<s>[INST] 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??
 [/INST] $d|a-b$,$d|a^2+ab+b^2$ 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 $(a^3,b^3)=1$) 
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