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deepkdeb@yahoo.com
science forum beginner

Joined: 29 Dec 2005
Posts: 38

Posted: Fri Jul 07, 2006 1:40 am    Post subject: Rational Points

The equation (1) represents an ellipse under the given conditions.

ax^2 + by^2 = r^2 (1) where r, b, a are integers > 0

Assertion: (1) cannot be satisfied for rational values of x, y.

Any comment upon the correctness of the assertion will be appreciated.
Rouben Rostamian

Joined: 01 May 2005
Posts: 85

Posted: Fri Jul 07, 2006 2:00 am    Post subject: Re: Rational Points

Deep <deepkdeb@yahoo.com> wrote:
 Quote: The equation (1) represents an ellipse under the given conditions. ax^2 + by^2 = r^2 (1) where r, b, a are integers > 0 Assertion: (1) cannot be satisfied for rational values of x, y. Any comment upon the correctness of the assertion will be appreciated.

Homework?
William Elliot
science forum Guru

Joined: 24 Mar 2005
Posts: 1906

Posted: Fri Jul 07, 2006 4:12 am    Post subject: Re: Rational Points

On Thu, 6 Jul 2006, Deep wrote:

 Quote: The equation (1) represents an ellipse under the given conditions. ax^2 + by^2 = r^2 (1) where r, b, a are integers > 0 Assertion: (1) cannot be satisfied for rational values of x, y. Any comment upon the correctness of the assertion will be appreciated. Look for counter examples.
science forum Guru

Joined: 24 Mar 2005
Posts: 790

Posted: Fri Jul 07, 2006 4:13 am    Post subject: Re: Rational Points

In article
"Deep" <deepkdeb@yahoo.com> wrote:

 Quote: The equation (1) represents an ellipse under the given conditions. ax^2 + by^2 = r^2 (1) where r, b, a are integers > 0 Assertion: (1) cannot be satisfied for rational values of x, y. Any comment upon the correctness of the assertion will be appreciated.

If a = b = 1 and r = 5, then x = 3, y = 4 gives a solution.
Robert B. Israel
science forum Guru

Joined: 24 Mar 2005
Posts: 2151

Posted: Fri Jul 07, 2006 6:25 am    Post subject: Re: Rational Points

Deep <deepkdeb@yahoo.com> wrote:
 Quote: The equation (1) represents an ellipse under the given conditions. ax^2 + by^2 = r^2 (1) where r, b, a are integers > 0 Assertion: (1) cannot be satisfied for rational values of x, y. Any comment upon the correctness of the assertion will be appreciated.

One of the most obviously false assertions you've made so far.
It's certainly false if a or b is a square. Thus if b = c^2, try
x = 2 r/(a+1), y = r (a - 1)/(c (a+1)).

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
deepkdeb@yahoo.com
science forum beginner

Joined: 29 Dec 2005
Posts: 38

Posted: Fri Jul 07, 2006 12:36 pm    Post subject: Re: Rational Points

Robert Israel wrote:
 Quote: In article <1152236444.887809.296900@m79g2000cwm.googlegroups.com>, Deep wrote: The equation (1) represents an ellipse under the given conditions. ax^2 + by^2 = r^2 (1) where r, b, a are integers > 0 Assertion: (1) cannot be satisfied for rational values of x, y. Any comment upon the correctness of the assertion will be appreciated. One of the most obviously false assertions you've made so far. It's certainly false if a or b is a square. Thus if b = c^2, try x = 2 r/(a+1), y = r (a - 1)/(c (a+1)). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada

Yes, you are right Professor Israel. Kindly note the corrections " none
of a or b is a perfect square"
Now kindly give a counter example.
The Qurqirish Dragon
science forum Guru Wannabe

Joined: 30 Apr 2005
Posts: 104

Posted: Fri Jul 07, 2006 2:35 pm    Post subject: Re: Rational Points

Deep wrote:
 Quote: Robert Israel wrote: In article <1152236444.887809.296900@m79g2000cwm.googlegroups.com>, Deep wrote: The equation (1) represents an ellipse under the given conditions. ax^2 + by^2 = r^2 (1) where r, b, a are integers > 0 Assertion: (1) cannot be satisfied for rational values of x, y. Any comment upon the correctness of the assertion will be appreciated. One of the most obviously false assertions you've made so far. It's certainly false if a or b is a square. Thus if b = c^2, try x = 2 r/(a+1), y = r (a - 1)/(c (a+1)). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada Yes, you are right Professor Israel. Kindly note the corrections " none of a or b is a perfect square" Now kindly give a counter example.

