Find a formula for all the points on the hyperbola $x^2 - y^2 = 1$? whose coordinates are rational numbers. So, I know that we first need to have an initial point. The answers I have say it's $(-1, 0)$ which makes sense because it satisfies the equation. But for example $(1, 0)$ satisfies it too. Why did we choose $(-1, 0)$?
In general, how do we go about choosing the initial point? 
 A: The constraint equation,
$$ x^2-y^2 = 1,$$
describes a hyperbola. 
We start by picking a rational point on the hyperbola, an easy one is $(1,0)$, lets call this point $A$. Then we draw a line with a rational slope, call it $l$, passing through the point $A$ and elsewhere on the parabola, lets call the point of intersection $B=(x_b,y_b)$.
The equation of the line is,
$$ y = m(x-x_a)+y_a \Rightarrow y = m(x-1), $$
sine we are interested in the intersection of this line with the hyperbola we will substitute it into,
$$ x^2-y^2 = 1,$$
obtaining, 
$$ x^2 - m^2(x-1)^2 = 1,$$
which can be simplified to,
$$ x^2 + \frac{2m^2}{1-m^2}x - \frac{m^2+1}{1-m^2} = 0,$$
now this is a quadratic equation for $x$ in terms of the rational number $m$. The points which satisfy this equation lie on both the line, $l$, and our hyperbola. There are two points we know of which do this, one is $A=(1,0)$ the other is $B=(x_b,y_b)$. This means we can factor this equation because we know one of the roots is $x=1$.  
Factoring $(x-1)$ out of the left hand side we get, 
$$ \left(x-1 \right)\left(x+\frac{m^2+1}{1-m^2} \right) = 0,$$
which gives two solutions, 
$$x=1 \qquad x=\frac{m^2+1}{m^2-1},$$
the first is just $x_a=1$ which we knew and the other is $x_b$ which we can see is rational.
$$ x_b =  \frac{m^2+1}{m^2-1} \qquad y_b = m \left( \frac{m^2+1}{m^2-1}-1\right)=\frac{2m}{m^2-1} \qquad m\in \mathbb{Q}$$
If we want the solution parameterized in terms of integers, say $p$ and $q$, we can let $m=p/q$ and make the appropriatie substitutions. 
A: Let $Y=a/b$
$$X=\sqrt { \frac {a^2 + b^2} {b^2}}=\frac 1 b \sqrt{a^2+b^2}$$
Now, we just need $a^2+b^2$ to be a square, i.e $a^2+b^2=c^2$ be pythagorean triples.
One nice parametrization is to let $a=p^2-q^2, b=2pq$, then $a^2+b^2=(p^2+q^2)^2$, so
$$X=\frac  {p^2+q^2}{2pq}$$
So (part of) our solutions are $(X,Y)=\left(\frac {p^2+q^2}{2pq},\frac {p^2-q^2}{2pq}\right)$.
All the solutions can be found if you parametrize all the pythagorean triples.
A: The choice $(-1,0)$ is good. So is $(1,0)$. Any rational point on the curve is a good choice. For instance we could choose $x=5/4$ and $y=3/4$. But simple is better, it makes the arithmetic nicer.
