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
So for example the following is an elliptical curve:
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!
What makes elliptical curves so useful is that we can create a group structure using them. Groups are very important mathematical structures – again 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).
Trying another example, y² = x³ -5x + 4 (above), with A = (1,0) and B = (0,2) we have C = (3,-4) and C’ = (3,4). Therefore (1,0) + (0,2) = (3,4).
We can also have elliptical curves as a collection of discrete points. We start with the curve y² = x³ +x+1, above, and just look at the positive integer solutions mod 7. For example, 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
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).
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.
If you enjoyed this post you might also like:
RSA Encryption – the encryption system which secures the internet.
Circular inversion – learn about some other geometrical transformations used in university level mathematics.