TITLE: If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$.
QUESTION [5 upvotes]: Hint:  $a^2 -ab +b^2 = (a+b)^2 -3ab.$
I know we can say that there exists an $x,y$ such that $ax + by = 1$.  So in this case,
$(a+b)x + ((a+b)^2 -3ab)y =1.$
I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$  Any suggestions?

REPLY [3 votes]: This is my solution:  
Lemma 1: $\forall a,b\in\mathbb{Z}$, if $\gcd(a,b)=1\Rightarrow\gcd(a+b,a)=\gcd(a+b,b)=1$.
Lemma 2: $\forall a,b,c\in\mathbb{Z}$, if $\gcd(a,b)=1\Rightarrow\gcd(a,bc)=\gcd(a,c)$.  
Proving lemma 1:
Let $d=\gcd(a+b,a)\Rightarrow d\mid(a+b)~\wedge~d\mid a\Rightarrow d\mid b$, but $\gcd(a,b)=1\Rightarrow d=1$.
Let $d=\gcd(a+b,b)\Rightarrow d\mid(a+b)~\wedge~d\mid b\Rightarrow d\mid a$, but $\gcd(a,b)=1\Rightarrow d=1$.  
Proving lemma 2:
Let $d=\gcd(a,bc)\Rightarrow d\mid a~\wedge~d\mid bc\Rightarrow d\mid ac~\wedge~d\mid bc\Rightarrow d\mid\gcd(ac,bc)\Rightarrow d\mid c\cdot\gcd(a,b)\Rightarrow d\mid c$.
Then $d\mid a~\wedge~d\mid c\Rightarrow d\mid\gcd(a,c)\Rightarrow d\le\gcd(a,c)$.  
Let $d'=\gcd(a,c)\Rightarrow d'\mid a~\wedge~d'\mid c\Rightarrow d'\mid a~\wedge~d'\mid bc\Rightarrow d'\mid\gcd(a,bc)\Rightarrow d'\le\gcd(a,bc)$.  
Then $\gcd(a,bc)\le\gcd(a,c)~\wedge~\gcd(a,c)\le\gcd(a,bc)\Rightarrow\gcd(a,bc)=\gcd(a,c)$.  
Proving the exercise:
Let's prove this: $\forall a,b\in\mathbb{Z},~\gcd(a,b)=1\Rightarrow\gcd(a+b,a^2+b^2-nab)\mid(n+2)$  
Let $d=\gcd(a+b,a^2+b^2-nab)$
$\Rightarrow d\mid(a+b)~\wedge~d\mid(a^2+b^2-nab)$
$\Rightarrow d\mid(a+b)^2~\wedge~d\mid(a^2+b^2-nab)$
$\Rightarrow d\mid\Big((a+b)^2-(a^2+b^2-nab)\Big)$
$\Rightarrow d\mid(a^2+b^2+2ab-a^2-b^2+nab)$
$\Rightarrow d\mid ab(n+2)$  
Suppose $d\mid a~\vee~d\mid b\Rightarrow d\mid ab\Rightarrow d\mid(a+b)~\wedge~d\mid ab\Rightarrow d\mid\gcd(a+b,ab)$
$\gcd(a+b,a)=1$ (Lemma 1)
$\gcd(a+b,ab)=\gcd(a+b,b)=1$ (Lemma 2, Lemma 1)
Then $d\mid1\Rightarrow d=1\Rightarrow d\mid(n+2)$.  
If $d\nmid ab\Rightarrow d\mid(n+2)$
Then the value of $d$ doesn't matter, $d\mid(n+2)$.  
In this case; $d=\gcd(a+b,a^2+b^2-ab),~n=1\Rightarrow d\mid(1+2)\Rightarrow d\mid3\Rightarrow d=1~\vee~d=3.$