Four Number Theorem : Let $a$, $b$, $c$, $d$ be integers such that $ab = cd$. 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.

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.
