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#1 2007-01-13 10:12:56

Stanley_Marsh
Member
Registered: 2006-12-13
Posts: 345

Elliptic...

Prove that  y^2=x^3 - 2 only has one solution.


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#2 2007-01-13 10:54:26

mathsyperson
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Registered: 2005-06-22
Posts: 4,900

Re: Elliptic...

I'm not sure what you mean. There are many solutions to this. Pick any value for x [as long as x > 2^(1/3)] and there will be 2 values of y that fit. Alternatively, pick any y-value and there will always be a value of x that makes the equation work.

Are there any restrictions on what x and y can be? For example, do they have to be integers?


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#3 2007-01-13 11:20:48

Patrick
Real Member
Registered: 2006-02-24
Posts: 1,005

Re: Elliptic...

mathsyperson - I'm not sure what it means, but maybe the name of the topic can give you a clue? "Elliptic"


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#4 2007-01-13 11:57:32

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Elliptic...

Note that y is even if and only if x is even.

Let x be even.  Then x contains 2 as a prime factor.  Thus, x^3 = 2^3 * m = 8m for some integer m.  8m - 2 = 2(4m - 1).  Also, it must be that y is even.  So y contains 2 as a prime factor.  Let y^2 = 4n for some integer n.  So:

4n = 2(4m-1)

or

2n = 4m - 1

Even = odd, contradiction.  Thus, x and y must be odd.

Halfway done...


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#5 2007-01-17 12:13:41

Stanley_Marsh
Member
Registered: 2006-12-13
Posts: 345

Re: Elliptic...

Sorrri , my bad,  The solutions must be integers ,  This kind of equations are called elliptic curves. Solve this can solve a problem which Fermat once challenged other two mathematicians , noted that  26 is between 25 , 27 , 25=5^2 , 27=3^3 ,  prove that 26 is the only number which lies between a square and a cube.


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#6 2007-01-17 12:17:45

Stanley_Marsh
Member
Registered: 2006-12-13
Posts: 345

Re: Elliptic...

I am working on it ,  I tried when x=k ,y=k+1, no solution , x=k y=k+2 , no soltution , when x=k,y=k+3 one solution ,  then use induction on n , assume x=k , y=k+n , n>3 contains no solution. then if y=k+n+1 can be proved to contain no solution ,it's done , but I have difficulty in working on the last step.


Numbers are the essence of the Universe

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#7 2007-01-17 12:54:07

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Elliptic...

No wonder I've been having so much difficulty.  But believe it or not, I'm still trying.  Hopefully I'll be getting some help from friends tonight.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#8 2007-01-17 13:10:05

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Elliptic...

Ok, here is another theorem.

Let p be a prime not equal to 2.  If p divides y, then p does not divide x.  Also, if p divides x, then p does not divide y.

The proof is simple:

p | y, so p | y^2, so p | x^3 - 2, p does not divide 2, so p does not divide x^3, so p does not divide x.

Just do it in reverse for the other way.

Maybe that helps?  x and y must be relatively prime.  That certainly restricts them quite a bit.

Edit:

Note that the stronger version of this is:

If p divides y, then p does not divide x.  Also, if p divides x, then p does not divide y.

With no restriction on what p is.  Simply combine my previous proof about even numbers with the one given above.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#9 2007-01-17 15:44:19

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Elliptic...

I found a solution.  It involves algebraic number theory.  Believe it or not, the solution has to do with factoring complex (as in imaginary) polynomials in a certain domain.  I take it you're looking for a purely number theory proof, right Stan?


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#10 2007-01-20 10:38:24

Stanley_Marsh
Member
Registered: 2006-12-13
Posts: 345

Re: Elliptic...

Yeah, I been reading books about number theory recently , but I am a newbie to it , still struggling with group theory . You can post it out , I'll try to comprehence it ,lol.  P.S another question , how to apply imaginary number to prove certain problem , I mean how to consider using it?


Numbers are the essence of the Universe

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#11 2007-01-20 10:58:01

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Elliptic...

The proof is here:

http://abstractnonsense.wordpress.com/2 … er-theory/

As to your second question, the answer is simply experience.  Knowing how polynomials act when factoring in the complex numbers and the various ways you can use this to arrive at a contradiction certainly helps.  But mostly, it's just experience.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#12 2007-01-20 11:00:34

Stanley_Marsh
Member
Registered: 2006-12-13
Posts: 345

Re: Elliptic...

Oh , Thank you anyway. I got to gain more exp lol


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#13 2007-01-20 14:06:43

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Elliptic...

And if you need any help with group theory, don't hesitate to ask.  Abstract Algebra is my favorite field at the current time.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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