TITLE: $a+b\phi$ and $a+b-b\phi$ are associates in $\mathbb{Z}+\phi \mathbb{Z}$, with $\phi=(1+\sqrt{5})/2$.
QUESTION [5 upvotes]: Let $a,b \in \mathbb{Z}$ such that $a^2+ab-b^2=a+b$. Show that $a+b\phi$ and $a+b-b\phi$ are associates in $\mathbb{Z}+\phi \mathbb{Z}$,  with $\phi=(1+\sqrt{5})/2$.
I have to find an unit $u$ such that $a+b\phi = u(a+b-b\phi)$. I've no idea how to begin, any hints? (I know that they both have the same norm, but that's just a necessary condition)

REPLY [4 votes]: Sometimes, you should just do the computation:
$$ \frac{a + b - b\phi}{a + b\phi} = \frac{(a + b - b\phi)(a + b - b\phi)}{a^2 + ab - b^2} = \frac{(a+b)^2 - 2(a+b)b \phi + b^2 (1 + \phi)}{a + b} $$
$$= a+b - 2b\phi + \frac{b^2}{a+b} (1 + \phi) = (2a+b-1) + (a-2b-1)\phi$$
You have already checked that these elements have the same norm (for example, because they are conjugates), so the norm of their ratio is $ 1 $, and it lies in $ \mathbf Z[\phi] $ by our computation. Hence, it is a unit in $ \mathbf Z[\phi] $, and the two elements are associates.