Proving the Existence of Rational Points on a Circle

[tehno]

x²+y²=c² is equation of a central circle,

so is that or is that not an elliptic curve?

http://mathworld.wolfram.com/EllipticCurve.html" .What do you think (eq (2) and (3))?

Elliptic curves I reffered to as an example.One power degree higher than conics but can have only finitely many rational points.
Circles and other conics ,in Cartesian plane have:
a) 0 rational points
or
b) infinitely many rational points.

For central circles case (a) or (b) depends exclusively on c.
For instance x²+y²=3 has no rational points while
x²+y²=1;x²+y²=2 have infinitely many of them.
Your only reasonable question ,as far as I can see, would be something like this:"What on Earth existence of circle rational points has to do with rational distances among them?".
As I already emphasized not much ,if the interest is only in rational distances.
But OP is to be blamed for the reason I brought this up.

He changed his mind on the formulation of the problem several times during the thread.
Reducing considerations to the unit circle and other things suggested to me he could be actually interested in situations where both points and distances are rational.That's interesting,but it doesn't turn to be case (unfortunately).
Again ,his concluding sentence prolongs that confusion,in the light of what he wrote before:

So the basic idea is just to choose rational points on the unit circle...

So,your basic idea is to choose:
-rational points on the unit circle
or
-points at the rational distance
or
-rational points at rational distance
?
Rational points means to me points with coordinates (p,q)\in\mathbb{Q}

 

[gonzo]

So,your basic idea is to choose:
-rational points on the unit circle
or
-points at the rational distance
or
-rational points at rational distance
?
Rational points means to me points with coordinates (p,q)\in\mathbb{Q}

Yes, I mean points with rational coordinates on the unit circle. It isn't hard to prove that if you pick two points with rational coordinates on the unit circle the distance between the points with double the angles is also rational.

 

[tehno]

Yes, I mean points with rational coordinates on the unit circle. It isn't hard to prove that if you pick two points with rational coordinates on the unit circle the distance between the points with double the angles is also rational.

Can't you (dis)prove it for x²+y²=3 ?

 

[gonzo]

The proof I have for the unit circle (and it's not my proof, I can't take credit for it) doesn't directly extend to circles with non-perfect square radii. However, there is another proof I've seen that might.

Why do you ask?

 

[matt grime]

I know what elliptic curves are, thank you very much. I also know what conics are, and the M-W theorem. The question is not about my knowledge but about your attempts at writing mathematics, tehno. Elliptic curves are abelian varieties - they have an addition of points. Do conics? Is there a Mordell Weil theorem (which is a statement about the finite generation of rational points for ab vars) for conics? None of what you wrote joins together well. And your rebuttal with a link to eq 2 and 3 is nonsense - they both point to

y^2=cubic in x (and you can't set the leading coefficient to 0).

Your only reasonable question ,as far as I can see, would be something like this:"What on Earth existence of circle rational points has to do with rational distances among them?".
As I already emphasized not much ,if the interest is only in rational distances.

but that is the only question i asked, and one you utterly failed to answer until now.

But OP is to be blamed for the reason I brought this up.

No, they are not.

He changed his mind on the formulation of the problem several times during the thread.

at no point did the OP require rationality of the points only of the distances between them.

 

[jing]

Techno - I agree with Matt Grime Gonzo never changed his mind about what he wanted proof of. However there was a lack of clarity (for me) about the statement.

For any positive integer k, you can find k points on a circle such that each point is a rational distance from every other point.

I took this to mean

For any positive integer k, you can find k points on any circle such that each point is a rational distance from every other point.

There followed a discussion with Gonzo about unit circles and such that led to an agreed restatement

So is the question now

'given k, a positive integer, is it possible to find a circle with k points on the circumference of the circle such that the distance between any two of these points is rational?' ?

which meant that I should have read the OP as

For any positive integer k, you can find k points on some circle such that each point is a rational distance from every other point.


Reading posts in general on the site has led me to conclude that:

Communicating in posts can be difficult as we know what we mean and so expect others to know exactly what we mean when we write the posts but often do not write with sufficient clarity to put over our exact meaning.

So extra care must be taken to read all posts carefully, to ask for clarification when needed and to give clarification when required.

Gonzo - I am sure everyone's posts have been helpful in clarifying your thoughts and in finding the proof you required, which I am pleased you have done.

 

[gonzo]

Thanks jing. But like I said, I can't take credit for it. Wish I could since it was really obvious and simple when I saw it. But I can't.

 

[tehno]

The proof I have for the unit circle (and it's not my proof, I can't take credit for it) doesn't directly extend to circles with non-perfect square radii. However, there is another proof I've seen that might.

Why do you ask?

Probably becouse I came up with solution for all circles.
OTOH,I'm not satisfied with it.
I'm almost certain there must be more elegant ,shorter,and
elementar proof of it.
It would be intereting to see other proof for c=1 case you saw.
Can you post it?Maybe it can be extended to the all cases.
Here I'll present an "ugly" proof.
Firstly ,about terminology and some definitions I will use.

For collection of points C \subset \mathbb{R}^2 I say to be at r-distance if the distance between every pair of points is rational.Let's call such collection of points
a rational distance set.Note that rationals form a rational
distance subset of the reals.Particularly,if C is a line
in \mathbb{R}^2,we say C is a dense
set of points at r-distance.
Prior to the proof two known lemmas I will reffer to (without proof).

