# Lucas primality test

(Redirected from Lucas–Lehmer test)
"Lucas–Lehmer test" redirects here. For the test for Mersenne numbers, see Lucas–Lehmer primality test. For the Lucas–Lehmer–Riesel test, see Lucas–Lehmer–Riesel 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.[1][2] 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 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 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 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:

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

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:

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 an-1 $\not\equiv$ 1 (mod n) then return composite
otherwise
LOOP2: for all prime factors q of n−1:
if a(n-1)/q $\not\equiv$ 1 (mod n)
if we did not check this equality for all prime factors of n−1
then do next LOOP2
otherwise return prime
otherwise do next LOOP1
return possibly composite.