Elliptical Curve Cryptography

Elliptical curves are a very important new area of mathematics which have been greatly explored over the past few decades.  They have shown tremendous potential as a tool for solving complicated number problems and also for use in cryptography.

Andrew Wiles, who solved one of the most famous maths problems of the last 400 years, Fermat’s Last Theorem, using elliptical curves.  In the last few decades there has also been a lot of research into using elliptical curves instead of RSA encryption to keep data transfer safe online.  So, what are elliptical curves?  On a simple level they can be regarded as curves of the form:

y² = x³ +ax + b


If we’re being a bit more accurate, we also need 4a³ + 27b² ≠ 0.  This stops the graph having “singular points” which cause problems with the calculations.  We also have a “point at infinity” which can be thought of as an extra point added on to the usual x,y plane representing infinity.  This also helps with calculations – though we don’t need to go into this in any more detail here!

Addition of two points A and B

What makes elliptical curves so useful is that we can create a group structure using them.  Groups are very important mathematical structures because of their usefulness in being applied to problem solving.  A group needs to have certain properties.  For example, we need to be able to combine 2 members of the group to create a 3rd member which is also in the group.  This is how it is done with elliptical curves:


Take 2 points A and B on y² = x³ -4x + 1.  In the example we have A = (2,1) and B = (-2,-1).   We now want to find an answer for A + B which also is on the elliptical curve.  If we add them as we might vectors we get (0,2) – but unfortunately this is not on the curve.  So, we define the addition A + B through the following geometric steps.


We join up the points A and B.  This line intersects the curve in one more place, C.


We then reflect the point C in the x axis. We then define this new point C’ = A + B.  In this case this means that (2,1) + (-2,-1) = (1/4, -1/8).

Addition of 2 points when A = B

We have to also be able to cope with the situation when the point A and B are the same.   Here we create the line through A which is the tangent to the curve at that point:

We then use the same transformation as before to say that A+B = C’.  For example with the curve y² = x³ -12x, if we start with the point A(-2,4) then this transformation tells us that A + A = (4,-4).

Elliptical curves over finite fields

For the purposes of cryptography we often work with elliptical curves over finite fields.  This means we (say) only consider integer coordinate solutions and work in modulo arithmetic (mod prime).

Say we start with the curve y² = x³ +x+1, and just look at the positive integer solutions mod 7.  (Plotted using the site here).

When x = 1,

y² = 1³ +1 + 1

y² = 3

So this has no integer solution.

Next, when x  = 2 we have:

y² = 2³ +2 +1 = 11.

However when we are working mod 7 we look at the remainder when 11 is divided by 7 (which is 4).  So:

y² = 4 (mod 7)

y = 2 or y = -2  = 5 (mod 7)

When x = 3 we have:

y² = 3³ +3 +1 = 31

y² = 3 (mod 7)

which has no integer solutions.

In fact, all the following coordinate points satisfy the equation (mod 7):

(2,2), (0,1), (0,6), (2,5).

Addition under modulo arithmetic

Let’s look at the coordinate points we calculated before for the elliptical curve y² = x³ +x+1 (integers solutions and mod 7) – they form a group under addition.  (Table generated here)

In order to calculate addition of points when dealing with elliptical curves with integer points mod prime we use the same idea as expressed above for general graphs.

The table tells us that (0,1) + (0,1) = (2,5).  If we were doing this from the graph we would draw the tangent to the curve at (0,1), find where it intersects the graph again, then reflect this point in the x axis.  We can do all this algebraically.

First we find the gradient of the tangent when x = 0:

Next we have to do division modulo 7 (you can use a calculator here, and you can also read more about division modulo p here).

Next we find the equation of the tangent through (0,1):

Next we find where this tangent intersects the curve again (I used Wolfram Alpha to solve this mod 7)

We then substitute the value x = 2 into the original curve to find the y coordinates:

(2,2) is the point where the tangent would touch the curve and (2,5) is the equivalent of the reflection transformation.  Therefore our answer is (2,5).  i.e (0,1) + (0,1) = (2,5) as required.

When adding points which are not the same we use the same idea – but have to find the gradient of the line joining the 2 points rather than the gradient of the tangent.  We can also note that when we try and add points such as (2,5) and (2,2) the line joining these does not intersect the graph again and hence we affix the point an infinity as (2,5) + (2,2).

Using elliptical codes for cryptography

Even though all this might seem very abstract, these methods of calculating points on elliptical curves form the basis of elliptical cryptography.  The basic idea is that it takes computers a very long time to make these sorts of calculations – and so they can be used very effectively to encrypt data.

Say for example two people wish to create an encryption key.

They decide on an elliptical curve and modulo. Let’s say they decide on y² = x³ +x+1 for integers, mod 7.

This creates the addition group

Next they choose a point of the curve.  Let’s say they choose P(1,1).

Person 1 chooses a secret number n and then sends nP (openly).  So say Person 1 chooses n = 2.  2(1,1) = (1,1) + (1,1) = (0,2).  Person 1 sends (0,2).

Person 2 chooses a secret number m and then sends mP (openly).  So say Person 2 chooses m = 3.  3(1,1) = (1,1) + (1,1) + (1,1) = (0,2) + (1,1) = (0,5).  Person 2 sends (0,5).

Both Person 1 and Person 2 can easily calculate mnP (the secret key).

Person 1 receives (0,5) and so does 2(0,5) = (0,5) + (0,5) = (1,1).  This is the secret key.

Person 2 receives (0,2) and so does 3(0,2) = (0,2) + (0,2) +(0,2) = (1,1).  This is the same secret key.

But for a person who can see mP and nP there is no quick method for working out mnP – with a brute force approach extremely time consuming. Therefore this method can be successfully used to encrypt data.