wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> hard >> Eliptic Curve!
        
(Message started by: Earendil on Mar 10^th, 2004, 3:59pm)

Title: Eliptic Curve!
Post by Earendil on Mar 10^th, 2004, 3:59pm

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

Title: Re: Eliptic Curve!
Post by Sir Col on Mar 10^th, 2004, 4:46pm

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=y³–2 has one solution; namely, 5²=3³–2. I believe they're called Mordell curves?

I look forward to seeing how we solve these Diophantine equations...

Title: Re: Eliptic Curve!
Post by Barukh on Mar 13^th, 2004, 5:29am

[smiley=blacksquare.gif][hide]
Rewrite the equation as follows: y³ = x² + 2. The idea is to factorize the RHS of this equation. Obviously, it’s 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 (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1077834292;start=6) 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.
[/hide][smiley=blacksquare.gif]

This proof was originally proposed by... Euler.

Title: Re: Eliptic Curve!
Post by Sir Col on Mar 13^th, 2004, 9:47am

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!