Author |
Topic: Eliptic Curve! (Read 592 times) |
|
Earendil
Newbie
 
Gender: 
Posts: 46
|
 |
Eliptic Curve!
« on: Mar 10th, 2004, 3:59pm » |
 Quote  Modify
|
Prove that 26 is the only natural number between a square (25) and a cube(27).
PS: Thx Sir Col for reminding me it is a natural number and not a integer
|
| « Last Edit: Mar 10th, 2004, 5:26pm by Earendil » |
 IP Logged |
|
|
|
Sir Col
Uberpuzzler
impudens simia et macrologus profundus fabulae
Gender:
Posts: 1825
|
|
Re: Eliptic Curve!
« Reply #1 on: Mar 10th, 2004, 4:46pm » |
Quote Modify
|
You obviously didn't mean x2=y3+2, otherwise 12=(-1)3+2 would be a solution too; that is, 0 is between the square, 1, and the cube, -1.
I know very little about generlisations regarding elliptic curves, but I know we're trying to prove that x2=y32 has one solution; namely, 52=332. I believe they're called Mordell curves?
I look forward to seeing how we solve these Diophantine equations...
|
|
IP Logged |
mathschallenge.net / projecteuler.net
|
|
|
Barukh
Uberpuzzler
Gender:
Posts: 2276
|
|
Re: Eliptic Curve!
« Reply #2 on: Mar 13th, 2004, 5:29am » |
Quote Modify
|
[smiley=blacksquare.gif]
Rewrite the equation as follows: y3 = x2 + 2. The idea is to factorize the RHS of this equation. Obviously, its not possible in [bbn], but it is possible in the ring R = [bbn][[sqrt](-2)], and fortunately the factorization is unique (see some discussion on this in y^2 = x^3 432 thread. There, we have (w stands for [sqrt](-2)): y3 = (x+w)(x-w).
The next step is to show that x+w, x-w are relatively prime in R. Let d = gcd(x+w, x-w), then d divides (x+w)-(x-w) = 2w = -w3. Since w is, of course, prime in R, d must be equal to one of 1, w, w2, w3. If d = w2, then x+w = w2(a+bw) = -2a 2bw, which is impossible. The case d = w3 is excluded similarly. d = w is a bit harder: in this case, (x+w)(x-w) = w2d, but this cannot be because it should be a cube of some number. So, d=1.
It follows then that both x+w and x-w are perfect cubes in R. Let x+w = (a+bw)3, a,b [in] [bbz]. Expanding and simplifying, we get: a3-6ab2 = x, 3a2b-2b3 = 1. Last equation factors (3a2-2b2)b = 1, which has the only solution a=1, b=1.
This proves the uniqueness of the solution.
[smiley=blacksquare.gif]
This proof was originally proposed by... Euler.
|
| « Last Edit: Mar 13th, 2004, 5:32am by Barukh » |
IP Logged |
|
|
|
Sir Col
Uberpuzzler
impudens simia et macrologus profundus fabulae
Gender:
Posts: 1825
|
|
Re: Eliptic Curve!
« Reply #3 on: Mar 13th, 2004, 9:47am » |
Quote Modify
|
I have the memory of a goldfish! When trying to solve NickH's problem I practised the method on exactly that equation! 
When I saw the title of the thread my mind went numb because I don't know much about elliptic curves, so I didn't contemplate for one moment that I knew how to solve it. Grr!
|
| « Last Edit: Mar 13th, 2004, 9:49am by Sir Col » |
IP Logged |
mathschallenge.net / projecteuler.net
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print