<s>[INST] Factoring an element in $\mathbb{Z}[\sqrt6]$ I want to know if $1+\sqrt6 $ factors in  $\mathbb{Z}[\sqrt6]$.
$1+\sqrt6=(a+b\sqrt6)(c+d\sqrt6)$ implies $a^2-6b^2\not=0,c=\frac{a-6b}{a^2-6b^2},a\not=0, d=\frac{a-b}{a^2-6b^2}$
How do I determine if there are integer solutions? 
 [/INST] Use the multiplicativity of the norm : if you define
$N(a+b\sqrt{6})=a^2-6b^2$, then $N(w_1w_2)=N(w_1)N(w_2)$ for any $w_1,w_2$.
So if $1+\sqrt{6}=(a+b\sqrt{6})(c+d\sqrt{6})$, then 
$-5=(a^2-6b^2)(c^2-6d^2)$, so that one of $a^2-6b^2$ or $c^2-6d^2$ is $\pm 1$.
If, say, $a^2-6b^2= \pm 1$, then $a+b\sqrt{6}$ is a unit in ${\mathbb Z}[\sqrt{6}]$.
So there are no interesting factorizations.
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