# Factor $55 – 88 \sqrt{-2}$ as a product of primes in $\mathbb{Z}$

To solve this problem, I let $K = \mathbb{Q}(\sqrt{-2})$, and I thought to take the norm $$N(55 – 88 \sqrt{-2}) = 55^2 + 2 \cdot 88^2 = 18513 = 3^2\cdot11^2 \cdot 17$$ If $a \in \mathbb{Z}[\sqrt{-2}]$ is irreducible, then $N(a) = p^f$, where $a$ lies over the prime $p$ and $f$ is the inertia degree of $K$ over $p$. Since the congruences $x^2 \equiv -2 \pmod 3$ and $x^2 \equiv -2 \pmod{11}$ are easily seen to be solvable (since $x^2 \equiv m \pmod p$ is solvable if and only if $m^{\frac{p-1}{2}} \equiv 1 \pmod p$), the primes $3$ and $11$ should split in $\mathbb{Z}[\sqrt{-2}]$. It was pretty easy to figure out that $$3 = (1 + \sqrt{-2})(1 – \sqrt{-2})$$ and $$11 = (3 + \sqrt{-2})(3 – \sqrt{-2})$$ and $17$ also splits as $(3 + 2 \sqrt{-2})(3 – 2 \sqrt{-2})$. So none of the primes $3, 11,$ or $17$ have inertia. Let $\sigma: K \rightarrow K$ be the unique nonidentity automorphism which is determined by $\sigma(\sqrt{-2}) = – \sqrt{-2}$.

If $\alpha_1 = 1 + \sqrt{-2}, \alpha_2 = 3 + \sqrt{-2}, \alpha_3 = 3 + 2\sqrt{-2}$, what I’m pretty sure should happen is that $55 – 88 \sqrt{-2}$ should be equal to some unit in $\mathbb{Z}[\sqrt{-2}]$, times either $\alpha_1^2$ or $\sigma \alpha_1^2$, times either $\alpha_2^2$ or $\sigma \alpha_2^2$, times either $\alpha_3$ or $\sigma \alpha_3$.

But how do I figure out which combination is right?

#### Solutions Collecting From Web of "Factor $55 – 88 \sqrt{-2}$ as a product of primes in $\mathbb{Z}$"

$55 – 88\sqrt{-2} = 11(5 – 8\sqrt{-2})$ so $11=\alpha_2\sigma(\alpha_2)$ is a factor, rather than $\alpha_2^2$ or $\sigma(\alpha_2)^2$, and we only need to factor $5 – 8\sqrt{-2}$.

I think the best thing to do now is to try dividing by the possible factors and check if the result is in $\Bbb{Z}[\sqrt{-2}].$

$$\frac{5 – 8\sqrt{-2}}{1 + \sqrt{-2}} = \frac{(5-8\sqrt{-2})(1 – \sqrt{-2})}{3} = \frac{-11 – 13\sqrt{-2}}{3} \notin \Bbb{Z}[\sqrt{2}]$$

$$\frac{5 – 8\sqrt{-2}}{1 – \sqrt{-2}} = \frac{(5-8\sqrt{-2})(1 + \sqrt{-2})}{3} = \frac{21 – 3\sqrt{-2}}{3} = 7 – \sqrt{-2} \in \Bbb{Z}[\sqrt{2}]$$

Hence $\sigma(\alpha_1)$ is a factor and $\alpha_1$ is not so $\sigma(\alpha_1)^2$ must be a factor. Dividing by $\sigma(\alpha_1)^2$ we obtain

$$\frac{5 – 8\sqrt{-2}}{(1 – \sqrt{-2})^2} = 3 + 2\sqrt{-2} = \alpha_3$$

so $55 – 88\sqrt{-2} = \sigma(\alpha_1)^2\alpha_2\sigma(\alpha_2)\alpha_3$.