Lucas primality test


In computational number theory, the Lucas test is a primality test for a natural number n; it requires that the prime factors of n − 1 be already known. It is the basis of the Pratt certificate that gives a concise verification that n is prime.

Concepts

Let n be a positive integer. If there exists an integer a, 1 < a < n, such that
and for every prime factor q of n − 1
then n is prime. If no such number a exists, then n is either 1, 2, or composite.
The reason for the correctness of this claim is as follows: if the first equivalence holds for a, we can deduce that a and n are coprime. If a also survives the second step, then the order of a in the group * is equal to n−1, which means that the order of that group is n−1, implying that n is prime. Conversely, if n is prime, then there exists a primitive root modulo n, or generator of the group *. Such a generator has order |*| = n−1 and both equivalences will hold for any such primitive root.
Note that if there exists an a < n such that the first equivalence fails, a is called a Fermat witness for the compositeness of n.

Example

For example, take n = 71. Then n − 1 = 70 and the prime factors of 70 are 2, 5 and 7.
We randomly select an a=17 < n. Now we compute:
For all integers a it is known that
Therefore, the multiplicative order of 17 is not necessarily 70 because some factor of 70 may also work above. So check 70 divided by its prime factors:
Unfortunately, we get that 1710≡1. So we still don't know if 71 is prime or not.
We try another random a, this time choosing a = 11. Now we compute:
Again, this does not show that the multiplicative order of 11 is 70 because some factor of 70 may also work. So check 70 divided by its prime factors:
So the multiplicative order of 11 is 70, and thus 71 is prime.
.

Algorithm

The algorithm can be written in pseudocode as follows:
algorithm lucas_primality_test is
input: n > 2, an odd integer to be tested for primality.
k, a parameter that determines the accuracy of the test.
output: prime if n is prime, otherwise composite or possibly composite.
determine the prime factors of n−1.
LOOP1: repeat k times:
pick a randomly in the range

return composite
else
LOOP2: for all prime factors q of n−1:

if we checked this equality for all prime factors of n−1 then
return prime
else
continue LOOP2
else
continue LOOP1
return possibly composite.