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

(4) Now (k x ) y �(k y ) x �m xy (mod p)�K (say) and K is the common key which
both X and Y can use, even though neither of them knows the other’s
secret key.
Before one can use the Diffie–Hellman system it is necessary to be able
to find a very large prime number, and this is a non-trivial task. We meet
this problem again in the RSA encryption system (where two large
primes are needed), where references to an interesting approach will be
found.
In a realistic example the prime p would be very large but the essence
of the method can be illustrated with a prime of moderate size.
Example 12.1
In the Diffie–Hellman system if p�59, m�3, x�7 and y�11 what will be
the values of k x , k y and K?
Solution
(1) We first note that ( p�1)�58�2�29 and so has no factor in common
with x or y.
(2) X computes 3 7 (mod 59) and Y computes 3 11 (mod 59). These calculations
can be done in various ways, some more efficient than others (M22).
In this case the numbers are sufficiently small that the powers can be
handled on a pocket calculator. Thus
3 7 �2187�37�59�4,
3 11 �177 147�59�3002�29
so k x �4 and k y �29.
(3) The value of the common key, K, is then given either by 4 11 (mod 59) or
by 29 7 (mod 59). These should give the same value; if they don’t we have
made a mistake. Therefore as a check we compute them both. The
numbers this time are somewhat bigger so we compute them by forming
powers and reducing (mod 59) at each stage thus:
(i) 4 5 �1024�17�59�21�21 (mod 59)
so
4 10 �441�7�59�28�28 (mod 59)
and therefore
4 11 �4�28�112�1�59�53�53 (mod 59).
We conclude that the common key, K, is 53.
(ii) We check this by computing 29 7 (mod 59).
29 2 �841�14�59�15�15 (mod 59)
Public key cryptography 167