I have a proof question that I'm struggling on for my abstract algebra class.
Suppose that $a, b, c, d$ are integers with $a\neq 0$. Suppose that $a$ and $b$ are relatively prime. Part (i) If $d|a$ then, $b$ & $d$ are relatively prime.
I've started off the question: $$ d|a\Rightarrow a=kd$$ Also $$ax+by=1$$ Therefore $$kdx+by=1$$ Now I'm not sure if this is enough to say $d$ & $b$ are relatively prime however I doubt it. I'm not sure how I can get rid of the $k$ factor in my expression. This only shows that $kd$ and $b$ are relatively prime. Then there is Part (ii):
Prove that $gcd(a,c)=gcd(a,bc)$
Here I have said if $a|bc$ then the equation holds as since $$ax+by=1\Rightarrow acx+bcy=c$$ Therefore if $a|bc$ then $a$ divides both terms of $acx+bcy$ therefore $a|c$. Therefore $gcd(a,c)=gcd(a,bc)=a$ However I'm not sure how to show the result for $a\not|bc$. It's trvial for $c=1$. Anyway any help would be much appreciated.