$(ab,ac)=a(b,c) , a >0$

Question

$(a,b)$ denotes G.C.D of $a$ and $b$

Let $(b,c)=d$. So, $d|b$ and $d|c$.

Claim : $(ab,ac)=ad$. As d|b and d|c , so $d|b.b...b$ ($a$ times) which means $d|ab$. similarly $d|ac$. So d is common divisor of $ab$ and $ac$.

Now let $f|ab$ and $f|ac$. To prove $f|d$

So we have $fq_1=ab$ and $fq_2=ac$. Also we have $dq_3=b$. So put in former equation i get $fq_1=adq_3$. Now $\frac{q_1}{q_3}$ may or maynot be an integer. How do i proceed ?

Answer 1

Let $(b,c)=d$. Now, we are to prove that $(ab,ac)=ad$. Assume the contrary, let $(ab,ac) = e$ where $e$ and $ad$ are distinct. Clearly, $ad$ divides $e$, hence $ad$ < $e$. Then, we have $(\frac{b}{d},\frac{c}{d}) = \frac{e}{ad}$. However, this shows that $\frac{e}{ad}=1$ since $(b,c)=d$. Contradiction.

Hence, $(ab,ac)=a(b,c)$