## Abstract

Computer assisted searches of solutions of the Diophantine equation $x^3 +y^3 +z^3 = k$ have been made since 1954. Thanks to some intelligent people, modern super computing facilities and a YouTube channel, we finally now 66 years later have solutions for all $k < 100$. Here we report an interesting solution when $k=2^3$.

Keywords: Sum of three cubes, Diophantine equation, taxicab number.

## Introduction

In 2015, on a Numberphile video “The uncracked problem with 33” Prof. Tim Browning discussed the Diophantine equation \begin{equation*} x^3+y^3+z^3=k \end{equation*} that has interested mathematicians a quite some time now. As explained in the video, it is know that this equation has no integer solutions for $k \equiv 4$ or $5\,(\mathrm{mod}\,9)$, i.e., $4,5,13,14, 22, 23, \ldots$ For other values of $k$, it has been conjectured that there are infinitely many solutions [1]. In 1953, for $k=3$, Prof. Louis J. Mordell now famously wrote [2]: “I do not know anything about the integer solutions of $x^3+y^3+z^3=3$ beyond the existence of the four sets $(1, 1, 1)$, $(4, 4, -5)$, etc.; and it must be very difficult indeed to find out anything about any other solutions.” This led to the first computer assisted search for $k < 100$ in 1954 [3].

Since those days, for values $k <1\,000$, several searches have been made and more effective algorithms proposed. For example, in 2007, Andreas-Stephan Elsenhans and Jörg Jahnel searched systematically for solutions where the positive integer $k < 100$ is neither a cube nor twice a cube and $|x|,|y|,|z| k < 10^{14}$ [4]. These values “New sums of three cubes” are tabulated here. At this point, they reported 14 unsolved values below $1\,000$: 33, 42, 74,114, 165, 390, 579, 627, 633, 732, 795, 906, 921, and 975.

In 2016, motivated by the original Numberphile video, Sander Huisman extended the search of Elsenhans and Jahnel up to $10^{15}$ [5]. In his report “Newer sums of three cubes,” he found 966 new solutions. The most exciting one was the discovery of the first solution for $k = 74$: \begin{equation*} 74 = (-284650292555885)^3 + 66229832190556^3 + 283450105697727^3. \end{equation*} This result was discussed by Browning on a follow-up Numberphile video “74 is cracked.

It was this follow-up video that got mathematician Andrew Booker hooked. In 2019, he proposed a new algorithm [6] and searched solutions for unsolved values below $1\,000$. He found the first known solution for $k = 33$: \begin{equation*} 33 = 8866128975287528^3 +(-8778405442862239)^3 +(-2736111468807040)^3. \end{equation*} This results was, of course, reported on a Numberphile video “42 is the new 33” indicating that after the discoveries by Huisman and Booker, the only unsolved value below 100 was $k=42$. In his paper, Booker also searched for solutions for $k = 3$, addressing a question of Prof. Mordell, but found none. He also reported the first known solution for $k = 795$: \begin{equation*} 795 = (14219049725358227)^3+(14197965759741571)^3+(2337348783323923)^3. \end{equation*} At this point, also I got interested.

## Methodology

Albeit having a PhD degree in applied mathematics, I don't have any formal education on number theory, let alone, have no experience of computer searches of this type. It was quite obvious, that an exhaustive search for the range $10^{16}$ was out of my reach. The only way I thought I could participate was to randomly sample large values of $x$, $y$ and $z$ and then test if the sum of their cubes gives a small solution, say less than $1\,000$. With a little bit of online research, I wrote the Python code used in this study.

import random

for x in range(10**13):
a = random.randint(10**14,10**18)
b = random.randint(10**14,10**18)
c = random.randint(10**14,10**18)

if a > b:
if a > c:
val = a**3 -b**3-c**3
else:
val = c**3 -b**3-a**3
elif b > c:
val = b**3 -a**3-c**3
else:
val = c**3-a**3-b**3

if abs(val) < 1000:
print(a)
print(b)
print(c)
print(val)
Outa = open("a.txt","w")
Outb = open("b.txt","w")
Outc = open("c.txt","w")
Outval = open("val.txt","w")
Outa.write(str(a))
Outa.close()
Outb.write(str(b))
Outb.close()
Outc.write(str(c))
Outc.close()
Outval.write(str(val))
Outval.close()
break
elif (x % 10**6) == 0:
print(x)

I knew that I would need to sample many many times in order to find a new solution. As noted by Huisman [5], the number of solutions for each decade up to search bounds $B$ in the range from $10^2$ to $10^{14}$ have been roughly $1\,000$ solutions per decade, which is in accordance with [1]. This gave me some initial hope, but I soon realized, after discovering a little error in my calculations, that the probability of finding a new solution was virtually non-existent. I still decided to give it a go.

## Results

On 10 July 2019 I set a computer search for finding a new solution. At the meantime, Andrew Booker had teamed up with Andrew Sutherland and with computer resources from Charity Engine they were searching for new solutions, too. On September 2019, while I was waiting my first solution to appear, they announced a solution for $k=42$: \begin{equation*} (-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3=42, \end{equation*} and again, a new Numberphile video “The Mystery of 42 is Solved” was made. This marked the end of the journey that was started in 1954, as we now had solutions for all $k < 100$. This left the original question of Mordell still unanswered, but three weeks later, they also found: \begin{equation*} 569936821221962380720^3 + (-569936821113563493509)^3 + (-472715493453327032)^3 = 3, \end{equation*} and yet again a new Numberphile video was made “3 as the sum of the 3 cubes.” Booker and Sutherland report these results and two other new solutions on a preprint [7].

