Re: a^(b^x) mod n?

Ο "richard miller" <richard@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> έγραψε στο μήνυμα
news:d9c8uj$sfm$1@xxxxxxxxxxxxxxxxxxxxxxx
[snip]

> It can be done efficiently with a computer. You use the fact that
>
> a^(b^x) = ( a^(b^(x-1) ) )^x

That's probably a typo for:

a^(b^x) = ( a^(b^(x-1) ) )^b

> And work in reverse, i.e. start with a^b mod n and keep raising to the
b'th
> power, see further. Denoting
> a_1 = a^b mod n,
>
> then
>
> a_2 = a_1^b mod n = a^(b^2)
>
> a_3 = a_2^b mod n = a^(b^3)
>
> and repeat x times to obtain
>
> a_x = a_x-1^b mod n = a^(b^x).
>
> E.g. if x is 1000, then repeat 1000 times.
>
> To raise a^b you use the binary method in the previous respondents reply.
>
> RJM
--
I. N. Galidakis
http://users.forthnet.gr/ath/jgal/
Eventually, _everything_ is understandable

.

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