New algorithm for cube (third power) computation

[Niels Möller]

nisse at lysator.liu.se (Niels Möller) writes:

> I've read the paper, and my first question is if it can be extended to
> fourth power (which I imagine would be a more useful buildingblock for
> high powers). But it looks difficult,

Maybe it's easier to do something useful for fifth power? E.g., using
that

  (x^3 + 10) (x^2 + 5x + 10) = x^5 + 5 x^4 + 10 x^3 + 10 x^2 + 50 x + 100

"masters"

  (x + 1)^5 = x^5 + 5 x^4 + 10 x^3 + 10 x^2 + 5x + 1

?

Regards,
/Niels

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