289. A remarkable number. Find such a number that its fractional part, its integer part, and the number itself form a geometric progression ${ }^{\star}$.
289. Given the conditions $x(x-[x])=[x]^2$, hence $x=$ $=[x] \frac{(1+\sqrt{5})}{2}, |x| \geqslant |[x]|$ and $x \geqslant 0$.

Further, $[x]+1 \geqslant [x] \frac{1+\sqrt{5}}{2}$ and $[x] \leqslant \frac{2}{\sqrt{5}-1}<2$. Therefore, $[x]=0$ or 1, and $x=0$ or $\frac{1+\sqrt{5}}{2}$.

[D. Silverman, M. M., 43, 56 (January 1970).]