Prove that $(ab,cd)=(a,c)(b,d)\left(\frac{a}{(a,c)},\frac{d}{(b,d)}\right)\left(\frac{c}{(a,c)},\frac{b}{(b,d)}\right)$ I'm working through Oystein Ore's Number Theory and its History.  On p. 109, I'm stuck on #2.

The question asks the reader to verify the following identity [Note: $(x,y)=\gcd(x,y)$]:
$$(ab,cd)=(a,c)(b,d)\left(\frac{a}{(a,c)},\frac{d}{(b,d)}\right)\left(\frac{c}{(a,c)},\frac{b}{(b,d)}\right)$$

I've tried numerous numeric examples and not found an exception.  I've tried a messy proof, substituting sample factors and exponents, but it's not very cohesive, clear, or robust.  Clearly, if $a,b,c,d$ are all relatively prime, the answer is clear.  I don't know how to concisely prove this if that's not the case though.
I've tried using the idea that $m(x,y)=(mx,my)$ to get rid of the denominators, but I still end up with some fractions.  I've tried to use the symmetry of the fractions to simplify things.
I also looked at this link without significant progress:
Is $\gcd(a,b)\gcd(c,d)=\gcd(ac,bd)$?
 A: We first prove that if $\alpha,\beta,\gamma,\delta$ are integers such that 
    $$(\alpha,\gamma)=(\alpha,\delta)=(\beta,\gamma)=(\beta,\delta)=1,$$ then $(\alpha\beta,\gamma\delta)=1.$
    Indeed, from $(\alpha,\gamma)=1=(\alpha,\delta)$ we deduce that $(\alpha,\gamma\delta)=1.$ Similarly, from
    $(\beta,\gamma)=1=(\beta,\delta)$ it follows that $(\beta,\gamma\delta)=1.$ Combining the foregoing two results,  we then conclude that $(\alpha\beta,\gamma\delta)=1.$ 
We   then show that,    if $(a,c)=1$ and $(b,d)=1,$  then $(ab,cd)=(a,d)(b,c).$ 
Indeed, because $(a,d)\mid a,$ $(b,c)\mid c,$ from $(a,c)=1$ we have $(a/(a,d), c/(b,c))=1.$ Similarly, from $(b,d)=1$ and $(b,d)\mid b, (a,d)\mid d,$ it follows that
$(b/(b,c), d/(a,d))=1.$  It is also clear that $(a/(a,d), d/(a,d))=1$ and $(b/(b,c), c/(b,c))=1.$ Thus, by the foregoing paragraph, we have
\begin{align*}
 (ab,cd)=&(a,d)(b,c)\left(\frac{a}{(a,d)}\frac{b}{(b,c)},\frac{d}{(a,d)}\frac{c}{(b,c)}\right)\\
 =&(a,d)(b,c)\cdot 1=(a,d)(b,c),
\end{align*}
provided $(a,c)=1=(b,d).$
Now consider the general case. Because 
    \begin{gather*}
  \left(\frac{a}{(a,c)},\frac{c}{(a,c)}\right)=1,\qquad \left(\frac{b}{(b,d)},\frac{d}{(b,d)}\right)=1,
 \end{gather*}
    we have 
    \begin{align*}
  (ab,cd)&=\left((a,c)(b,d)\frac{a}{(a,c)}\frac{b}{(b,d)}, (a,c)(b,d)\frac{c}{(a,c)}\frac{d}{(b,d)}\right)\\
  &=(a,c)(b,d)\left(\frac{a}{(a,c)}\frac{b}{(b,d)},\frac{c}{(a,c)}\frac{d}{(b,d)}\right)\\
  &=(a,c)(b,d)\left(\frac{a}{(a,c)},\frac{d}{(b,d)}\right)\left(\frac{b}{(b,d)},\frac{c}{(a,c)}\right)\\
  &=(a,c)(b,d)\left(\frac{a}{(a,c)},\frac{d}{(b,d)}\right)\left(\frac{c}{(a,c)},\frac{b}{(b,d)}\right).
