<s>[INST] The only natural number $x$ for which $x+\sqrt{-2}$ is a cube in $\mathbb{Z}[\sqrt{-2}]$ is $x=5$ Let $A =  \mathbb{Z}[\sqrt{-2}]= \{a+b\sqrt{-2} \ : a, b \in \mathbb{Z}\}$. Show that the only natural number $x$ for which $x+\sqrt{-2}$ is a cube in $A$ is $x=5$. 
So I have to show that there exists $c,d \in \mathbb{Z}$ such that $(x+\sqrt{-2})^3= c + d \sqrt{-2}$. From the last equation, I obtain $x(x^2-6) = c \in \mathbb{Z}$ and $3x^2-2= b \in \mathbb{Z}$, and here I blocked. Could I conclude from both equations? If not, is there exists another way to proceed (hint)?
Thanks!
 [/INST] Directly
$$(a+b\sqrt{-2})^2(a+b\sqrt{-2})=(a^2-2b^2+2ab\sqrt{-2})(a+b\sqrt{-2})=$$
$$=(a^3-6ab^2)+(3a^2b-2b^2)\sqrt{-2}$$
We get that it must be
$$3a^2b-2b^2=1\iff b(3a^2-2b)=1\iff b,\,3a^2-2b=\pm 1$$
But $b=-1\implies 3a^2-2(-1)=-1\implies3a^2=-3$ , impossible, so it must
$$\;b=1\implies3a^2-2=1\iff 3a^2=3\implies a^2=1\iff a=\pm1\;$$
and thus we want $\;\pm1+\sqrt{-2}\;$ , so that for example
$$(1+\sqrt{-2})^3=1+3\sqrt{-2}-6-2\sqrt{-2}=-5+\sqrt{-2}$$
But $\;-5\notin\Bbb N\;$ , so let us try the other option with $\;a=-1\,,\,\,b=1\;$ :
$$(-1+\sqrt{-2})^3=-1+3\sqrt{-2}+6-2\sqrt{-2}=5+\sqrt{-2}$$
and this time $\;x=5\in\Bbb N\;$ .
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