I'm reading Stillwell's marvelous book Elements of number theory. This is not a homework question, I'm studying it for myself. I solved the question below and want to ask if my solution is correct.
if $\:y^3=x^2+11=\left(x+\sqrt{-11}\right)\left(x-\sqrt{11}\right)$,use unique prime factorization in $\:\mathbb{Z}\left[\frac{1+\sqrt{-11}}{2}\right]$ to show that : $$ x+\sqrt{-11}=\left(a^3-33ab^2\right)+\left(3a^2b-11b^3\right)\sqrt{-11}$$ where $a,b$ are both integers or else both half-integers that are not integers
We know that $\mathbb{Z}\big[\frac{1+\sqrt{-11}}{2}\big]$ is UFD. Thus, if we could prove that $\operatorname{gcd}(x+\sqrt{-11},x-\sqrt{-11})=1$ in $\mathbb{Z}\big[\frac{1+\sqrt{-11}}{2}\big]$ when $x$ and $y$ satisfy $y^3=x^2+11$, then it will follow from UFD property of $\mathbb{Z}\big[\frac{1+\sqrt{-11}}{2}\big]$ and the fact that $x+\sqrt{-11}$ divides a cube, that $x+\sqrt{-11}$ is a cube $$ x+\sqrt{-11}=(a+b\sqrt{-11})^3=a^3-33ab^2+\sqrt{-11}(3a^2b-11b^3). $$
If integers $x$ and $y$ satisfy $y^3=x^2+11$, then $x$ can not be odd, otherwise, $y$ would be even and we would get $$ 8m^3=4n^2+4n+1+11=8\frac{n^2+n}{2}+12\Rightarrow 0=4 \, (\mod 8), $$ contradiction.
$d=\operatorname{gcd}(x+\sqrt{-11},x-\sqrt{-11})$ would divide the norm $x^2+11$ and the difference $x+\sqrt{-11}-(x-\sqrt{-11})=2\sqrt{-11}$, whose norm is $44$. However $$ \operatorname{gcd}(x^2+11,44)=1\quad \textrm{in}\quad \mathbb{Z}. $$ This is because $x^2+11$ is odd and $x$ can not be divisible by $11$ otherwise $y$ would be divisible by 11 and $$ 11^3m^3=11^2n^2+11\Rightarrow 0=11 (\mod 11^2), $$ contradiction.
Thus $d=1$.
Is this reasoning correct?