Re: modular exponentiation

Jeroen Demeyer on Fri, 25 Nov 2005 14:01:14 +0100

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Re: modular exponentiation

To: Henk Karssenberg <henk.karssenberg@hu.nl>
Subject: Re: modular exponentiation
From: Jeroen Demeyer <J.Demeyer@UGent.be>
Date: Fri, 25 Nov 2005 13:54:30 +0100
Cc: pari-users@list.cr.yp.to
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Henk Karssenberg wrote:

Dear M.,
In PARI I try to calculate k = Mod(6682632708227277^28345232917,72057594037927889) but this gives an overflow. Is there any option tocalculate Mod(a^m,n) or a^m % n with huge numbers ?
Thank you & kind regards.


Henk,

You should do k = Mod(a,n)^m.
This works because Mod(a,n) creates an object in Z/nZ.

References:
- modular exponentiation
  - From: "Henk Karssenberg" <henk.karssenberg@hu.nl>

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