# If $\gcd(a, b) = 1$, then $\gcd(ab, c) = \gcd(a, c) \cdot\gcd(b, c)$

How can I prove that if $\gcd(a, b) = 1$, then $\gcd(ab, c) = \gcd(a, c) \times \gcd(b, c)$?

By eea there exists $ax+by=1$ from $\gcd(a,b)=1$ so a and be are co-primes there also exists $dk=a$ and $dj= b$ where $d=\gcd(a,b)=1$ this is all the information I have gathered from the question but I dont know how to approach and solve it. Can anyone help explain to me how to arrive at the answer? Thanks!

• Hint: do you know the Fundamental Theorem of arithmetic? Jun 25 '14 at 1:35
• I haven't learned that yet. =( Jun 25 '14 at 1:36
• Perhaps if you share how gcd is defined for you, we can help you better. Jun 25 '14 at 2:04

Without using primes. We show that $(ab,c) \mid (a,c)(b,c)$ and that $(a,c)(b,c)\mid (ab,c)$.

We have $ax+by=1$ multiplying by $c$ we have $acx+bcy=c$

Now $$(a,c)(b,c)\left[\frac{a}{(a,c)}\frac{c}{(b,c)}x+\frac{b}{(b,c)}\frac{c}{(a,c)}y\right]=c$$ where of course $\frac{a}{(a,c)}$ etc are integers. So $(a,c)(b,c)\mid c$. It is clear that $(a,c)(b,c)\mid ab$ since $(a,c)\mid a$ and $(b,c)\mid b$. And therefore we have $(a,c)(b,c)\mid (ab,c)$.

To show the other direction note that there are $p,q,r,s$ such that

$$ap+qc=(a,c)$$ and $$br+cs=(b,c)$$ thus $$(a,c)(b,c)=abpr +(aps+brq+qsc)c$$ and this latter is divisible by $(ab,c)$

• Remark that Bezout is not needed, it is true in any gcd domain - see my answer. Jun 25 '14 at 2:31

Hint $$\$$ By basic gcd laws (associative, commutative, distributive)

$$\quad (a,c)(b,c) = (ab,ac,bc,cc) = (ab,\color{#c00}{(a,b,c)}c) = (ab,c),\,\$$ by $$\ \,(a,b)=1\,\Rightarrow\,\color{#c00}{(a,b,c)=1}$$