# 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