Problem 2084

a) Let $\alpha = \frac{p}{q}$, where $p, q \in \mathbb{N}$, $p < q$. Consider $y = pq + \frac{1}{q}$. Then all $x = y + m$, where $m = q^n$, $n \in \mathbb{N}$, $n \ge 2$, have equal fractional parts and satisfy the given equation. Indeed, then we have:
$$\{x\} = \{y\}, \quad m\{y\} \in \mathbb{N}, \quad my\{y\} \in \mathbb{N},$$

$$\begin{align*} x\{x\} &= (y+m)\{y\} = y\{y\} + m\{y\}, & [x\{x\}] &= [y\{y\}] + m\{y\}, \\ x[x\{x\}] &= (y+m)([y\{y\}] + m\{y\}) = y[y\{y\}] + m[y\{y\}] + my\{y\} + m^2\{y\}, & \\ \{x[x\{x\}]\} &= \{y[y\{y\}]\} = \frac{p}{q} = \alpha. \end{align*}$$
$$b) For \$\alpha = \frac{p}{q}\$, \$p, q \in \mathbb{N}\$, \$p < q\$, consider \$x = pqn^2 + \frac{1}{qn}\$, \$n \in \mathbb{N}\$. Then$$
$$\begin{align*} \{x\} &= \frac{1}{qn}, \quad x\{x\} = \left( pqn^2 + \frac{1}{qn} \right) \frac{1}{qn} = pn + \frac{1}{q^2 n^2}, \quad [x\{x\}] = pn, \\ x[x\{x\}] &= \left( pqn^2 + \frac{1}{qn} \right) pn = p^2 q n^3 + \frac{p}{q}, \quad \{x[x\{x\}]\} = \frac{p}{q} = \alpha. \end{align*}$$
$$$$