Taxicab Numbers and the Eisenstein Integers

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?

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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: $a^3+b^3=(a+b)(a+b\omega )(a+b\bar{\omega })$, 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:

• $\bar{\omega }={\omega }^2$
• $\omega +\bar{\omega }=-1$
• $\omega \bar{\omega }=1$.

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:

• $(10+9)(10+9\omega )(10+9\bar{\omega })$
• $(9+10)(9+10\omega )(9+10\bar{\omega })$
• $(12+1)(12+\omega )(12+\bar{\omega })$
• $(1+12)(1+12\omega )(1+12\bar{\omega })$

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.