Show that $n,1+\sqrt{11}\in\Bbb{Z}[\sqrt{11}]$ generate the whole ring if $n$ ends in $7$. I am starting to learn about rings and ideals in abstract algebra. I came across a textbook problem that I am having a lot of trouble solving:

Prove that for any positive integer $n$ ending in $7$, the ideal generated by $n$ and $1+\sqrt{11}$ in $\mathbb{Z}[\sqrt{11}]$ is trivial. In other words, that it is the same as the whole ring.

I would appreciate any help on this!
 A: $\!\!\bmod\overbrace{ 1\!+\!\sqrt{11}}^{\textstyle {\rm ideal}\ I}\!:\,\ \sqrt{11}\equiv -1\Rightarrow \overbrace{11\!\equiv\! \sqrt{11}^2\!\equiv 1}^{\textstyle \color{#80f}{10\equiv 0}}\ $ so $\ \overbrace{\color{#0a0}0\equiv 7\!+\!\color{#90f}{10}j}^{\textstyle n\in I}\equiv \color{#0a0}7\Rightarrow \overbrace{0\equiv 3(\color{#0a0}7)\!-\!2(\color{#90f}{10})\equiv\color{#c00}{\bf 1}}^{\textstyle \color{#c00}{\bf 1}\in I}$
Generally $\ w\in I\Rightarrow$ $\, \underbrace{N :=w\bar w \in I}_{\textstyle \text{Norm}(w)\in I},\,$ so $\,n\in I\Rightarrow I\supseteq (n,N) = (\color{#c00}{d}), \ \color{#c00}d=\gcd(n,N)$
i.e. the generator $\,w\,$ has a multiple (its norm $N$) that is simpler (an integer) so we can adjoin $N$ as a generator of $I,\,$ then we can combine all integer generators by taking their $\rm\color{#c00}{gcd}$, i,e, $$(w,n) = (w,w\bar w,n) = (w,N,n) = (w,(N,n)) = (w,\color{#c00}d)$$
This is a special case of the method of simpler multiples.  OP is special case $\,\color{#c00}{d = 1}$.
A: Hint: Note that the ideal also contains the element
$$(1+\sqrt{11})(1-\sqrt{11})=1-11^2=-10.$$
A: Assume $n$ is positive. Then we have that $n=10k+7$ for some $k \in \mathbb{Z}$ and $(1-\sqrt{11})(1+\sqrt{11})=-10$.
Thus it follows that $\gcd(n, 10)=1$, thus $1 \in I$.
