# Sums of three cubes

Unsolved problem in mathematics:

Is there a number that is not 4 or 5 modulo 9 and that cannot be expressed as a sum of three cubes?

In the mathematics of sums of powers, it is an open problem to characterize the numbers that can be expressed as a sum of three cubes of integers, allowing both positive and negative cubes in the sum. A necessary condition for ${\displaystyle n}$ to equal such a sum is that ${\displaystyle n}$ cannot equal 4 or 5 modulo 9, because the cubes modulo 9 are 0, 1, and −1, and no three of these numbers can sum to 4 or 5 modulo 9.[1] It is unknown whether this necessary condition is sufficient.

Semi-log plot of solutions of ${\displaystyle x^{3}+y^{3}+z^{3}=n}$ for integer ${\displaystyle x}$, ${\displaystyle y}$, and ${\displaystyle z}$, and ${\displaystyle 0\leq n\leq 100}$. Green bands denote values of ${\displaystyle n}$ proven not to have a solution.

Variations of the problem include sums of non-negative cubes and sums of rational cubes. All integers have a representation as a sum of rational cubes, but it is unknown whether the sums of non-negative cubes form a set with non-zero natural density.

## Small cases

A nontrivial representation of 0 as a sum of three cubes would give a counterexample to Fermat's Last Theorem for the exponent three, as one of the three cubes would have the opposite sign as the other two and its negation would equal the sum of the other two. Therefore, by Leonhard Euler's proof of that case of Fermat's last theorem,[2] there are only the trivial solutions

${\displaystyle a^{3}+(-a)^{3}+0^{3}=0.}$

For representations of 1 and 2, there are infinite families of solutions

${\displaystyle (9b^{4})^{3}+(3b-9b^{4})^{3}+(1-9b^{3})^{3}=1}$  (discovered[3] by K. Mahler in 1936)

and

${\displaystyle (1+6c^{3})^{3}+(1-6c^{3})^{3}+(-6c^{2})^{3}=2.}$  (discovered[4] by A.S. Verebrusov in 1908, quoted by L.J. Mordell[5])

These can be scaled to obtain representations for any cube or any number that is twice a cube.[5] There exist other representations, and other parameterized families of representations, for 1.[6] For 2, the other known representations are[6][7]

${\displaystyle 1\ 214\ 928^{3}+3\ 480\ 205^{3}+(-3\ 528\ 875)^{3}=2,}$
${\displaystyle 37\ 404\ 275\ 617^{3}+(-25\ 282\ 289\ 375)^{3}+(-33\ 071\ 554\ 596)^{3}=2,}$
${\displaystyle 3\ 737\ 830\ 626\ 090^{3}+1\ 490\ 220\ 318\ 001^{3}+(-3\ 815\ 176\ 160\ 999)^{3}=2.}$

However, 1 and 2 are the only numbers with representations that can be parameterized by quartic polynomials in this way.[5] Even in the case of representations of 3, Louis J. Mordell wrote in 1953 "I do not know anything" more than its small solutions

${\displaystyle 1^{3}+1^{3}+1^{3}=4^{3}+4^{3}+(-5)^{3}=3,}$

other than the fact that in this case each of the three cubed numbers must be equal modulo 9.[8][9]

Analogous to the first Pythagorean triple is an example of a special case: the cube sum of the cubes of three integers

${\displaystyle 3^{3}+4^{3}+5^{3}=6^{3}}$

## Computational results

Since 1955, and starting with the instigation of Mordell, many authors have implemented computational searches for these representations.[10][11][7][12][13][14][15][16][17][18]Elsenhans & Jahnel (2009) used a method of Noam Elkies (2000) involving lattice reduction to search for all solutions to the Diophantine equation

${\displaystyle x^{3}+y^{3}+z^{3}=n}$

for positive ${\displaystyle n}$  at most 1000 and for ${\displaystyle \max(|x|,|y|,|z|)<10^{14}}$ ,[17] leaving only 33, 42, 74, 114, 165, 390, 579, 627, 633, 732, 795, 906, 921, and 975 as open problems for ${\displaystyle n\leq 1000}$ , and 192, 375, and 600 remain with no primitive solutions (i.e. ${\displaystyle \gcd(x,y,z)=1}$ ). After Timothy Browning covered the problem on Numberphile, Huisman (2016) extended these searches to ${\displaystyle \max(|x|,|y|,|z|)<10^{15}}$  solving the case of 74, with solution

${\displaystyle 74=(-284\ 650\ 292\ 555\ 885)^{3}+66\ 229\ 832\ 190\ 556^{3}+283\ 450\ 105\ 697\ 727^{3}.}$

Through these searches, it was discovered that all ${\displaystyle n<100}$  that are unequal to 4 or 5 modulo 9 have a solution, with at most two exceptions, 33 and 42.[18]

In 2019, Andrew Booker settled the case ${\displaystyle n=33}$  by discovering that

${\displaystyle 33=8\ 866\ 128\ 975\ 287\ 528^{3}+(-8\ 778\ 405\ 442\ 862\ 239)^{3}+(-2\ 736\ 111\ 468\ 807\ 040)^{3}.}$

In order to achieve this, Booker exploited an alternative search strategy with running time proportional to ${\displaystyle \min(|x|,|y|,|z|)}$  rather than to their maximum,[19] an approach originally suggested by Heath-Brown et al.[20] He also found that

${\displaystyle 795=(-14\ 219\ 049\ 725\ 358\ 227)^{3}+14\ 197\ 965\ 759\ 741\ 571^{3}+2\ 337\ 348\ 783\ 323\ 923^{3},}$

and established that there are no solutions for ${\displaystyle n=42}$  or any of the other unresolved ${\displaystyle n\leq 1000}$  with ${\displaystyle |z|\leq 10^{16}}$ .

