Prime Numbers

**page929** · Sep 20th 2010, 08:52 AM

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?

**roninpro** · Sep 20th 2010, 09:42 AM

You could try playing around with Bezout's identity.

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

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.)

**Unbeatable0** · Sep 20th 2010, 10:28 AM

Alternatively:

$<br /> \gcd(a+b,ab) = 1 \Leftrightarrow \gcd(a+b,a) = \gcd(a+b,b) = 1 \Leftrightarrow \gcd(a,b)=1<br />$

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