$\gcd( x+y, x^2 - xy + y^2) $ is 1 or 3 for coprime $x$, $y$

Question

How does one show that: if $\gcd( x, y) = 1$, then $\gcd( x+y, x^2 - xy + y^2) = 1\,{\rm or }\,3$.

Answer 1

Let $d=(x+y,x^2-xy+y^2)$

$$d \mid x+y \Rightarrow d \mid (x+y)^2$$ $$ d \mid x^2-xy+y^2 \Rightarrow d \mid (x+y)^2-3xy$$

$$d \mid (x+y)^2-((x+y)^2-3xy) \Rightarrow d \mid 3xy (*)$$

Let $(d,x)=m>1$,then it has a prime factor,let $q$.

$$q \mid x \\ q \mid d,d \mid x+y \Rightarrow q \mid x+y \\ \text{So, } q \mid x+y-x=y \\ \text{As } q \mid x \text{ and } q \mid y \Rightarrow q \mid (x,y)=1 \ \text{ ,that is a contradiction.} \\ \text{So,we conclude that } (d,x)=1$$

Similarly,we show that $(d,y)=1$

$$(*) \xrightarrow[(d,y)=1]{(d,x)=1} d \mid 3$$

Therefore,: $ d=1 \text{ or } d=3$.