Sums of cubes: multiple representations

I’m continuing a short series of posts on representing numbers as a sum of three cubes; previous posts are 33 is the sum of three cubes and More sums of three cubes.

We now know that every number less than $100$ which is not equivalent to $(\pm 4) \pmod 9$ can be written as a sum of three cubes. But for some of them we only know of one such representation, which can involve very large numbers. Is it possible that there are yet more ways, using even larger numbers?

Tricubic sums for 0

$0$ can of course be written as $0^3 + 0^3 + 0^3$ , but in fact it can be written as a sum of three cubes in infinitely many ways: pick any number $c$ , then $0 = 0^3 + c^3 + (-c)^3$ . In fact, we know these are must be the only tricubic sums for $0$ . For suppose there were some nonzero $x$ , $y$ , and $z$ such that $0 = x^3 + y^3 + z^3$ . In order to get a sum of zero, one of them has to be negative and two positive, or vice versa. If two are negative, then just negate the entire equation so that two are positive and one negative. Then move the negative one (say, $z$ ) to the other side of the equation, giving us $x^3 + y^3 = z^3$ —but this would contradict Fermat’s Last Theorem!¹

Tricubic sums for 1

$1$ can of course be written in infinitely many ways as $1^3 + c^3 + (-c)^3$ . However, there are also other ways. For example, as you can easily check for yourself,

$1 = 9^3 + (-6)^3 + (-8)^3.$

In fact, for any integer value of $b$ , it turns out that

$1 = (9b^4)^3+(3b-9b^4)^3+(1-9b^3)^3$ .

Let’s prove it! Remember that by the Binomial Theorem, $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$ . There are a lot of exponents and such to keep track of, but ultimately the algebra is not hard, just somewhat tedious:

$\begin{array}{cl} & (9b^4)^3+(3b-9b^4)^3+(1-9b^3)^3 \\[1em] =& (9b^4)^3 + \big[(3b)^3 + 3 (3b)^2 (-9b^4) + 3 (3b) (-9b^4)^2 + (-9b^4)^3\big] + \big[1^3 + 3(1^2)(-9b^3) + 3(1)(-9b^3)^2 + (-9b^3)^3\big] \\[1em] =& 3^6 b^{12} + 3^3 b^3 - 3^5 b^6 + 3^6 b^9 - 3^6 b^{12} + 1 - 3^3 b^3 + 3^5 b^6 - 3^6 b^9 \\[1em] =& 1\end{array}$

If you look carefully, you’ll see that everything cancels in the last step except for the single $1$ !

I got the example $1 = 9^3 + (-6)^3 + (-8)^3$ by choosing $b = 1$ . As another example, if we pick $b = 47$ , then we get

$1 = 43917129^3 + (-43916988)^3 + (-934406)^3$

which you can verify for yourself!

Tricubic sums for 2

There are also infinitely many tricubic sums for $2$ : for any value of $c$ ,

$2 = (1 + 6c^3)^3 + (1 - 6c^3)^3 + (-6c^2)^3$

I’ll let you prove this one yourself. Just expand using the Binomial Theorem and watch everything magically cancel!

Tricubic sums for 3 and beyond

It turns out that as of this writing (September 2019), we only know of two tricubic sums for 3, both rather small:

$3 = 1^3 + 1^3 + 1^3 = 4^3 + 4^3 + (-5)^3$ .

Are there any more? No one knows. However, in 1992, Roger Heath-Brown published a paper in which he conjectured that every number not of the form $\pm 4 \pmod 9$ has not just one but infinitely many tricubic sums! If true, this means there are infinitely many more tricubic sums for $3$ in particular—but if so they involve rather large numbers. So rather than tackling $114$ (the current smallest number for which we don’t know a tricubic sum), the next big prize might be to find another tricubic sum for the humble $3$ . Indeed, Andrew Booker says as much in his Numberphile interview.

We don’t actually need to invoke Fermat’s Last Theorem in full generality—the $n=3$ case of Fermat’s Last Theorem was first proved by Euler in 1770.↩

About Brent

Associate Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.

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