Lucas primality test
# 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

    $a^{n-1}\ \equiv \ 1{\pmod {n}}\,$

and for every prime factor q of n − 1

    $a^{({n-1})/q}\ \not \equiv \ 1{\pmod {n}}\,$

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:

    $17^{70}\ \equiv \ 1{\pmod {71}}.$

For all integers a it is known that

    $a^{n-1}\equiv 1{\pmod {n}}\ {\text{ if and only if }}{\text{ ord}}(a)|(n-1).$

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:

    $17^{35}\ \equiv \ 70\ \not \equiv \ 1{\pmod {71}}$
    $17^{14}\ \equiv \ 25\ \not \equiv \ 1{\pmod {71}}$
    $17^{10}\ \equiv \ 1\ \equiv \ 1{\pmod {71}}.$

Unfortunately, we get that $17^{10}≡1$ (mod 71). 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:

    $11^{70}\ \equiv \ 1{\pmod {71}}.$

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:

    $11^{35}\ \equiv \ 70\ \not \equiv \ 1{\pmod {71}}$
    $11^{14}\ \equiv \ 54\ \not \equiv \ 1{\pmod {71}}$
    $11^{10}\ \equiv \ 32\ \not \equiv \ 1{\pmod {71}}.$

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:

