Originally Posted by

**GoldendoodleMom** I would just like some feedback on the work I have done for the following proof.

Prove that if gcd(a,b)=1, then gcd(a+b, ab)=1.

Proof: Let gcd(a,b) = 1. From an earlier exercise it is known that for any positive integer n and any integer a, gcd(a, a+n) divides n. Let d = gcd(a, a+n). This implies that d|a, and thus a = dx for some integer x, and also that d|(a+n), and thus a+n = dy for some integer y. So n = dy - a => n = dy - dx => n = d(y-x) => d|n. Since gcd(a,b)=1, let b=n. This implies d = gcd(a, a+b) but d|b since gcd(a,b)=1, so d=1. Let a=n. Then d=gcd(b,b+a) which implies d|b and d|(a+b), but d|a since gcd(a,b)=1, so d=1. Now, 1=ax+(a+b)y and 1=bx+(a+b)y so (ax+(a+b)y)*(bx+(a+b)y)=1. {expand and simplify} (a+b)(an integer)+(ab)(an integer) = 1. Therefore by theorem 2.4{a and b are relatively prime iff there exist integers x andy such that 1=ax+by}, gcd(ab, a+b)=1.

Any constructive criticism would be appreciated!