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

key, d, we use the Euclidean Algorithm which is explained in M25, but the
method is illustrated by the following:
so
72�2�29�14,
29�2�14�1,
1�29�2�14,
but we have seen, in the first line above, that 14�72�2�29 and so
or
1�29�2�(72�2�29)�5�29�2�72, that is:
5�29�2�72�1,
5�29�1 (mod 72)
(i.e. 145�144�1) which means that d, the decipherment key, is 5.
This may seem confusing, but the method is to use the Euclidean
Algorithm to find the highest common factor of the encipherment key, e,
and the number ( p�1)(q�1). Since e and ( p�1)(q�1) have no common
factor the highest common factor is 1. If we now ‘work backwards’
through the Euclidean Algorithm from the last line to the first, replacing
the last number on the right of each identity by the other numbers
involved, we will eventually obtain the decipherment key, d, as given by
the congruence in (3) above. A formal description, and an alternative
approach, are described in M25.
Example 13.2 (A slightly more realistic case, which we shall
use below)
Find the decipherment key, d, when n�3127 and the encipherment key, e,
is 17.
Solution
We note that n�3127�53�59 and 53 and 59 are both primes. It follows
that n is a valid modulus for the RSA and that (p�1)(q�1)�(52)(58)�
3016. Furthermore since 17 has no factor in common with 3127 or with
3016 it is a valid encipherment key. Proceeding as before, that is, as in the
Euclidean Algorithm:
3016�177�17�7,
1117�2�7�3,
1117�2�3�1,
Encipherment and the internet 177