Proposition 1   Let $D$ be an integral domain. Then $D$ is a lcm domain iff it is a gcd domain.

Proposition 2   Let $D$ be an integral domain and $a,b\in D$ . Then the following are equivalent:

1. $a,b$ have an lcm,
2. for any $r\in D$ , $ra,rb$ have a gcd.

Proof. For arbitrary $x,y \in D$ , denote $\LCM(x,y)$ and $\GCD(x,y)$ the sets of all lcm's and all gcd's of $x$ and $y$ , respectively.

$(1\Rightarrow 2)$ . Let $c\in \LCM(a,b)$ . Write $a=mc$ , $b=nc$ for some $m,n\in D$ . Set $d=mnc$ . We claim that $d\in \GCD(a,b)$ . Since $d=mnc=an=bm$ , we get $d\mid a$ and $d\mid b$ . Next, suppose $s\mid a$ and $s\mid b$ . Write $a=sf$ and $b=sg$ . Let $h=sfg$ . Then $h=ag=bf$ , so that $a\mid h$ and $b\mid h$ , which implies that $c\mid h$ , or $cs\mid hs$ . Since $hs = sfgs = ab=cd$ , we get $cs\mid cd$ , or $s\mid d$ (since $c\ne 0$ and $D$ is an integral domain). Therefore, $d\in \GCD(a,b)$ .

Now, for any $r\in D$ , we want to show that $rd\in \GCD(ra,rb)$ . Clearly $rd\mid ra$ and $rd\mid rb$ . Now, suppose $s\mid ra$ and $s\mid rb$ . Then $sp=ra$ for some $p\in D$ . Since $a=mc$ (in the previous paragraph), $sp=rmc$ , so that $spn=rmnc=rd$ , showing that $s\mid rd$ . Hence $rd\in \GCD(ra,rb)$ .

$(2\Rightarrow 1)$ . Suppose $k\in \GCD(a,b)$ . Write $ki=a$ , $kj=b$ for some $i,j\in D$ . Set $\ell = kij$ , so that $ab=k\ell$ . We want to show that $\ell \in \LCM(a,b)$ . First, notice that $\ell = aj=bi$ , so that $a\mid \ell$ and $b\mid \ell$ . Now, suppose $a\mid t$ and $b\mid t$ , we want to show that $\ell \mid t$ as well. Write $t=ax=by$ . Then $ta=aby$ and $tb=abx$ , so that $ab\mid ta$ and $ab\mid tb$ . Since $\GCD(ta,tb)\ne \varnothing$ , we have $tk\in \GCD(ta,tb)$ (see proof of this here), implying $ab\mid tk$ . In other words $tk=abz$ for some $z\in D$ . As a result, $tk=abz=k\ell z$ , or $t=\ell z$ . In other words, $\ell \mid t$ , as desired.