On Fri, 22 Jul 2011 07:22:30 -0700 (PDT), Simon Roberts <2assume@gmail.com> wrote:
>This is a trivial except that it shows x^(2^i) mod n for all >i>1 and all x!=0 mod n forms the same set as x^(2^j) mod n >j>1 i!=j. But this may be trivial as well. Maybe someone can >show me this is trivial as well (better than I can).
You need to quote the relevant parts of any message you are replying to.
As it stands, I have no idea what it is that you're claiming is trivial.
