# Number is Sum of Five Cubes

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## Theorem

Let $n \in \Z$ be an integer.

Then $n$ can be expressed as the sum of $5$ cubes (either positive or negative) in an infinite number of ways.

## Proof

We have for any $m, n \in \Z$:

\(\displaystyle \paren {6 m + n}^3\) | \(\equiv\) | \(\displaystyle n^3\) | \(\displaystyle \pmod 6\) | Congruence of Powers | |||||||||

\(\displaystyle \) | \(\equiv\) | \(\displaystyle n\) | \(\displaystyle \pmod 6\) | Euler's Theorem: $\map \phi 6 = 2$ |

By definition of modulo arithmetic:

- $\exists k \in \Z: \paren {6 m + n}^3 = n + 6 k$

We also have:

\(\displaystyle \paren {k + 1}^3 + \paren {k - 1}^3 - k^3 - k^3\) | \(=\) | \(\displaystyle \paren {k^3 + 3 k^2 + 3 k + 1} + \paren {k^3 - 3 k^2 + 3 k - 1} - k^3 - k^3\) | Cube of Sum | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 6 k\) |

Thus $n = \paren {6 m + n}^3 + k^3 + k^3 - \paren {k + 1}^3 - \paren {k - 1}^3$ is an expression of $n$ as a sum of $5$ cubes.

In the equation above, $m$ is arbitrary and $k$ depends on both $m$ and $n$.

As there is an infinite number of choices for $m$, there is an infinite number of such expressions.

$\blacksquare$

## Remark

From the result above we can also write each term of the sum explicitly:

\(\displaystyle k\) | \(=\) | \(\displaystyle \frac {\paren {6 m + n}^3 - n} 6\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {216 m^3 + 108 m^2 n + 54 m n^2 + n^3 - n} 6\) | Cube of Sum | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 36 m^3 + 18 m^2 n + 9 m n^2 + \frac {n^3 - n} 6\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle n\) | \(=\) | \(\displaystyle \paren {6 m + n}^3 + \paren {36 m^3 + 18 m^2 n + 9 m n^2 + \frac {n^3 - n} 6}^3 + \paren {36 m^3 + 18 m^2 n + 9 m n^2 + \frac {n^3 - n} 6}^3\) | ||||||||||

\(\displaystyle \) | \(\) | \(\, \displaystyle - \, \) | \(\displaystyle \paren {36 m^3 + 18 m^2 n + 9 m n^2 + \frac {n^3 - n} 6 + 1}^3 - \paren {36 m^3 + 18 m^2 n + 9 m n^2 + \frac {n^3 - n} 6 - 1}^3\) |

However, because of the complexity of this expression it is recommended to first calculate $k$.

## Sources

- 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $5$ - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $5$