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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 $ ?

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$\, r \!=\! \frac{ac}g,\ s\! =\! \frac{bd}g$ 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

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Let us take g to be odd then if g divides 2ac then g divides ac. Now consider g to be even we have to show that ac contains 2 in its prime factorisation. Since g divides ab then a or b have to be even, in the worst case if b and d are even and a and c are odd then g still divides ab and cd however g cannot divide (ac+bd) since ac is odd and (ac+bd) is also odd arriving at a contradiction. Thus ac has to be even and g divides ac.

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