Sum of four cubes problem

Unsolved problem in mathematics:

Is every integer the sum of four perfect cubes?

(more unsolved problems in mathematics)

The sum of four cubes problem asks whether every integer is the sum of four cubes of integers. It is conjectured the answer is affirmative, but this conjecture has been neither proven nor disproven. Some of the cubes may be negative numbers, in contrast to Waring's problem on sums of cubes, where they are required to be positive.

Partial results

The substitutions $X=T$, $Y=T$, and $Z=-T+1$ in the identity $$(X+Y+Z)^{3}-X^{3}-Y^{3}-Z^{3}=3(X+Y)(X+Z)(Y+Z)$$ lead to the identity $$(T+1)^{3}+(-T)^{3}+(-T)^{3}+(T-1)^{3}=6T,$$ which shows that every integer multiple of 6 is the sum of four cubes. (More generally, the same proof shows that every multiple of 6 in every ring is the sum of four cubes.)

Since every integer is congruent to its own cube modulo 6, it follows that every integer is the sum of five cubes of integers.

In 1966, V. A. Demjanenko proved that any integer that is congruent neither to 4 nor to −4 modulo 9 is the sum of four cubes of integers. For this, he used the following identities: $${\begin{aligned}6x&=(x+1)^{3}+(x-1)^{3}-x^{3}-x^{3}\\6x+3&=x^{3}+(-x+4)^{3}+(2x-5)^{3}+(-2x+4)^{3}\\18x+1&=(2x+14)^{3}+(-2x-23)^{3}+(-3x-26)^{3}+(3x+30)^{3}\\18x+7&=(x+2)^{3}+(6x-1)^{3}+(8x-2)^{3}+(-9x+2)^{3}\\18x+8&=(x-5)^{3}+(-x+14)^{3}+(-3x+29)^{3}+(3x-30)^{3}\ .\end{aligned}}$$ These identities (and those derived from them by passing to opposites) immediately show that any integer which is congruent neither to 4 nor to −4 modulo 9 and is congruent neither to 2 nor to −2 modulo 18 is a sum of four cubes of integers. Using more subtle reasonings, Demjanenko proved that integers congruent to 2 or to −2 modulo 18 are also sums of four cubes of integers.

The problem therefore only arises for integers congruent to 4 or to −4 modulo 9. One example is $$13=10^{3}+7^{3}+1^{3}+(-11)^{3},$$ but it is not known if every such integer can be written as a sum of four cubes.

18x±2 case

According to Henri Cohen's translation of Demjanenko's paper, these identities

$${\begin{aligned}54x+2&=(29484x^{2}+2211x+43)^{3}+(-29484x^{2}-2157x-41)^{3}+(9828x^{2}+485x+4)^{3}+(-9828x^{2}-971x-22)^{3}\\54x+20&=(3x-11)^{3}+(-3x+10)^{3}+(x+2)^{3}+(-x+7)^{3}\\216x-16&=(14742x^{2}-2157x+82)^{3}+(-14742x^{2}+2211x-86)^{3}+(4914x^{2}-971x+44)^{3}+(-4914x^{2}+485x-8)^{3}\\216x+92&=(3x-164)^{3}+(-3x+160)^{3}+(x-35)^{3}+(-x+71)^{3}\end{aligned}}$$ together with their complementary identities leave the 108x±38 case, proving the proposition. He also proves the 108x±38 case in his paper.

See also

• Sums of three cubes