Here is my problem:

Let a,b be nonzero integers. Prove that (a,b) = 1 if and only if (a+b, ab) = 1.

Any suggestion on where to start?

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- Sep 20th 2010, 08:52 AMpage929Prime Numbers
Here is my problem:

Let a,b be nonzero integers. Prove that (a,b) = 1 if and only if (a+b, ab) = 1.

Any suggestion on where to start? - Sep 20th 2010, 09:42 AMroninpro
You could try playing around with Bezout's identity.

(From right to left) Suppose that $\displaystyle \gcd(a+b,ab)=1$. There exist integers $\displaystyle x,y$ such that $\displaystyle (a+b)x+(ab)y=1$. Distributing the terms, $\displaystyle ax+bx+aby=1$, and $\displaystyle a(x+by)+bx=1$. This shows that $\displaystyle \gcd(a,b)$ must divide 1. Therefore, $\displaystyle \gcd(a,b)=1$.

You can try a similar argument for the other direction. (You may need to insert some terms into the equation by adding and subtracting the same thing.) - Sep 20th 2010, 10:28 AMUnbeatable0
Alternatively:

$\displaystyle

\gcd(a+b,ab) = 1 \Leftrightarrow \gcd(a+b,a) = \gcd(a+b,b) = 1 \Leftrightarrow \gcd(a,b)=1

$