Find all rational numbers $x$ such that there exists an integer $a \geq 1$ such that $x^{\lfloor x\rfloor}=a / 2$
We note $x=u+\frac{b}{c}$ where $u, b, c$ are integers such that $b \wedge c=1$ and $b \leqslant c$. Therefore, $: 2 \cdot(b+u \cdot c)^{u}=$ $a \cdot c^{u}$. Thus $2 \mid a$ or $2 \mid c$. If $2 \mid a$, then $c \mid b+u \cdot c$, so $c \mid b$ and thus $x$ is an integer, and therefore $a=2 \cdot x^{x}$. Otherwise, $a \wedge 2=1$, thus, $2 \mid c^{u}$, hence, $2 \mid c, c=2 d$. Therefore, $(b+u c)^{u}=a \cdot 2^{u-1} \cdot d^{u}$. Thus, $2^{u-1} \mid(b+2 u d)^{u}$, but $b \wedge 2 d=1$, so $b \wedge 2=1$, thus, necessarily $u=1$. Hence $b+2 d=a d$, so $d \mid b$. Thus $d \mid b \wedge c=1$, hence $d=1$. Therefore, $b \leqslant 2$, so $b=0$ or $b=1$. But, $a \wedge 2=1$, so $b=1$. This gives $x=\frac{3}{2}$. It is easily verified that the integers and $\frac{3}{2}$ are indeed solutions.