 \end{align*}
A: Theorem 1. Let $x$ and $y$ be two integers. Then, there exist integers $p$
and $q$ such that $px+qy=\gcd\left(  x,y\right)  $.
Theorem 1 is Bezout's theorem, and we assume it to be known. Notice that the
integers $x$ and $y$ are allowed to be $0$ (even both of them, in which case
we use the convention $\gcd\left(  0,0\right)  =0$).
Proposition 2. Let $n$ and $m$ be two nonnegative integers such that
$n\mid m$ and $m\mid n$. Then, $m=n$.
Proposition 2 is obvious. Equalities between gcd's are usually proven with the
help of Proposition 2.
Lemma 3. Let $x$, $y$, $z$ and $w$ be four integers such that $\gcd\left(
x,z\right)  =1$ and $\gcd\left(  y,w\right)  =1$. Then, $\gcd\left(
xy,zw\right)  =\gcd\left(  y,z\right)  \cdot\gcd\left(  x,w\right)  $.
Proof of Lemma 3. Theorem 1 (applied to $w$ instead of $y$) shows that there
exist integers $p$ and $q$ such that $px+qw=\gcd\left(  x,w\right)  $. Let us
denote these $p$ and $q$ by $p_{1}$ and $q_{1}$. Thus, $p_{1}$ and $q_{1}$ are
integers satisfying $p_{1}x+q_{1}w=\gcd\left(  x,w\right)  $.
Theorem 1 (applied to $y$ and $z$ instead of $x$ and $y$) shows that there
exist integers $p$ and $q$ such that $py+qz=\gcd\left(  y,z\right)  $. Let us
denote these $p$ and $q$ by $p_{2}$ and $q_{2}$. Thus, $p_{2}$ and $q_{2}$ are
integers satisfying $p_{2}y+q_{2}z=\gcd\left(  y,z\right)  $.
Theorem 1 (applied to $z$ instead of $y$) shows that there exist integers $p$
and $q$ such that $px+qz=\gcd\left(  x,z\right)  $. Let us denote these $p$
and $q$ by $g$ and $h$. Thus, $g$ and $h$ are integers satisfying
$gx+hz=\gcd\left(  x,z\right)  $. Hence, $gx+hz=\gcd\left(  x,z\right)  =1$.
Theorem 1 (applied to $y$ and $w$ instead of $x$ and $y$) shows that there
exist integers $p$ and $q$ such that $py+qw=\gcd\left(  y,w\right)  $. Let us
denote these $p$ and $q$ by $g^{\prime}$ and $h^{\prime}$. Thus, $g^{\prime}$
and $h^{\prime}$ are integers satisfying $g^{\prime}y+h^{\prime}w=\gcd\left(
x,z\right)  $. Hence, $g^{\prime}y+h^{\prime}w=\gcd\left(  x,z\right)  =1$.
Now,
$\underbrace{\gcd\left(  y,z\right)  }_{=p_{2}y+q_{2}z}\cdot\underbrace{\gcd
\left(  x,w\right)  }_{=p_{1}x+q_{1}w}$
$=\left(  p_{2}y+q_{2}z\right)  \cdot\left(  p_{1}x+q_{1}w\right)  $
$=p_{1}p_{2}xy+q_{1}p_{2}\underbrace{yw}_{=yw1}+q_{2}p_{1}\underbrace{xz}
_{=xz1}+q_{1}q_{2}zw$
$=p_{1}p_{2}xy+q_{1}p_{2}yw\underbrace{1}_{=gx+hz}+q_{2}p_{1}xz\underbrace{1}
_{=g^{\prime}y+h^{\prime}w}+q_{1}q_{2}zw$
$=p_{1}p_{2}xy+q_{1}p_{2}yw\left(  gx+hz\right)  +q_{2}p_{1}xz\left(
g^{\prime}y+h^{\prime}w\right)  +q_{1}q_{2}zw$
$=p_{1}p_{2}xy+q_{1}p_{2}ywgx+q_{1}p_{2}ywhz+q_{2}p_{1}xzg^{\prime}
y+q_{2}p_{1}xzh^{\prime}w+q_{1}q_{2}zw$
$=\left(  p_{1}p_{2}+q_{1}p_{2}wg+q_{2}p_{1}zg^{\prime}\right)  xy+\left(
q_{1}p_{2}yh+q_{2}p_{1}xh^{\prime}+q_{1}q_{2}\right)  zw$ (by a
straightforward computation)
is a $\mathbb{Z}$-linear combination of $xy$ and $zw$, and therefore divisible
by $\gcd\left(  xy,zw\right)  $ (since both $xy$ and $zw$ are divisible by
$\gcd\left(  xy,zw\right)  $). In other words,
(1) $\gcd\left(  xy,zw\right)  \mid\gcd\left(  y,z\right)  \cdot
\gcd\left(  x,w\right)  $.
