**TI-84 Plus CE: Simple Elliptic Curves Determined by Two Points**

**Introduction**

The program ECURVE determines the coefficients of a simple elliptic curve:

y^2 = x^3 + G * x + H

from two points P:(A,B) and Q:(C,D). It also computes the point R:(E,F) by the following equations:

E = θ^2 - A - C

F = θ * (A + C) - B

Determining G and H.

Using the two points P:(A,B) and Q:(C,D):

A^3 + A * G + H = B^2

C^3 + C * G + H = D^2

A * G + H = B^2 - A^3

C * G + H = D^2 - C^3

Subtracting the bottom equation from the top gets:

( A - C ) * G = (B^2 - A^3) - (D^2 - C^3)

G = ( (B^2 - A^3) - (D^2 - C^3) ) / (A - C)

H can be determined one of two ways:

H = B^2 - A^3 - A * G

H = D^2 - C^3 - C * G

**TI-84 Plus CE Program: ECURVE**

Func

FnOff

"EWS 2019-10-20"

Disp "P:(A,B), Q:(C,D)"

Prompt A,B,C,D

(D-B)/(C-A)→θ

θ^2-A-C→E

θ*(A+E)-B→F

Disp "R:(E,F)",E,F

Pause

((B^2-A^3)-(D^2-C^3))/(A-C)→G

B^2-A^3-A*G→H

Disp "Y^2=X^3+G*X+H",G,H

Pause

"√(X^3+G*X+H)"→Y_1

"-Y_1"→Y_2

ZoomFit

Y_1 and Y_2 are the Y1 and Y2 from the Vars, Y-Vars, Function menu.

**Example 1**

P:(-2, 3) (A = -2, B = 3)

Q:(-1, 0) (C = -1, D = 0)

Results:

R:(12, -33) (E = 12, F = -33)

G = -16

H = -15

Equation: y^2 = x^3 - 16 x - 15

**Example 2**

P:(-1, 1) (A = -1, B = 1)

Q:(2, 4) (C = 2, D = 4)

Results:

R:(0, -2) (E = 0, F = -2)

G = 2

H = 4

Equation: y^2 = x^3 + 2 x + 4

Source:

Rosing, Michael.

__Implementing Elliptic Curve Cryptography__Manning Publishing Co: Greenwich, CT 1999 ISBN 10: 1884777694

Eddie

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