# 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?(more unsolved problems in mathematics)
Semilog plot of solutions of x³ + y³ + z³ = n for integer x, y and z, and n in [0, 100]. Green bands denote where no solution is proven to exist. The purple line denotes 42, the remaining n < 100 for which a solution has not been found.

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. An obvious 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.

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. For, 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

${\displaystyle (9b^{4})^{3}+(3b-9b^{4})^{3}+(1-9b^{3})^{3}=1}$

and

${\displaystyle (1+6c^{3})^{3}+(1-6c^{3})^{3}+(-6c^{2})^{3}=2.}$

These can be scaled to obtain representations for any cube or any number that is twice a cube.[3][4] There exist other representations, and other parameterized families of representations, for 1.[5] For 2, the other known representations are[5][6]

${\displaystyle 1214928^{3}+3480205^{3}+(-3528875)^{3}=2,}$
${\displaystyle 37404275617^{3}+(-25282289375)^{3}+(-33071554596)^{3}=2,}$
${\displaystyle 3737830626090^{3}+1490220318001^{3}+(-3815176160999)^{3}=2.}$

However, 1 and 2 are the only numbers with representations that can be parameterized by quartic polynomials in this way.[7] Even in the case of representations of 3, Louis J. Mordell wrote in 1953 "I do not know anything" more than its small solutions

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

and more than the fact that in this case each of the three cubed numbers must be equal modulo 9.[8][9] As of March 2019 these remain the only known representations of 3.[10]

## Computational results

Since 1955, and starting with the instigation of Mordell, many authors have implemented computational searches for these representations.[11][12][6][13][14][15][16][17][18][19] 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 ${\displaystyle \max(|x|,|y|,|z|)<10^{14}}$,[18] and Huisman (2016) extended these searches to ${\displaystyle \max(|x|,|y|,|z|)<10^{15}}$. Through these searches, it was discovered that all ${\displaystyle n<100}$ that are unequal to 4 or 5 modulo 9 have a solution, with two exceptions, 33 and 42.[19]

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

${\displaystyle 33=8866128975287528^{3}+(-8778405442862239)^{3}+(-2736111468807040)^{3}.}$

In order to achieve this, Booker developed an alternative search strategy with running time proportional to ${\displaystyle \min(|x|,|y|,|z|)}$ rather than to their maximum. After this discovery, the only remaining two-digit value of ${\displaystyle n}$ for which the problem remains unknown is ${\displaystyle n=42}$.[20][10]

## 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.[21] 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.[22] However, 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 easily 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.[21]

## 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.[23] It is conjectured that the representable numbers have positive natural density.[24][25] This, also, 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.[26][27][28] 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).[29][30]

## References

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2. ^ Machis, Yu. Yu. (2007), "On Euler's hypothetical proof", Mathematical Notes, 82 (3): 352–356, doi:10.1134/S0001434607090088, MR 2364600
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