IMO 2006 Shortlisted Problems

First note that if $a_0\ge 0$, then all $a_i\ge 0$. For $a_i\ge 1$ we have (in view of $⟨a_i⟩<1$ and $\lfloor a_i\rfloor >0$ ) $$\lfloor a_{i+1}\rfloor \le a_{i+1}=\lfloor a_i\rfloor \cdot ⟨a_i⟩<\lfloor a_i\rfloor$$ the sequence $\lfloor a_i\rfloor$ is strictly decreasing as long as its terms are in $[1,\infty )$. Eventually there appears a number from the interval $[0,1)$ and all subsequent terms are $0$.

Now pass to the more interesting situation where $a_0<0$; then all $a_i\le 0$. Suppose the sequence never hits $0$. Then we have $\lfloor a_i\rfloor \le -1$ for all $i$, and so $$1+\lfloor a_{i+1}\rfloor >a_{i+1}=\lfloor a_i\rfloor \cdot ⟨a_i⟩>\lfloor a_i\rfloor$$ this means that the sequence $\lfloor a_i\rfloor$ is nondecreasing. And since all its terms are integers from $(-\infty ,-1]$, this sequence must be constant from some term on: $$\lfloor a_i\rfloor =c\text{ for }i\ge i_0;c\text{ a negative integer. }$$ The defining formula becomes $$a_{i+1}=c\cdot ⟨a_i⟩=c\left(a_i-c\right)=c{a}_i-c^2$$ Consider the sequence $$\begin{array}{c}{b}_i=a_i-\frac{c^2}{c-1}\end{array}\text{\hspace{1em}(1)}$$ It satisfies the recursion rule $$b_{i+1}=a_{i+1}-\frac{c^2}{c-1}=c{a}_i-c^2-\frac{c^2}{c-1}=c{b}_i$$ implying $$\begin{array}{c}{b}_i=c^{i-i_0}{b}_{i_0}\text{ for }i\ge i_0\end{array}\text{\hspace{1em}(2)}$$ Since all the numbers $a_i$ (for $i\ge i_0$ ) lie in $[c,c+1)$, the sequence $\left(b_i\right)$ is bounded. The equation (2) can be satisfied only if either $b_{i_0}=0$ or $|c|=1$, i.e., $c=-1$.

In the first case, $b_i=0$ for all $i\ge i_0$, so that $$a_i=\frac{c^2}{c-1}\text{ for }i\ge i_0.$$ In the second case, $c=-1$, equations (1) and (2) say that $$a_i=-\frac{1}{2}+(-1{)}^{i-i_0}{b}_{i_0}=\{\begin{array}{ll}{a}_{i_0}& \text{ for }i=i_0,i_0+2,i_0+4,\dots ,\\ 1-a_{i_0}& \text{ for }i=i_0+1,i_0+3,i_0+5,\dots .\end{array}$$ Summarising, we see that (from some point on) the sequence $\left(a_i\right)$ either is constant or takes alternately two values from the interval $(-1,0)$. The result follows.