GCD domain is LCM domain

On this Wiki page it is written:

A GCD domain is an integral domain $R$ with the property that any two non-zero elements have a greatest common divisor (GCD). Equivalently, any two non-zero elements of $R$ have a least common multiple (LCM).

How to prove last statement that is equivalence of GCD and LCM for all elements? (I am able to prove in Bezout ring but I am not able to prove in general GCD ring.)

Does the existence of gcd of two elements implies existence of lcm and conversely in any integral domain?

The result that is given in Wikipedia without proof is an immediate corollary of the following:

Theorem: Let $D$ be a domain and $a,b\in D$. TFAE $\colon$

i) $\text{lcm}(a,b)$ exists.

ii) For all $r\in D\setminus\{0\}$, $\gcd(ra,rb)$ exists.

Proof: i)$\implies$ ii) Let's call $\text{lcm}(a,b)=m$. We have $a\mid ab$ and $b\mid ab$, then $m\mid ab$. We claim that $\gcd(a,b)=ab/m$. Indeed, $$a=\frac{ab}{m}\frac{m}{b}\implies \frac{ab}{m}\mid a,$$ $$b=\frac{ab}{m}\frac{m}{a}\implies \frac{ab}{m}\mid b.$$

Thus, $ab/m$ is a common divisor of $a$ and $b$.

Let $e\in D$ such that $e\mid a$ and $e\mid b$. Then, $e \mid ab$, so that $ab/e$ is an integer. Moreover, $a \mid ab/e$ (since $e \mid b$) and $b \mid ab/e$ (likewise), whence $m \mid ab/e$. That is, $em\mid ab$. Hence $e\mid ab/m$, and this shows that $\gcd(a,b)=ab/m$.

Set $\gcd(a,b)=ab/m=d$. Given $r\neq 0$, we claim that $\gcd(ra,rb)=rd$. Indeed, $d\mid a$ and $d\mid b$ implies that $rd\mid ra$ and $rd\mid rb$. Let $s\in D$ such that $s\mid ra$ and $s\mid rb$. Then, $rab/s$ is an integer and satisfies $a \mid rab/s$ (since $s \mid rb$) and $b \mid rab/s$ (likewise), so that $m \mid rab/s$. Hence, $sm\mid rab$. Therefore $s\mid rab/m=rd$. This proves that $\gcd(ra,rb)=rd$.

ii)$\implies$ i) We claim that $$\text{lcm}(a,b)=\frac{ab}{\gcd(a,b)}.$$

Indeed, set $d=\gcd(a,b)$, then $$\frac{ab}{d}=a\frac{b}{d},$$ $$\frac{ab}{d}=b\frac{a}{d}.$$

So $a\mid ab/d$ and $b\mid ab/d$. Let $n\in D$ such that $a\mid n$ and $b\mid n$, thus $ab\mid nb$ and $ab\mid na$. It follows that $ab\mid \gcd(na,nb)=n\gcd(a,b)$, i.e., $ab\mid nd$ and then $\frac{ab}{d}\mid n$. Hence, $\text{lcm}(a,b)=ab/d$.

The above theorem is used in this paper (Dinesh Khurana, On GCD and LCM in domains -- A conjecture of Gauss, Resonance, 8(6), 72–79) by D. Khurana where he proves that for every $n\ge 3$ non-square $\Bbb{Z}[\sqrt{-n}]$ is not a GCD-domain, and hence not a UFD. More exactly, he shows that if $n+1=pk$ for some prime $p$ and $k\ge 2$, then $\text{lcm}(p,1+\sqrt{-n})$ doesn't exist; and if $n+1$ is prime then $\text{lcm}(2,2+\sqrt{-n})$ doesn't exist. So we have an alternative proof to the fact that $\Bbb{Z}[\sqrt{-n}]$ is not a UFD for $n\ge 3$.

I want to make a few supplements to the above good answer by Xam. If $R$ is an integral domain and $gcd(a,b)$ exists, then $lcm(a,b)$may not exist.

Consider $Z[\sqrt{-3}]$. Let $a = 2$ and $b = 1+\sqrt{-3}$. If $lcm(2,1+\sqrt{-3})$ exists, then $lcm(2,1+\sqrt{-3})$ $\mid$ $2(1+\sqrt{-3})$ and $lcm(2,1+\sqrt{-3})$ $\mid$ $2(1-\sqrt{-3})$(because $2(1-\sqrt{-3}) = -(1+\sqrt{-3})(1+\sqrt{-3})$). However, the factors of $2(1+\sqrt{-3})$ are $2, (1+\sqrt{-3})$ and $(1-\sqrt{-3})$.

Thus $lcm(2,1+\sqrt{-3})$ must be equal to $2(1+\sqrt{-3})$. In the same way, according to $lcm(2,1+\sqrt{-3})$ $\mid$ $2(1-\sqrt{-3})$, we have $lcm(2,1+\sqrt{-3})$ must be equal to $2(1-\sqrt{-3})$. This is a contradiction.

Actually, $gcd(4,2(1+\sqrt{-3}))$ doesn't exist，but$gcd(2,1+\sqrt{-3}) =1$