number theory $\gcd(a,bc)=\gcd(a,c)$

Question

Suppose $\forall (a;b;c;d) \in \mathbb{Z}^4$ as $\gcd(a,b)=\gcd(c,d)=1$:

How can I prove that $\gcd(a,bc)=\gcd(a,c)$?

Also, how to prove that $\gcd(ac,bd) = (\gcd(a,d))*(\gcd(b,c))$?

What is $d$? Why is it $\forall (a;b;c)\in\mathbb Z^3$ without $d$? — Thomas Andrews, Mar 26 '13 at 20:49
Title makes it look like it might eventually be a duplicate or maybe the stepchild of a duplicate of this. — rschwieb, Mar 26 '13 at 21:02
@user68578 $\mathbb{Z}^3$ is the set of all three-tuples of integers. — apnorton, Mar 26 '13 at 21:05
@user68578 Do you know what $\mathbb{R}^3$ is? Well, replace $\mathbb{R}$ by $\mathbb{Z}$ to get $\mathbb{Z}^3$:
$$\mathbb{Z}^3 = { (p,q,r) : p,q,r \in \mathbb{Z} } . $$ — Fly by Night, Mar 26 '13 at 21:05

Math Gems · Accepted Answer · 2013-03-26T21:15:10.020

4

$\rm\ \color{#C00}{(a,bc)} = (a,ac,bc) = (a,(ac,bc)) \color{#C00}{\stackrel{(1)}{=} (a,(a,b)c)}\ [\, = (a,c)\ \ if\ \ (a,b) = 1].\ $ So applying $\,\color{#C00}{\stackrel{(1)}{=}}$

$\rm\qquad \color{#0A0}{(d,ac)} \color{#C00}{\stackrel{(1)}{=}} (d,a(c,d)) = (d,a)\ $ and $\rm\ \color{blue}{(b,ac)} \color{#C00}{\stackrel{(1)}{=}} (b,(b,a)c) = (b,c),\, $ by $\rm\,(a,b)\!=\!1\!=\!(c,d)$

$\rm \Rightarrow\ (ac,db) \color{#C00}{\stackrel{(1)}{=}} (ac,\color{#0A0}{(d,ac)}\color{blue}{(b,ac)}) = (ac,(d,a)(b,c)) = (d,a)(b,c)\:$ by $\rm\:(d,a)\mid a,\ (b,c)\mid c$

edited Mar 26 '13 at 21:15

answered Mar 26 '13 at 21:06

Math Gems

20,090

I couldn't understand this step
$\rm (ac,db) \color{#C00}{\stackrel{(1)}{=}} (ac,\color{#0A0}{(d,ac)}\color{blue}{(b,ac)})$

Can you please show me how did u pass from right hand to the left?
– pourjour Mar 26 '13 at 22:58
I applied the top red gcd law $(1)$ twice, first to map d into the green term, then to map b into the blue term. – Math Gems Mar 26 '13 at 23:09

score 3 · Answer 2 · answered Mar 26 '13 at 20:59

3

Hint: Let $GCD(a,c)=x, a=a'x, c=c'x. $

Then $GCD(a',c')=1,$

so $GCD(a,bc)=GCD(a'x,bc'x)=xGCD(a',bc')=x$.

answered Mar 26 '13 at 20:59

Boris Novikov

17,754

Cameron Buie · Answer 3 · 2013-03-26T21:30:11.117

Note that if a number divides $a$ and $c$, then it divides $a$ and $bc$, so $\gcd(a,bc)\geq\gcd(a,c).$ On the other hand, $\gcd(a,bc)$ divides $a,$ and since $a,b$ have no common divisors, then $\gcd(a,bc)$ does not divide $b$. But $\gcd(a,bc)$ does divide $bc$, so it must divide $c$, and hence, $\gcd(a,bc)\leq\gcd(a,c)$ (why?).

We can make similar arguments in the second problem. One direction of the inequality should be clear. You'll need to use the facts that (i) $a,b$ have no common factors and (ii) $c,d$ have no common factors to get the other direction.

score 1 · Answer 4 · answered Mar 26 '13 at 21:29

Theorem $(a,b)=1$ implies $(a,bc) = (a,c)$.
We need to show that $g=(a,c)$ is the universal $g$ such that $g\mid a$ and $g\mid bc$. Certainly $(a,c)\mid a$ and $(a,c)\mid bc$ holds, so we just need to show universality. To that end let $g\mid a$ and $g\mid bc$, we will show that $g|(a,c)$. By $(a,b)=1$ if $g\mid b$ then $g=1$, so suppose $g \not \mid b$, then $g \mid c$ and we are done.

number theory $\gcd(a,bc)=\gcd(a,c)$

4 Answers4

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