# $g\mid ab, g\mid cd$ and $g\mid (ac+bd)$. Prove that $g \mid ac$ and $g \mid bd$ , $a,b,c,d \in \mathbb{Z}$

If $$g$$ is a divisor of $$ab,cd$$ and $$ac+bd$$ prove that it is also a divisor of $$ac$$ and $$bd$$, where $$a,b,c,d$$ are integers.

There are several existing solutions of this problem on this site, but I approached this problem in a different way, consider $$(ac-bd)^2 = (ac+bd)^2-4abcd$$ , since $$g^2 \mid \left[ (ac+bd)^2-4abcd \right]$$, this implies $$g^2 \mid (ac-bd)^2$$ and hence $$g \mid (ac-bd)$$. This further imples that $$g \mid 2ac$$ and $$g \mid 2bd$$.

Now i am stuck at this point, how do I show from this that $$g \mid ac$$ and $$g \mid bd$$ ?

$$\, r \!=\! \frac{ac}g,\ s\! =\! \frac{bd}g$$ are roots of $$\,\overbrace{x^2\!-\!(r\!+\!s) x\! +\! rs}^{\textstyle (x-r)(x-s)}\,$$ with integer coef's, so $$\,r,s\in\Bbb Z\,$$ by Rational Root Test