The Nagell-Lutz Theorem

peccavi_2006 · **Posted:** Thu, 23 Feb 2012 16:18:44 UTC

My Supervisor wants me to give a talk on part of my project, and decided that the Nagell-Lutz theorem was the most interesting, yet easiest to convey, piece of reading I had done.
The problem is, I know that the theorem isn't an "iff" statement - having $(x,y) \in \mathbb{Z}$ with y|D

(where D is the discriminant of f(x)), does not mean that $P=(x,y) \in C : y^{2} = f(x) = x^{3} + ax^{2} + bx + c;$ where $a,b,c \in \mathbb{Z}$ is a point of finite order. Alas, I'm having trouble finding such a point.
I want to use the curve $y^{2} = x^{3} + 17$ , and I thought that the point P=(8,9)

would be a good candidate (as it satisfies all of the conditions of the converse of the theorem). However, trying to calculate 2P I keep getting $(\frac{880}{9}, - \frac{26099}{27})$ , which is not on the curve ( $x^{3} + 17 > y^{2}$ ). I can't work out what I'm doing wrong - the formulae I'm using is correct because I've used it plenty of other times to duplicate points, and once I work out the kinks with this stupid point, it will clearly be contradiction to the converse (as P will satisfy all of the conditions, but have infinite order). So can someone tell me where I'm going wrong? Because this is starting to irritate me.
The formulae I'm using are:
$x(2P) = \lambda ^{2} - a - 2x_{1}$ where $$ \lambda = \frac{f'(x_{1})}{2y_{1}}$ and here a =0

and
$y(2P) = -(\lambda x_{3} + \nu)$ , where $\nu = y_{1} - \lambda x_{1}$

outermeasure · **Posted:** Thu, 23 Feb 2012 16:43:08 UTC

peccavi_2006 wrote:

My Supervisor wants me to give a talk on part of my project, and decided that the Nagell-Lutz theorem was the most interesting, yet easiest to convey, piece of reading I had done.
The problem is, I know that the theorem isn't an "iff" statement - having $(x,y) \in \mathbb{Z}$ with y|D

(where D is the discriminant of f(x)), does not mean that $P=(x,y) \in C : y^{2} = f(x) = x^{3} + ax^{2} + bx + c;$ where $a,b,c \in \mathbb{Z}$ is a point of finite order. Alas, I'm having trouble finding such a point.
I want to use the curve $y^{2} = x^{3} + 17$ , and I thought that the point P=(8,9)

would be a good candidate (as it satisfies all of the conditions of the converse of the theorem). However, trying to calculate 2P I keep getting $(\frac{880}{9}, - \frac{26099}{27})$ , which is not on the curve ( $x^{3} + 17 > y^{2}$ ). I can't work out what I'm doing wrong - the formulae I'm using is correct because I've used it plenty of other times to duplicate points, and once I work out the kinks with this stupid point, it will clearly be contradiction to the converse (as P will satisfy all of the conditions, but have infinite order). So can someone tell me where I'm going wrong? Because this is starting to irritate me.
The formulae I'm using are:
$x(2P) = \lambda ^{2} - a - 2x_{1}$ where $$ \lambda = \frac{f'(x_{1})}{2y_{1}}$ and here a =0

and
$y(2P) = -(\lambda x_{3} + \nu)$ , where $\nu = y_{1} - \lambda x_{1}$

The point (8,9) does not lie on the curve (8^3+17=529, not 81).

Try the point P=(-1,4)

instead, for example. Alternatively, use the curve y^2=x^3+3

and

, which is possibly the simplest (counter)example.

peccavi_2006 · **Posted:** Thu, 23 Feb 2012 18:21:44 UTC

ugh, this is what happens when people spring talks on me - I lose my head completely.

I was going to use (-1,4), as at some point I duplicate it for some reason (I think just to show how the formulas work), but I can't use it as a contradiction to the converse of the theorem because 4 doesn't divide the discriminant

I will totally use the other curve and your suggestion for P - thanks outermeasure

Edit: Oh! I know why I thought (8,9) was on the curve - I had $x^{3} = 64$ for the x coordinate...and then square-rooted both sides because algebra is just that hard >.<

outermeasure · **Posted:** Fri, 24 Feb 2012 05:15:32 UTC

peccavi_2006 wrote:

ugh, this is what happens when people spring talks on me - I lose my head completely.

I was going to use (-1,4), as at some point I duplicate it for some reason (I think just to show how the formulas work), but I can't use it as a contradiction to the converse of the theorem because 4 doesn't divide the discriminant

I will totally use the other curve and your suggestion for P - thanks outermeasure

Edit: Oh! I know why I thought (8,9) was on the curve - I had $x^{3} = 64$ for the x coordinate...and then square-rooted both sides because algebra is just that hard >.<

Why? For an elliptic curve y^2=x^3+Ax+B

, the discriminant is $\Delta=-16(4A^3+27B^2)$ , so obviously 4 divides the discriminant.

Hmm... maybe you are using the version with modular discriminant instead?

peccavi_2006 · **Posted:** Fri, 24 Feb 2012 14:09:29 UTC

yep you're totally right - I stupidly copied down the equation for the discriminant as $... - 27b^{3}$ :cry:

Oh well - no one will notice it on the poster, I'm sure...

S.O.S. Mathematics CyberBoard

The Nagell-Lutz Theorem

Who is online