Number is Sum of Five Cubes

From ProofWiki
Jump to navigation Jump to search

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


This page has been identified as a candidate for refactoring of basic complexity.
In particular: Find a way to incorporate this as a result without implementing it as a "remark"
Until this has been finished, please leave {{Refactor}} in the code.

New contributors: Refactoring is a task which is expected to be undertaken by experienced editors only.

Because of the underlying complexity of the work needed, it is recommended that you do not embark on a refactoring task until you have become familiar with the structural nature of pages of $\mathsf{Pr} \infty \mathsf{fWiki}$.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Refactor}} from the code.


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 $k$.


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Number_is_Sum_of_Five_Cubes&oldid=569070"