# Is $\gcd(a,b)\gcd(c,d)=\gcd(ac,bd)$?

2,164

## 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, 2020

• 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.

• It is not true. Search for a counterexample. It is small.
• Consider $(3,4)$ and $(4,3)$.
• Even smaller: $(2,1)$ and $(1,2)$.
• Did you try any example before asking?
• Why try it yourself when you can get it done for you here? (within 10 minutes)
• @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.
• It is easy to prove that $\gcd(a,b)\gcd(c,d) | \gcd(ac,bd)$
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