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Main Question or Discussion Point
As the title says, for any a>1 and n>0
You cannot say that for sure, or I am wrong?Set [tex] a^n=1+M[/tex]. Consequently [tex]a^n\equiv 1ModM[/tex] and is the smallest n for a>1, n>0. It also follows from the last equation that a belongs to the reduced residue group of order [tex]\phi{M}[/tex]. Consequently by LaGrange's Theorem, n is a divisor of the order of the group.