Let $a\mid c$ and $b\mid c$ such that $\gcd(a,b)=1$, Show that $ab\mid c$

Question

Let $a\mid c$ and $b\mid c$ such that greatest common divisor (gcd) $\gcd(a,b)=1$, Show that $ab\mid c$.

Hint: Write a,b,c in their prime factorizations and use the given information. — Fredrik Meyer, Sep 05 '11 at 18:28
The factorization theorem for the integers says that every integer can be uniquely written as a product of primes. — Fredrik Meyer, Sep 05 '11 at 18:32
@Alvou: Please don't post in a commanding tone (Show that...), please accept satisfactory answers to your questions (this one and previous ones). What have you tried? — Tyler, Sep 05 '11 at 19:08
To all who repeatedly suggest using prime factorizations: this is frequently used towards the proof of the FTA. So it would be best if that were not used. In particular, there are proofs that do not use FTA. (I know Rosen's book develops this before FTA, for example). — davidlowryduda, Sep 05 '11 at 19:40
Some related posts: Show that if $a | c$ and $b | c$, then $ab | c$ when $a$ is coprime to $b$., Prove: If $a|m$ and $b|m$ and $\gcd(a,b)=1$ then $ab|m$, If $\gcd(a,b)=1$ and $a$ and $b$ divide $c$, then so does $ab$. — Martin Sleziak, May 03 '19 at 12:30

Bill Dubuque · Accepted Answer · 2020-10-03T20:05:25.667

Hint $\rm\quad\ \ a,b\ |\ c\ \Rightarrow\ ab\ |\ ac,bc\ \Rightarrow\ ab\ |\ (ac,bc) = (a,b)c = c\ $ via $\rm\:(a,b)= 1\:.$

Note $\ $ This proof works in every domain where GCDs exist, since it doesn't use Bezout's identity. Instead it uses the GCD distributive law $\rm\ (ac,bc) = (a,b)c\:,\ $ true in every GCD domain, viz.

Lemma $\rm\quad (a,b)\ =\ (ac,bc)/c\ \ $ if $\rm\ (ac,bc)\ $ exists.

Proof $\rm\quad\ d\ |\ a,b\ \iff\ dc\ |\ ac,bc\ \iff\ dc\ |\ (ac,bc)\ \iff\ d\ |\ (ac,bc)/c$

The above proofs use the universal definitions of GCD, LCM, which often serve to simplify proofs, e.g. see this proof of the GCD * LCM law.

score 4 · Answer 2 · answered Sep 05 '11 at 18:31

4

If $\operatorname{gcd}(a,b)=1$ then there exist $x,y$ such that $ax+by=1$. Now let $a,b|c$ and note that $c = cax+cby$. Can you show that both terms on the right hand side are divisible by $ab$?

answered Sep 05 '11 at 18:31

Matthew Towers

10,854

why do you divide both terms on the right hand by ab? – Sep 05 '11 at 18:38
@alvoutila, because you want to show that $ab | c$? – Sep 05 '11 at 18:44
2

thanks that helped. So if a|c ~ c=aq and b|c ~ c=bq' then if syt(a,b)=xa+yb=1 => c=xca+ybc = x bq'a+yaqb=ab(xq'+yq)=abq'' => ab|c. Isn't it – Sep 05 '11 at 18:57
First sentence is Bézout's identity. – anon Sep 05 '11 at 19:12
@alvoutila, you are right. :) – Sep 05 '11 at 19:13

score 0 · Answer 3 · edited Jul 30 '12 at 20:04

Here's a solution using set theory.

Let the set $A$ be the prime factors of $a$,

set $B$ be the set of prime factors of $b$,

and set $C$ be the prime factors of $c$.

$a\;|\;c\Longrightarrow A\subset C$
$b\;|\;c\Longrightarrow B\subset C$
$\gcd(a,b) = 1 \Longrightarrow A\cap B\subset \varnothing$

We want to show that $ab\;|\;c \Longrightarrow A \cup B \subseteq C$.

Do you see why this is given the $3$ points above?

Let $a\mid c$ and $b\mid c$ such that $\gcd(a,b)=1$, Show that $ab\mid c$

3 Answers3

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