Sums of four cubes problem

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The sums of four cubes problem[1] is to ask whether any rational integer is the sum of four cubes of rational integers. By putting X = T, Y = T, Z = - T + 1 in the identity

[math]\displaystyle{ \qquad (X + Y + Z)^{3} - X^{3} - Y^{3} - Z^{3} = 3 (X + Y) (X + Z) (Y + Z), }[/math]

we get the identity

[math]\displaystyle{ \qquad (T + 1)^{3} + (- T)^{3} + (- T)^{3} + (T - 1)^{3} = 6 T , }[/math]

which shows that in any ring, any multiple of 6 (i.e. any element of this ring of the form 6a, a being itself an element of the ring) is sum of four cubes.

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

According to a conjecture that is still open,[2] any rational integer would be the sum of four cubes of rational integers.

In 1966, V. A. Demjanenko proved that any rational integer that is congruent neither to 4 nor to - 4 modulo 9 is the sum of four cubes of rational integers. For this, he used in particular the following identities:

[math]\displaystyle{ \qquad 6x = (x+1)^{3}+(x-1)^{3}-x^{3}-x^{3}, }[/math]
[math]\displaystyle{ \qquad 6x+3 = x^3+(-x+4)^3+(2x-5)^3+(-2x+4)^3, }[/math]
[math]\displaystyle{ \qquad 18x+1 = (2x+14)^3+(-2x-23)^3+(-3x-26)^3+(3x+30)^3, }[/math]
[math]\displaystyle{ \qquad 18x+7 = (x+2)^3+(6x-1)^3+(8x-2)^3+(-9x+2)^3, }[/math]
[math]\displaystyle{ \qquad 18x+8 = (x-5)^3+(-x+14)^3+(-3x+29)^3+(3x-30)^3. }[/math]

These identities (and those derived from them by passing to opposites) immediately show that any rational 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 rational integers. Using more subtle reasonings, Demjanenko proved that rational integers congruent to 2 or to - 2 modulo 18 are also sums of four cubes of rational integers.[3]

The problem therefore only arises for rational integers congruent to 4 or to -4 modulo 9. We have for example

[math]\displaystyle{ \qquad 13 = 10^3 + 7^3 + 1^3 + (-11)^3. }[/math]

Notes and references

  1. ↑ Referred to as the “four cube problem” in H. Davenport, The Higher Arithmetic: An Introduction to the Theory of Numbers, Cambridge University Press, 7th edition, 1999, p. 173, 177.
  2. ↑ At least in 1982. See Philippe Revoy, “Sur les sommes de quatre cubes”, L’Enseignement Mathématique, t. 29, 1983, p. 209-220, online here or here, p. 209 on the point in question.
  3. ↑ V.A. Demjanenko, “On sums of four cubes”, Izvestiya Vysshikh Uchebnykh Zavedenii. Matematika, vol. 54, no. 5, 1966, p. 63-69, available online at the site Math-Net.Ru. For a demonstration in French, see Philippe Revoy, “Sur les sommes de quatre cubes”, L’Enseignement Mathématique, t. 29, 1983, p. 209-220, online here or here.

See also




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