1. For a real number $x$, the following are defined: the greatest integer of $x$, $\lfloor x\rfloor$, is the greatest integer not greater than $x$, and the fractional part of $x$, $\{x\}=x-\lfloor x\rfloor$.

For example, $\lfloor 3.14\rfloor=3,\{3.14\}=0.14,\lfloor-2.4\rfloor=-3,\{-2.4\}=0.6$. Determine all positive numbers $x$ for which $\{x\},\lfloor x\rfloor$, and $x$ are three consecutive terms of a geometric sequence.
First solution.

It must be $\{x\} \cdot x=\lfloor x\rfloor^{2}$.

For $\lfloor x\rfloor=0$, we would get the geometric sequence $0,0,0$.

For $\lfloor x\rfloor=1$ we have

$x \in[1,2),\{x\}=x-1 \in[0,1) \text{ and }\{x\} \cdot x=1, \text{ i.e. } x(x-1)=1$.

The solutions to the quadratic equation $x^{2}-x-1=0$ are $x=\frac{1 \pm \sqrt{5}}{2}$, so because $x>0$ we have $x=\frac{1+\sqrt{5}}{2}$. (Then $\{x\}=\frac{1-\sqrt{5}}{2}$.)

(8 points)

For $\lfloor x\rfloor \geq 2$ it holds that

$$
\lfloor x\rfloor^{2} \geq 2\lfloor x\rfloor
$$

and since $\{x\} \cdot x < 2\lfloor x\rfloor$, there are no other solutions.

(8 points)

Second solution.

From $\frac{\{x\}}{\lfloor x\rfloor}=\frac{\lfloor x\rfloor}{x}$ it follows that $\frac{\lfloor x\rfloor}{\{x\}}=\frac{x}{\lfloor x\rfloor}=\frac{\lfloor x\rfloor+\{x\}}{\lfloor x\rfloor}=1+\frac{\{x\}}{\lfloor x\rfloor}$ If $\lfloor x\rfloor \geq 2$, the number on the left side of this equation is greater than 2, and the number on the right side is less
than 2. Therefore, $\lfloor x\rfloor=1,(\lfloor x\rfloor \neq 0)$. (5 points)

(5 points)

Now $1+\{x\}=\frac{1}{\{x\}}$, i.e. $\{x\}^{2}+\{x\}-1=0$. The solutions to this quadratic equation are $\frac{-1 \pm \sqrt{5}}{2}$. Because $0 \leq\{x\}<1$, it follows that $\{x\}=\frac{-1 + \sqrt{5}}{2}$

Thus, $x=\frac{1}{\{x\}}=\frac{\sqrt{5}+1}{2}$