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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.

Bill Dubuque
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Hawaiian Rolls
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    Is (a,b) the gcd? If yes, then try out some examples. How can it maybe go wrong? – user2103480 Oct 19 '18 at 10:52
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    $a|bc$ does not imply that $a|b$ or $a|c$. For example, $6|2 \cdot 3$, but $6$ divides neither $2$ nor $3$. The implication holds if $a$ is prime though. More generally, $a|bc$ implies $a|c$ if $\textrm{gcd}\left(a,b\right) = 1$. – Sam Streeter Oct 19 '18 at 10:55
  • @Amjad (a,b) is the gcd. I used $gcd(a,b) = gcd(15,20) = 5$. I inserted it into $a|(a,b)c$ and it turned into $5c = 15q$. The only way this works out is if $c = 3q$, but if it doesn't this doesn't work out, I believe. – Hawaiian Rolls Oct 19 '18 at 11:07
  • @SamStreeter My wording was off. I didn't mean to say that a|b or b|c was implied. I just knew that a|b can be a property of $a|bc$. So if $a$ is relatively prime is $a|(a,b)c$ able to be proved by just using $a|bc => a|c$ if $gcd(a,b)=1$ then rewriting as $a|(a,b)c = a|1c. $? – Hawaiian Rolls Oct 19 '18 at 11:07
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    @HawaiianRolls Sure, the result you're after follows from the result I have stated in the special case where $a$ is relatively prime to $b$, but this doesn't help with the general solution. However, the answers below, as you have seen, give a nice and simple proof by Bézout's identity. – Sam Streeter Oct 19 '18 at 11:16

5 Answers5

5

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 $a\mid kac$. Also $a\mid bc$ which implies $a\mid lbc$. And hence $a$ divides the sum $kac+lbc=(a,b)c$.

J.G.
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Mark
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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?

Michael Burr
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  • Thank you for your input. I think Mark explained the different reasons. I got to where you were on my notebook, I just couldn't figure out why $acx + bcy$ were divisible by $a$. I did not know you use the fact that $a$ divides the sums as proof that $a|(a,b)c$. – Hawaiian Rolls Oct 19 '18 at 11:15
  • Note that if you haven't proved it yet, you might need to prove the claim that if $a\mid b$ and $a\mid c$, then $a\mid(b+c)$. – Michael Burr Oct 19 '18 at 12:34
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$ a\mid bc\!\iff\!\! a\mid ac,bc\!\!\! \overset{\rm\color{#0a0}U\!\!}{\iff}\! a\mid (ac,bc)\! \overset{\rm\color{#c00} D_{\phantom{|}}\!}{=}\!(a,b)c\:\!$ by $\rm \color{#0a0}U\! =\! $ Universal Property, $\rm\color{#c00}D\! =\! $ Distributive Law.

Or $\!\bmod a\!:\ ac,bc\equiv 0\,$ so $\:\!a,b\:\!$ periods of $\:\!c\:\!\Rightarrow (a,b)\:\!$ is too, by periods closed under gcd.

Beware $ $ Other proofs using Bezout essentially replace above use of the gcd distributive law by its Bezout-based proof, so they are less general - e.g. they fail in UFDs 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, we'd have: $\,x\!f\!+\!2g\!=\!1\underset{x\ =\ 0}\Longrightarrow 2g(0)=1\Rightarrow 2\mid 1\,$ in $\Bbb Z\Rightarrow\!\Leftarrow\,;\,$ similarly in $\,\Bbb Q[y,z]\,$ for $(y,z)=1$ by eval at $\,y\!=\!0\!=\!z$.

Bill Dubuque
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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 $dA=a$

lab bhattacharjee
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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:

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

Maged Saeed
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    W/o proving "this result" that's a circular proof, since $,a\mid (a,b)c\iff a/(a,b)\mid c,$ are trivially equivalent (to $, (a,b)c/a\in\Bbb Z)\ \ $ – Bill Dubuque Sep 05 '25 at 02:34