On the other hand, multiplying the relations
$\gcd\left(  y,z\right)  \mid y$ and $\gcd\left(
x,w\right)  \mid x$, we obtain $\gcd\left(  y,z\right)  \cdot\gcd\left(
x,w\right)  \mid yx=xy$. Also, multiplying the relations
$\gcd\left(  y,z\right)  \mid z$ and
$\gcd\left(  x,w\right)  \mid w$, we obtain $\gcd\left(  y,z\right)  \cdot
\gcd\left(  x,w\right)  \mid zw$. We thus know that both $xy$ and $zw$ are
divisible by $\gcd\left(  y,z\right)  \cdot\gcd\left(  x,w\right)  $.
Therefore, the greatest common divisor of $xy$ and $zw$ is also divisible by
$\gcd\left(  y,z\right)  \cdot\gcd\left(  x,w\right)  $. In other words, we have
(2) $\gcd\left(  y,z\right)  \cdot\gcd\left(  x,w\right)  \mid\gcd\left(
xy,zw\right)  $.
Now, we have proven (1) and (2). Thus, we can apply Proposition 2 to
$n=\gcd\left(  y,z\right)  \cdot\gcd\left(  x,w\right)  $ and $m=\gcd\left(
xy,zw\right)  $. We thus obtain $\gcd\left(  xy,zw\right)  =\gcd\left(
y,z\right)  \cdot\gcd\left(  x,w\right)  $. This proves Lemma 3.
Theorem 4. Let $a$, $b$, $c$ and $d$ be four integers. Let $n=\gcd\left(
a,c\right)  $ and $m=\gcd\left(  b,d\right)  $; assume that $n\neq0$ and
$m\neq0$. Then,
$\gcd\left(  ab,cd\right)  =\gcd\left(  a,c\right)  \cdot\gcd\left(
b,d\right)  \cdot\gcd\left(  \dfrac{a}{n},\dfrac{d}{m}\right)  \cdot
\gcd\left(  \dfrac{c}{n},\dfrac{b}{m}\right)  $.
Proof of Theorem 4. Let $x=\dfrac{n}{a}$, $y=\dfrac{m}{b}$, $z=\dfrac{n}{c}$
and $w=\dfrac{n}{d}$. Then, $a=nx$, $b=my$, $c=nz$ and $d=nw$. Also,
$x=\dfrac{n}{a}$ is an integer (since $n=\gcd\left(  a,c\right)  \mid a$), and
similarly $y$, $z$ and $w$ are integers.
Now, $n=\gcd\left(  \underbrace{a}_{=nx},\underbrace{c}_{=nz}\right)
=\gcd\left(  nx,nz\right)  =n\gcd\left(  x,z\right)  $. Since $n\neq0$, we can
divide this equality by $n$, and obtain $1=\gcd\left(  x,z\right)  $. The same
argument (using $m,b,d,y,w$ instead of $n,a,c,x,z$) shows that $1=\gcd\left(
y,w\right)  $. Thus, Lemma 3 yields
$\gcd\left(  xy,zw\right)  =\underbrace{\gcd\left(  y,z\right)  }
_{=\gcd\left(  z,y\right)  }\cdot\gcd\left(  x,w\right)  =\gcd\left(
z,y\right)  \cdot\gcd\left(  x,w\right)  $
$=\gcd\left(  x,w\right)  \cdot\gcd\left(  z,y\right)  $.
