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

212
appendix
For the RSA encipherment system we need only the special case when
N�pq, where p and q are different primes. In this case �(N)�( p�1)(q�1).
M24 Finding numbers which are ‘probably’ primes
The ‘sieve of Eratosthenes’ will find all the primes below any given
number N and, if a list of all the primes is what is required, this is the standard
method. If, however, we only want to know if a particular integer is a
prime then finding a list of all the primes below it is not necessary and, if
the number is very large, likely to be very time-consuming. Unfortunately
there is no very fast general method for testing if any given large
number, N, is a prime and if N is large enough to be considered for the RSA
method, say about 10 50 , the time required to establish its primality
beyond doubt is likely to be prohibitive. In view of this a different
approach was proposed by Rabin in 1976 [12.8] and, in a different form,
by Solovay and Strassen in 1977 [12.9]. The idea behind this is to use a
test, which involves some number less than N, that
(1) will always fail if the number, N, is a prime,
(2) will succeed more often than not if N is not a prime.
When N is not a prime the test proposed by Rabin will succeed at least 75%
of the time. If we use many numbers less than N, m say, and apply the test
with each of these m and find that the test never succeeds then the probability
that N is not prime is estimated to be (0.25) m and by taking m large
enough the probability that N is prime can be made arbitrarily close to 1.
A description of Rabin’s test can be found in [1.2], Chapter 9. For a bibliography
of some relevant papers see [13.12].
M25 The Euclidean Algorithm
This algorithm is used to find the highest common factor (h.c.f.) of two
integers, x 1 and x 2 . If the h.c.f is denoted by h the algorithm can also be
used to find integers m, n such that
mx 1 �nx 2 �h
which is relevant to the RSA encryption/decryption system.
The Euclidean Algorithm is carried out as follows.
We may suppose that both x 1 and x 2 are positive and that x 1 is greater
than x 2; if not, interchange them.
Divide x 1 by x 2 to give a remainder x 3 :
x 1 �a 1 x 2 �x 3 where a 1 is an integer and 0�x 3 � x 2.