Find all rational points where $x^2 - y^2 = 1$ (need help simplifying quadratic formula)

Question

The original problem is to find all rational points where $x^2 - y^2 = 1$ I know how to go about the problem, but whenever I get to the point of simplifying my equation, I keep having problems. This is what I have now:

choosing point $(-1, 0)$

So we have:

$x^2 - (m (x+1))^2 = 1$

$= x^2 (1-m^2) + 2xm^2 - m - 1 = 0 $

I need to simplify the quadratic equation where :

$a = (1-m^2), b = 2m^2, c = -m-1 $

How can I simplify the part under the square root? Namely this part: $ \sqrt{(2m)^2 - 4 (1-m^2) (-m-1)}$

when I simplify I get this:

$- 4m^3 + 4m -4$

but that doesn't help with the square root. Can anyone point out the right direction for this?

thank you!

Answer 1

Just a remark. The quadratic equation has been incorrectly exapnded. It should be $$ x^2( - m^2 + 1) - 2xm^2 - (m^2 + 1)=0, $$ with solutions $$ x_{1,2}=-\frac{m^2 + 1}{m^2 - 1}, -1, $$ For the question itself, it has been answered here.

Answer 2

How about writing the equation as $1=\left(\frac{1}{x}\right)^2+\left(\frac{y}{x}\right)^2$? You know all rational solutions $(u,v)$ to $u^2+v^2=1$, right?

Answer 3

A little late to the party. The way to do this that avoids computational problems is to write it as parametrized by some new letter, $t.$ Given integers $(p,q)$ we take $$ (-1,0) + t (p,q), $$ or $$ (x,y) = (-1 + tp, tq). $$ So far, $t$ is rational, $p,q$ are integers, and $x,y$ are rational. So, when does $x^2 - y^2 = 1?$ We have $x = -1 + tp,$ $y = tq.$ $$ 1 = x^2 - y^2 = 1 - 2tp + p^2 t^2 - q^2 t^2, $$ $$ 0 = -2tp + (p^2 - q^2)t^2, $$ $$ 2tp = (p^2 - q^2)t^2. $$ This is obviously true when $t=0.$ When $t \neq 0$ and $p \neq \pm q,$ we have $$ 2p = (p^2 - q^2)t, $$ or $$ t = \frac{2p}{p^2 - q^2}. $$ No quadratic formula.Since $$ -1 = \frac{-p^2 + q^2}{p^2 - q^2}, $$ $$ x = -1 + tp = \frac{-p^2 + q^2}{p^2 - q^2} + \frac{2p^2}{p^2 - q^2}, $$ $$ x = \frac{p^2 + q^2}{p^2 - q^2}, $$ $$ y = \frac{2pq}{p^2 - q^2}. $$ An answer at your earlier question got this, but it is not necessary to know anything about Pythagorean triples to find it.