But
$\gcd\left(  \underbrace{a}_{=nx}\underbrace{b}_{=my},\underbrace{c}
_{=nz}\underbrace{d}_{=mw}\right)  =\gcd\left(  nxmy,nzmw\right)  =\gcd\left(
nm\cdot xy,nm\cdot zw\right)  $
$=nm\cdot\underbrace{\gcd\left(  xy,zw\right)  }_{=\gcd\left(  w,x\right)
\cdot\gcd\left(  z,y\right)  }=\underbrace{n}_{=\gcd\left(  a,c\right)
}\underbrace{m}_{=\gcd\left(  b,d\right)  }\cdot\gcd\left(  \underbrace{x}
_{=\dfrac{a}{n}},\underbrace{w}_{=\dfrac{d}{m}}\right)  \cdot\gcd\left(
\underbrace{z}_{=\dfrac{c}{n}},\underbrace{y}_{=\dfrac{b}{m}}\right)  $
$=\gcd\left(  a,c\right)  \cdot\gcd\left(  b,d\right)  \cdot\gcd\left(
\dfrac{a}{n},\dfrac{d}{m}\right)  \cdot\gcd\left(  \dfrac{c}{n},\dfrac{b}
{m}\right)  $.
Theorem 4 is proven.
This is probably not the simplest or shortest proof, but was the easiest one
to write (it took me almost no focus and very little editing, just a lot of
copy & paste). The annoying computations in the proof of Lemma 3 could have
been simplified using ideal notation, but I don't know if you have this
background. There is certainly an alternative proof by comparing exponents of primes, but my kind of argument generalizes better. For example, Lemma 3 above can be straightforwardly generalized to the following result:
Lemma 5. Let $A$ be a commutative ring. Let $X$, $Y$, $Z$ and $W$ be four ideals of $A$ such that $X+Z=A$ and $Y+W=A$. Then, $XY+ZW = \left(Y+Z\right)\left(X+W\right)$.
Lemma 3 can be recovered from Lemma 5 by setting $A = \mathbb Z$, $X = x \mathbb Z$, $Y = y \mathbb Z$, $Z = z \mathbb Z$ and $W = w \mathbb Z$. The proof I gave for Lemma 3 is essentially a proof for Lemma 5, artificially restricted to the case of principal ideals in $\mathbb Z$. Theorem 4 is harder to generalize, since it is not clear what the analogue of (for example) $\dfrac{a}{n}$ is for ideals; but given that it is a corollary of Lemma 3, a point could be made in favor of regarding Lemma 3 as the main theorem.
A: $$(ab,cd)=(a,c)(b,d)\left(\frac{a}{(a,c)},\frac{d}{(b,d)}\right)\left(\frac{c}{(a,c)},\frac{b}{(b,d)}\right) \tag{T}$$
Using $\pi_p(n) = k \iff p^k \mid n \text{ and } p^{k+1} \not \mid n$ (that is, $\pi_p(n)$ is the largest power of $p$ that divides $n$), and using $A=\pi_p(a), B=\pi_p(b)\dots$
Then (T) is equivalent to, for all primes $p$,
$$\begin{array}{rcl}\min(A+B, C+D) 
&=& \min(A,C)\\
&+& \min(B,D)\\
&+& \min(A-\min(A,C),D-\min(B,D)) \\
&+& \min(C-\min(A,C),B-\min(B,D)) \\
\end{array}$$
The $A>C,B>D$ (and the symmetric case) is trivial.  The remaining case, $A<C, B>D$ (and the symmetric case is equivalent):
$$\begin{array}{rcl}\min(A+B, C+D) 
&=& A\\
&+& D\\
&+& \min(0,0) \\
&+& \min(C-A,B-D) \\ \\
&=& \min(A + D + C - A, A + D + B-D) \\ \\
&=& \min(D + C, A + B) \\
\end{array}$$
Due to the isomorphic relation of min/max, this proof also works for proving the theorem when $(x,y)$ is used to mean ${\rm lcm}(x,y)$.
