Ideal class group of $K = \mathbb{Q}[x]/ (x^{3} + 11x + 21)$ In this question, I asked about some non-cyclic and hand-doable example of ideal class group, and Yong Hao Ng the example with minimal discriminant is $K = \mathbb{Q}(\alpha)$ with $\alpha^{3} + 11\alpha + 21 = 0$. With help of SAGE and some tedious computations, I proved that the class group is generated by two ideal classes $[(3, \alpha-1)]$ and $[(3, \alpha)]$ and both have order 2. 
To show that $(3, \alpha - 1)$ is not a principal ideal, I proved that $\epsilon = \alpha^{2}- \alpha - 4$ is a fundamental unit of $K$ by using Artin's inequality. Then if $(3, \alpha - 1)=(\beta)$ for some $\beta\in \mathcal{O}_{K} = \mathbb{Z}[\alpha]$, $(\alpha+2) = (3, \alpha - 1)^{2} = (\beta^{2})$ implies that $v(\alpha + 2) = \beta^{2}$ for some unit $v\in U_{K} =\mathcal{O}_{K}^{\times}$. 
We can assume $v = 1$ or $v = \epsilon$, and for example, if $\alpha + 2 = \beta^{2} = (a\alpha^{2} + b\alpha +c)^{2}$, then by expanding this, we should have $c^{2} + 42ab = 2$, which is impossible by viewing the equation mod 3. Similarly, we can show that there's no $\beta\in \mathbb{Z}[\alpha]$ s.t. $\beta^{2} = \epsilon(\alpha + 2)= \alpha^{2}- 17\alpha - 29$ (by mod 7).
I can also prove that $(3, \alpha)$ is not a principal ideal since $(3, \alpha)(13, \alpha + 3) = (\alpha + 3)$ and $(13, \alpha + 3)$ is not a principal ideal by the similar method. 
To prove that the class group is isomorphic to Klein 4-group, the only thing we have to do now is $[(3, \alpha -1)] \neq [(3, \alpha)]$, which is equivalent to that $\mathfrak{p}_{3}\mathfrak{p}_{3}' = (3, \alpha -1)(3, \alpha)$ is not a principal ideal. 
However, I can't prove this by the above method, since I can't find any contradiction.
We have $(\mathfrak{p}_{3}\mathfrak{p}_{3}')^{2} = (-\alpha-1)$, so we only need to show that $(-\alpha - 1)v =\beta^{2}$ is not possible for any $v\in U_{K}$ and $\beta\in \mathcal{O}_{K}$. 
As before, $-\alpha - 1 = \beta^{2}$ doesn't have a solution by observing the coefficient of $\beta^{2}$. (-1 is not a square mod 3). 
However, I can' prove that $(\alpha^{2} - \alpha - 4)(-\alpha - 1) = 16\alpha + 25$ is not a square in $\mathcal{O}_{K}$. This is equivalent to show that the following system of Diophantine equation
$$
b^{2} + 2ac - 11a^{2} = 0 \\
2bc + 22ab - 21a^{2} = 16\\
c^{2} + 42ab = 25
$$
doesn't have any integer solution $(a, b, c)$. Clealy, the last equation doesn't give any contradiction modulo any prime $p$. 
So I try to find other nontrivial relation among ideal classes using SAGE, but I failed to find any one of them that help me to prove that the above ideal is not principal. Also, I tried $v = \epsilon^{2k+1}$, instead of $\epsilon$, and nothing gives a contradiction. 
Also, it is enough to show that one of the following (prime) ideals
$$
\mathfrak{p}_{3}'' = (3, \alpha + 1) \\
\mathfrak{p}_{17} = (17, \alpha - 2) \\
\mathfrak{p}_{23} = (23, \alpha - 8) \\
\mathfrak{p}_{29} = (29, \alpha + 4) \\
\mathfrak{p}_{29} = (29, \alpha  - 10)
$$
are not principal. But I failed for every above ideal. Do you have any idea? Thanks in advance. 

Edit: I just find a way that I can't do without SAGE. If $\sqrt{16\alpha+25} = \beta\in K$, then $\beta$
$$
\left(\frac{\beta^{2}-25}{16}\right)^{3} + 11\left(\frac{\beta^{2} - 25}{16}\right) + 21 = \frac{\beta^{6} - 75\beta^{4} + 4691\beta^{2} - 9}{4096} = 0.
$$
However, according to SAGE, the polynomial $x^{6} - 75x^{4} + 4691x^{2} - 9$ is irreducible over $\mathbb{Q}$, so the degree of $\beta$ is 6, which contradicts to $\beta\in \mathbb{Q}(\alpha)$.  However, I think I can't show that the degree 6 polynomial is irreducible (and even it is hard to compute it by hands.)
 A: As you observed, $(\mathfrak{p}_{3}\mathfrak{p}_{3}')^{2} = (\alpha+1)$, so to prove $\mathfrak{p}_{3}\mathfrak{p}_{3}'$ is not principal, it suffices to show none of $\pm(\alpha+1)$ and $\pm \epsilon(\alpha+1)$ is a square in $\mathcal{O}_K$, where $\epsilon = \alpha^{2}- \alpha - 4$ is the fundamental unit you identified. There is a very efficient way to do this:
Let $f(x)=x^3+11x+21$, then $101\mid f(6)$. Since the ring of integer is monogenic, we have a homomorphism: $$\mathcal{O}_K \to \mathbb{F}_{101} \qquad \alpha \mapsto 6$$ under this, $\pm(\alpha+1)$ is mapped to $\pm 7$, but it is easy to check $\pm 7$ is not a quadratic residue modulo $101$. Hence $\pm(\alpha+1)$ is not a square in $\mathcal{O}_K$. The same prime $101$ does not work for $\pm \epsilon(\alpha+1)$, because they are mapped to $\pm (6^2-6-4)(6+1)$, both are quadratic residue of $101$.
Therefore we try another prime. Note that $29\mid f(10)$, use the homomorphism: $\mathcal{O}_K \to \mathbb{F}_{29}, \alpha \mapsto 10$, $\pm \epsilon(\alpha+1)$ is mapped to $\pm (10 + 1) (10^2 - 10 - 4)$ which are quadratic nonresidue modulo $29$. 
A: I think you can show directly that $\alpha\not\in (3,\alpha-1)$ so $(3,\alpha)$ and $(3,\alpha-1)$ cannot be equal.  
Suppose instead that $\alpha \in (3,\alpha-1)$, such that
$$
\alpha = 3\cdot g(\alpha) + (\alpha-1) \cdot h(\alpha)
$$
for some $g(\alpha),h(\alpha) \in \mathcal O_K = \mathbb Z[\alpha]$. Then letting $h(\alpha) = a + b\alpha + c\alpha^2$ and $a,b,c \in\mathbb Z$ we have
$$
\begin{align}
\alpha &= 3\cdot g(\alpha) + (\alpha-1)(a+b\alpha + c\alpha^2)\\
&= (-a - 21 c) + (a - b - 11 c)\alpha + (b - c) \alpha^2 +3\cdot g(\alpha)\\
\alpha &\equiv -a + (a-b-11c)\alpha + (b-c)\alpha^2 \pmod 3
\end{align} 
$$
Therefore
$$
c \equiv b \pmod 3,\quad a\equiv 0 \pmod 3,
$$
giving
$$
\alpha \equiv (-b-11b)\alpha \equiv 0 \pmod 3
$$
which is a contradiction.  
Hence we cannot have
$$
(3,\alpha) = (3,\alpha-1)
$$
