0
$\begingroup$

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!

$\endgroup$

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ 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. $\endgroup$ – André Nicolas Jan 25 '16 at 16:21
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Jan 25 '16 at 16:27
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Lana Jan 25 '16 at 16:14
  • $\begingroup$ Both questions were made by the same person. $\endgroup$ – YoTengoUnLCD Jan 25 '16 at 16:14
  • $\begingroup$ But the comments and answers there solve this, don't they ? $\endgroup$ – Dietrich Burde Jan 25 '16 at 16:16
  • $\begingroup$ No they don't. Like I mentioned, I am trying to solve using the quadratic formula specifically. $\endgroup$ – Lana Jan 25 '16 at 16:19
  • $\begingroup$ Yup, I see what I was doing wrong now. The square root simplifies to 4. $\endgroup$ – Lana Jan 25 '16 at 16:41
0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.