# Math Help - proof about gcd

I have to prove that if $a,b,c\in\mathbb{Z}$ Show that $gcd(a,b)=1 and gcd(a,c)=1$ iff $gcd(a,lcm[b,c])=1$ I've tried proving the forward direction and im stuck. Here is what I have so far.

proof: $\rightarrow$: Assume $gcd(a,b)=1 and gcd(a,c)=1$. Assume further that $gcd(a,[b,c])\neq 1$. Let $d=(a,[b,c])$ which implies that $d|a and d|[b,c] \implies d|b or d|c$
Case 1: Suppose $d|b and d\nmid c$. Since $d|a and d|b$ and since $gcd(a,b)=1$ then d is a common multiple of $gcd(a,b)$ thus $d=1$ A contradiction.

Case 2: Suppose $d\nmid b and d|c$. Same here as case 1 and d=1. A contradiction.
Case 3: Suppose $d|b and d|c$ Since $gcd(a,b)=1 and gcd(a,c)=1 and d|a$, $d|b, and d|c$ then $d=1$ A contradiction.\

Would this be right?

2. ## Re: proof about gcd

Originally Posted by bonfire09
I have to prove that if $a,b,c,d \in\mathbb{Z}$ Show that $gcd(a,b)=1 and gcd(a,c)=1$ iff $gcd(a,lcm[b,c])=1$ I've tried proving the forward direction and im stuck. Here is what I have so far.
Please review what you have posted. What does $d$ have to do with anything in the given?

3. ## Re: proof about gcd

ok ive fixed it. The part that that I think needs to be explained further is the part where I say since $d|[b,c]\implies d|b or d|c$.

4. ## Re: proof about gcd

Actually I take that back. Its wrong. The claim that $d|[a,b]\implies d|a pr d|b$ is wrong.