Addition of relatively prime numbers

**I-Think** · October 31st 2010, 08:10 PM

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

Proof
Note $1=as+bt$
Assume $gcd(a, (a+b))=k$
Then
$am+(a+b)n=a(m+n)+bn=k*1$
$a(m+n)+bn=aks+bkt$
Thus
$m+n=ks,n=kt$
So
$m+kt=ks,m=ks-kt$
So
$a(ks-kt)+bkt=aks+bkt$
$-akt=0$
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?

**I-Think** · October 31st 2010, 08:12 PM

This thread is a repost. Sorry. Marking it as solved and requesting that it be removed please

**Drexel28** · October 31st 2010, 08:35 PM

Originally Posted by I-Think

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

Proof
Note $1=as+bt$
Assume $gcd(a, (a+b))=k$
Then
$am+(a+b)n=a(m+n)+bn=k*1$
$a(m+n)+bn=aks+bkt$
Thus
$m+n=ks,n=kt$
So
$m+kt=ks,m=ks-kt$
So
$a(ks-kt)+bkt=aks+bkt$
$-akt=0$
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?

Originally Posted by I-Think

This thread is a repost. Sorry. Marking it as solved and requesting that it be removed please

I can't find your other thread. Your proof looks fine from what I see. Yes, you're theorem generalizes. One merely needs note that if $\text{gcd}(a,a+b)\mid \text{gcd}(a,b)$ since $z\mid a\text{ and }z\mid a+b\implies z\mid a\text{ and }z\mid b$ . Also, $\text{gcd}(a,b)\mid \text{gcd}(a,b)$ since $z\mid a\text{ and }z\mid b\implies z\mid a\text{ and }z\mid a+b$

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