Let $a$, $b$, $c$, $d$ be integers such that $ab = cd$. Then there exist integers $x$, $y$, $z$, $w$ such that $xy = a$, $zw = b$, $xz = c$, $yw = d$.
My Progress: I tried playing with $\gcd(a,c)$, $\gcd(a,d)$,$\gcd(b,c)$, $\gcd(b,d)$. But I am not able to proceed.
Please post hints rather than solution, it helps me a lot.
Thanks in advance.
Euler's Four Number Theorem ($\rm 4NT$) shows how to derive a common refinement of two factorizations $\,ab = cd,\,$ see $(3)$ below. It is a fundamental result in divisor theory, very closely related to uniqueness of prime factorizations, so we give a few proofs below to help lend insight.
$(1)\ $ By here $\,a\mid cd\Rightarrow \color{#c00}{a = xy},\ x\mid c,\ y\mid d,\,$ so $\, c = xz,\ \color{#0a0}{d = yw},\,$ for some $\,z,w\in\Bbb Z,\,$ hence using these values to solve for $\,b\,$ we obtain $\,b = c/\color{#0a0}d\color{#c00}a = xz\color{#0a0}{yw}/(\color{#c00}{xy}) = zw$.
$(2)\ $ equal fractions are scalings of a reduced fraction $\,\large \color{#c00}{\frac{x}w},\,$ by unique fractionization, so
$$\dfrac{a}d = \dfrac{c}b\ \Rightarrow\ \begin{align} &a = y\,\color{#c00} x,\ \ c = z\, \color{#c00}x\\ &d = y\:\! \color{#c00} w,\ \ b = z\:\! \color{#c00}w \end{align}\qquad$$
$(3)\ \,{\rm let}\, \ \color{#90f}{g :=(a,b,c,d)}.\,$ Cancelling $\,g^2\,$ from $\,\color{#0a0}{ab}=\color{#c00}{cd}\,$ reduces to case $\,\color{#90f}{g = 1}\,$ with solution
$$ \overbrace{\underbrace{(a,c)}_{\textstyle x}\underbrace{(a,d)}_{\textstyle y}}^{\textstyle \color{#0a0}a}\,\overbrace{\underbrace{(b,c)}_{\textstyle z}\underbrace{(b,d)}_{\textstyle w}}^{\textstyle\color{#0a0} b}\, =\, \overbrace{\underbrace{(c,a)}_{\textstyle x}\underbrace{(c,b)}_{\textstyle z}}^{\textstyle\color{#c00}c} \overbrace{\underbrace{(d,a)}_{\textstyle y}\underbrace{(d,b)}_{\textstyle w}}^{\textstyle\color{#c00}d}\qquad$$
by $\ (a,c)(a,d) = (a(a,c,d),\color{#c00}{cd}) = (a(a,c,d),\color{#c00}{ab}) = a(\color{#90f}{a,c,d,b}) = a,\,$ and similarly for the other products (by symmetry). Here $\,(a,c)(a,d)(b,c)(b,d)\,$ is a common refinement of the two factorizations $\,ab = n = cd.\,$ See here for more details on such gcd arithmetic (& ideal arithmetic).
Remark $ $ The solution is summarized by the following Shreier refinement matrix formulation of Euler's Four Number Theorem for the proofs $(2)$ and $(3)$ above
$$\begin{array}{c | c c} (2) & c & d\\ \hline \color{#c00}a& \color{#c00}x & \color{#c00}y\\ b& z & w \end{array}\qquad \begin{array}{c | c c} (3)&\ c & d\\ \hline \color{#c00}a&\color{#c00}{(a,c)} & \color{#c00}{(a,d)}\\ b& (b,c) & (b,d) \end{array}\qquad$$
where the row label is the product of the row elements, $ $ e.g. $\, \color{#c00}{xy = a = (a,c)(a,d)},\,$ and likewise the column label is the product of the column elements, e.g. $\, xz = c = (a,c)(b,c).\,$ Analogous refinement matrices can display the common refinements of two (any length) factorizations of the same element in a UFD or gcd domain, e.g. see this answer, which also explains how this is equivalent to uniqueness of prime factorizations (and many well-known equivalent properties).