Lm1.

A triple (x,y,z) of naturals is a primitive pythagorean triple
if x^2+y^2=z^2 holds and if x,y,z are coprime.
All primitive pythagorean triplets where y is even are given by:
x=m^2-n^2,y=2mn,z=m^2+n^2;m>n
m,n are relatively prime and one of them is even.

Lm2.

Affine-rational solutions to \alpha^2 =\beta^2 + 1

are parameterized by:

\alpha=\frac{m^2+1}{2m},\beta=\frac{m^2-1}{2m};m\in\mathbb{Q}
.

_____________________________________________________________
With terminology given above your proposition can be formulated now in this way:

Any circle contains a dense set of points at r-distance.
_____________________________________________________________
Proof:

Let's identify the complex plane,as is often done,with \mathbb{R}^2.
Two points in the complex plane z_{1},z_{2} we can always choose that
they are at r-distance and have rational lenght.Then,becouse of the identity:
\frac{||z_{1}-z_{2}||}{||z_{1}||\cdot ||z_{2}||}=||\frac{1}{z_{1}}-\frac{1}{z_{2}}||

Points 1/z_{1}[/tex] and 1/z_{2} are at r-distance too.<br /> <br /> Consider a vertical line <b>I</b> in the complex plane.<br /> By Lm1 &amp; Lm2 ascertain that we can always parameterize all points on <b>I</b> whose<br /> lenghts and imaginary parts are rational.This set of points is dense in <b>I</b>.<br /> Furthery ,this set points is mapped, as long as <b>I</b> isn&#039;t imaginary axis,<br /> by the complex mapf(z)=z^{-1} to a dense ,rational distance subset of a<br /> circle.Since <b>I</b> can be translated at will through all vertical lines,we can obtain<br /> circless of all radii.<br /> <b>QED</b><br /> <br /> That&#039;s a cumbersome proof of something geometrically obvious (to me).<br /> Ugly and quite probably overkill to the problem.<br /> But it&#039;s a proof.<br /> [EDIT: That <b>I</b> can be parameterized in described manner make sure like this<br /> Vertical line is x=a.We seek points (a,y) on it such that y is rational and also<br /> (a^2+y^2)^(1/2}=z is rational.We get pythagorean equation a^2+y^2=z^2 or 1+(y/a)^2=(z/a)^2.By Lm2 we have y= a[(m^2-1)/(2m)],for m\in\mathbb{Q} .The parametrization occurs ]<br /> If need be I&#039;ll post the proof for lemma 2.It&#039;s not long.<br /> But that&#039;s is only thing what I&#039;m willing to do more ( lack of time and nerves for a chit-chat on this subject).<br /> <br /> <blockquote data-attributes="" data-quote="matt grimme" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> matt grimme said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> but that is the only question i asked, and one you utterly failed to answer until now. </div> </div> </blockquote>Ok.How about posting your solution instead?<br /> <br /> <blockquote data-attributes="" data-quote="jing" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> jing said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Communicating in posts can be difficult as we know what we mean and so expect others to know exactly what we mean when we write the posts but often do not write with sufficient clarity to put over our exact meaning.<br /> <br /> So extra care must be taken to read all posts carefully, to ask for clarification when needed and to give clarification when required. </div> </div> </blockquote>Well said.<br /> Quite sometimes,I read to quickly.Somewhere around half of this thread I saw the title<br /> &quot;Rational points on the circle&quot; .Didn&#039;t care before that about rational points,and I don&#039;t know why I thought he was interested in both distances and points being rational.<br /> he wanted both.

 

[gonzo]

I already posted the basis of my proof, the rest is trivial.

Pick two rational points (points with rational coordinates) on the unit circle:

e^{i \theta_1} and e^{i \theta_2}

Then it is really easy to show that the distance between the points with double the angles e^{2i \theta_1} and e^{2i \theta_2} is rational.

Since there are infinite rational points, you can always find what you need.

 

[matt grime]

l
Ok.How about posting your solution instead?

did I say I had one? I merely have very strong objections to your misuse of mathematics (M-W theorems, and invoking elliptic curves: x^2+y^2=-1 is a perfectly good conic (over C - birational, isomorphic as a curve over C, in fact, to x^2+y^2=1) and has precisely one Q-rational point contradicting your previous assertions, for example).

 

[tehno]

I pleadge gulity as charged for invoking elliptic curves (and for misreading problem conditions as the thread developed as I already said).
I reffered to elliptics only in comparative purposes to conics,but that was completely unnecessary complication to the problem I agree. I accept your criticism as concerns that matter.
I never said x²+y²=-1 has infinitely many rational points. x²+y²=1 in Cartesian plane has that property.

 

[matt grime]

I never said x²+y²=-1 has infinitely many rational points. x²+y²=1 in Cartesian plane has that property.

I meant x^2+y^2=0, not -1, a degnerate case - but what differentiates it from x^2-y^2=0? Or for that matter, what about x^4=3y^4? That has 1 Q-rational point, and is a quadratic, so what's special about conics? Or x^2-1=sqrt(3)y^2? Another curve with only one Q-rational point, although it isn't defined over Q.