Prove there exist $a, b$ coprime such that $a|r, b|s$ and $lcm(r,s)=ab$ with $r, s\in\Bbb{N}$

Question

I just made a proof and I would like to verify it.

Statement

If $r, s \in\Bbb{N} \implies \exists$ $a,b \in\Bbb{Z}$ coprime such that $a|r, b|s$ and $lcm(r,s)=ab$.

My proof

If $r$ and $s$ are coprime, with $a=r$ and $b=s$ we have:

• $a$ and $b$ are coprime integers
• $a|r$
• $b|s$
• $lcm(r,s)=rs=ab$

If $r$ and $s$ are not coprime, since both are greater than $1$ (otherwise they would be coprime) we have that $gcd(r,s)=d\ne 1$.

We can express $r=dq$, $s=dq'$ with $q$ and $q'$ coprime integers. Using the relationship between $lcm$ and $gcd$ we have that $lcm(r,s)=\frac{|rs|}{gcd(r,s)} \implies lcm(r,s)=\frac{dqdq'}{d}=dqq'$.

If $q$ and $d$ are coprime, with $a=q$ and $b=dq'$ we have:

• $a$ and $b$ are coprime integers
• $a|r$ since $q|dq$
• $b|s$ since $q'd|q'd$
• $lcm(r,s)=dqq'=ab$

The following part is not correct

If $q$ and $d$ are not coprime, let be $gcd(q,d)=l$. With $a=ql$ and $b=\frac{dq'}{l}$ we have:

• $a$ and $b$ are coprime integers
• $a|r$ since $ql|dq$
• $b|s$ since $\frac{q'd}{l}|q'd$
• $lcm(r,s)=dqq'=\frac{qlq'd}{l}=ab$

This is the corrected (I think) version of the last part

If $q$ and $d$ are not coprime: let be $q = \prod_i p_i^{q_i}$ and $d = \prod_i p_i^{d_i}$ the prime factorization of $q$ and $d$ respectively. We define $D=\{i:q_i>0,d_i>0\}$.

Let $l = \prod_{i\in D} p_i^{d_i}$

Then with $a=ql$ and $b=\frac{dq'}{l}$ we have:

• $a$ and $b$ are coprime integers by construction
• $a|r$ since $ql|dq$
• $b|s$ since $\frac{q'd}{l}|q'd$
• $lcm(r,s)=dqq'=\frac{qlq'd}{l}=ab$

$\square$

I wanted to know if my proof is ok and since this exercise belongs to a 'groups - primitive roots' sheet of exercises, if there is another proof related with those concepts.