Proving gcd(r,s)=1 with gcd(r^2-s^2, r^2+s^2)

[awesome220]

Can anyone help me with this?

If gcd(r,s)=1 then prove that gcd(r^2-s^2, r^2+s^2)=1 or 2.

i'm so confused.

 

[CRGreathouse]

Can anyone help me with this?

If gcd(r,s)=1 then prove that gcd(r^2-s^2, r^2+s^2)=1 or 2.

i'm so confused.

Suppose $n|(r^2-s^2)$ and $n|(r^2+s^2)$. (This would be the case for the gcd of the two expressions.) Then there are some integers a, b with
$$an=r^2-s^2$$ and $$bn=r^2+s^2$$.
Then $(a+b)n=2r^2$ and so n divides $2r^2$. Does this help?

 

[awesome220]

I understand, but how does that give us that gcd (r^2-s^2, r^2+s^2) = 1 or 2?

 

[awesome220]

nevermind, I think i see it! Thanks!