There’s a famous, and famously tedious, story about the number 1729 and how it became known as a taxicab number. You can look it up if you’re that interested.

What’s interesting to me is the numbers themselves: numbers that are the sum of two cubes in two (or more) different ways:

  • $1729 = 10^3 + 9^3$
  • $1729 = 12^3 + 1^3$

You might recall that the sum of two cubes factorises: $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$, which means that 1729 has two ‘obvious’ factors – 19 and 13. It also has a third factor, 7, and my question is is there a simple way to find that?

This is a question to which I don’t know the answer, but which has led me down an interesting path.

Recalling that $a^2 + b^2$ factors over the Gaussian integers as $(a+bi)(a-bi)$, I wondered if there was a similar thing for taxicab numbers and of course there is: $a^3 + b^3 = (a+b)(a+b\omega)(a+b\overline\omega)$, where $\omega = \frac{-1+i \sqrt{3}}{2}$, one of the complex cube roots of 1. Numbers in this form, the field $\mathbb{Z}[i\sqrt{3}]$ if you prefer, are called the Eisenstein integers.

A few handy observations:

  • $\overline\omega = \omega^2$
  • $\omega + \overline \omega = -1$
  • $\omega \overline\omega = 1$.

The upshot of the second of those is that Eisenstein integers can all be written in a unique way as $p + q\omega$.

This means that 1729 readily factorises in four different ways:

  • $(10+9)(10+9\omega)(10+9\overline\omega)$
  • $(9+10)(9+10\omega)(9+10\overline\omega)$
  • $(12+1)(12+\omega)(12+\overline\omega)$
  • $(1+12)(1+12\omega)(1+12\overline\omega)$

Or, if we write the final factor of each in the conventional form:

  • $(10+9)(10+9\omega)(1 - 9\omega)$
  • $(9+10)(9+10\omega)(-1 - 10\omega)$
  • $(12+1)(12+\omega)(11 - \omega)$
  • $(1+12)(1+12\omega)(-11 - 12\omega)$

And these final factors are especially interesting – applying the original logic in reverse, (for example) $1-9\omega$ is a factor of $1^3 - 9^3$, or -728.

Similarly, the final factors are also factors of -1001, 1330 and -3059 – all of which are also multiples of 7.

However, I’ve not (yet) been able to come up with a simple method for pulling out the factor of 7 that’s any neater than just “divide 1729 by 19 and 13, and see what’s left over”. I thought it would be a good idea to share my work so far and see if any of my dear readers have an insight to share.