Problem 4. Determine the real numbers $\mathrm{x}$, which satisfy the relation:

$$
|[x-a]|+[|x-a|]=1 \text {, where } a \text { is a real parameter. }
$$

(We denote $[t]$ as the integer part of the real number $t$.)
## Solution.

Let $y=x-a$ and since $|[y]| \geq 0 \Rightarrow[|y|] \in\{0,1\}$

If $[|y|]=0 \Rightarrow|y| \in[0,1) \Rightarrow y \in(-1,1)$ and since, from the equation, $|[y]|=1 \Leftrightarrow[y]= \pm 1$, we have the solution $y \in(-1,0)$

If $[|y|]=1 \Rightarrow|y| \in[1,2) \Rightarrow y \in(-2,-1] \cup[1,2)$ and since, from the equation $|[y]|=0 \Leftrightarrow[y]=0 \Leftrightarrow y \in[0,1)$ in this case there is no solution