For $\gcd(p,q) = 1$ and $\gcd(r,s) = 1$ show $\gcd(ps,rq) = 1$.

Question

For $\gcd(p.q)=1$ and $\gcd(r,s)=1$ show $\gcd(ps,rq)=1$.

I tried via contradiction. Let $\gcd(ps,rq) \neq 1$, so let $d > 1$ a common divisor of $ps$ and $rq$, i.e. $ps = dx$ and $rq = dy$ for some integers $x,y$. So $ p = \frac{dx}{s} $ and $q = \frac{dy}{r}$. It follows $\gcd(p,q) = \gcd(\frac{dx}{s}, \frac{dy}{r}) \geq d > 1$. That should do it, but I don't think the last but one inequality holds. If $s \nmid x$ or $r \nmid y$ this won't work... Might be the wrong way, but I would be very thankful, if someone could help me!

Answer 1

This statement is simply untrue. As, Arturo notes, if you have $p=2,q=3,r=2,s=3$, we have that $gcd(2,3) =1$ for both number sets, but $gcd(2*3,3*2) = 6$, thus proving your statement wrong. How about adding the extra condition that all $p,q,r,s$ must be distinct? However, we can have $p=4,q=7,r=1,s=21$ and have $gcd(4,7) =1, gcd(1,21)=1$, but $gcd(4*21,1*7) = gcd(84,7) = 7.$ Thus, this statement is untrue.