Proof a number is Fibonacci number

Question

I have a question regarding the proof that a number n is a Fibonacci number if and only if $5n^2-4$ or $5n^2+4$ is a perfect square. I don't understand the second part of the proof: knowing that $5n^2-4$ or $5n^2+4$ is a perfect square, prove that n is a Fibonacci number.

I attach the solution I found in this Google Book - Fibonacci and Lucas Numbers with Applications by Thomas Koshy:

I don't understand the part of the proof starting from "Since m and n...".

Basically:

1. Why must $(m + n \sqrt5)/2$ and $(m - n \sqrt5)/2$ be integers in the given extension field?
2. Why must they be units in this field if their product is -1 (what is actually "a unit in an extension field" and why does it have that form?)

Thank you in advance.

Answer 1

They are algebraic integers in the extensions field as $m$ and $n$ have the same parity; more precislely, as $m$ and $n$ have the same parity the numbers you consider are of the form $a + b (1+ \sqrt{5})/2$ with (usual) integers $a,b$, which means precisely they are algebraic integers in that extension field.

By units in the extension field it is meant invertible elements in the ring of algebraic integers of the field. If the product is $1$, they are clearly invertible in that ring (both elements are in the ring and the product is $1$); if it is $-1$ just changing the sign of one of them also yields a product of $1$ and thus shows they are invertible. That units have that form is also a fact about rings of algebraic integers, up to signe the group is cyclic of infinite order and $\alpha$ is a generator; see Dirichlet's unit theorem for context.

Answer 2

The ring of integers $\mathcal{O}_K$ in the quadratic number field $K=\mathbb{Q}(\sqrt{5})$ is given by $\mathbb{Z}+\frac{1+\sqrt{5}}{2}\mathbb{Z}$. This is a basic result of algebraic number theory. So the author means that these elements are in the ring of integers of the extension field. Furthermore the ring $\mathcal{O}_K$ has a unit group, i.e., its invertible elements. The product equal $\pm 1$ equation says that they are invertible. Finally we use a famous theorem on the unit group of number fields, i.e., Dirichlet's unit theorem. It says in our case that the unit group of $\mathcal{O}_K$ is infinite cyclic times $\{\pm 1\}$.