Lucas primality test

Let n be a positive integer. If there exists an integer a, 1 < a < n, such that

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

and for every prime factor q of n − 1

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

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.

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:

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

For all integers a it is known that

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

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:

${\displaystyle 17^{35}\ \equiv \ 70\ \not \equiv \ 1{\pmod {71}}}$

${\displaystyle 17^{14}\ \equiv \ 25\ \not \equiv \ 1{\pmod {71}}}$

${\displaystyle 17^{10}\ \equiv \ 1\ \equiv \ 1{\pmod {71}}.}$

Unfortunately, we get that 17¹⁰≡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:

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

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:

${\displaystyle 11^{35}\ \equiv \ 70\ \not \equiv \ 1{\pmod {71}}}$

${\displaystyle 11^{14}\ \equiv \ 54\ \not \equiv \ 1{\pmod {71}}}$

${\displaystyle 11^{10}\ \equiv \ 32\ \not \equiv \ 1{\pmod {71}}.}$

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).

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]
            if ${\displaystyle a^{n-1}\not \equiv 1{\pmod {n}}}$ then
                return composite
            else # ${\displaystyle \color {Gray}{a^{n-1}\equiv 1{\pmod {n}}}}$
                LOOP2: for all prime factors q of n−1:
                    if ${\displaystyle a^{\frac {n-1}{q}}\not \equiv 1{\pmod {n}}}$ then
                        if we checked this equality for all prime factors of n−1 then
                            return prime
                        else
                            continue LOOP2
                    else # ${\displaystyle \color {Gray}{a^{\frac {n-1}{q}}\equiv 1{\pmod {n}}}}$
                        continue LOOP1

    return possibly composite.

• 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