Five Cube Theorem

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Theorem

Every integer can be expressed as a sum of five cube numbers.


Proof

Let $r$ be an integer.

Then $r$ can be expressed in exactly one the following forms:

$\paren {6 m + 0}$
$\paren {6 m + 1}$
$\paren {6 m + 2}$
$\paren {6 m + 3}$
$\paren {6 m + 4}$
$\paren {6 m + 5}$

for some $m \in \Z$.


$\paren {6 m + 0} = \paren {m + 1}^3 + \paren {m - 1}^3 + \paren {- m}^3 + \paren {- m}^3 + 0^3$
$\paren {6 m + 1} = \paren {m + 1}^3 + \paren {m - 1}^3 + \paren {- m}^3 + \paren {- m}^3 + 1^3$
$\paren {6 m + 2} = \paren {m}^3 + \paren {m - 2}^3 + \paren {1 - m}^3 + \paren {1 - m}^3 + 2^3$
$\paren {6 m + 3} = \paren {m - 3}^3 + \paren {m - 5}^3 + \paren {- m + 4}^3 + \paren {- m + 4}^3 + 3^3$
$\paren {6 m + 4} = \paren {m + 3}^3 + \paren {m + 1}^3 + \paren {- m - 2}^3 + \paren {- m - 2}^3 + \paren {-2}^3$
$\paren {6 m + 5} = \paren {m + 2}^3 + \paren {m}^3 + \paren {- m - 1}^3 + \paren {- m - 1}^3 + \paren {-1}^3$

$\blacksquare$


Also see