Exercise 1.2.50* (Value of $(a+b, a^2 -ab + b^2)$)

Proof. Write $d=a+b\wedge a^2-\mathit{ab}+b^2$.

Then $d\mid a+b$ and $d\mid a^2-\mathit{ab}+b^2$. Therefore $b\equiv -a(\mathit{mod}d)$. Hence

so

By hypothesis, $a\wedge b=1$, thus $a\wedge a+b=1$, and $a^2\wedge a+b=1$.

Since $d\mid a+b$, then $d\wedge a^2=1$ (for every integer $c$, if $c\mid d$ and $c\mid a^2$, then $c\mid a+b$, so $c\mid a^2\wedge a+b=1$).

From $d\mid 3a^2$ and $d\wedge a^2=1$, we deduce $d\mid 3$, where $d\ge 0$.

This shows that $d=1$ or $d=3$. □