Distinct integers with $a=\text{lcm}(|a-b|,|a-c|)$ and permutations Do there exist three pairwise different integers $a,b,c$ such that $$a=\text{lcm}(|a-b|,|a-c|), b=\text{lcm}(|b-a|,|b-c|), c=\text{lcm}(|c-a|,|c-b|)?$$
None of the integers can be $0$, because the lcm is never $0$. So we know that $|a-b|<\max(a,b)$ (and same with $|b-c|,|c-a|$.) But this is still plausible, because lcm is greater than (or equal to) each of the two numbers.

It seems the following. 
All greatest common divisors considered below are positive.  Put $d=\text{gcd}(a,b,c)$. Let $d’=\text{gcd}(a-b,a-c)$ Then $d|d’$. From the other side, $d’|a$, so $d’|b=a-(a-b)$ and $d’|c=a-(a-c)$. So $d’|d$ an therefore $d’=d$. Hence $|a|=|a-b||a-c|/d$. Similarly, $|b|=|b-a||b-c|/d$ and $|c|=|c-a||c-b|/d$. Put $a=a’d$, $b=b’d$, and $c=c’d$. Then $\text{gcd}(a’,b’,c’)=1$. Moreover, $|a’|=|a’-b’||a’-c’|$. Hence $a’|b’c’$ so $|a’|=\text{gcd}(a’,b’)\text{gcd}(a’,c’)$. Similarly, $|b’|=\text{gcd}(b’,a’) \text{gcd}(b’,c’)$, and $|c’|=\text{gcd}(c’,b’) \text{gcd}(c’,a’)$. Put $x=\text{gcd}(b’,c’)$, $y=\text{gcd}(a’,c’)$, and $z= \text{gcd}(a’,b’)$. Then $|a’|=yz$, $|b’|=xz$, and $|c’|=xy$. So $yz=|yz-xz||yz-xy|$ and therefore $1=|y-x||z-x|$. Similarly $1=|y-x||z-x|$ and $1=|x-z||y-z|$. Without loss of generality we may assume that $x>y$ and $x>z$. Then $|x-y||x-z|=1$ implies $y=z=x-1$. Then $y-z=0,$ a contradiction. 
