Is $\gcd(a,b)\gcd(c,d)=\gcd(ac,bd)$?
Solution 1
No, let $a= 2, b=3, c=3, d= 2$, then $\gcd(a,b) = 1 = \gcd(c,d) = \gcd(a,c) = \gcd(b,d)$, but $\gcd(ac, bd) = 6$.
Solution 2
Even if you demand that the numbers $a, b, c, d$ are all different, it is trivial to find a counterexample:
$$\begin{align} \gcd(1,6) &= 1; \\ \gcd(2,3) &= 1; \\ \gcd(1,2) \cdot \gcd(6,3) &\neq \gcd(1 \cdot 6, 2 \cdot 3). \end{align}$$
Solution 3
It is not true generally. By using simple gcd arithmetic, employing only basic universal gcd laws (associative, commutative, distributive laws), we can determine precisely when it holds true and, hence, easily construct counterexamples.
Theorem $\ $ If $\rm\:(a,c)=1=(b,d)\:$ then $\rm\:(ac,bd) = (a,b)(c,d)\!\iff\! (a,d) = 1 = (b,c) $
Proof $\ $ We apply the Lemma below a few times to compute gcd products.
Notice $\rm\: (ac,bd) = (a,bd)(c,bd)\ $ by $\rm\:(a,c)=1\:\Rightarrow (a,c,bd)=1$
Further $\rm\:(a,bd) = (a,b)(a,d)\ $ since $\rm\ (b,d) = 1\:\Rightarrow (a,b,d) = 1$
Further $\rm\:(c,bd) = (c,b)(c,d)\ $ since $\rm\ (b,d) = 1\:\Rightarrow (c,b,d) = 1$
Hence $\rm\: (ac,bd) = (a,\!bd)(c,\!bd) = (a,b)(a,d)(c,b)(c,d)\ $ by combining the above.
Hence $\rm\: (ac,bd) = (a,\:b)\:(c,\:d)\ \iff\ (a,d)\:(c,b) = 1\ $ by comparing with prior. $\ $ QED
Lemma $\rm\ (x,y)(x,z) = (x,yz)\ \ if\ \ (x,y,z) = 1$
Proof $\rm\quad (x,y)(x,z) = (xx,xy,xz,yz) = (x(x,y,z),yz) = (x,yz)\ \ \ $ QED
Solution 4
gcd is a multiplicative function , so:
If $\gcd(a,c)=1$ then $\gcd(ac,bd)=\gcd(a,bd)\cdot \gcd(c,bd)$
and :
If $\gcd(b,d)=1$ then $\gcd(ac,bd)=\gcd(b,ac)\cdot \gcd(d,ac)$
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Sayantan
Updated on April 30, 2020Comments

Sayantan about 3 years
Let $a$,$b$,$c$ and $d$ be four natural numbers such that $\gcd(a,c)=1$ and $\gcd(b,d)=1$. Then is it true that,$$\gcd(a,b)\gcd(c,d)=\gcd(ac,bd)$$ I'm awfully weak in number theory. Can anyone please help? Thank you.

sdcvvc about 11 yearsIt is not true. Search for a counterexample. It is small.

wxu about 11 yearsConsider $(3,4)$ and $(4,3)$.

Asaf Karagila about 11 yearsEven smaller: $(2,1)$ and $(1,2)$.

Did about 11 yearsDid you try any example before asking?

GEdgar about 11 yearsWhy try it yourself when you can get it done for you here? (within 10 minutes)

Sayantan about 11 years@GEdgar : I admit that I did not look for any counterexample and when I was posting this question I was thinking something similar to what you have said in your comment. Actually, this came from some different problem. Anyway, I apologize for the whole mess. Regrads.

N. S. about 11 yearsIt is easy to prove that $\gcd(a,b)\gcd(c,d)  \gcd(ac,bd)$

Asaf Karagila about 11 years@Sayantan: Here's a good advice for any field of mathematics which you study, if you don't meddle with things by hand and chew on the problems you will have a hard time to grasp the material. Most of the introductory level courses give problems in order to exercise the use of definitions; theorems; and practice writing correct proofs. When approaching a problem you should have all your definitions and theorems at hand and you need to read again all the definitions of things appearing in your exercise and all the theorems about them which seem relevant. The answer should appear from that.


TMM about 11 yearsI don't see what the first two equations have to do with the third.

TMM about 11 years(1) This all looks irrelevant to the question, which you have not answered.

Gaute about 11 yearsThe first two equations are conditions for a,b,c,d to fulfill in the original question.

Bill Dubuque over 2 yearsSee also here