A question related to the concept of being "relatively prime"

Question

Suppose that I have $a, b, c, d \in \mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers.

If I have the equation $ab = 2cd$ and I know that $\gcd(a,c)=\gcd(c,d)=1$, then it follows that I have the following divisibility constraints: $$a \mid 2d,$$ and $$c \mid b.$$

Is this correct?

Answer 1

Your results are correct (but the second assumption $\gcd(c,d)=1$ is not used), For this sort of problems, the theorem to keep in mind is Euler's Lemma: $$\left(x|yz \text{ and } \gcd(x,y)=1\right)\Rightarrow x|z \tag 1$$

Now return to your question $ab=2cd$,using $(1)$ for $x=a$, $y=c$ and $z=2d$ we conclude that $a|2d$ so we can write $2d=qa$ and hence $$b=qc$$

and from here follows that $c|b$;

Answer 2

Yes, by Euler's Lemma $\,(c,a)=1,\ c\mid ab\,\Rightarrow\,c\mid b,\ $ so $\ a(b/c) = 2d,\,$ so $\,a\mid 2d$

Remark $\ $ Note that $\, ab = 2cd\iff \dfrac{b}c = \dfrac{2d}a\,\ $ so $\,\ c\mid b\iff a\mid 2d$