Prove that $\gcd(ac, bd)=1$ if $\gcd(a, b)=1$ and $\gcd(c, d)=1.$

Question

Prove:

$\gcd(a,b)=1$ and $\gcd(c,d)=1$ implies $\gcd(ac,bd)=1.$

method wich seems like it should work is simplifying the product that we got by Bézout:

$$(ua+vb)^2(u'c+v'd)^2=1.$$

but it doesn't.

It’s not true. $a=d=2,b=c=1.$ – Thomas Andrews Oct 27 '21 at 00:50 — Thomas Andrews, Oct 27 '21 at 00:50

score 2 · Accepted Answer · answered Oct 27 '21 at 01:01

2

It’s not true. If $c=b,d=a,$ and $\gcd(a,b)=1,$ then $\gcd(c,d)=1,$ but $cd=ab,$ so $$\gcd(ab,cd)=ab.$$

answered Oct 27 '21 at 01:01

Thomas Andrews

Please strive not to add more dupe answers to dupes of FAQs. – Bill Dubuque Oct 27 '21 at 20:59

1 Answers1