I am trying to see whether that sentence holds.
I did this :
Let $\gcd(ab, cd) = g$ :
$$\exists s, t:\; s(ab) + t(cd) = g$$
Now we have :
$$
\begin{cases}
s'a + t'c =1\\
s(ab) + t(cd) = g
\end{cases}
\implies (s'a+t'c)(s(ab) + t(cd))= g
\implies ss'a^2b + s'atcd + t'csab + tt'c^2d= g \\
\implies (ss'a^2+t'csa)b+(s'atc+tt'c^2)d=g
$$
Now $z=(ss'a^2+t'csa), \; r = (s'atc+tt'c^2)$ .
So they exist $z,r \in N:\; zb + rd = g$
But, does that imply that $\gcd(b, d) = g$ ?
Because we know that $\gcd(b, d)$ is the minimum linear combination of $b, d$.
We know that $\gcd(s,t) = 1 \wedge \gcd(s',t') = 1$ but I am not sure how to proceed further.
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N. F. Taussig
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tonythestark
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2Let $d=a$ and $b=c$, then $\gcd(ab,cd)=ab$, but $\gcd(b,d)=1$. – Christian E. Ramirez Nov 21 '22 at 19:20
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you are right, thank you – tonythestark Nov 21 '22 at 19:23
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1@tonythestark FYI, using an Approach0 search, I found the fairly related Is $\gcd(a,b)\gcd(c,d)=\gcd(ac,bd)$?. – John Omielan Nov 21 '22 at 19:27
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1@tonythestark Also FYI, from your first display equation, we have $s(ab)+t(cd)=g ; \to ; (sa)b+(tc)d=g$. This is of the same form as your later $zb+rd=g$, with $z = sa$ and $r=tc$. Also, regarding your question of "But, does that imply that $\gcd(b,d)=g$", note it only just implies that $\gcd(b,d)\mid g$. – John Omielan Nov 21 '22 at 19:37
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1The answer should be clear by examining the general formula in the linked dupe (the method used there is very useful for problems like this - see the link in the Remark there). – Bill Dubuque Nov 21 '22 at 20:30
1 Answers
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Hint. So, $a$ and $c$ share no factors. But what about $a$ with $d$ or $c$ with $b$? What if $a$ shares factors with $d$ that $b$ doesn't, or $c$ shares factors with $b$ that $d$ doesn't? Specifically, how would that affect the values of $\gcd(ab,cd)$ versus $\gcd(b,d)$?
anon
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okay I see what you mean. I am basically throwing away a, c when in fact a could increase the common factor (if for example $\gcd(a, d) >1$).. – tonythestark Nov 21 '22 at 19:22