The ideal $I=(3,1+\sqrt{-23})\subseteq \mathcal O_{\mathbb{Q}(\sqrt{-23})}$ Intro:
$K=\mathbb Q(\sqrt{-23})$ be a number field with obvious minimum polynomial.
$\mathcal O_K$ be its ring of integers which is determined as $$\mathcal O_K=\mathbb Z\left[\frac{1+\sqrt{-23}}{2}\right]$$
since $-23\equiv 1 \mod 4$
I want to determine if $I,I^2,I^3$ are principal or not.
I was able to calculate norm of $I$ as: $N(I)=3$.
Then using basic ideas I can show that $I$ is not principal.
$N(I^2)=9=\left(x+y/2\right)^2+\frac{23}4y^2=3^2$
Only solution is $x=\pm 3,y=0$
Why we can't say that $I^2$ is generated by the ideal $(3)$?
I want to show $I^3$ is principal ideal.
Norm is multiplicative so $N(I^3)=27$
So $N(I^3)=27=\left(x+y/2\right)^2+\frac{23}4y^2=3^3$
Has following integer solutions $(x,y)=(-3,2),(-1,2),(1,2),(3,-2)$
And I am stuct to find which one is appropriate candidate, which kind of theorem says/guarantees that?
 A: $w\!=\!\sqrt{-23},\ I\! =\! \overbrace{(3,w\!-\!2)}^{\textstyle (3,w\!+\!1)}$ $\Rightarrow \color{darkorange}{3^3}\!\in I^3$ $\Rightarrow\overbrace{\color{c00}{(\color{#c00}{w\!-\!2})(\color{#0a0}{w\!+\!2})}\in I^3}^{\textstyle w^2\!-\!4 = \color{darkorange}{-3^3\ \ \ [*]}},\:$ $\overbrace{(\color{#0a0}{w\!+\!2},I^3)\!=\!1}^{{\textstyle  (w\!+\!2,I)\!=\!1}}$ $\overbrace{\Rightarrow \color{#c00}{w\!-\!2\in I^3}}^{\color{#0af}{\rm EL=}\text{Euclid's Lemma}\!\!}$ using EL. $ $ Hence $\,I^3=((3,w\!-\!2)^3,\color{#c00}{w\!-\!2})^{\phantom{|^{|^|}}}\!\!\! = (3^3,w\!-\!2) \overset{\color{darkorange}{[*]}}= (w\!-\!2)\,$ [so $I^2$ is not principal, else $I^3 = I I^2\Rightarrow I^{\phantom{|^{|}}}\!\!\!$ principal]. $\ \small\bf QED$
A: This is a very famous question. Repeated many times. Every time I see it, I think that I don't know the solution! I am an amateur number theorist. This is neither a complete nor systematic solution. If you see mistakes please comment. I wonder if there is a systematic solution for such questions.
Let $x=\frac{1+\sqrt{-23}}{2}$ then, $x^2=x-6$.
Let $I=(3,1+\sqrt{-23})$. Then,
$$I=(3,2x)=(3,-2x+3x)=(3,x).$$
The norm of $I$ is $N(I)=|\Bbb{Z}[x]/I|=|\Bbb{Z}_3|=3$.
For any number $a+bx\in\Bbb{Z}[x]$, $N(a+bx)=a^2+ab+6b^2$. If $I=(\alpha)$ were principal then $N(I)=N(\alpha)=3.$ But this is impossible as the possible norms are:
$$0,1,6,8,9,12,16,18,23,24,25,26,...$$
So $I$ is not principal as OP told. Next,
$$I^2=(9,3x,x^2)=(9,3x,x+3)=(9,x+3).$$
Then, $N(I^2)=9.$ Norm of ideal trick doesn't work. Suppose $I^2=(\alpha)$ is principal. Then $(a+bx)\alpha=9$ and $(c+dx)\alpha=x+3$ for some integers $a,b,c,d.$ Hence, $(a^2+ab+6b^2)N(\alpha)=81$ and $(c^2+cd+6d^2)=18.$ From these equations, it is easy to deduce that $N(\alpha)=1.$ So, without loss of generality $\alpha=1$. But then, for some integers $a,b,c,d$, $(a+bx)9+(c+dx)(x+3)=1$ which gives
$9a+3c-6d=1$ which is impossible. So, $I^2$ is not principle. Next,
$$I^3=I.I^2=(27,3x+9,9x,x^2+3x)=(27,3x+9,4x-6)=(27,3x+9,x+12)=(27,x+12)=(2x-3).$$
For the last equality, we observe that
$$(-2x-1)(2x-3)=27,$$
$$(-x)(2x-3)=x+12,$$
$$2(x+12)-27=2x-3.$$
So $I^3$ is principal with norm $N(I^3)=N(2x-3)=27$.
A: $I^3=(b)\,$ by below: put $\,a\!=\!3,\, b= 2\!-\!\sqrt{-23},\,$ so $ \,b\!+\!\bar b=\color{#0af}4\,\Rightarrow\, \color{#0a0}{(a,b,\bar b)} = (3,b,\bar b,\color{#0af}4)=(1)$.
Lemma $\, $ If $\ \color{#c00}{(a^3) = (b\bar b)}\ $ then $\ (a,b)^3 = (b)\!\iff\!  \color{#0a0}{(a,b,\bar b)}=(1),\,$ by $\rm\color{#c00}{EL}$ = Euclid's Lemma.
Proof $\ \ (\color{#c00}a,b)^{\color{#c00}3}\! = \color{#c00}{(b)}(\color{#c00}{(\bar b)}\!+\!(a,b)^2) = (b)\!\iff\!$ $(\bar b)\!+\!(a,b)^2\! = (1)\!\!\!\overset{\color{#c00}{\rm EL}\!\!}\iff\!$
$\color{#0a0}{(\bar b)\!+\!(a,b)}=(1)$
