Code and ciphers: Julius Caesar, the Enigma and the internet

210
appendix
(2) In using modular arithmetic it pays to remove the highest possible
power of 2 from the exponent, then find the remainder of the other (odd)
factor of the exponent, and finally use repeated squaring to get to the
original exponent. So, for example, since 96�3�(32) if we form (59) 3
(mod 97) and repeatedly square it five times, reducing (mod 97) at each
stage, we will obtain the required result; the details are as follows:
so
so
so
so
so
59�59�3481�35�97�86,
(59) 3 �86�59�5074�52�97�30�30 (mod 97),
(59) 6 �(30) 2 �900�9�97�27�27 (mod 97),
(59) 12 �(27) 2 �729�7�97�50�50 (mod 97),
(59) 24 �(50) 2 �2500�25�97�75�75 (mod 97),
(59) 48 �(75) 2 �5625�57�97�96�96 (mod 97)��1 (mod 97),
and so, finally,
(59) 96 �(�1) 2 �1 (mod 97),
i.e. (59) 96 leaves remainder 1 when divided by 97, as asserted.
M23 Proof of the Fermat–Euler Theorem
It is helpful to begin with a proof of Fermat’s Little Theorem; the generalistion
to the Fermat–Euler Theorem is then almost obvious.
Fermat’s Little Theorem asserts that
If p is a prime and M is any integer not divisible by p then
M ( p�1) �1 (mod p).
Proof
A complete set of residues (‘remainders’) (mod p) of numbers not divisible
by p is
1, 2, 3, ..., ( p�1).