Yep, looking at Integer.__pow__
def __pow__(self, n, modulus):
if modulus is not None:
from sage.rings.finite_rings.integer_mod import Mod
return Mod(self, modulus) ** n
so that should be exactly the same thing.
On Thu, Mar 17, 2011 at 10:26 AM, Graham Enos <graham.e...@gmail.com> wrote:
> Another option is to use the pow() function, as in pow(a, b, c).
>
> On Mar 17, 1:58 am, Robert Bradshaw <rober...@math.washington.edu>
> wrote:
>> mod(a, c)^b
>>
>> On Wed, Mar 16, 2011 at 10:44 PM, Santanu Sarkar
>>
>>
>>
>>
>>
>>
>>
>> <sarkar.santanu....@gmail.com> wrote:
>> > How one can calculate a^b mod c in Sage for large b?
>>
>> > --
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