How to prove that GCD(2K+2,4k)=2 and if GCD(a,b)=1, GCD(a,c)=1 -> GCD(a,bc)=1

Question

How to prove that GCD(4k+2,4k) = 2

And

if GCD(a,b)=1, GCD(a,c) = 1

Then GCD(a,bc)=1

$k,a,b,c \in \mathbb{Z}$

Answer 1

Hint: For the first part, $2$ clearly divides both $4k + 2 = 2(2k + 1)$ and $4k = 2(2k)$. Can you show that $\gcd(2k+1,2k) = 1$? For the second, if $\gcd(a,b) = \gcd(a,c) = 1$, then there exist $u,v,u',v'\in\Bbb Z$ such that $au + bv = au' + cv' = 1$. Try to use these equations find $s,t\in\Bbb Z$ (in terms of $a,b,c,u,v,u',$ and $v'$) such that $as + bct = 1$.