Mathematical Reflections 2012, Issue 4 - Problem S235

Problem:
Solve the equation $$\dfrac{8}{\{x\}}=\dfrac{9}{x}+\dfrac{10}{[x]},$$ where $[x]$ and $\{x\}$ denote the greatest integer less or equal than $x$ and the fractional part of $x$, respectively.

Proposed by Titu Andreescu.

Solution:

Clearly $x \notin \mathbb{Z}, x \not \in (0,1)$ and from $\dfrac{8}{\{x\}} > 0$, we gather $x > 1$. Since $0 < \{x\} < 1$ and $0 < [x] \leq x$, we have $$8 < \dfrac{8}{\{x\}} \leq \dfrac{19}{[x]},$$ so $[x]=1,2$. Clearing the denominators and using the fact that $\{x\}=x-[x]$ we obtain the equation $$10x^2-9x[x]-9[x]^2=0.$$
(i) If $[x]=1$, we get $10x^2-9x-9=0$, which gives $x=\dfrac{3}{2},-\dfrac{3}{5}$, i.e. $x=\dfrac{3}{2}$. 

(ii) If $[x]=2$, we get $10x^2-18x-36=0$, which gives $x=3,-\dfrac{6}{5}$, i.e. no solution.

So, $x=\dfrac{3}{2}$ is the only solution to the given equation.