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

Multiply each of these by M:
M, 2M, 3M, ......, ( p�1)M.
No two of these numbers produce the same residue (mod p) for if, say,
aM�bM (mod p)
then M(a�b) is divisible by p; but M is not divisible by p and a and b are
both less than p. Hence the ( p�1) multiples of M are all different (mod p);
they must therefore be
1, 2, 3, ..., ( p�1)
in some order. So
(M)(2M)(3M)...(( p�1)M)�(1)(2)(3)...( p�1) (mod p)�( p�1)! (mod p).
Since ( p�1)! has no factor in common with p we can divide it out of both
sides to give
M ( p�1) �1 (mod p)
which proves Fermat’s Little Theorem.
Proof of the Fermat–Euler Theorem
We are now dealing with a composite modulus N. The proof follows along
the same lines as above but now, instead of using all of the residues (mod
p) we now consider only those residues which have no factor in common
with N. If we denote these residues by
a 1 , a 2 , ..., a k
then k��(N), where �(N) is Euler’s function, which was defined in M11. If
we multiply each of the k residues by M then, as before, they are all different
since if
M(a r )�M(a s ) (mod N)
then M(a r �a s ) is divisible by N, but this is impossible since M has no factor
in common with N and (a r �a s ) is less than N. We have therefore proved
that
if M has no factors in common with N then
M �(N) �1 (mod N)
which is the Fermat–Euler Theorem.
Mathematical aspects 211