elliptic curves - cemc · 2020. 8. 24. · elliptic curves have in nitely many real points. as an...

Elliptic Curves

Dr. Carmen Bruni

University of Waterloo

November 4th, 2015

Dr. Carmen Bruni Elliptic Curves

Revisit the Congruent Number Problem

Congruent Number Problem

Determine which positive integers N can be expressed as the areaof a right angled triangle with side lengths all rational.

For example 6 is a congruent number since it is the area of the3− 4− 5 right triangle.


From Triangles to Curves

Now, we’re going to take the information about our triangleand get a new equation which will turn out to represent acurve in the real plane.

Let x2 + y2 = z2 and xy = 2N for rationals x , y , z and somecongruent number N.

Adding and subtracting 2xy = 4N to the first equation gives

x2 + 2xy + y2 = z2 + 4N x2 − 2xy + y2 = z2 − 4N

Factoring and dividing by 4 gives the two equations(x + y

2

)2

= (z/2)2 + N

(x − y

2

)2

= (z/2)2 − N



With the equations(x + y

2

)2

= (z/2)2 + N

(x − y

2

)2

= (z/2)2 − N

we multiply these two equations together gives(x + y

2

)2

·(x − y

2

)2

= ((z/2)2 + N)((z/2)2 − N)((x + y

2

)·(x − y

2

))2

= (z/2)4 + N(z/2)2 − N(z/2)2 − N2

((x − y)(x + y)

4

)2

= (z/2)4 − N2

(x2 − y2

4

)2

= (z/2)4 − N2



Letting u = z/2 and v = (x2 − y2)/4, the previous equationbecomes

v2 = u4 − N2

Multiplying by u2 gives

(uv)2 = (u2)3 − N2u2

Finally, we let y = uv and x = u2 which gives us the equation

y2 = x3 − N2x

We call such curves where y2 equals a cubic in x an EllipticCurve (provided the discriminant is nonzero; this is the casefor cubics associated to the Congruent Number Problem).


Examples of an Elliptic Curve

Let’s look at examples of elliptic curves. What do they looklike on the real plane?

Let’s try to draw y2 = x3 − x first by drawing y = x3 − x andthen trying to draw the elliptic curve.


Drawing y = x3 − x

First, note that y = x3 − x = x(x − 1)(x + 1) and so theequation has three zeroes at x = 0,±1.

Now let’s break this curve into four intervals and see whathappens in each interval y = x3 − x = x(x − 1)(x + 1).

Between −∞ and −1, the function is negative.

Between −1 and 0, the function is positive.

Between 0 and 1, the function is negative.

Between 1 and ∞, the function is positive.

Lastly, the curve should look smooth with no breaks.


The Cubic Curve y = x3 − x

Here is the picture (Using Desmos.com)


The Elliptic Curve y 2 = x3 − x

What changes when we make the left hand side y2 instead ofy?

For almost all values of x , we will get not 1 but 2 outputvalues (the exceptions are the roots).

This means that we no longer have a function, rather a curve.

The cubic must be positive to have a real root! So all theareas where the picture is negative are gone.

The curve still has no breaks and is symmetric about thex-axis, that is, if I reflect the top half of the picture, it shouldmatch the bottom half.

The function should still be smooth (even at 1).


The Cubic Curve y = x3 − x

Here is the picture (All graphs courtesy of Desmos.com)

Notice that the curve has two connected components!


Connected Components

Note: In general, not all elliptic curves have two components.Some have one like y2 = x3 − 1:

However, the elliptic curves associated to the Congruent NumberProblem always have two connect components.


Points on an elliptic curve

Elliptic curves have infinitely many real points.

As an example, y2 = x3 − x has infinitely many real points bynoticing that the cubic on the right is always positive whenx > 1 and hence we can find a y value by taking the squareroot.

So if we take x = 2, then we see that y2 = 23 − 2 = 6 and sothe point P = (2,

√6) and Q = (2,−

√6) are on the curve.


Points on an elliptic curve

From the perspective of Diophantine equations, it isinteresting to ask: How many integer points are on ellipticcurves?

For the example y2 = x3 − x , it turns out that (±1, 0) and(0, 0) are the only integer points, though this is hardlyobvious.

How many rational points are on elliptic curves?

Above, the only rational points are also the integral ones.

More on this later.


Group Law of an Elliptic Curve

With an elliptic curve, wecan actually describe a wayto, given two rational pointsP and Q, create a thirdrational point R.

Let’s begin with the ellipticcurve y2 = x3 − x + 1 forillustrative purposes.

y2 = x3 − x + 1



Let’s take the pointsP = (−1.324, 0) andQ = (0, 1) (correct to threedecimal places).

y2 = x3 − x + 1



Draw the line between Pand Q. It intersects thecurve in a third point asshown in the picture atcoordinates (1.895, 2.43).

y2 = x3 − x + 1



Draw the vertical linethrough the point whichmust intersect the curve in athird point, in our case,R = (1.895,−2.43) (this isthe same as reflecting aboutthe x-axis).

