Let $ab=cd$ and $\gcd(a,c)=1$. Then $a | d$ and $c | b$.

Question

Let $ab=cd$ and $\gcd(a,c)=1$. Then $a | d$ and $c | b$.

$ab=cd \implies a|cd,$ but $\gcd(a,c)=1 \implies a \nmid c$, so $a | d$.

$ab=cd \implies c|ab,$ but $\gcd(a,c)=1 \implies c \nmid a$, so $c | b$.

Is the solution correct?

Answer 1

If

$\gcd(a, c) = 1, \tag 1$

then there are $m, n \in \Bbb Z$ with

$ma + nc = 1; \tag 2$

so,

$mab + nbc = b; \tag 3$

now $ab = cd$ implies $c \mid ab$, whence $c \mid b$; likewise,

$mad + ncd = d, \tag 4$

and $ab = cd$ yields $a \mid cd$, so $a \mid d$ follows.

Answer 2

The first implication is the consequence of the definition of divisibility

and second $a|cd$ and $\gcd(a,c)=1 \implies a|d $

is true because of Euclid lemma. Similarly for the second one.

For Euclid lemma: page 24 Elementary Number Theory 6th edition by David M. Burton

https://en.wikipedia.org/wiki/Euclid%27s_lemma

Answer 3

Considering the equation $ab=cd$ modulo $a$, we have $cd\equiv 0 \mod a$. Since $\gcd(a,c)=1$, $c$ is invertible modulo $a$, and thus $d\equiv 0 \mod a$. That is, $a\mid d$. We can similarly conclude that $c\mid b$ by considering $ab=cd$ modulo $c$.