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 $\,c+d=y^2z$

Question

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!

Answer 1

You have the right idea with using the $\gcd$, but I don't know how to appropriately finish what you've done using the $\operatorname{lcm}$ as well. Instead, let

$$z = \gcd(a + b, c + d), \; \; a + b = ez, \; \; c + d = fz, \; \; \gcd(e, f) = 1 \tag{1}\label{eq1A}$$

For simpler algebra, let $j = ad - bc$. Then substitute \eqref{eq1A} into the equation that you gave to be satisfied to get

$$j^2 = (ez)(fz) = z^2(ef) \tag{2}\label{eq2A}$$

Thus, $z \mid j \; \; \to \; \; j = kz$, which substituting into \eqref{eq2A} and dividing by $z^2$ gives

$$k^2 = ef \tag{3}\label{eq3A}$$

As proven in multiple posts here, such as Show if a product of coprime numbers is a perfect square, so are the numbers - without FTA, a product of coprime integers being a square mean each of those integers is a square themselves. Thus, since \eqref{eq1A} gives that $\gcd(e, f) = 1$, then \eqref{eq3A} means that for some positive integers $x$ and $y$, coprime to each other, we have

$$e = x^2, \; \; f = y^2 \tag{4}\label{eq4A}$$

Using \eqref{eq1A}, this gives what you were asked to prove, i.e., $a + b = x^{2}z$ and $c + d = y^{2}z$.

Answer 2

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 $w=y^2$.