Number Theory: Quadratic Fields

Question

Suppose $32 =\alpha \beta $ for $\alpha,\beta $ relatively prime quadratic integers in Q[i]. Show that $\alpha= \varepsilon \gamma^2 $ some unit $\varepsilon $ and some quadratic integer $ \gamma $ in Q[i].

So i or -i is a unit in the field Q[i] but, I can't figure out two quadratic integers that multiply to $32$. I feel like the rest of the problem will become a lot easier if I can figure this out. I'm not even sure if that is the right direction to approach the problem but it's the path that I'm trying to pursue with it. If someone could help me out with this or help me figure out another direction to approach the problem that would be great! Thanks :)

Answer 1

It seems that the only admissible answer to this problem is a relatively trivial one.

The domain of the Gaussian integers bears a unique factorization theorem. According to this theorem, the unique factorization of $32$ is

$32=(-i)(1+i)^{10}$

and since this is composed by a single prime factor that means if $\alpha,\beta$ are to be relatively prime, then

$\alpha=\epsilon_1, \beta=\epsilon_2(1+i)^{10} ~~\text{or}~~\beta=\epsilon_1, \alpha=\epsilon_2(1+i)^{10}$

where $\epsilon_1,\epsilon_2$ are units endowed with the constraint $\epsilon_1\epsilon_2=-i$ which admits the solutions

$$(\epsilon_1,\epsilon_2)=\{(1,-i), (-1,i), (-i,1), (i,-1)\}$$

In any case one can trivially check that always $\alpha=\epsilon \gamma^2$ with the only possibilities for the quadratic integer being $\gamma=1,(1+i)^5$.

Answer 2

$\alpha=a+bi$ and $\beta=c+di$ give taking norms $$(a^2+b^2)(c^2+d^2)=32^2=2^{10}$$ It follows $$a^2+b^2=2^x\\c^2+d^2=2^{10-x}$$ Some easy calculation gives $$\alpha=2+2i\\\beta=8+8i$$ possibly with a unit factor but it is a known fact that the only units here are $\pm1,\pm i$. Also you have to consider possibly other obvious signes adding other solutions basically the same as given above.