Decomposition of numbers: Show that there are positive integers $x, y$ and $z$, with $x$ and $y$ coprime, such that $a+b=x^2z\,$ and $\,c+d=y^2z$ The positive integers $a$, $b$, $c$ and $d$ satisfy the equation $$(ad-bc)^2=(a+b)(c+d).$$ Show that there are positive integers $x$, $y$ and $z$, with $x$ and $y$ coprime, such that $$a+b=x^2z~\text{ and }~c+d=y^2z.$$
I tried using the fact that $$(a+b)(c+d)=\text{lcm}(a+b,c+d)\gcd(a+b,c+d)$$ and by letting $m=\text{lcm}(a+b,c+d)$ and $z=\gcd(a+b,c+d)$, I got $$a+b=z\Big(\frac{a+b}{z}\Big)=z(a+b)\Big(\frac{m}{(ad-bc)^2}\Big)=z\Big(\frac{k(a+b)^2}{(ad-bc)^2}\Big),$$ since $m=k(a+b)$ for some integer $k$. However, I hit a road block in terms of simplifying further. Any help is appreciated! Thank you!

Write $a+b=vz,c+d=wz$. By the fundamental theirem of arithmatic, $v=p_1^{e_1}p_2^{e_2}..., w=q_1^{f_1}q_2^{f_2}...$. Since $v$ and $w$ are coprime, $p_i\neq q_j\forall i,j$. Finally,
$$(ad-bc)^2=(a+b)(c+d)=vwz^2$$
$$l^2:=\left(\frac{ad-bc}{z}\right)^2=vw$$
Now $l$ has its own prime factorization $r_1^{g_1}r_2^{g_2}...$, so we get
$$r_1^{2g_1}r_2^{2g_2}...=p_1^{e_1}p_2^{e_2}...q_1^{f_1}q_2^{f_2}...$$
Since each prime number appears at most once in $p_i, q_j$, we get that each $r_k$ is matched to exactly one $p_i$ or $q_i$. This means that each $e_i,f_j$ is matched to a $2g_k$, and is thus even. It follows that both $v$ and $w$ are square, that is, $v=x^2$ and $w=y^2$.