Define P + Q = R for pointson an elliptic curve (notethat this isn’t just addingthe coordinates!)

y2 = x3 − x + 1



If P = Q, then we can stilladd points.

Here, we use the tangentline to find a third point ofintersection.

To the right, we start withthe point P = (−1, 1) onthe same elliptic curve.

y2 = x3 − x + 1



Using calculus, we cancalculate the tangent line atP to be y = x + 2.

This intersects the ellipticcurve at the point (3, 5).

y2 = x3 − x + 1



Reflecting as before gives usthat 2P = P + P = (3,−5).

y2 = x3 − x + 1



What about if the linebetween P and Q is vertical?

We define a “point atinfinity” and call it R = O.This point intersects allvertical lines.

In this case, we also callQ = −P (this is thereflection of P about thex-axis).

ThusP − P = P + Q = R = O

y2 = x3 − x + 1


Your Turn!

Try an example. Add thepoints P = (0, 1) andQ = (3, 5).

y2 = x3 − x + 1


Your Turn!

The slope of the linebetween P and Q is

m =5− 1

3− 0=

4

3

and the y intercept is b = 1since P = (0, 1) is on theline y = 4

3x + 1.

Thus the equation of theline between P and Q isy = 4

3x + 1.

y2 = x3 − x + 1


Your Turn!

Where does the line y = 43x + 1 intersect y2 = x3 − x + 1?

Plug the equation of the line into the elliptic curve to get:

(43x + 1)2 = x3 − x + 1169 x

2 + 83x + 1 = x3 − x + 1

x3 − 169 x

2 − 113 x = 0

x(x2 − 169 x −

113 ) = 0

The last quadratic must have x = 3 as a root since we knowthe line intersects at the points P = (0, 1) and Q = (3, 5). Sofactoring the above gives

x(x − 3)(x + 119 ) = 0

Thus the other point of intersection occurs when x = −119 .

The corresponding y value is

y = 43(−11

9 ) + 1 = −4427 + 1 = −17

27


Your Turn!

This line intersects theelliptic curve at the point(−11

9 , −1727 ).

Then finally, reflecting(negating the y -coordinate)gives the pointR = (−11

9 , 1727)

y2 = x3 − x + 1


Formulas For Adding Points

Let’s summarize the above for adding two points P = (x1, y1)and Q = (x2, y2) on the elliptic curve y2 = x3 + Cx + D.

Let ` be the line connecting P and Q and suppose ` is definedby

y = mx + b

We can describe the slope m and the y -intercept b via

m =

{y2−y1x2−x1

If P 6= Q3x21+C2y1

If P = Qand b = y1 −mx1

where again we used calculus to compute the tangent line inthe case when P = Q.



As in our example, we can find the intersection of

y2 = x3 + Cx + D and y = mx + b

by solving

(mx + b)2 = x3 + Cx + D

m2x2 + 2mxb + b2 = x3 + Cx + D

0 = x3 −m2x2 + (C − 2mb)x + D − b2



This new polynomial has x1 and x2 as solutions since P and Q areon both the line and the curve. Hence,

0 = x3 −m2x2 + (C − 2mb)x + D − b2

= (x − x1)(x − x2)(x − x3)

= x3 − (x1 + x2 + x3)x2 + (x1x2 + x1x3 + x2x3)x − x1x2x3

which must hold for all values of x . Hence the coefficients oneither side match up. Thus, comparing the x2 coefficients oneither side gives

−m2 = −(x1 + x2 + x3) ⇒ x3 = m2 − x1 − x2

and y3 = mx3 + b. Hence reflecting gives

P + Q = (x3,−y3)


Formulas For Adding Points on y 2 = x3 − N2x

When we add P = (x , y) = Q on the elliptic curvey2 = x3 − N2x with N squarefree, the formula for thex-coordinate of P + P becomes:

(x2 − N2)2

(2y)2

(see the problem set). Notice here that the x-coordinate is asquare, has an even denominator and the numerator shares nocommon factor with N provided P 6= (0, 0) or (±N, 0) (seethe problem set).


Revisit the Congruent Number Problem

Congruent Number Problem

Determine which positive integers N can be expressed as the areaof a right angled triangle with side lengths all rational.


Key Theorem 1

Theorem 1.

Let (x , y) be a point with rational coordinates on the elliptic curvey2 = x3 − N2x where N is a positive squarefree integer. Supposethat x satisfies three conditions:

1 x is the square of a rational number

2 x has an even denominator

3 x has a numerator that shares no common factor with N

Then there exists a right angle triangle with rational sides and areaN, that is, N is congruent.


Key Theorem 2

Theorem 2.

A number N is congruent if and only if the elliptic curvey2 = x3 − N2x has a rational point P = (x , y) distinct from (0, 0)and (±N, 0).

Thus, determining congruent numbers can be reduced to findingrational points on elliptic curves!


Next Time

We prove these theorems.

We figure out how to go from a rational point on an ellipticcurve to a rational right triangle with area N.

We revisit Don Zagier’s example.

We discuss some tricks for finding rational points on ellipticcurves.


elliptic curves - cemc · 2020. 8. 24. · elliptic curves have in nitely many real points. as an...

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