### Taxicabs

A while back, Lieven Le Bruyn linked to an article on the story of Ramanujan and Hardy and the number 1729. In 1919, Ramanujan knew instantly that 1729 was an interesting number, because it may be expressed in two ways as

$1^{3} + 12^{3} = 9^{3} + 10^{3}$.

On being pressed further, Ramanujan did not know the smallest number that may be expressed as a sum of two fourth powers, although Euler had solved this problem. But there is a good reason why 1729 is more natural. Recall that Pythagorean triples solve equations of the form

$x^{2} + y^{2} = z^{2}$

because they form three sides of an irregular triangle, whose lengths are expressed as areas of squares. So the simplest cubic analogue should be about volumes on the faces of a tetrahedron. And as every category theorist knows, tetrahedra are naturally marked by face pairs, with a three dimensional arrow going between face compositions.

Let's try to match the Ramanujan quadruple $(1,12,9,10)$ to volumes associated to the faces of a tetrahedron. Each edge of the tetrahedron must correspond to a distinct pair of numbers, like $(9,12)$. The opposite edge corresponds to the conjugate pair, which for $(9,12)$ is $(1,10)$. We could choose the tetrahedron with edge lengths equal to, say, the average of the numbers in the edge pair, but there ought to be a

$(1,3, \sqrt{10})$

$(3,4,5)$

$(5, \sqrt{10}, \sqrt{17})$

$(1,4, \sqrt{17})$

which is kind of cute, since $10 - 1 = 9$, $17 - 16 = 1$, $10 + 17 - 25 = 2$ and so on.

$1^{3} + 12^{3} = 9^{3} + 10^{3}$.

On being pressed further, Ramanujan did not know the smallest number that may be expressed as a sum of two fourth powers, although Euler had solved this problem. But there is a good reason why 1729 is more natural. Recall that Pythagorean triples solve equations of the form

$x^{2} + y^{2} = z^{2}$

because they form three sides of an irregular triangle, whose lengths are expressed as areas of squares. So the simplest cubic analogue should be about volumes on the faces of a tetrahedron. And as every category theorist knows, tetrahedra are naturally marked by face pairs, with a three dimensional arrow going between face compositions.

Let's try to match the Ramanujan quadruple $(1,12,9,10)$ to volumes associated to the faces of a tetrahedron. Each edge of the tetrahedron must correspond to a distinct pair of numbers, like $(9,12)$. The opposite edge corresponds to the conjugate pair, which for $(9,12)$ is $(1,10)$. We could choose the tetrahedron with edge lengths equal to, say, the average of the numbers in the edge pair, but there ought to be a

*right angled*tetrahedron for the Ramanujan numbers. This tetrahedron would have three Pythagorean faces and one skew face. Now the smallest Pythagorean triple $(3,4,5)$ provides a right angled tetrahedron with edge lengths $(1,3,4,5, \sqrt{10} , \sqrt{17})$. The Ramanujan triangles would have to be$(1,3, \sqrt{10})$

$(3,4,5)$

$(5, \sqrt{10}, \sqrt{17})$

$(1,4, \sqrt{17})$

which is kind of cute, since $10 - 1 = 9$, $17 - 16 = 1$, $10 + 17 - 25 = 2$ and so on.

## 5 Comments:

oh i thought there was going to be an interesting anecdote on taxis :( I'll go back to my knitting

Hi Kerie. The first link has the taxi story. I hope the knitting is progressing well, since winter is on the way.

Its difficult to get a good geometric interpretation of the Ramanujan quadruple and perhaps your description uses the tetrahedron in more of a combinatorical sense than a geometric one.

In fact the areas of the faces can be related using their squares. I.e. if A, B, C, D are the areas of the faces with A,B and C being for the right angled triangles and D the acute traingle, then the relation is

D^2 = A^2 + B^2 + C^2

(use heron's formula for easy algebraic proof)

You can also make a tetrahedron where all four face triangles are right angled and the area relationship comes out as

A^2 + B^2 = C^2 + D^2

Hi Phil

Duh! The completely right angled tetrahedron must be the right one. Then we can take 9 to the 2/3 etc to get numbers from the Ramanujan set. Nice.

LOL! That's the way to do it.

Actually it is quite useful to map number theory problems to polytopes or graphs to help you think about their symmetries in that way, even if the equations themselves dont have a perfect geometric interpretation like the pythagorean triples do.

Many years ago I got interested in a number theory problem that dates back to Diophantus. He was looking for sets of four numbers such that the product of any two is one less than a square. You can see how that problem maps very nicely to a tetrahedron. After a long time I realised that the problem has a natural generalisation where the geometric structure is a cube with the embedded tetrahedron giving the special case. You then find that the 2x2x2 hyperdeterminant is key to solving the problem.

I put a paper about it on the arxiv back in the days when they accepted submissions from nonentities like me. It got cited by Duff et al when they found the relationship between hyperdterminants and string dualities. I found that hilariously ironic because I had always been miffed that string theorists never cited my work on string theory!

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