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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.

J. M. ain't a mathematician
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Sayantan
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    It is not true. Search for a counterexample. It is small. – sdcvvc Apr 28 '12 at 11:36
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    Consider $(3,4)$ and $(4,3)$. – wxu Apr 28 '12 at 11:37
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    Even smaller: $(2,1)$ and $(1,2)$. – Asaf Karagila Apr 28 '12 at 11:43
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    Did you try any example before asking? – Did Apr 28 '12 at 11:43
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    Why try it yourself when you can get it done for you here? (within 10 minutes) – GEdgar Apr 28 '12 at 12:45
  • @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. – Sayantan Apr 28 '12 at 13:54
  • It is easy to prove that $\gcd(a,b)\gcd(c,d) | \gcd(ac,bd)$ – N. S. Apr 28 '12 at 17:23
  • @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. – Asaf Karagila Apr 28 '12 at 19:29

4 Answers4

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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 $\gcd(ac, bd) = 6$.

TMM
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Justin Young
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Even if you demand that the numbers $a, b, c, d$ are all different, it is trivial to find a counterexample:

$$\begin{align} \gcd(1,6) &= 1; \\ \gcd(2,3) &= 1; \\ \gcd(1,2) \cdot \gcd(6,3) &\neq \gcd(1 \cdot 6, 2 \cdot 3). \end{align}$$

TMM
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Gaute
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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 $\gcd(ac,bd)=\gcd(b,ac)\cdot \gcd(d,ac)$

Pedja
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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

Bill Dubuque
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