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
:<math>a^{n-1}\ \equiv\ 1 \pmod n \, </math>
and for every prime factor q of n − 1
:<math>a^{({n-1})/q}\ \not\equiv\ 1 \pmod n \, </math>
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 (Z/nZ)* is equal to n − 1, which means that the order of that group is n − 1 (because the order of every element of a group divides the order of the group), implying that n is prime. Conversely, if n is prime, then there exists a primitive root modulo n, or generator of the group (Z/nZ)*. Such a generator has order |(Z/nZ)*| = 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:
:<math>17^{70}\ \equiv\ 1 \pmod {71}.</math>
For all integers a it is known that
:<math>a^{n - 1}\equiv 1 \pmod{n}\ \text{ if and only if } \text{ ord}(a)\mid(n-1).</math>
Therefore, the multiplicative order of 17 (mod 71) is not necessarily 70 because some factor of 70 may also work above. So check 70 divided by its prime factors:
:<math>17^{35}\ \equiv\ 70\ \not\equiv\ 1 \pmod {71}</math>
:<math>17^{14}\ \equiv\ 25\ \not\equiv\ 1 \pmod {71}</math>
:<math>17^{10}\ \equiv\ 1\ \equiv\ 1 \pmod {71}.</math>
Unfortunately, we get that 17<sup>10</sup> ≡ 1 (mod 71). So we still do not know if 71 is prime or not.
We try another random a, this time choosing a = 11. Now we compute:
:<math>11^{70}\ \equiv\ 1 \pmod {71}.</math>
Again, this does not show that the multiplicative order of 11 (mod 71) is 70 because some factor of 70 may also work. So check 70 divided by its prime factors:
:<math>11^{35}\ \equiv\ 70\ \not\equiv\ 1 \pmod {71}</math>
:<math>11^{14}\ \equiv\ 54\ \not\equiv\ 1 \pmod {71}</math>
:<math>11^{10}\ \equiv\ 32\ \not\equiv\ 1 \pmod {71}.</math>
So the multiplicative order of 11 (mod 71) is 70, and thus 71 is prime.
(To carry out these modular exponentiations, one could use a fast exponentiation algorithm like binary or addition-chain exponentiation).
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 [2, n − 1]
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.
See also
- Édouard Lucas, for whom this test is named
- Fermat's little theorem
- Pocklington primality test, an improved version of this test which only requires a partial factorization of n − 1
- Primality certificate