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.

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    $\begingroup$ Having done a computer check, I highly suspect not. $\endgroup$ – Edward Jiang Dec 20 '14 at 4:46

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.


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