# Number is Sum of Five Cubes

## 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 $m, n \in \Z$:

 $\ds \paren {6 m + n}^3$ $\equiv$ $\ds n^3$ $\ds \pmod 6$ Congruence of Powers $\ds$ $\equiv$ $\ds n$ $\ds \pmod 6$ 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:

 $\ds \paren {k + 1}^3 + \paren {k - 1}^3 - k^3 - k^3$ $=$ $\ds \paren {k^3 + 3 k^2 + 3 k + 1} + \paren {k^3 - 3 k^2 + 3 k - 1} - k^3 - k^3$ Cube of Sum $\ds$ $=$ $\ds 6 k$

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:

 $\ds k$ $=$ $\ds \frac {\paren {6 m + n}^3 - n} 6$ $\ds$ $=$ $\ds \frac {216 m^3 + 108 m^2 n + 54 m n^2 + n^3 - n} 6$ Cube of Sum $\ds$ $=$ $\ds 36 m^3 + 18 m^2 n + 9 m n^2 + \frac {n^3 - n} 6$ $\ds \leadsto \ \$ $\ds n$ $=$ $\ds \paren {6 m + n}^3 + \paren {36 m^3 + 18 m^2 n + 9 m n^2 + \frac {n^3 - n} 6}^3 + \paren {36 m^3 + 18 m^2 n + 9 m n^2 + \frac {n^3 - n} 6}^3$ $\ds$  $\, \ds - \,$ $\ds \paren {36 m^3 + 18 m^2 n + 9 m n^2 + \frac {n^3 - n} 6 + 1}^3 - \paren {36 m^3 + 18 m^2 n + 9 m n^2 + \frac {n^3 - n} 6 - 1}^3$

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