Multiplicative inverse of a gaussian integer

Question

So, I had to prove that Gaussian integers had an identity element and ended up with it being $(1 + 0i)$. Now I have to see if any $(a+bi)$ (except $(0 + 0i)$)has a multiplicative inverse. Then I ended up with this: \begin{align*} ac-bd &= 1\\ ad + bc &= 0\\ \end{align*} But I don't really know if I can prove that for any given $a,b\in\mathbb{Z}$ I can find $c,d\in \mathbb{Z}$ such that that happens. Am I on the wrong path? Can I really prove it?

Answer 1

Hint :

1. Consider the application $N : a+ib \mapsto a^2+b^2$

2. Show that for every $a+ib$ and $c+id$, one has $N((a+ib)(c+id))=N(a+ib)N(c+id)$.

3. Deduce that if $a+ib$ has a multiplicative inverse, then $N(a+ib)=\pm 1$.

4. Deduce all the Gaussian integers which have a multiplicative inverse.