Let $a\mid bc $ then prove or disprove $a\mid (a,b)c$

Question

Prove or disprove:

Let $a\mid bc$ then $a\mid (a,b)c$

Here is my approach, but I am not sure if I am doing this correctly or efficiently.

Let $a\mid bc$. It follows that either

1. $a\mid b$ Proof: $b=ar, a\mid bc => (ar)c = a \rightarrow a(rc)=a \rightarrow a|a(rc)$

2. $a\mid c$.

Since $rc$ is an integer. $a\mid bc$. Similar for $(2). a|c $

Let $a\mid (a,b)c$.

Using: the definition of $\gcd(a,b)=1=ax+by$ if $\exists x,y \in Z$

then we can rewrite it as $a\mid dc$. This is as far as I go. I can't manipulate it so that I show that $a\mid (a,b)c$. Does this mean that I would have to disprove $a\mid bc$ then $a\mid (a,b)c$?

Any help would be appreciated.

Answer 1

I'm really confused about your solution. $a\mid bc$ doesn't imply that either $a\mid b$ or $a\mid c$ unless $a$ is a prime number. Anyway, the statement you try to prove/disprove is true. You can write $(a,b)=ka+lb$ when $k,l\in\mathbb{Z}$. Then $(a,b)c=kac+lbc$. $a$ divides $a$ and so $a\mid $. Also $a\mid bc$ which implies $a\mid lbc$. And hence $a$ divides the sum $kac+lbc=(a,b)c$.

Answer 2

Try $\dfrac aA=\dfrac bB=(a,b)\implies(A,B)=1\ \ \ \ (1)$

$a|bc\implies A|Bc\implies A|c$ by $(1)$

Now $(a,b)c=dc$ is divisible by $dA=a$

Answer 3

The main problem in your proof is that knowing that $a|bc$ does not imply that $a|b$ or $a|c$.

Hint: Use the fact that the $\gcd(a,b)$ can be written as follows: $$ \gcd(a,b)=ax+by $$ for some integers $x$ and $y$. By substituting this into the expression $\gcd(a,b)c$, you get $$ \gcd(a,b)c=acx+bcy. $$ Both of the terms $acx$ and $bcy$ are divisible by $a$ (but for different reasons). Can you work from here?

Answer 4

first of all, you need to pay attention when you said if $a|bc$ then $a|b$ or $a|c$ this is not true unless $a$ is prime. Consider this counter example: $6|2 \times 3$.

For the proposition you claimed, you may need to be familiar with this result:

$a|bc \to \frac{a}{(a,b)}|c$

Using this fact, consider the following:

$\frac{a}{(a,b)}|c \to c = \frac{a}{(a,b)} k \to c(a,b) = ak \to a|(a,b)c$

Now, can you prove this generalized version of your result:

$a|bc \to a|(a,b)(a,c)$

Answer 5

$a\:\!\:\!|\:\!\:\!(a,b)c\color{#c00}{\overset{\rm D}{=}}(ac,bc)\!\!\!\!\color{#0c0}{\overset{\rm U\!\!}{\iff}}\!\! a\:\!|\:\!ac,bc\!\!\iff\!\! a\:\!|\:\!bc\,$ by $\rm \color{#0c0}U\! =\! $ Universal Property, $\rm\color{#c00}D\! =\! $ Distributive Law.

Remark $ $ Other proofs using Bezout are essentially Bezout-based proofs of the gcd distributive law so they are less general - they don't work in domains like $\:\!\Bbb Z[x]\:\!$ and $\:\!\Bbb Q[y,z]$ where Bezout fails, e.g. $\,(x,2) = 1 = (y,z)\,$ but these gcds cannot be written as linear combinations, else, e.g. $x\!f\!+\!2g\!=\!1\overset{x\ =\ 0}\Longrightarrow 2g(0)=1\Rightarrow 2\mid 1\,$ in $\Bbb Z\Rightarrow\!\Leftarrow\,;\,$ ditto for $(y,z)=1$ by eval at $\,y\!=\!0\!=\!z$.