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.
