Basic Facts of Division This is rather embarrassing, but I haven't had much acquaintance with elementary facts of division. I am working on part $3$ of link.
Here are my questions:
Let $q,p,n\in \Bbb Z^+$ such that $q\mid n$ and $p\mid n$. Denote $d=\operatorname{gcd}(q,p)$.  Then let $q=kd$ and $p=ld$.


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*Why is it that $kld = \operatorname{lcm}(q,p)$?

*Why does $\operatorname{lcm}(q,p) \mid n$?  I can see how $\operatorname{gcd}(q,p)$ would divide $n$, but not $\operatorname{lcm}(q,p)$.

 A: By definition $a\mid b$ means there is some (positive, since we are dealing with positive integers, here) integer $c$ such that $b = ca$.
Note that $kld = (kd)l = ql$ and $kld = k(ld) = (ld)k = pk$.
Thus $kld$ is a multiple of both $p$ and $q$, i.e., a common multiple.
Now suppose $n$, let's say, is any common multiple of $p$ and $q$ (this is what we mean when we say $p\mid n$ and $q\mid n$), so let's say:
$n = pu = qv$.
Since $p = ld$, we have $n = ldu$, and similarly, $n = kdv$.
Clearly, $\dfrac{n}{kld} = \dfrac{ldu}{ldk} = \dfrac{u}{k}$.
Similarly, $\dfrac{n}{kld} = \dfrac{kdv}{kdl} = \dfrac{v}{l}$.
Thus $\dfrac{u}{k} = \dfrac{v}{l}$, or, put another way: $ul = vk$.
To finish, we need to prove an observation about $k$ and $l$, namely $\gcd(k,l) = 1$.
For suppose not, suppose $f\mid k$ and $f\mid l$, with $f > 1$. This means $k = fs$, and $f = lt$ for some (positive) integers $s,t$. Going back to $p$ and $q$ we have:
$p = ld = ftd = (df)t$, and $q = kd = fsd = (df)s$.
Thus $df$ is a common divisor of $p$ and $q$, but $df > d$ (since we assumed $f > 1$), and $d$ is the greatest common divisor, which is a contradiction.
So the only actual possibility for a common divisor of $k$ and $l$ is $1$. In particular this means that any prime occuring in the factorization of $k$ does not occur in $l$.
Now back to $ul = vk$. By equality, since clearly $k \mid vk$, we have $k \mid ul$. But none of the primes occuring in the factorization of $k$ occur in $l$, so they must all occur in $u$ (make sure you are clear on this). See if you can figure out why this means that $k \mid u$.
If $k \mid u$, that means $\dfrac{u}{k}$ is an integer, that is: $kld \mid n$.
(My apologies for all the extra leters)
