MATH 216T
Homework 10
Solutions
1. Consider the elliptic curve
y 2 + 4y = x3 − 3x2 − 7x + 6
(a) Find a transformation of the form
X =x+a
Y =y+b
such that the equation of the elliptic curve is of the form
Y 2 = X 3 + AX + B
What are A and B?
(b) Find a point P (X, Y ) on this new elliptic curve with integral coordinates (so X and Y are integers). Find 2P .
Solution : (a) The transformation for Y is easy to see: we complete the square by adding 4 to both sides. We get
y 2 + 4y + 4 = x3 − 3x2 − 7x + 6 + 4
and so
(y + 2)2 = x3 − 3x2 − 7x + 10
Hence the substitution is
Y =y+2
The transformation for X may not be so obvious. We illustrate two methods.
If X = x + a then x = X − a. Substituting this into
x3 − 3x2 − 7x + 10
we find
(X − a)3 − 3(X − a)2 − 7(X − a) + 10
Working this out and simplifying, we get
X 3 − (3 + 3a)X 2 + (3a2 + 6a − 7)X + 10 + 7a − 3a2 − a3
To eliminate the term in X 2 , we must have that
3 + 3a = 0
So a = −1 and we get
X 3 − 10X + 1
In general, if we have two terms
xn + an−1 xn−1
and we make a substitution
X =x+a
the terms in X n and X n−1 are
X n + (an−1 − na)X n−1
Hence we must choose a =
an−1
and the substitution becomes
n
X =x+
1
an−1
n
In our example, the substitution is
X =x−1
3
Note that there is an ‘easy way’ to write f (x) = x − 3x2 − 7x + 10 in terms of powers of x − 1: Taylor series! We get
that
f 00 (1)
f 000 (1)
f (x) = f (1) + f 0 (1)(x − 1) +
(x − 1)2 +
(x − 1)3
2
3!
We easily get that
f (x) = x3 − 3x2 − 7x + 10 so f (1) = 1
f 0 (x) = 3x2 − 6x − 7
so f 0 (1) = −10
00
f (x) = 6x − 6
so f 00 (1) = 0
000
f (x) = 6
so f 000 (1) = 6
Hence
x3 − 3x2 − 7x + 10 = 1 − 10(x − 1) + (x − 1)3
The substitution
X =x−1
transforms y 2 + 4y = x3 − 3x2 − 7x + 6 into Y 2 = X 3 − 10X + 1
Y =y+2
(b) It easily follows from (a) that P = (0, 1) lies on the curve C : Y 2 = X 3 − 10X + 1. To find 2P , we consider the
tangent line l to C at P . Using implicit differentiation, we get
2Y dY = 3X 2 dX − 10 dX
Substituting (X, Y ) = (0, 1), we find that the slope of l is
dY
= −5
dX
So the equation of l is
l : Y = −5X + 1
The points of intersection between l and C are the solutions of
(
Y = −5X + 1
2
3
Y = X − 10X + 1
(1)
(2)
Substituting (1) into (2), we get
(−5X + 1)2 = X 3 − 10X + 1
and so
X 3 − 25X 2 = 0
Since the sum of the solutions is 25 and X = 0 is a solution of multiplicity 2, we get that X = 25 and so Y =
−5 · 25 + 1 = −124. Hence
P = (0, 1) lies on Y 2 = X 3 − 10X + 1 and 2P = (25, 124).
2. Suppose we have an elliptic curve
y 2 = x3 + ax2 + bx + c
Then we can still define the addition of two points on the curve. As usual, e denotes the point at infinity (the ‘vertical’
direction). If P is a point on the curve, we define P + e = P .
So let P1 and P2 be two points on the curve with P1 6= e 6= P2 . Then the line through P1 and P2 (or the tangent
line at P1 if P1 = P2 ) intersects the curve in a third point P . We put P1 + P2 = −P where −P = e if P = e and
−P = (x, −y) if P = (x, y).
Consider the elliptic curve
C : y 2 = x3 − 2x2 + 5x
Then P (1, 2) and Q(0, 0) are points on this curve.
2
(a) Find P + Q.
(b) Find 2P .
(c) What is the order of P ?
Solution : (a) We easily get that the line l through P and Q has equation
l : y = 2x
So the points of intersection between l and C are the solutions of
(
y = 2x
(1)
y 2 = x3 − 2x2 + 5x (2)
Substituting (1) into (2), we get
(2x)2 = x3 − 2x2 + 5x
and so
x3 − 6x2 + 5x
Since the sum of the solutions is 6 and x = 1 and x = 0 are solutions, we get that x = 5 and so y = 2 · 5 = 10.
(1, 2) + (0, 0) = (5, −10)
(b) Let l be the tangent line to C at P . Using implicit differentiation, we get
2y dy = 3x2 dx − 4x dx + 5 dx
Substituting (x, y) = (1, 2), we find that the slope of l is
dy
=1
dx
So the equation of l is
l :y =x+1
The points of intersection between l and C are the solutions of
(
y =x+1
2
3
(1)
2
y = x − 2x + 5x (2)
Substituting (1) into (2), we get
(x + 1)2 = x3 − 2x2 + 5x
and so
x3 − 3x2 + 3x − 1 = 0
Since the sum of the solutions is 3 and x = 1 is a solution of multiplicity 2, we get that x = 1 and so y = 1 + 1 = 2.
(1, 2) + (1, 2) = (1, −2)
(c) From (b), we get that
2P = P + P = (1, 2) + (1, 2) = (1, −2) = −P
So 3P = e. Since P 6= e, we have that P has order 3.
(1, 2) has order 3.
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