If gcd(aa,b)=1, then gcd(a-b,a+b)=1 or 2 for all integers a,b

Show that if then or for all integers .

I don't know how to start.

Re: If gcd(aa,b)=1, then gcd(a-b,a+b)=1 or 2 for all integers a,b

Re: If gcd(aa,b)=1, then gcd(a-b,a+b)=1 or 2 for all integers a,b

Hello.

You could do following:

Let be such that , then we're goning to show that .

If , and , so . On the other hand, .

Hence , but , then therefore or

Best regards.

Re: If gcd(aa,b)=1, then gcd(a-b,a+b)=1 or 2 for all integers a,b

Thank you! This is the part that I could not prove.

Re: If gcd(aa,b)=1, then gcd(a-b,a+b)=1 or 2 for all integers a,b

Quote:

Originally Posted by

**Fernando** Hello.

You could do following:

Let

be such that

, then we're goning to show that

.

If

,

and

, so

. On the other hand,

.

Hence

, but

, then

therefore

or

Best regards.

I don't fully understand the logic in your proof. Do you mean when you say ? Why does and imply ?

Re: If gcd(aa,b)=1, then gcd(a-b,a+b)=1 or 2 for all integers a,b

Hi, yeah ! it's my notation.

OK. There is a theorem states that: Let and be integers differents of zero. Then iff holds following:

and

If and then .

So, by the theorem We can assert if and then and I guess you could continue...

Best regards.

Re: If gcd(aa,b)=1, then gcd(a-b,a+b)=1 or 2 for all integers a,b

Quote:

Originally Posted by

**Fernando** Hi, yeah !

it's my notation.

OK. There is a theorem states that: Let

and

be integers differents of zero. Then

iff

holds following:

and

If

and

then

.

So, by the theorem We can assert if

and

then

and I guess you could continue...

Best regards.

Do you mean the following theorem?

Let and not both zero. Then a positive integer is the greatest common divisor of and if and only if the following equivalence holds:

.

This is different to what you stated.

Re: If gcd(aa,b)=1, then gcd(a-b,a+b)=1 or 2 for all integers a,b

Hi

I don't know, what's the wrong? Do you think the first theorem is wrong?

Re: If gcd(aa,b)=1, then gcd(a-b,a+b)=1 or 2 for all integers a,b

Quote:

Originally Posted by

**Fernando** Hi

I don't know, what's the wrong? Do you think the first theorem is wrong?

I think the first theorem is correct after thinking through it. Sorry I had not seen it before and confused it with the one that I posted.