RTP: If $\displaystyle gcd(a,b)=1$, then $\displaystyle gcd(a, (a+b))=1$, where a,b are positive integers

Proof

Note $\displaystyle 1=as+bt$

Assume $\displaystyle gcd(a, (a+b))=k$

Then

$\displaystyle am+(a+b)n=a(m+n)+bn=k*1$

$\displaystyle a(m+n)+bn=aks+bkt$

Thus

$\displaystyle m+n=ks,n=kt$

So

$\displaystyle m+kt=ks,m=ks-kt$

So

$\displaystyle a(ks-kt)+bkt=aks+bkt$

$\displaystyle -akt=0$

This is impossible, so our assumption is wrong

Thus our statement is true

Questions

1. Is this proof valid?

2. Can I generalize to :If gcd(a,b)=k, then gcd(a, (a+b))=k

3. Is there a simpler non-contradiction, non-FTA proof?