How do we conclude that the relation is equal to $1$ ?

Question

We have a curve of the form $$s^2-\alpha t^2=1 \tag 1$$ (in $ts$-coordinates).

If $(a,b)$ and $(c,d)$ are points of $(1)$, then $(ac+\alpha cd, ad+bc)$ is point of $(1)$.

$(a,b)$ is a point of $(1)$ $\Rightarrow a^2-\alpha b^2=1$

$(c,d)$ is a point of $(1)$ $\Rightarrow c^2-\alpha d^2=1$

So that $(ac+\alpha cd, ad+bc)$ is apoint of $(1)$ it must be $(ac+\alpha cd)^2-\alpha (ad+bc)^2=1$.

$(ac+\alpha cd)^2-\alpha (ad+bc)^2=a^2c^2+\alpha^2c^2d^2+2ac^2d\alpha-\alpha a^2d^2-\alpha b^2c^2-2\alpha a bcd$

But how is this equal to $1$ ?

$$$$

We define the additive law $+$ on $(1)$ by $$(a,b)+(c,d)=(ac+\alpha bd, ad+bc)$$ We observe that $$-(a,b)=(a,-b)$$ and the identity is $(1,0)$.

How have we concluded that $-(a,b)=(a,-b)$ ?

Answer 1

Since $a^2-\alpha b^2=1$ and $c^2-\alpha d^2=1$, by multiplication we have $$a^2c^2+\alpha^2 b^2d^2 -\alpha(a^2d^2+b^2c^2)=1.\tag{1}$$

Now compare with $(ac+\alpha bd)^2 -\alpha(ad+bc)^2$. The "middle" terms cancel, and we get the right-hand side of (1).

As to $-(a,b)=(a,-b)$, calculate $(a,b)+(a,-b)$ using the addition formula. You will get $(1,0)$.

Answer 2

Is it not $(ac+\alpha bd, ad+bc)$ a point of $(1)$?

Then you can show it by simple computation:

\begin{align*} &(ac+\alpha bd)^2-\alpha (ad+bc)^2=a^2c^2+2\alpha abcd +\alpha^2b^2d^2-\alpha a^2d^2-2\alpha abcd -\alpha b^2c^2\\& = a^2(c^2-\alpha d^2) -\alpha b^2 ( c^2-\alpha d^2)=1 \end{align*} by the two rules.

The second part $-(a,b)=(a,-b)$ can be proved by adding $(a,b)+(a,-b)$ by the given rule to get $(0,0)$.

Similarly you can show they are equal to $(1,0)$.