# Infinite Number of Integers which are Sum of 3 Sixth Powers in 2 Ways

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## Theorem

There exist an infinite number of positive integers which can be expressed as the sum of $3$ sixth powers in $2$ different ways.

## Proof

There are many parametric solutions to $x^6 + y^6 + z^6 = u^6 + v^6 + w^6$. One is given by:

\(\ds x\) | \(=\) | \(\ds 2 m^4 + 4 m^3 n - 5 m^2 n^2 - 12 m n^3 - 9 n^4\) | ||||||||||||

\(\ds y\) | \(=\) | \(\ds 3 m^4 + 9 m^3 n + 18 m^2 n^2 + 21 m n^3 + 9 n^4\) | ||||||||||||

\(\ds z\) | \(=\) | \(\ds -m^4 - 10 m^3 n - 17 m^2 n^2 - 12 m n^3\) | ||||||||||||

\(\ds u\) | \(=\) | \(\ds m^4 - 3 m^3 n - 14 m^2 n^2 - 15 m n^3 - 9 n^4\) | ||||||||||||

\(\ds v\) | \(=\) | \(\ds 3 m^4 + 8 m^3 n + 9 m^2 n^2\) | ||||||||||||

\(\ds w\) | \(=\) | \(\ds 2 m^4 + 12 m^3 n + 19 m^2 n^2 + 18 m n^3 + 9 n^4\) |

This set of solutions also satisfy:

\(\ds x^2 + y^2 + z^2\) | \(=\) | \(\ds u^2 + v^2 + w^2\) | ||||||||||||

\(\ds 3 x + y + z\) | \(=\) | \(\ds 3 u + v + w\) |

This needs considerable tedious hard slog to complete it.In particular: For people who enjoy degree $24$ polynomials in $2$ unknownsTo 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 `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## Sources

- 1976: Simcha Brudno:
*Triples of Sixth Powers with Equal Sums*(*Math. Comp.***Vol. 30**,*no. 135*: pp. 646 – 648) www.jstor.org/stable/2005335 - 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $160,426,514$ - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $160,426,514$