Elements with norm $1$ in a quadratic field

Question

Consider the number field $K=\mathbb{Q}(\sqrt{5})$ and the norm map $N:\mathbb{Q}(\sqrt{5}) \to \mathbb{Q}^*$ given by $N(a+b \sqrt{5})=a^2-5b^2$.

Is there an explicit description of the kernel of this map, i.e of the elements having norm $=1$?

Answer 1

Hilbert's Theorem 90 implies that any such element may be written in the form $\frac{a-b\sqrt{5}}{a+b\sqrt{5}}$ for $a,b \in \Bbb Q$. You can achieve $a,b \in \Bbb Z$ by multiplying with a common denominator.

Answer 2

in lowest terms, the ones with even denominator are $$ a = \frac{2u^2 + 2uv + 3 v^2}{2 u^2 + 2uv - 2 v^2} \; , \; \; b = \frac{ 2uv + v^2 }{2 u^2 + 2uv - 2 v^2} $$

Odd $$ a = \frac{u^2 + 5 v^2}{ u^2 - 5 v^2} \; , \; \; b = \frac{ 2uv }{ u^2 - 5 v^2} $$

there should be conditions mod 2,5 to enforce no common factors

If you just want integers $a_n^2 - 5 b_n^2 = 1$ in sequence $a_{n+2} = 18 a_{n+1} - a_n$ and $b_{n+2} = 18 b_{n+1} - b_n$ beginning with pairs $(1,0)$ and $(9,4)$ and $(161,72)$