Interesting modulus

We'll split the solution into cases in order to observe the effects of each individual absolute value sign.

Case 1: $x\le1$

$(x-1)$ , $x$ , and $(x+1)$ are negative or zero. The absolute value signs will negate these expressions. This gives us the equation:

$-x+1-x-x-1=x+2$

Solving this equation yields $x=-\frac{1}{2}$ , but this contradicts the condition of this case.

Case 2: $-1<x\le0$

$(x-1)$ and $x$ are negative or zero, while $(x+1)$ is positive or zero. The absolute value signs will negate $(x-1)$ and $x$ , and the absolute value sign will do nothing to $(x+1)$ . This gives us the equation:

$-x+1-x+x+1=x+2$

Solving this equation yields $x=0$ .

Case 3: $0<x\le1$

$(x-1)$ is negative or zero, while $x$ and $(x+1)$ are positive or zero. The absolute value sign will negate $(x-1)$ , and the absolute value signs will do nothing to $x$ and $(x+1)$ . This gives us the equation:

$-x+1+x+x+1=x+2$

This equation is true for all values of $x$ . Therefore, the solution to this case is $0<x\le1$ .

Case 4: $x>1$

$(x-1)$ , $x$ , and $(x+1)$ are positive or zero. The absolute value signs will do nothing to these expressions. This gives us the equation:

$x-1+x+x+1=x+2$

Solving this equation yields $x=1$ , but this contradicts the condition of this case.

The union of all the solutions of all four cases is $0\le x\le1$ . There are two integers in this interval: $0$ and $1$ . The sum of these integers is $\boxed{1}$ .

This problem is interesting to me because it is a linear absolute value equation, but there are infinite solutions on a closed interval.

Note that the LHS is positive since $x$ is an integer and $|x| \geq 0 \Rightarrow |x-1|+|x|+|x+1|>0$ . Thus, the the value of the RHS is positive, which means $x \geq -1$ . We also have that if $x$ is positive, we get $3x$ as the sum on the LHS.

$3x = x+2$

$2x=2$

$x=1$

Therefore, $x=1$ is the only positive solution. Since $x \geq -1$ , we only need to check $-1$ and $0$ for the other solutions. We find that $0$ also satisfies. Thus the sum of all integral solutions is $1+0=1$ .