If $a,b\in R$ are distinct numbers satisfying $|a-1|+|b-1|=|a|+|b|=|a+1|+|b+1|$,then find the minimum value of $|a-b|$ If $a,b\in R$ are distinct numbers satisfying $|a-1|+|b-1|=|a|+|b|=|a+1|+|b+1|$,then find the minimum value of $|a-b|$

I squared both sides of $|a-1|+|b-1|=|a|+|b|$ and $|a|+|b|=|a+1|+|b+1|$ and equated them but the equation obtained is messy and i am stuck.Please help.

Consider the $16$ cases:
$\hspace{3cm}$
$$\begin{align}1) \ &a,b\ge 1 \Rightarrow \\
& \ \ \ \ \ a-1+b-1=a+b=a+1+b+1 \Rightarrow \emptyset;\\
2) \ &0\le a\le 1, b\ge 1 \Rightarrow \\
& \ \ \ \ \ -a+1+b-1=a+b=a+1+b+1 \Rightarrow \emptyset;\\
3) \ &-1\le a\le 0, b\ge 1 \Rightarrow \\
& \ \ \ \ \ -a+1+b-1=-a+b=a+1+b+1 \Rightarrow \\
& \ \ \ \ \ a=-1 \Rightarrow b=1 \Rightarrow (a,b)=(-1,1) \Rightarrow \\
& \ \ \ \ \ |a-b|=2 \ \text{(min)};\\
4, 7, 8) \ &(a,b)=(-1,1);\\
5, 6, 11, 12,15,16) \ &\emptyset;\\
9,10,13,14) \ &(a,b)=(1,-1). \end{align}$$
