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I am trying to find rational points on this surface

$$ \left( \left( 1-x \right) ^{2}+{y}^{2} \right) \left( \left( 1+x \right) ^{2}+{y}^{2} \right) ={z}^{2}$$

I am actually only interested in points where both $x$ and $y$ are in the range of $\left(0..1\right)$.

There are trivial rational points at $\left(0,0,1\right)$, $\left(0,1,2\right)$, $\left(1,0,0\right)$, but I cannot figure out how to generalize a parameterization. Of particular interest are rational points in the interior of the surface and not the boundary anyways. It may also be the case that there actually are not any rational points on the interior of the surface in the range of interest, as well.

Is it possible to actually compute a solution to this, or even possible (tractable) to just prove that no rational solutions exist in the domain of interest?

I've taken a first stab at tagging this under algebraic geometry and rational numbers, but if anyone who can edit tags knows of better ones to use for this kind of problem, please feel free to update that information.

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I've flagged one of the answers below as helpful, because it most definitely is, and I will have to investigate it more deeply to determine if it helps me find out if using the formulation can ever reveal any solutions to the original rational problem where both $x$ and $y$ are in the $(0..1)$ range. Since that range is part of my original problem, I cannot actually say at this time that the helpful answer is my accepted answer until I have either found such an answer, or used that formulation to determine that no answers in the range of interest exist. I haven't investigated it very thoroughly yet, and it may take me a few hours to do so (I will provide an update on this question once I have done so), but I wanted to give an explanation for why I have not flagged what I must agree is the most informative answer as an accepted one at this time. Just to clear things up in advance, however, if no solutions in the range of interest can be found with the diophantine representation provided, is it possible that there may be another form that does have such solutions?

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The good news is that using the diophantine formula seems good for solving the problem as I had phrased it... the bad news is that it turns out that there's a single additional algebraic requirement in the problem which I had completely forgotten to mention.

The single additional requirement is that $x^2+y^2$ must be the square of a rational number. I imagine this makes the problem probably many times harder than how I had originally phrased it and I apologize profusely for forgetting to mention this requirement initially. I can offer no excuse for this oversight beyond that this requirement was simply not at the forefront of my mind while I was typing a description of the problem here.

So given the additional requirement, is it still possible to compute a general rational expression for each of x, y, and z, or is it somehow possible (tractable) to determine that no rational points exist other than on the aforementioned boundary?

That is definitely all of the requirements, by the way... there really aren't any more. I feel genuinely embarrassed for not fully describing all of the algebraic requirements initially, and I again apologize for forgetting to mention this detail initially.

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  • $\begingroup$ You need to solve this Diophantine equation? $\endgroup$ – individ Jun 24 '14 at 5:37
  • $\begingroup$ Do I NEED to? No... but I'd like to. I'm weird that way.... sometimes my brain goes on weird math trajectories and then I spend days or sometimes even weeks thinking about how to solve something that admittedly probably doesn't have a single practical purpose. My desire for an answer is no less genuine, however... and if anyone knows how to help, I'd appreciate it most sincerely. $\endgroup$ – Mark Jun 24 '14 at 5:42
  • $\begingroup$ So try to be specific. You can rewrite the equation in different ways. So what the equation must be solved in whole or rational? $\endgroup$ – individ Jun 24 '14 at 5:48
  • $\begingroup$ Rational... it could be rewritten as diophantine, but I think it would need an additional unknown. $\endgroup$ – Mark Jun 24 '14 at 6:18
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    $\begingroup$ It is better to solve this equation in integers. $((x-d)^2+y^2)((x+d)^2+y^2)=z^2$ He might be the solution. $\endgroup$ – individ Jun 24 '14 at 11:40
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A.

For each Pythagorean triple $(a,b,c)$ one can denote $$ x=\dfrac{a}{c}, \qquad y=\dfrac{b}{c}, $$ so $$ x^2+y^2=1, $$ and $$ \left( (1-x)^2+y^2 \right) \left( (1+x)^2+y^2 \right) \\ =\left( 1-2x+x^2+y^2 \right) \left( 1+2x+x^2+y^2 \right) \\ =\left( 2-2x\right) \left( 2+2x\right) = 4(1-x^2) = (2y)^2, $$ hence $z=2y$ is rational.

Following by the link above, you can find parameterizations.

B.

There are many rational solutions of equation $$ \left( (1-x)^2+y^2 \right) \left( (1+x)^2+y^2 \right) =z^2,\tag{1} $$ where $$ x^2+y^2=r^2, \quad x,y\in (0,1), \tag{2} $$ $x,y,z,r\in\mathbb{Q}$ (all are rationals), and $r\ne 1$.

Here are a few examples:

$(x,y,z,r) = \left( \dfrac{8}{28}, \dfrac{15}{28}, \dfrac{975}{28^2},\dfrac{17}{28}\right)$; (small denominator);

$(x,y,z,r) = \left( \dfrac{153}{37^2}, \dfrac{104}{37^2}, \dfrac{1360}{37^2},\dfrac{185}{37^2}\right)=\left( \dfrac{153}{1369}, \dfrac{104}{1369}, \dfrac{1360}{1369},\dfrac{185}{1369}\right)$; (small $r$);

$(x,y,z,r) = \left( \dfrac{1020}{1038}, \dfrac{931}{1038}, \dfrac{2103325}{1038^2},\dfrac{1381}{1038}\right)$; (large $r$);

$\ldots$


To illustrate which changes come with condition $x^2+y^2=r^2$, where $x,y,r\in\mathbb{Q}$ and to show that circle $x^2+y^2=1$ contains the most of points (for limited common denominator of $x,y$) I drawed $2$ images (where denominator $c$ is $\le 2000$):

Rational points $x,y$ ($0<x,y<1$), for which $z\in\mathbb{Q}$ (see $(1)$):

enter image description here

Rational points $x,y$ ($0<x,y<1$), for which $z,r\in\mathbb{Q}$ (see $(1), (2)$):

enter image description here

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The formula looks quite cumbersome, but it has simplified and will write here is simple.

Equation:

$$((x+d)^2+y^2)((x-d)^2+y^2)=z^2$$

Has the solution:

$$x=(k+n)^2p^2+(k-n)^2s^2$$

$$d=(k^2-n^2)(p^2+s^2)$$

$$y=4knps$$

$$z=4kn((k+n)^2p^4+2(k^2+n^2)p^2s^2+(k-n)^2s^4)$$

$k,n,p,s$ - any integers, any sign.

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Is it possible to actually compute a solution to this? Maybe - it would be done by going through very small increments of x and for each one going through very small increments of y and calculating z as the square root of the LHS. This would be given to a fixed precision and so by default would be rational. However, if the decimal expansion terminated or repeated you would have found a rational point. If it failed you could do it again with smaller increments. Project Euler 26 is relevant to this.

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    $\begingroup$ That would only approximate the solution with rational answers, however, and would not actually properly be an exact solution. $\endgroup$ – Mark Jun 24 '14 at 6:21
  • $\begingroup$ Unless you found one with an expansion that terminated or repeated. I don't know how likely this approach would be to succeed. Interesting question - by looking at the data generated by this approach would it be possible to determine that we were 'near' a rational solution? $\endgroup$ – user117644 Jun 24 '14 at 7:19
  • $\begingroup$ Now I know - unlikely - I got screen fulls of obviously irrational numbers. If any are 'near' a rational it's not obvious. $\endgroup$ – user117644 Jun 24 '14 at 8:51

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