**Elliptical Curve Cryptography**

*This post builds on some of the ideas in the previous post on elliptical curves. This blog originally appeared in a Plus Maths article I wrote here. The excellent Numberphile video above expands on some of the ideas below.
*

On a (slightly simplified) level elliptical curves they can be regarded as curves of the form:

y² = x³ +ax + b

So for example the curve below is an elliptical curve. This curve also has an added point at infinity though we don’t need to worry about that here. Elliptical curve cryptography is based on the difficulty in solving arithmetic problems on these curves. If you remember from the last post, we have a special way of defining the addition of 2 points.

Let’s say 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 (below), 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 also need to have a definition when A and B define the same point on the curve. This will give us the definition of 2A. In this case we take the tangent to the curve at that point, and then as before find the intersection of this line and the curve, before reflecting the point. This probably is easier to understand with another graph:

Here we used the graph y² = x³ -5x + 4 again. This time point A = B = (-1.2, 2.88) and we have drawn the tangent to the curve at this point, which gives point D, which is then reflected in the x axis to give D’. D’ = (2.41, -2.43). Therefore we can say 2A = D’, or 2(-1.2, 2.88) = (2.41, -2.43).

Now addition of points is defined we can see how elliptical curve cryptography works. The basic idea is that given 2 points on the curve, say A and B, it takes a huge amount of computing power to work out the value a such that aA = B. For example, say I use the curve y² = x³ -25x to encrypt, and the 2 points on the curve are A = (-4,6) and B = (1681/144 , -62279/1728). Someone who wanted to break my encryption would need to find the value a such that a(-4,6) = (1681/144 , -62279/1728). The actual answer is a =2 which we can show graphically. As we want to show that 2(-4,6) = (1681/144 , -62279/1728) , we can use the previous method of finding the tangent at the point (-4,6):

We can then check with Geogebra which shows that B’ is indeed (1681/144 , -62279/1728). When a is chosen so that it is very large, this calculation becomes very difficult to attack using brute force methods – which would require checking 2(4,-6), 3(4,-6), 4(4,-6)… until the solution (1681/144 , -62279/1728) was found.

**NSA and hacking data**

Elliptical curve cryptography has some advantages over RSA cryptography – which is based on the difficulty of factorising large primes – as less digits are required to create a problem of equal difficulty. Therefore data can be encoded more efficiently (and thus more rapidly) than using RSA encryption. Currently the digital currency Bitcoin uses elliptical curve cryptography, and it is likely that its use will become more widespread as more and more data is digitalised. However, it’s worth noting that as yet no-one has proved that it* has* to be difficult to crack elliptical curves – there may be a novel approach which is able to solve the problem in a much shorter time. Indeed many mathematicians and computer scientists are working in this field.

Government digital spy agencies like the NSA and GCHQ are also very interested in such encryption techniques. If there was a method of solving this problem quickly then overnight large amounts of encrypted data would be accessible – and for example Bitcoin currency exchange would no longer be secure. It also recently transpired that the NSA has built “backdoor” entries into some elliptical curve cryptography algorithms which have allowed them to access data that the people sending it thought was secure. Mathematics is at the heart of this new digital arms race.

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.

**IB Revision**

If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

The Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!

The Practice Exams section takes you to ready made exams on each topic – again with worked solutions. This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.

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