If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!

**Ramanujan’s Taxi Cabs and the Sum of 2 Cubes **

The Indian mathematician Ramanujan (picture cite: Wikipedia) is renowned as one of great self-taught mathematical prodigies. His correspondence with the renowned mathematician G. H Hardy led him to being invited to study in England, though whilst there he fell sick. Visiting him in hospital, Hardy remarked that the taxi that had brought him to the hospital had a very “rather dull number” – number 1729. Ramanujan remarked in reply, ” No Hardy, it’s a very interesting number! It’s the smallest number expressible as the sum of 2 cubes in 2 different ways!”

Ramanujan was profoundly interested in number theory – the study of integers and patterns inherent within them. The general problem referenced above is finding integer solutions to the below equation for given values of A:

In the case that A = 1729, we have 2 possible ways of finding distinct integer solutions:

The smallest number which can be formed through 3 distinct (positive) integer solutions to the equation is A = 87, 539, 319.

Although this began as a number theory problem it has close links with both graphs and group theory – and it is from these fields that mathematicians have gained a deeper understanding as to the nature of its solutions. The modern field of elliptical curve cryptography is closely related to the ideas below and provides a very secure method of encrypting data.

We start by sketching the graph of:

For some given integer value of A. We will notice that the graph has a line of symmetry around y = x and also an asymptote at y = -x. If we plot:

We can see that both our integer solutions to this problem (1,12) and (9,10) lie on the curve:

**Group theory**

Groups can be considered as sets which follow a set number of rules with regards to operations like multiplication, addition etc. Establishing that a set is a group then allows certain properties to be inferred. If we can establish the following rules hold then we can create an Abelian group. If we start with a set A and and operation Θ.

1) **Identity.** For an element e in A, we have a Θ e = a for all a in A.

(for example 0 is the identity element for the addition operation for the set of integers numbers. a+0 = a for all a in the real numbers).

2) **Closure**. For all elements a,b in A, a Θ b = c, where c is also in A.

(For example with the addition operation, the addition of 2 integers numbers is still an integer)

3) **Associativity**. For all elements a,b,c in A, (a Θ b) Θ c = a Θ (b Θ c)

(For example with the addition operation, (1+2) + 3 = 1 + (2+3) )

4) **Inverse**. For each a in A there exists a b in A such that a Θ b = b Θ a = e. Where e is the identity.

(For example with the addition operation, 4+-4 = -4+4 = 0. 0 is the identity element for addition)

5) **Commutativity**. For all elements a,b in A, a Θ b = b Θ a

(For example with the addition operation 1+2 = 2+1).

As we have seen, the set of integers under the operation addition forms an abelian group.

**Establishing a group**

So, let’s see if we can establish a Abelian group based around the rational coordinates on our graph. We can demonstrate with the graph:

We then take 2 coordinate points with rational coordinates (i.e coordinates that can be written as a fraction of integers). In this case A (1,12) and B (9,10).

We then draw the line through A and B. This will intersect the graph in a 3rd point, C (except in a special case to be looked at in a minute).

We then reflect this new point C in the line y = x, giving us C’.

In this case C’ is the point (46/3, -37/3)

We therefore define *addition* (our operation Θ) in this group as:

A + B = C’.

(1,12) + (9,10) = (46/3, -37/3).

We now need to deal with the special case when a line joining 2 points on the curve does not intersect the curve again. This will happen whenever the gradient of this line is -1, which will make it parallel to the graph’s asymptote y = -x.

In this case we affix a special point at infinity to the Cartesian (x,y) plane. We define this point as the point through which all lines with gradient -1 intersect. Therefore in our expanded geometry, the line through AB *will* intersect the curve at this point at infinity. Let’s call our special point Φ. Now we have a new geometry, the (x,y) plane affixed with Φ.

We can now create an Abelian group. For any 2 rational points P(x,y), Q(x,y) we will have:

1) **Identity.** P + Φ = Φ + P = P

2) **Closure**. P + Q = R. (Where R(x,y) is also a rational point on the curve)

3) **Associativity**. (P+Q) + R = P+(Q+R)

4) **Inverse**. P + (-P) = Φ

5) **Commutativity**. P+Q = Q+P

**Understanding the identity**

Let’s see if we can understand some of these. For the identity, if we have a point A on the line and the point at infinity then this will contain the line with gradient -1. Therefore the line between the point at infinity and A will intersect the curve again at B. Our new point, B’ will be created by reflecting this point in the line y = x. This gets us back to point A. Therefore P + Φ = P as required.

**Understanding the inverse**

With the inverse of our point P(x,y) given as -P = (-x,-y) we can see that this is the reflection in the line y = x. We can see that we we join up the 2 points reflected in the line y = x we will have a line with slope -1, which will intersect with the curve at our point at infinity. Therefore P + (-P) = Φ.

Through our graphical understanding the commutativity rule also follows immediately, It doesn’t matter which of the 2 points come first when we draw a line that connects them, therefore P+Q = Q+P.

**Understanding associativity and closure**

Neither associativity nor closure are obvious from our graph. We could check individual points to show that (P+Q) + R = P+(Q+R), but it would be harder to explain why this always held. Equally whilst it’s clear that P+Q will always create a point on the curve it’s not obvious that this will be a *rational* point.

In fact we do have both associativity and closure for our group as we have the following algebraic definition for our addition operation:

The addition of 2 points is given by:

In the case of our curve:

If we take P = (1,12). P + P will be given by:

We can check this result graphically. If P and Q are the same point, then the line that passes through both P and Q has to be the tangent to the curve at that point. Therefore we would have:

Here the tangent at A does indeed meet the curve again – at point C, which does reflect in y = x to give us the coordinates above.

We could also find this intersection point algebraically. If we differentiate the original curve to find the gradient when x = 1 we can find the equation of the tangent when x=1 and then substitute this back into the equation of the curve to find the intersection point. This would give us:

We would then reverse the x and y coordinates to reflect in the line y = x. This also gives us the same coordinates.

More generally if we have the 2 rational coordinates on the curve:

We have the algebraic formula for addition as:

If P = (1,12) and Q = (9,10), P + Q would give (after much tedious substitution!):

This agrees with the coordinates we found earlier using the much easier geometrical approach. As we can see from this formula, both coordinate points will always be rational – as they will be composed of combinations of our original rational coordinates. For any given curve there will be a generator set of coordinates through which we can generate all other rational coordinates on the curve through our addition operation.

So, we seem to have come a long way from our original goal – finding integer solutions to an algebraic equation. Instead we seem to have got sidetracked into studying graphs and establishing groups. However by reinterpreting this problem as one in group theory then this then opens up many new mathematical techniques to help us understand the solutions to this problem.

A fuller introduction to this topic is the very readable, “Taxicabs and the Sum of Two Cubes” by Joseph Silverman (from which the 2 general equations were taken) .

**Essential Resources for IB Teachers**

If you are a **teacher** then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over **2000 pages of pdf content** for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:

**Original pdf worksheets**(with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.**Original Paper 3 investigations**(with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.- Over 150 pages of
**Coursework Guides**to introduce students to the essentials behind getting an excellent mark on their exploration coursework. - A large number of
**enrichment activities**such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.

There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!

**Essential Resources for both IB teachers and IB students**

1) Exploration Guides and Paper 3 Resources

I’ve put together a **168 page** Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made **Paper 3 packs** for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.

## Leave a comment

Comments feed for this article