Exercise 5.7.15 (Part of Nagell-Lutz Theorem)

Show that 27 ( x 3 Ax + B ) ( x 3 Ax B ) ( 3 x 2 4 A ) ( 3 x 2 A ) 2 = 4 A 4 27 B 2 . Deduce that if an elliptic curve is given by y 2 = x 3 Ax B , with A and B integers, and if P 1 = ( x 1 , y 1 ) and 2 P 1 are points with integral coordinates, 2 P 1 O , then y 1 2 ( 4 A 3 27 B 2 ) .

Answers

Proof. To verify the identity, it is sufficient to expand the left member. With Sagemath

var(’x,A,B’)
u = 27*(x^3 - A*x + B)*(x^3 - A*x -B) - (3*x^2 - 4*A)*(3*x^2 - A)^2; u

( 3 x 2 A ) 2 ( 3 x 2 4 A ) + 27 ( x 3 Ax + B ) ( x 3 Ax B )

u.expand()

4 A 3 27 B 2

So

27 ( x 3 Ax + B ) ( x 3 Ax B ) ( 3 x 2 4 A ) ( 3 x 2 A ) 2 = 4 A 4 27 B 2 . (1)

Let f ( x , y ) = y 2 p ( x ) , where p ( x ) = x 3 Ax B , and A and B are integers. By the explicit formulas (5.53) (where a = 0 ) may be rewritten as

x 3 = x ( 2 P 1 ) = ( p ( x 1 ) 2 y 1 ) 2 2 x 1 , where  y 1 2 = p ( x 1 ) = x 1 3 A x 1 B .

Since x 1 , y 1 , x 3 , p ( x 1 ) are integers, this shows that y 1 2 p ( x 1 ) 2 . Since y 1 2 = p ( x 1 ) , we obtain

y 1 2 27 ( x 1 3 A x 1 + B ) p ( x 1 ) ( 3 x 1 2 4 A ) p ( x 1 ) 2 .

By identity (1), this gives

y 1 2 4 A 3 27 B 2 .

If P 1 = ( x 1 , y 1 ) and 2 P 1 are points with integral coordinates of the curve y 2 = x 3 Ax B (where A , B are integers), 2 P 1 O , then y 1 2 4 A 3 27 B 2 . □

Note: That is the easy part of the (strong form of) Theorem of Nagell-Lutz Theorem: A point ( x 1 , y 1 ) with finite order of E f ( ) has integral coordinates and satisfies y 1 = 0 or y 1 2 D . (See Silverman, Tate “Rational points on elliptic curves, Chapter 2, and Exercise 2.11). This allows us to find the torsion group of E f ( ) .

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2025-05-26 10:49
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