Finding the inverse elements of $\mathbb{Z}[i] = \{ a + ib | a,b \in \mathbb{Z} \}$

Question

So I am asked to find the inverse elements of this set $\mathbb{Z}[i] = \{ a + ib | a,b \in \mathbb{Z} \}$ (I know that this is the set of Gaussian integers).

I was pretty much doing the same thing the correction suggested. Suppose $x = a+ib \in \mathbb{Z}[i]$ and $y = a' + ib' \in \mathbb{Z}[i]$. We suppose that $y$ is the inverse of $x$, that is $xy=1 \iff |x|^2 |y|^2 =1$ thus $(a^2 + b^2)(a'^2 + b'^2)=1$. At this point I got stuck a bit, and read the correction, which stated that $a^2 + b^2$ has an inverse in $\mathbb{N}$, and I am unable to understand why is that the case?

Answer 1

Let $c=a^2+b^2$ and $c'=a'^2+b'^2$. Then $c$ and $c'$ are both non-negative integers, and $cc'=1$. So the non-negative integer $c$ has an "inverse" (reciprocal) in $\Bbb N$ namely $c'$. What must $c$ (and $c'$) be?

Answer 2

Hint

Since $(a^2+b^2)(a'^2+b'^2)>1$ if $a,b,c,d\notin \{0,1\}$ it should be easy to find the solutions...