a) Prove that for any rational number there exists an infinite set of real numbers that satisfy the equation and any two of them have the same fractional part.
b) Prove that for any rational number there exists an infinite set of real numbers that satisfy the equation and any two of them have different fractional parts.
(The fractional part of a real number is given by , where is its integer part, i.e., the greatest integer that does not exceed .)
a) Let , where , . Consider . Then all , where , , , have equal fractional parts and satisfy the given equation. Indeed, then we have:
\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*}