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.