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Is it true that $\gcd(a,bc)=\gcd(a,b)\gcd({a\over\gcd(a,b)}, c)$?

It is true in quite a few examples that I came up with, e.g.

$a = 18, b = 21, c = 33$
$\gcd(18,21)\gcd({18\over\gcd(18,21)}, 33) = 3 \gcd({18 \over 3}, 33) = 3 \times 3 = 9$
$\gcd(18,693) = 9$

Here is my (probably not very good) attempt at a proof:

For a $b$ where $\gcd(a, b) = 1$
For a $c$ where $\gcd(a, c) = 1$

$\gcd(k_1a, k_1b) = k_1$
$\gcd(k_2a, k_2c) = k_2$

$\gcd(k_1k_2a, k_1k_2bc) = k_1k_2$

Hence, $\gcd(k_1k_2a, k_1b) \gcd({k_1k_2a \over \gcd(k_1k_2a, k_1b)}, k_2b) = k_1 \gcd({k_1k_2a \over k_1}, k_2c) = k_1k_2 = \gcd(k_1k_2a, k_1k_2bc)$

Is this statement true, or did I just stumble across several lucky coincidences?

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Greg
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    You can work on this one prime at a time. Assume $p$ is prime, $a=p^r$, $b=p^s$, $c=p^t$, evaluate both sides under all possible order relations among $r$, $s$, and $t$. – Gerry Myerson May 27 '13 at 12:45

2 Answers2

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Yes, by basic gcd laws (distributive, commutative, associative, etc) we have $$(a,b)\left(\frac{a}{(a,b)},c\right) = (a,(a,b)c) = (a,ac,bc) = (a(1,c),bc) = (a,bc) $$

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It is generally true (for positive arguments) that $\gcd(dx,dy)=d\gcd(x,y)$. Now $d=\gcd(a,b)$ divides both $a$ and $b$, so you get from this that $$ \gcd(a,bc)=d\gcd(a/d,(b/d)c). $$ Moreover $a/d$ and $b/d$ are relatively prime (a common divisor, multiplied by $d$, would divide both $a$ and $b$, and this can only happen for $1$). So what you are asking boils down to: if $a'$ and $b'$ are relatively prime, does it hold for all $c$ that $\gcd(a',b'c)=\gcd(a',c)$? The answer is yes, this is a well known fact attributed to Gauss.

Marc van Leeuwen
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