[CST-2] advanced algorithms 1996/9/6 (fwd)

[Matej Pfajfar]

> I think you can use that to say that 12^x where x = y*4 is 1 mod 65
> Work it form there?
>
> answer is:
>
> 12^(4*16) mod 65
>
> 1
>
That's the thing if you can use that (which I think you can but i am not
thinking straight) then the algorithm I described works anyway...
The reason I think you can do that is becase Miller-Rabin does it ..

--
Matej Pfajfar
St John's College, University of Cambridge, UK

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