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

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

Should $K = k$ in the first statement? Also, what are your thoughts? — Stahl, Nov 10 '16 at 07:33
@stahl Yes! I think the first one is obvious, but I can't think of a way to formulate it. The 2nd one probably has to do with the fact that GCD(a,b) = au +bv, and is equal to 1 only if the GCD is 1. — Tony, Nov 10 '16 at 07:39

score 0 · Accepted Answer · answered Nov 10 '16 at 07:46

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 $as + bct = 1$.

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

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