IDLE 2.6.4
>>> # these 4 test are all that is needed to find the order of 2 in Z/101Z*
>>> pow(2, 25, 101)
10
>>> pow(2, 50, 101)
100
>>> pow(2, 4, 101)
16
>>> pow(2, 20, 101)
95
>>> # The fillwoing are given jsut illustating Fermat's last theorem
>>> # - sure to get these results
>>> pow(2, 100, 101)
1
>>> pow(2, 100, 101)
1
>>> # Hence both factors of 2 and both factors of 5 are needed for the order
>>> # of the element 2 +101Z.
>>>
>>> # I made an error in class suggesting that the base 5
>>> # had something to do with this problem.
>>> # The book example dealt with the generator *2* only.
>>> # Another annoyance with the book example is the use of 2
>>> # in several different ways: as generator and as group order factor.
>>> # We could test the order of the group generated by 7 similarly:
>>> pow(7, 25, 101) # remove max factors of 2 from power 100
10
>>> pow(7, 50, 101) # remove one less factor of 2
100
>>> pow(7, 4, 101) # remove max factors of 5 from 100
78
>>> pow(7, 20, 101) # remove one less factor of 5
84
>>> # hence need all factors of 2 and of 5:
>>> # 7 has order 100 and generates all of Z/101Z*