslant[2 x]=2 x>x$. Therefore, $a<2 .[a]=1$. The equation becomes
$$
x=[a x]=x+[\{a\} x],
$$

which means
$$
\{a\} x<1 \text{. }
$$

If $\{a\} x<1, x^{\prime}<x$, then it is obvious that $\{a\} x^{\prime}<1$. Therefore, the $n-1$ positive integer solutions of $\{a\} x<1$ must be $x=1,2, \cdots, n-1$. Thus, $\frac{1}{n} \leqslant\{a\}<\frac{1}{n-1}$. Hence, $1+\frac{1}{n} \leqslant a<1+\frac{1}{n-1}$.