Thread: gcd(a,b) proof - need help

1. gcd(a,b) proof - need help

Hello and thanks for looking

I have to prove the following :Prove that gcd(a,b) = 1 if and only if gcd(ab,a+b)=1

Thanks a lot in advance for the help

2. Originally Posted by octagonreturns
Hello and thanks for looking

I have to prove the following :Prove that gcd(a,b) = 1 if and only if gcd(ab,a+b)=1

Thanks a lot in advance for the help
Let $d=\gcd(ab,a+b)$ then $d|ab$ so $d|a$ or $d|b$. WLOG say $d|a$ then since $d|(a+b)$ it means $d|(a+b-a)\implies d|b$. And so $d=1$.

3. Hello,

Proving this implication : $gcd(ab,a+b)=1 \implies gcd(a,b)=1$

Let $d=gcd(a,b)$

d divides a and d divides b. Therefore, d (and even d²) divides ab and d divides (a+b).

But we know that gcd(ab,a+b)=1.

--> d divides 1.

So d=1.

4. Thank you all so very much for your time and help!