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I can't prove this statement:

Let $a_1,a_2,...,a_n$ and $b_1,b_2,...,b_n$ be non zero elements of an integral domain $R$ such that $a_1b_1=a_2b_2=\cdots=a_nb_n=x$

If $gcd(ra_1,ra_2,...,ra_n)$ exists for all $0\neq r\in R$, then $lcm(b_1,b_2,...,b_n)$ also exists and satisfies $$gcd(a_1,a_2,...,a_n)lcm(b_1b_2,...,b_n)=x$$

Any hints?

I use the following lemma:

Let $a_1,a_2,...,a_n$ and $r$ be nonzero elements of an integral domain $R$.

$1)$ if $lcm(a_1,a_2,..a_n)$ exists, then $lcm(ra_1,ra_2,...,ra_n)$ also exists and $$lcm(ra_1,ra_2,...ra_n)=rlcm(a_1,a_2,...,a_n)$$

$2)$ if $gcd(ra_1,ra_2,...,ra_n)$ exists, then $gcd(a_1,a_2,...a_n)$ also exists and $$gcd(ra_1,ra_2,...,ra_n)=rgcd(a_1,a_2,...,a_n).$$

My attemp: We suppose that $gcd(ra_1,ra_2,...,ra_n)$ exists, therefore $a:=gcd(a_1,a_2,...,a_n)$ exists for the lemma, and therefore $a| a_i$ for each $i$. Since $x=a_ib_i$ we obtain that $ab_i |x$. We easily can show that $x=lcm(ab_1,ab_2,...,ab_n)$ but i don't know how procede then.

Any hints?

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