If $(a,b) = 1$, show that $(a-b,a^2+ab+b^2)=1 \text{ or } 3$ I have a question...We know that $(a,b)=1$ and we want to show that $(a-b,a^2+ab+b^2)=1 \text{ or } 3$.How can I show this??
I thought that we could suppose that $(a-b,a^2+ab+b^2)=d$.Then we know that $d|a-b$ and $d|,a^2+ab+b^2$.But how can I continue??
 A: Hint $\,\ (a\!-\!b,a^2\!+ab+b^2)\overset{\color{#c00}{(1)}} = (a\!-\!b,\color{#c0f}{3b^2})\overset{\color{#0a0}{(2)}} = (a\!-\!b,3),\ $ by $\ \color{#0a0}{(2)\!:\ }(a\!-\!b,b) = (a,b) = 1,\ $ and by 
$\, \color{#c00}{(1)\!:}\, \ (a\!-\!b,c) = (a\!-\!b,d) \,$ if $\,c\equiv d\pmod{\!a\!-\!b},\,$ and $\,{\rm mod}\ a\!-\!b\!:\ a\equiv b\,\Rightarrow\, a^2\!+ab+b^2\equiv \color{#c0f}{3b^2}$
A: Sometimes it is useful to try dividing these things out. So here (treating the quadratic as a polynomial in $a$ and using the division algorithm for polynomials) $$a^2+ab+b^2=(a-b)(a+2b)+3b^2$$
you should be able to make some progress from there ...
A: $d|a-b$,$d|a^2+ab+b^2$ and $a^2+ab+b^2=(a-b)^2+3ab$ imply that
$$d|3ab,d|3a(a^2+ab+b^2)=3a^3+3a^2b+3ab^2,d|3ab(a-b)=3a^2b-3ab^2,d|2a(3ab).$$
Hence, $$d|3a^3+3a^2b+3ab^2+3a^2b-3ab^2-6a^2b=3a^3$$ and similarly $d|3b^3$ and now 
$$d\le(3a^3,3b^3)=3(a^3,b^3)=3.$$ so $d=1\,\text{or}\,2\,\text{or}\,3$, but $d$ can't be $2$, since in this case $a^3$ and $b^3$ must be two even numbers and $(a^3,b^3)\neq1$. (Note that $(a,b)=1$ implies $(a^3,b^3)=1$) 
