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

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!

marked as duplicate by Dietrich Burde, jameselmore, SchrodingersCat, Em., DanJan 25 '16 at 21:46

• The simplest way is not to do it. First note that the quadratic has been incorrectly expanded. After correcting, we know one of the roots is $-1$. From the Vieta formula we know the product of the roots. Thus the missing root is the negative of that product. – André Nicolas Jan 25 '16 at 16:21
• But if you really want to use the quadratic formula, start from the correct expansion of $x^2-(m(x+1))^2=1$. It will work out. – André Nicolas Jan 25 '16 at 16:27

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.

• Hello, thanks for your answer. This is my own question. But the method for solving is different. I want to know how to solve this, specifically using the quadratic formula. – Lana Jan 25 '16 at 16:14
• Both questions were made by the same person. – YoTengoUnLCD Jan 25 '16 at 16:14
• But the comments and answers there solve this, don't they ? – Dietrich Burde Jan 25 '16 at 16:16
• No they don't. Like I mentioned, I am trying to solve using the quadratic formula specifically. – Lana Jan 25 '16 at 16:19
• Yup, I see what I was doing wrong now. The square root simplifies to 4. – Lana Jan 25 '16 at 16:41

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?

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.

• This is a much better way to go about it. Thank you so much. – Lana Jan 27 '16 at 2:46