In September 2019, Booker and Andrew Sutherland settled the ${\displaystyle n=42}$  case, using 1.3 million hours of computing on the Charity Engine global grid to discover that

${\displaystyle 42=(-80\ 538\ 738\ 812\ 075\ 974)^{3}+80\ 435\ 758\ 145\ 817\ 515^{3}+12\ 602\ 123\ 297\ 335\ 631^{3},}$

as well as solutions for several other previously unknown cases.[21]

Booker and Sutherland also found a third representation of 3 using a further 4 million compute-hours on Charity Engine:

${\displaystyle 3=569\ 936\ 821\ 221\ 962\ 380\ 720^{3}+(-569\ 936\ 821\ 113\ 563\ 493\ 509)^{3}+(-472\ 715\ 493\ 453\ 327\ 032)^{3}.}$ [21][22]

This discovery settled a 65-year old question of Louis J. Mordell that has stimulated much of the research on this problem.[8]

While presenting the third representation of 3 during his appearance in a video on the Youtube channel Numberphile, Booker also presented a representation for 906:

${\displaystyle 906=(-74\ 924\ 259\ 395\ 610\ 397)^{3}+72\ 054\ 821\ 089\ 679\ 353\ 378^{3}+35\ 961\ 979\ 615\ 356\ 503^{3}.}$ [23]

The only remaining unsolved cases up to 1,000 are 114, 390, 627, 633, 732, 921, and 975, and there are no known primitive solutions (i.e. ${\displaystyle \gcd(x,y,z)=1}$ ) for 192, 375, and 600.[21][24]

## Popular interest

The sums of three cubes problem has been popularized in recent years by Brady Haran, creator of the YouTube channel Numberphile, beginning with the 2015 video "The Uncracked Problem with 33" featuring an interview with Timothy Browning.[25] This was followed six months later by the video "74 is Cracked" with Browning, discussing Huisman's 2016 discovery of a solution for 74.[26] In 2019, Numberphile published three related videos, "42 is the new 33", "The mystery of 42 is solved", and "3 as the sum of 3 cubes", to commemorate the discovery of solutions for 33, 42, and the new solution for 3.[27][28][23]

Booker's solution for 33 was featured in articles appearing in Quanta Magazine[29] and New Scientist[30], as well as an article in Newsweek in which Booker's collaboration with Sutherland was announced: "...the mathematician is now working with Andrew Sutherland of MIT in an attempt to find the solution for the final unsolved number below a hundred: 42".[31] The number 42 has additional popular interest due to its appearance in the 1979 Douglas Adams science fiction novel The Hitchhiker's Guide to the Galaxy as the answer to The Ultimate Question of Life, the Universe, and Everything.

Booker and Sutherland's announcements[32][33] of a solution for 42 received international press coverage, including articles in New Scientist,[34] Scientific American,[35] Popular Mechanics,[36] The Register,[37] Die Zeit,[38] Der Tagesspiegel,[39] Helsingin Sanomat,[40] Der Spiegel,[41] New Zealand Herald,[42] Indian Express,[43] Der Standard,[44] Las Provincias,[45] Nettavisen,[46] Digi24,[47] and BBC World Service.[48] Popular Mechanics named the solution for 42 as one of the "10 Biggest Math Breakthroughs of 2019".[49]

The resolution of Mordell's question by Booker and Sutherland a few weeks later sparked another round of news coverage.[22][50][51][52][53][54][55]

In Booker's invited talk at the fourteenth Algorithmic Number Theory Symposium he discusses some of the popular interest in this problem and the public reaction to the announcement of solutions for 33 and 42.[56]

## Solvability and decidability

In 1992, Roger Heath-Brown conjectured that every ${\displaystyle n}$  unequal to 4 or 5 modulo 9 has infinitely many representations as sums of three cubes.[57] The case ${\displaystyle n=33}$  of this problem was used by Bjorn Poonen as the opening example in a survey on undecidable problems in number theory, of which Hilbert's tenth problem is the most famous example.[58] Although this particular case has since been resolved, it is unknown whether representing numbers as sums of cubes is decidable. That is, it is not known whether an algorithm can, for every input, test in finite time whether a given number has such a representation. If Heath-Brown's conjecture is true, the problem is decidable. In this case, an algorithm could correctly solve the problem by computing ${\displaystyle n}$  modulo 9, returning false when this is 4 or 5, and otherwise returning true. Heath-Brown's research also includes more precise conjectures on how far an algorithm would have to search to find an explicit representation rather than merely determining whether one exists.[57]

## Variations

A variant of this problem related to Waring's problem asks for representations as sums of three cubes of non-negative integers. In the 19th century, Carl Gustav Jacob Jacobi and collaborators compiled tables of solutions to this problem.[59] It is conjectured that the representable numbers have positive natural density.[60][61] This remains unknown, but Trevor Wooley has shown that ${\displaystyle \Omega (n^{0.917})}$  of the numbers from ${\displaystyle 1}$  to ${\displaystyle n}$  have such representations.[62][63][64] The density is at most ${\displaystyle \Gamma (4/3)^{3}/6\approx 0.119}$ .[1]

Every integer can be represented as a sum of three cubes of rational numbers (rather than as a sum of cubes of integers).[65][66]

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