
Problem 1. For which integer values of the parameter $a$ the equation $|2 x+1|+|x-2|=a$ has an integer solutions?


Solution. We shall consider the following cases:
I. $x>2$. Then $2 x+1>0, x-2>0$ and the equation is equivalent to $2 x+1+x-2=a$. Thus, $x=\frac{1+a}{3}$, which is a solution when $\frac{1+a}{3}>2$, i.e. when $a>5$.

II. $-\frac{1}{2} \leq x \leq 2$. Then $2 x+1 \geq 0, x-2 \leq 0$ and the equation has the form $2 x+1-(x-2)=a$. Thus, $x=a-3$, which is a solution when $-\frac{1}{2} \leq a-3 \leq 2$, i.e. $\frac{5}{2} \leq a \leq 5$. In this interval the integers are $a=3,4,5$ and respectively we get $x=0,1,2$, which are integer solutions.

III. $x\frac{5}{2}$.

According to the considered cases the given equation has solution when $a>\frac{5}{2}$. When $a=3$ the only integer solution is $x=0$. When $a=4$ there are two integer solutions $x=1$ and $x=-1$. When $a=5, x=2$ is the only integer solution. When $a>5$ the equation has two solutions $x_{1}=\frac{1+a}{3}$ and $x_{2}=\frac{1-a}{3}$. For these solutions we have: if $a=3 k-1$, then $x_{1}$ is integer only; if $a=3 k+1$, then $x_{2}$ is integer only; if $a=3 k$, there is no integer solution.

As a result, we get integer solutions when $a=3$ or $a=3 k \pm 1$, where $k$ is a positive integer, greater than 1.
