Problem 2. Solve the equation $[x] \cdot\{x\}=x-1$, where $[x]$ represents the integer part, and $\{x\}$ is the fractional part of the real number $x$.
Solution: Since $x=\{x\}+[x], \forall x \in \square$ we have

$[x] \llbracket\{x\}=\{x\}+[x]-1$ thus $(\{x\}-1)([x]-1)=0$ $\qquad$ $4 p$

But $\{x\} \neq 1$ so $[x]=1$ from which $x \in[1,2)$ $\qquad$ $3 p$