Let $a$ and $b$ be relatively prime integers. Show that $\gcd(a+b,a^2-3ab+b^2)= 1$ or $5$ Let $a$ and b be relatively prime integers. Show that $\gcd(a+b,a^2-3ab+b^2)=1$ or $5$
Proof: $s\mid (a+b)$ and $s\mid (a^2-3ab+b^2)$ implies $s\mid(a+b)^2=a^2+b^2+2ab$ and $s\mid (a^2-3ab+b^2)$
implies $s\mid((a+b)^2-(a^2-3ab+b^2))=5ab$. 
From a previous question I posted I figure that the I am on the wrong track. Can someone help me with the next step?

Note that $$a^2-3ab+b^2=(a+b)^2-5ab$$ So if $p$ is a prime, and such that $p|(a+b)$ and $p|(a+b)^2-5ab$ then $p|5ab$. Now $p\not | ab$ since if $p|a$, say then from $p|a+b$ follows $p|b$ contradiction the fact the $a$ and $b$ are relatively prime therefore $p|5$. The same argument shows that $gcd =5$. 
