How to find examples of non-unique euclidean division in $\Bbb Z[i]$?

Question

The euclidean division of $1+8i$ by $2-4i$ can be one of two things:

$$1+8i=(2-4i)(i-1)-1+2i\tag{1}$$ $$\underbrace{1+8i}_\text{a}=\underbrace{(2-4i)}_\text{b}(\underbrace{-2+i}_\text{q})+\underbrace{1-2i}_\text{r}\tag{2}$$

And we have that $N(r)<N(b)$ where $N(a+bi)=a^2+b^2$ and the $b$'s aren't associated so those are really two different euclid divisions. (do the $r$'s have to be associated? here one is $(-1)$ times the other...)

What's the mindset in which we have to be to look for and find such examples?

Answer 1

You even have non-unique Euclidean division in the integers. Dividing $9$ by $4$ could be either:

$$ 9 = 4 \cdot 2 + 1 $$ $$ 9 = 4 \cdot 3 + (-3) $$

You pretty much always have these examples, except for the special case that the remainder is zero. In cases like $\mathbb{Z}$ and $\mathbb{Z}[i]$, given any one quotient-remainder pair, you can find others by adjusting the quotient by testing out the other numbers near the quotient.

I don't recall for sure how this all works out, but I think for $\mathbb{Z}[i]$, that on average you should expect there to be $\pi$ different remainders satisfying $N(r) < N(b)$.

Answer 2

$\mathbb{Z}[i]$ is a UFD, i.e. a unique factorisation domain. Therefore, whatever elements from $\mathbb{Z}[i]$ we divide, the factorisation is unique. Also, if you may be interested: wikipedia says that $\mathbb{Z}[e^{\frac{2\pi i}{n}}]$ is a unique factorisation domain for all $1\leqslant n \leqslant 22$ and not for $n=23$ (strange, isn't it?). We can compute $\mathbb{Z}[i]$ by taking $n=4$ (write exponent as cosine and sine). BTW, all euclidean domain are UFD's.