Three, 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 $a$. ( $[x]$ denotes the greatest integer less than or equal to $x$)
Three, 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 becomes
$$
x=[a x]=[a] x+[\{a\} x] .
$$
$\because[a] \geqslant 1, \therefore(1)$ holds if and only if
$$
[a]=1 \text { and }\{a\} x<1 \text {. }
$$
$\because x=0$ clearly satisfies the equation,
$\therefore[a x]=x$ holds only for $n-1$ positive integers.
If $\{a\} x<1, x^{\prime}<x$, then obviously
$\{a\} x^{\prime}<1$.
Therefore, the positive integer solutions of $\{a\} x<1$ must be $x=1,2$, $\cdots, n-1$.
At this time, $\{a\} x<1$, so $\{a\}<\frac{1}{n-1}$.
When $x \geqslant n, x$ is not a solution of (1) (otherwise there would be at least $n+1$ solutions), which means $\{a\} x=1$ must hold.

From $\{a\} x \geqslant 1$ and $x \geqslant n$, it follows that $\{a\} n \geqslant 1$, so $\{a\} \geqslant \frac{1}{n}$.
Therefore, $1+\frac{1}{n} \leqslant a<1+\frac{1}{n-1}$.