THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

It is clear that $x\notin \mathbb{Z}$ and as $a^{[x]}>0$, ${\log }_a\{x\}>0$, we should have $x>0$. If $x\in (0,1)$, then $0<{\log }_{a}x=x-1<0$, which is absurd. Hence, $x\in (0,\infty )\setminus \mathbb{N}$. Let $f:\mathbb{R}\to (0,+\infty )$, $f(x)=a^x$, which is of course a bijective and strictly decreasing function. Now the equation could be written as $f([x])+f^{-1}(\{x\})=[x]+\{x\}$. Let us denote $f^{-1}(\{x\})=y$. Then $\{x\}=f(y)$ and the previous relation becomes $f([x])+y=[x]+f(y)\iff f([x])-[x]=f(y)-y$. As the function $g(t)=f(t)-t$ is strictly decreasing, and thus injective, we obtain that $[x]=y$, that is $f([x])=\{x\}$.

Finally, if $[x]=n\in \mathbb{N}$, we get $a^n=x-n$, and thus, the solutions of the equation are the numbers $x_n=n+a^n$, for $n\in {\mathbb{N}}^{\ast }$.