I am stuck with a number theory problem which needs the following lemma.
Lemma- Let $x$ and $y$ be to positive integers such that $g.c.d(x,y)=1$,then the we also have $g.c.d(x+y,x^2-xy+y^2)=1$. My question is whether the lemma is true or not and if it is true then I need the proof.Thanks in advance.
It is not true. Take $x=5,y=4$. $\gcd(x+y,x^2-xy+y^2)=\gcd(9,21)=3$
I found it by subtracting convenient multiples of $x+y$ from the second term. $\gcd(x+y,x^2-xy+y^2)=\gcd(x+y,-2xy+y^2)=\gcd(x+y,-3xy)$ The constant $3$ suggests that we want to find an $x+y$ which has a factor of $3$. We could also have used $x=2,y=1$