Factor $55 - 88 \sqrt{-2}$ as a product of primes in $\mathbb{Z}[\sqrt{-2}]$ 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?  

I think that only after doing several problems like this one do certain "obvious" basic facts really begin to feel truly obvious and basic.
The first such fact, which Zoe already mentioned, is that if for $a + b \sqrt d$ with $a$ and $b \in \mathbb Z$ we have $\gcd(a, b) > 1$, then $a + b \sqrt d$ is divisible by some purely real rational integer. In this case that would be 11. Then, like Zoe already said, $$\frac{55 - 88 \sqrt{-2}}{11} = 5 - 8 \sqrt{-2}.$$
The second "obvious" fact is that since we're working in an imaginary quadratic ring, smaller $a$ and $b$ guarantee a smaller norm. So $N(5 - 8 \sqrt{-2}) = 153$, which is more manageable than $N(55 - 88 \sqrt{-2}) = 18513$.
The two important "obvious" facts specific to $\mathbb Z[\sqrt{-2}]$ are first that there are only two units, 1 and $-1$, and that purely real positive integers congruent to 1 or $3 \pmod 8$ are composite in $\mathbb Z[\sqrt{-2}]$ even if prime in $\mathbb Z$.
So, as you already figured out, 3, 11, 17 all split. Obviously, going by the norms, $$(1 - \sqrt{-2})^2 (1 + \sqrt{-2})^2 (3 - 2 \sqrt{-2})(3 + 2 \sqrt{-2}) = 153.$$ This means six prime factors (with some repetition).
The problem now is: which of those factors do you delete to get $5 - 8 \sqrt{-2}$? Dividing, we get $$\frac{153}{5 - 8 \sqrt{-2}} = 5 + 8 \sqrt{-2}.$$ Oops, sorry, that was no help.
One of the "obvious" general facts is telling us that the combination we're looking for can't have conjugates. So we must remove either $(1 - \sqrt{-2})^2$ or $(1 + \sqrt{-2})^2$, and either $3 - 2 \sqrt{-2}$ or $3 + 2 \sqrt{-2}$.
Instead of complicating your life with Greek letters, since we're dealing with numbers of small norm, why not just try multiplying a few combinations and see what you get?
$$(1 - \sqrt{-2})^2 (3 - 2 \sqrt{-2}) = -11 - 4 \sqrt{-2}$$
Wrong quadrant of the complex plane.
$$(1 - \sqrt{-2})^2 (3 + 2 \sqrt{-2}) = 5 - 8 \sqrt{-2}$$
Ding, ding, ding!
Of course it could very well have happened that I tried all the wrong combinations before hitting on the right one. Then I would have looked even less clever. But looking clever has never been my primary concern.