A: After being away from the problem for more than a year, Leox's comment reminded me of the problem, so I looked at it again.  I think I solved it! (EDIT: Per the comments, this answer is incomplete and has an error.) I use just the basic GCD (a,b) and LCM [a,b] identities presented in the book to that point:
$$ab=(a,b)[a,b]$$ and $$(ma,mb)=m(a,b)$$
It's a bit involved, but I begin with the right side of the given identity and work to yield the left side.
$$(a,c)(b,d)\left(\frac{a}{(a,c)},\frac{d}{(b,d)}\right)\left(\frac{c}{(a,c)},\frac{b}{(b,d)}\right)$$
Reorganize terms:
$$=(a,c)\left(\frac{a}{(a,c)},\frac{d}{(b,d)}\right)(b,d)\left(\frac{c}{(a,c)},\frac{b}{(b,d)}\right)$$
Mulltiply first big parenthesis by $(a,c)$ and second by $(b,d)$:
$$=\left(\frac{a(a,c)}{(a,c)},\frac{d(a,c)}{(b,d)}\right)\left(\frac{c(b,d)}{(a,c)},\frac{b(b,d)}{(b,d)}\right)$$
Simplify:
$$=\left(a,\frac{d(a,c)}{(b,d)}\right)\left(\frac{c(b,d)}{(a,c)},b\right)$$
Substitute $1/(b,d)=[b,d]/bd$ and $1/(a,c)=[a,c]/ac$:
$$=\left(a,d(a,c)\frac{[b,d]}{bd}\right)\left(c(b,d)\frac{[a,c]}{ac},b\right)$$
Cancel $d$'s from first term and $c$'s from second term and rewrite:
$$=\left(a,\frac{(a,c)[b,d]}{b}\right)\left(\frac{(b,d)[a,c]}{a},b\right)$$
Treat the whole first term like $m$ in $m(a,b)=(ma,mb)$ and multiply it into the second term:
$$=\left(\left(a,\frac{(a,c)[b,d]}{b}\right)\frac{(b,d)[a,c]}{a},\left(a,\frac{(a,c)[b,d]}{b}\right)b\right)$$
Multiply:
$$=\left(\left(\frac{a(b,d)[a,c]}{a},\frac{(a,c)[b,d](b,d)[a,c]}{ba}\right),\left(ab,\frac{b(a,c)[b,d]}{b}\right)\right)$$
Simplify fractions, and note that $(a,c)[a,c]=ac$ and $[b,d](b,d)=bd$:
$$=\left(\left((b,d)[a,c],\frac{acbd}{ba}\right),\left(ab,\frac{(a,c)[b,d]}{1}\right)\right)$$
$$=\bigg(\Big((b,d)[a,c],cd\Big),\Big(ab,(a,c)[b,d]\Big)\bigg)$$
Again note that $(a,c)[a,c]=ac$ and $[b,d](b,d)=bd$ to rewrite the individual GCD and LCM terms:
$$=\bigg(\Big(\frac{bd}{[b,d]}\frac{ac}{(a,c)},cd\Big),\Big(ab,\frac{ac}{[a,c]}\frac{bd}{(b,d)}\Big)\bigg)$$
$$=\bigg(\Big(\frac{abcd}{[b,d](a,c)},cd\Big),\Big(ab,\frac{abcd}{[a,c](b,d)}\Big)\bigg)$$
Factor, using the idea $(ma,mb)=m(a,b)$:
$$=\bigg(cd\Big(\frac{ab}{[b,d](a,c)},1\Big),ab\Big(1,\frac{cd}{[a,c](b,d)}\Big)\bigg)$$
I don't have a good answer why these fractions have to work out to be natural numbers, but algebraically that's the result when factored. 
The GCD of 1 and any other natural number is 1:
$$=\Big(cd*1,ab*1\Big)=\Big(cd,ab\Big)=\Big(ab,cd\Big)$$
