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Here's how to find all factorizations of $-134 + 26\sqrt{-5}$ in $\mathbf{Z}[\sqrt{-5}]$, using some more theory.

$\mathbf{Z}[\sqrt{-5}]$ is the ring of integers of $\mathbf{Q}(\sqrt{-5})$ and hence it is a Dedekind domain.

We begin by factoring the principal ideal $I = (-134 + 26\sqrt{-5})$ into primes.

Note that $I$ contains the integer $N(-134 + 26\sqrt{-5}) = 21336 = 2^3\cdot 3 \cdot 7 \cdot 127$, so $(21336) = IJ$ for some $J$, and hence the prime factors of $I$ are a subset of those of $(21336)$.

By the Kummer-Dedekind theorem, the primes $2, 3, 7$ and $127$ split into primes in $\mathbf{Z}[\sqrt{-5}]$ as follows:

$2 = (2, 1 + \sqrt{-5})^2$,

$3 = (3, 1 + \sqrt{-5})(3, 2 + \sqrt{-5})$,

$7 = (7, 3 + \sqrt{-5})(7, 4+\sqrt{-5})$, and

$127 = (127, 54 + \sqrt{-5})(127, 73 + \sqrt{-5})$.

To determine if one of these primes $\mathfrak{p} = (p, r + \sqrt{-5})$ divides $I$, we check whether $I \subset \mathfrak{p}$. That is, whether $-134 + 26\sqrt{-5} = px + (r + \sqrt{-5})y$ has any solutions $x,y \in \mathbf{Z}[\sqrt{-5}]$. Writing $x = a + b\sqrt{-5}$ and $y = c + d\sqrt{-5}$, the equation becomes $$-134 = pa + rc - 5d, \quad 26 = pb + rd + c,$$ which reduces to $-134 - 26r = pa - prb - (r^2 + 5)d$ (using $c = 26 - pb - rd$). This equation has solutions if and only if $134 + 26r \in (p, r^2 + 5) = (p)$. The last equality holds because $r$ was obtained from the Kummer-Dedekind theorem. In conclusion, we have to check whether $134 + 26r \equiv 0 \pmod p$.

For example, $(3, 1 + \sqrt{-5})$ does not divide $I$ because $160 \equiv 1 \pmod 3$, and $(3, 2 + \sqrt{-5})$ does divide $I$ because $186 = 3\cdot 62$. In this way we find that the primes dividing $I$ are $(2, 1 + \sqrt{-5})$, $(3, 2 + \sqrt{-5})$, $(7, 4 + \sqrt{-5})$ and $(127, 73 + \sqrt{-5})$.

Now we compute $\operatorname{ord}_\mathfrak{p}(I)$ for these primes. From the equality $(21336) = IJ$ it follows that for $\mathfrak{p}$ one of $(3, 2 + \sqrt{-5}), (7, 4 + \sqrt{-5})$ and $(127, 73 + \sqrt{-5})$ we have $\operatorname{ord}_\mathfrak{p}(I) = 1$. For $\mathfrak{p} = (2, 1 + \sqrt{-5})$ we know that $1 \leq \operatorname{ord}_\mathfrak{p}(I) \leq 3$, and since $\mathfrak{p}^2 = (2)$, we can see that $\mathfrak{p}^3 = (4, 2 + 2\sqrt{-5})$ divides $I$, as $-134 + 26\sqrt{-5} = 13\cdot(2+2\sqrt{-5}) - 40\cdot 4$.

Hence the factorization of the ideal $I = (-134 + 26\sqrt{-5})$ into primes in $\mathbf{Z}[\sqrt{-5}]$ is given by $$I = (2, 1 + \sqrt{-5})^3(3, 2 + \sqrt{-5})(7, 4 + \sqrt{-5})(127, 73 + \sqrt{-5}).$$

Since all these primes are non-principal and $\mathbf{Q}(\sqrt{-5})$ has class number $2$, any factorization of $I$ into principal ideals must be such that every principal ideal is a product of an even number of these primes.

The proper partitions of $6$ (the number of primes counted with multiplicity) into even numbers are $6 = 2 + 2 + 2$ and $6 = 4 + 2$. The latter partition corresponds to factorizations into reducible elements, so we ignore it.

You can compute the ${4 \choose 2} + 1 = 7$ different products of two of these primes; they are

$\mathfrak{p}_2\mathfrak{p}_{3} = (1 - \sqrt{-5})$,

$\mathfrak{p}_2\mathfrak{p}_{7} = (3 - \sqrt{-5})$,

$\mathfrak{p}_2\mathfrak{p}_{127} = (3 + 7\sqrt{-5})$,

$\mathfrak{p}_3\mathfrak{p}_{7} = (1 + 2\sqrt{-5})$

$\mathfrak{p}_3\mathfrak{p}_{127} = (19 + 2\sqrt{-5})$,

$\mathfrak{p}_7\mathfrak{p}_{127} = (22 + 9\sqrt{-5})$, and

$\mathfrak{p}_2^2 = (2)$.

Observe (by looking at $\mathfrak{p}_2$) that any factorization of $I$ into principal ideals must either be of the form $\mathfrak{p}_2^2 (\mathfrak{p}_2\mathfrak{p}_i) (\mathfrak{p}_j\mathfrak{p}_k)$ or of the form $(\mathfrak{p}_2\mathfrak{p}_i)(\mathfrak{p}_2\mathfrak{p}_j)(\mathfrak{p}_2\mathfrak{p}_k)$. Hence all the factorizations of $I$ into principal ideals are $\mathfrak{p}_2^2 (\mathfrak{p}_2\mathfrak{p}_3) (\mathfrak{p}_7\mathfrak{p}_{127}), \mathfrak{p}_2^2 (\mathfrak{p}_2\mathfrak{p}_7) (\mathfrak{p}_3\mathfrak{p}_{127}), \mathfrak{p}_2^2 (\mathfrak{p}_2\mathfrak{p}_{127}) (\mathfrak{p}_3\mathfrak{p}_7)$ and $(\mathfrak{p}_2\mathfrak{p}_3)(\mathfrak{p}_2\mathfrak{p}_7)(\mathfrak{p}_2\mathfrak{p}_{127})$.

Writing this out, we get four factorizations

$I = (2)(1 - \sqrt{-5})(22 + 9\sqrt{-5})$, $I = (2)(3-\sqrt{-5})(19 + 2\sqrt{-5})$, $I = (2)(3 + 7\sqrt{-5})(1+2\sqrt{-5})$, $I = (1-\sqrt{-5})(3-\sqrt{-5})(3 + 7\sqrt{-5})$.

As the only units in $\mathbf{Z}[\sqrt{-5}]$ are $\pm 1$, this leads (by choosing the right signs) to four factorizations of $-134 + 26\sqrt{-5}$ in $\mathbf{Z}[\sqrt{-5}]$:

$-134 + 26\sqrt{-5} = -2(1 - \sqrt{-5})(22 + 9\sqrt{-5})$, $-134 + 26\sqrt{-5} = -2(3-\sqrt{-5})(19 + 2\sqrt{-5})$, $-134 + 26\sqrt{-5} = 2(3 + 7\sqrt{-5})(1+2\sqrt{-5})$, $-134 + 26\sqrt{-5} = -(1-\sqrt{-5})(3-\sqrt{-5})(3 + 7\sqrt{-5})$.

Factorization number 3 is the one from the question. The factorizations into reducible elements (corresponding to the partition $6 = 4 + 2$) are obtained by multiplying out two of the three factors in the products above.