Show that if $a$ and $b$ are positive integers then there are divisors $c$ of $a$ and $d$ of $b$ with $(c, d) = 1$ and $cd = [a, b]$
Since $a$ and $b$ are positive, surely both will have a prime power factorization.
So initially I thought I would have to show $c$ is product of primes where the the power $e$ of $p$ would be the max for that prime $p$ between $a$ and $b$, but that doesn't seem right.
Does anyone have a better solution to this problem?
Best Answer
Let $a,b\in\mathbb N$ such that $a\neq b$. Consider $p,q\in\mathbb N$ such that $a=(a,b)p$ and $b=(a,b)q$. Then $(p,q)=1$ and $(p,(a,b))=1$. Hence $(p,q(a,b))=1$ If we define $c=p$ and $d=q(a,b)$ we have what we wanted.
CORRECTION: Let $a,b\in\mathbb N$ such that $a\neq b$. Consider $p,q\in\mathbb N$ such that $a=(a,b)p$ and $b=(a,b)q$. Then $(p,q)=1$ and $[a,b]=(a,b)pq$. Now we want to factorize $(a,b)$ in a suitable way, say $(a,b)=xy$, in order to get $c=xp$ and $d=yq$ coprime.
Let $s\in\mathbb{N}$ such that $(a,b)=(p,(a,b))\cdot(q,(a,b))\cdot s$. Notice that $(p,q)=1$ implies that $(p,(a,b))$ and $(q,(a,b))$ are coprime. Notice also that $s$ is coprime to p and q.
Now consider $c=p\cdot (p,(a,b))\cdot s$ and $d=q\cdot (q,(a,b))$. We have that $cd=[a,b]$, $c|a$ and $d|b$. The remaining part to show is that $c$ and $d$ are coprime.
Since $p$ and $q$ are coprime and $(q,(a,b))$ divides $q$, we conclude that $p$ is prime to $(q,(a,b))$ and then it is prime to $d$ as well.
Now, since $(p,(a,b))$ divides $p$, and $(p,d)=1$ then $(p,(a,b))$ is prime to $d$.
Similarly, since $s$ and $q$ are coprime, $s$ and $(q,(a,b))$ are coprime. Thus $s$ is prime to $d$.
Therefore $c=p\cdot (p,(a,b))\cdot s$ is prime to $d$.