If a+b=r^2, then x=y=1 is a solution.
Denis Feldmann2

Joined: 23 Apr 2006
Posts: 87

Posted: Fri Jul 07, 2006 3:58 pm    Post subject: Re: Rational Points

The Qurqirish Dragon a écrit :
 Quote: Deep wrote: Robert Israel wrote: In article <1152236444.887809.296900@m79g2000cwm.googlegroups.com>, Deep wrote: The equation (1) represents an ellipse under the given conditions. ax^2 + by^2 = r^2 (1) where r, b, a are integers > 0 Assertion: (1) cannot be satisfied for rational values of x, y. Any comment upon the correctness of the assertion will be appreciated. One of the most obviously false assertions you've made so far. It's certainly false if a or b is a square. Thus if b = c^2, try x = 2 r/(a+1), y = r (a - 1)/(c (a+1)). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada Yes, you are right Professor Israel. Kindly note the corrections " none of a or b is a perfect square" Now kindly give a counter example. If a+b=r^2, then x=y=1 is a solution.

for a=2, b= 7 , r=1, x=y=1/3 is a solution (and also x=3/5, y=1/5 ; etc.)

>
OwlHoot
science forum beginner

Joined: 25 Jun 2006
Posts: 6

Posted: Sat Jul 08, 2006 1:27 pm    Post subject: Re: Rational Points

Deep wrote:
 Quote: The equation (1) represents an ellipse under the given conditions. ax^2 + by^2 = r^2 (1) where r, b, a are integers > 0 Assertion: (1) cannot be satisfied for rational values of x, y. Any comment upon the correctness of the assertion will be appreciated.

It's wrong.

By homogeneity there are non-zero (assumed hereafter) rational
solutions
if and only if there integer solutions. So it suffices to find
conditions for the
latter.

We can assume a and b are squarefree (each a product of distinct
primes),
because if say c^2 divides b we can consider an equation of the same
form

a.x^2 + (b/c^2).(c.y)^2 = r^2.

Then denoting a, b = e.f, e.g resp where GCD(f, g) = 1, clearly e is
also squarefree and thus divides r. So it suffices to consider:

f.x^2 + g.y^2 = e.(r/e)^2

Similarly, noting that GCD(GCD(e,f), g) = 1 and GCD(GCD(e,g), f) = 1,
it can be seen that GCD(e, f) (also squarefree) divides y and GCD(e,g)
divides x, and upon dividing these out it turns out that integer
solutions
to the original imply and are implied by integer solutions of:

A.X^2 + B.Y^2 + C.R^2 = 0

in which the integers A, B, C are coprime (relatively prime in pairs).

It isn't hard to prove that when A, B, C are coprime, then the latter
has non-zero integer solutions if and only if:

* A, B, C are not all of the same sign (true in your case,
because C < 0 < A, B)

* -B.C, -C.A, -A.B are quadratic residues mod |A|, |B|, |C| resp.

For example 3.x^2 + 5.y^2 = 7.z^2 has no non-zero integer
solutions because, with A, B, C = 3, 5, -7 resp, 5.7 is not a
quadratic residue mod 3 (because 5.7 == 2 mod 3, whereas
a square is == 0 or 1 mod 3).

Apologies if this post come out with crabby-looking "dot and carry one"
lines, but I'm typing in Google groups in a vile proportional-spacing
font
and Google has an extremely irritating habit of rearranging lines at
random!
OwlHoot
science forum beginner

Joined: 25 Jun 2006
Posts: 6

Posted: Sat Jul 08, 2006 1:35 pm    Post subject: Re: Rational Points

Deep wrote:
 Quote: The equation (1) represents an ellipse under the given conditions. ax^2 + by^2 = r^2 (1) where r, b, a are integers > 0 Assertion: (1) cannot be satisfied for rational values of x, y. Any comment upon the correctness of the assertion will be appreciated.

<html>
<body>
<pre>
By homogeneity there are non-zero (assumed hereafter) rational
solutions
if and only if there integer solutions. So it suffices to find
conditions for the
latter.
</pre>
</body>
</html>

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