A question on non-principalness of ideal $\langle 3, 1 + \sqrt{223} \rangle \subset \textbf Z[\sqrt{223}]$ Consider the ideal $I = \langle 3, 1 + \sqrt{223} \rangle \subset \textbf Z[\sqrt{223}]$. 
Q: How do I show $I$ is non-principal? This is related to an exercise in Number Fields by Daniel Marcus. I do not think there should be any advanced machinery showing up. 
I tried the following. Consider $I = \langle a \rangle$ for some $a \in \textbf Z[\sqrt{223}]$. The ideal norm gives $3$ for both. Then $x^2 - 223y^2 = -3$ has a solution by $a$. Quadratic residue by modding out $223$ or $3$ does not help. I am aware there is the following links. However $-3$ is never showing up in a trivial way.
Show that $\mathbb{Z}[\sqrt{223}]$ has three ideal classes.
Or by continued partial fractions. (This is not used in Marcus's Number Fields for sure. I guess there should be a much naïve way of solving this.)
How to prove there are no solutions to $a^2 - 223 b^2 = -3$.
Or A diophantine equation
 A: Here is an algorithm for solving this type of problem. It is not in Marcus, but then I think Marcus makes some assumptions as to background, I dont know what he has in mind as a solution, but at least this method is simple and general. It is related to continued fractions and ultimately quadratic forms. 
${\bf Theorem.}$The solutions of $x^2-dy^2=N$ where $|N|<\sqrt{d}$ are given by 
$$p_n^2-dq_n^2=(-1)^{n-1}a_{n+1}$$ where $\frac{p_n}{q_n}$ are the continued fraction approximations. 
To find the continued fraction expansion Build a sequence of triples $(a_n,b_n,c_n)$ such that $b_n^2-4a_nc_n=D$. 
In the case of $d=223$, we have $D=4\cdot 223=892$. The $a_n$ will give the possible values for $N$.
Further 
1)$a_{n+1}=-c_n$
2)$2c_n|b_n+b_{n+1}$
3) $\sqrt{D}-2|c_n|<b_{n+1}<\sqrt{D}$
So in the case of $d=223$ here is the calculation.
$\sqrt{892}=29.8\cdots$
Start with
$$(a_0,b_0,c_0)=(223,0,-1).$$
To get the next triple, set, 
$$(223,0,-1)(1,b_1,c_1)$$
we must have $2|0+b_1$ and $27.8<b_1<29.8$ so $b_1=28$ and since $D=b_1^2-4a_1c_1$, $c_1=27$.
To get the next one, 
$$(-1,28,27)(-27,b_2,c_2)$$
 and $2\cdot 27| 28+b_2$ with $0\leq b_2<29.8$ , thus $b_2=26$.
The complete sequence is 
$$(223,0,-1)$$
$$(-1,28,27)$$
$$(-27,26,2)$$
$$(-2,26,27)$$
$$(-27,28,1)$$
$$(-1,28,27)$$
thus the last is the same as the second, so we have repeat. Thus we see that the only numbers $\leq 14$ that are represented are $223, 1, 27, 2$ so 
only $1,2$ amongst the candidate numbers are represented and split into principal ideals. Thus $3,11,13$ split into non principal ideals.
If further you calculate the differences $\delta_n=\frac{b_n+b_{n+1}}{2|c_n|}$ you get the continued fraction expansion, here we have 
$$\delta_0=14$$ 
$$\delta_1=1$$ 
$$\delta_2=13$$ 
$$\delta_3=1$$
$$\delta_4=28$$ giving 
$$\sqrt{223}=[14,\overline{1,13,1,28}]$$  
${\bf Additional \ Explanation:}$ If $\sqrt{d}=[k_0, k_1, k_2, \cdots ]$ is a continued fraction expansion, and let 
$$\tau_n=[k_n, k_{n+1}, k_{n+2}, \cdots ]$$
Then $$\tau_n=\frac{b_n+\sqrt{D}}{2c_n}$$ where $c_n,b_n$ are the numbers above. 
A: Assume there exists $\alpha$ such that $\alpha \cdot \bar{\alpha} =\pm 3$. 
Consider a unit  $\omega = 224 + 15 \sqrt{223}$, (  a fundamental unit ). There exists an integer $k$ such that 
$|\omega^k|< |\alpha| < |\omega|^{k+1}$. Substituting $\alpha$ with $\frac{\alpha}{\omega^k}$ we may assume $1< |\alpha| < |\omega|$. We conclude $\frac{3}{\omega}<|\bar{\alpha}|<3$. If $\alpha = a + b \sqrt{223}$ we get 
$$|a + b \sqrt{223}| < \omega\\
|a- b\sqrt{223}|  <  3$$
From the above we get $|a|,\sqrt{223}|b| < \frac{\omega+3}{2}$. This reduces the problem to a finite check, that can done by computer. 
Note: This method uses the different embeddings of the field $\mathbb{Q}(\sqrt{223})$ into $\mathbb{R}$. In general, for a number field $K$ and an order of $K$ one can find in this way elements of a prescribed norm, if we have a fundamental system of units ( or at least a system of maximal rank). This is treated nicely in the book Number Theory by Borevich and Shafarevich. 