Firstly note that it is sufficient to prove the theorem when $a,b,c,d,x,y,z,w$ are all natural numbers. For if any of the given numbers is $0$ then the solution tuple $(x,y,z,w)$ is trivial and if there are negatives involved you can look for $x,y,z,w$ for $\lvert a \rvert,\lvert b \rvert, \lvert c \rvert, \lvert d \rvert$ and then adjust for signs.
If $b = 1$ you can take $(x,y,z,w) = (c,d,1,1)$, say the result holds for all $a,b,c,d$ when $1 \leq b < n$ and say $an = cd$ for some $a,c,d$. Let $p$ be a prime divisor of $n$ then $p \vert c$ or $p \vert d$. Say $p \vert c$, then we'll have an equation of the form $am = c'd$ where $n=mp,c=pc'$ and $1 \leq m<n$ so by hypothesis there exists $(r,s,t,u)$ all naturals such that $a = rs, m = tu, c' = rt, d = su$ that gives $n = (pt)u$ and $c = r(pt)$, therefore $ (r,s,pt,u)$ is the tuple corresponding to $an = cd$, similarly one can find the tuple if $p \vert d$. This proves the theorem for natural numbers by induction.
Here are some cases to consider:
If the products are equal to $0$, WLOG, if $a=0$, then $c$ or $d$ must be $0$.
If $a=0$ and $c=0$, let $x=0$. If $d=0$, then we let $y=0$ and choosing $w$ and $z$ should be easy. If $d\ ne 0$, we let $w=1$ and you can choose your $y$ and $z$ accordingly.
Now consider the cases where the product is non-zero.
$$\frac{a}{c}=\frac{d}{b}=\frac{y}{z}$$ where $y$ and $z$ are chosen to satisfy $\gcd(y,z)=1$. Try to argue how to determine $w$ and $x$ from here.
Whenever , one of $a,b,c,d$ is $0$, it goes trivial. Without loss of any generality, take all of $a,b,c,d$ are positive. Now as $ a $ is the divisor of $ cd $ , so, we can find such positive integers $x,y$ with $a=xy$ that $x$ is divisor of $ c $ and $y$ is divisor of $d$ (either trivially by $1$ or properly or improperly). Now , also, $ b $ is divisor of $\frac{cd}{a} =\frac{c}{x} \frac{d}{y} $ . (Remind, $\frac{c}{x} $, $\frac{d}{y} $ are integers), Also, we can find such positive integers $z,w$ with $b=zw$ that that $z$ is divisor of $\frac{c}{x} $ and $w$ is divisor of $\frac{d}{y} $ (either trivially by $1$ or improperly ). Now, you can continue.
Okay, this is my answer (which I got by the hints everyone provided), I will really be grateful if someone proof reads it.
Since $ab=cd$, we have $d=a\cdot\frac {b}{c} \\ b=c\cdot\frac {b}{c}$
Now, let $\alpha =\frac {p}{q}=\frac {b}{c}$, where gcd$(p,q)=1$
So we have $a\cdot \alpha=d$ and $\\c\cdot \alpha=b$.
Now, let gcd$(a,c)=m$. So, let $a=me$ and $c=mf$, where gcd$(e,f)=1.$
Note that $m\cdot \alpha$ is always an integer.
So we have $a=m\cdot e$, $b=(m\alpha)\cdot f$ , $c=m\cdot f$ and $d=(m\alpha)\cdot e$
So we can take $x=m$,$y=e$,$z=f$ and $w=m\alpha$.
And we are done.
Since $a\mid cd$ we can write $a=xy$ where one factor divide $c$ and other $d$, say $x|c$ and $y|d$ Then $c=xz$ for some $z$ and $d=yt$ for some $t$. Then $b=zt$: $$ xyb= xzyt \implies b=zt$$