I left my computer search to go on for some months, but at some point the computer was rebooted or I stopped the script for running, I don't quite remember anymore which one was it. Anyways, no new solutions were found.

On 15 October 2020, during a Finnish Autumn school break, I decided to revisit this question. It was pretty obvious again, that the original code was not going to give me a solution, so I decided to check what kind of solution I would get if I would limit the search to $10^4$ and would also be satisfied with a solution of the same size. After running the script, it printed the numbers $7\,576, 4\,112, 7\,960$ and $4\,096$ which I arranged to a solution \begin{equation*} -7576^3-4112^3+7960^3 = 4096. \end{equation*} I immediately recognized $4\,096$ as a power of 2, i.e, $2^{12}$, so I decided to check the prime factors of the other numbers, too. I found that $7\,576 = 2^3 \times 947$, $4\,112 = 2^4 \times 257$ and $7\,960 = 2^3 \times 5 \times 199$, so the original solution given by the algorithm could be re-arranged to \begin{equation*} -947^3-514^3+995^3 = 2^3. \end{equation*} I found this solution quite fascinating, as also $k=2^3$ is a cube and the other numbers are relatively large. I decided to right away communicate this mesmerizing solution to the mathematical community via Twitter. At the time of writing this text, a day later, this tweet has already gotten zero likes and retweets combined.

The question that rises with $k$ being a cube is: Are there any others? At first glance, it would seem that requiring $k$ being a cube would make things more complicated, but this is not the case. For example, if we think about it a little bit (and even if we don't), the equations \begin{equation*} a^3+0^3+0^3 = a^3 \end{equation*} and \begin{equation*} a^3+a^3+0^3 = 2 a^3 \end{equation*} always give solutions. This is quite likely the reason why cubes and twice the cubes were not included in the original dataset of Elsenhans and Jahnel: here.

Moreover, if we read the article “Diophantine Equation--3rd Powers” on Wolfram MathWorld, we find that the general rational solution to $x^3+y^3+z^3=k^3$ exists. For example, famously, Plato's number $216 = 6^3$ is the sum of the cubes for the Pythagorean triple $(3, 4, 5)$ \begin{equation*} 3^3 + 4^3 + 5^3 = 6^3 \end{equation*} and is also the case $a=1, b=0$ of Ramanujan's formula: \begin{equation*} (3a^2+5ab-5b^2)^3 + (4a^2-4ab+6b^2)^3 + (5a^2-5ab-3b^2)^3 = (6a^2-4ab+4b^2)^3. \end{equation*} In fact, when $k$ is a cube the question is related to the famous story of Ramanujan and G. H. Hardy, and how the taxicab number $1\,729$ is the smallest integer that can be expressed as a sum of two positive integer cubes in two distinct ways: \begin{equation*} 1729 = 1^3 + 12^3 = 9^3 + 10^3. \end{equation*} If we allow also negative numbers, we can re-arrange the earlier example with Plato's number and have even smaller sum: \begin{equation*} 91 = 6^3 - 5^3 = 4^3 + 3^3. \end{equation*} The smallest sum that our re-arranged solution would give is the following: \begin{equation*} 135\,796\,752 = 995^3-947^3 = 514^3 + 2^3. \end{equation*} On the other hand, Euler conjectured that there were no positive integral solutions to \begin{equation*} a^4 + b^4 + c^4 = d^4. \end{equation*} In 1988, the smallest possible counterexample was found [8]: \begin{equation*} 95800^4 + 217519^4 + 414560^4 = 422481^4. \end{equation*}

## Acknowledgements

Most of my knowledge on this topic comes from the series of Numberphile videos “Sums of three cubes.” Do yourself a favor and go find the originals.

## References

[1] D. R. Heath-Brown, The density of zeros of forms for which weak approximation fails, Math. Comp. 59 (1992), no. 200, 613–623.
[2] L. J. Mordell, On the integer solutions of the equation x2 + y2 + z2 + 2xyz = n, J. London Math. Soc. 28 (1953), 500–510.
[3] J. C. P. Miller and M. F. C. Woollett, Solutions of the Diophantine equation x3 +y3 +z3 = k, J. London Math. Soc. 30 (1955), 101–110.
[4] Andreas-Stephan Elsenhans and J ̈org Jahnel, New sums of three cubes, Math. Comp. 78 (2009), no. 266, 1227–1230.
[5] Sander G. Huisman, Newer sums of three cubes, arXiv:1604.07746, 2016.
[6] A. R. Booker, Cracking the problem with 33, Res. Number Theory 5 (2019), no. 3, 5:26.
[7] Andrew R. Booker and Andrew V. Sutherland, On a question of Mordell, arXiv:2007.01209, 2020.
[8] N. D. Elkies, On A4 + B4 + C4 = D4, Math. of Comp. 51 (1988), 825–835.