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This article is about the Lucas–Lehmer test (LLT), that only applies to Mersenne numbers. There is also a Lucas-Lehmer-Riesel test for numbers of the form N = k2n−1, with 2n > k, based on the LLT: see Lucas-Lehmer-Riesel test. There is also a Lucas-Lehmer-Reix test for Fermat numbers Fn = 22n + 1, with seed = 5, based on the LLT: see the "External links".

In mathematics, the Lucas–Lehmer test is a primality test for Mersenne numbers. The test was originally developed by Edouard Lucas in 1856 [1][2], and subsequently improved by Lucas in 1878 and Derrick Henry Lehmer in the 1930s.

The test

The Lucas-Lehmer test works as follows. Let Mp = 2p − 1 be the Mersenne number to test with p an odd prime (because p is exponentially smaller than Mp, we can use a simple algorithm like trial division for establishing its primality). Define a sequence {s i} for all i ≥ 0 by

$s_i= \begin{cases} 4 & \mbox{if }i=0; \ s_{i-1}^2-2 & \mbox{otherwise.} \end{cases}$

The first few terms of this sequence are 4, 14, 194, 37634, ... (sequence A003010 in OEIS). Then Mp is prime iff

$s_{p-2}\equiv0\pmod{M_p}.$

The number sp − 2 mod M p is called the Lucas–Lehmer residue of p. (Some authors equivalently set s1 = 4 and test sp−1 mod Mp). In pseudocode, the test might be written:

```// Determine if Mp = 2p − 1 is prime
Lucas-Lehmer(p)
    var s ← 4
    var M ← 2p − 1
    repeat p − 2 times:
        s ← ((s × s) − 2) mod M
    if s = 0 return PRIME else return COMPOSITE
```

By performing the `mod M` at each iteration, we ensure that all intermediate results are at most p bits (otherwise the number of bits would double each iteration). It is exactly the same strategy employed in modular exponentiation.

Time
complexity

In the algorithm as written above, there are two expensive operations during each iteration: the multiplication `s × s`, and the `mod M` operation. The `mod M` operation can be made particularly efficient on standard binary computers by observing the following simple property:

$k \equiv (k \hbox{ mod } 2^n) + \lfloor k/2^n \rfloor \pmod{2^n - 1}$.

In other words, if we take the least significant n bits of k, and add the remaining bits of k, and then do this repeatedly until at most n bits remain, we can compute the remainder after dividing k by the Mersenne number 2n−1 without using division. For example:

|              |    |                          |
|--------------|----|--------------------------|
| 916 mod 25−1 | =  | 11100101002 mod 25−1     |
|              | =  | 111002 + 101002 mod 25−1 |
|              | =  | 1100002 mod 25−1         |
|              | =  | 12 + 100002 mod 25−1     |
|              | =  | 100012 mod 25−1          |
|              | =  | 100012                   |
|              | =  | 17.                      |

Moreover, since `s × s` will never exceed M2 < 22p, this simple technique converges in at most 2 p-bit additions, which can be done in linear time. As a small exceptional case, the above algorithm may produce 2n−1 for a multiple of the modulus, rather than the correct value of zero; this should be accounted for.

With the modulus out of the way, the asymptotic complexity of the algorithm depends only on the multiplication algorithm used to square s at each step. The simple "grade-school" algorithm for multiplication requires O(p2) bit-level or word-level operations to square a p-bit number, and since we do this O(p) times, the total time complexity is O(p3). The most efficient known multiplication method, the Schönhage-Strassen algorithm based on the Fast Fourier transform, requires O(p log p log log p) time to square a p-bit number, reducing the complexity to O(p2 log p log log p) or Õ(p2).[1]

By comparison, the most efficient randomized primality test for general integers, the Miller-Rabin primality test, takes O(k p2 log p log log p) bit operations using FFT multiplication, where k is the number of iterations and is related to the error rate. This is a constant factor difference for constant k, but in practice the cost of doing many iterations and other differences lead to worse performance for Miller-Rabin. The most efficient deterministic primality test for general integers, the AKS primality test, requires Õ(p6) bit operations in its best known variant and is dramatically slower in practice.

Examples

Suppose we wish to verify that M3 = 7 is prime using the Lucas-Lehmer test. We start out with s set to 4 and then update it 3−2 = 1 time, taking the results mod 7:

• s ← ((4 × 4) − 2) mod 7 = 0

Because we end with s set to zero, M3 is prime.

On the other hand, M11 = 2047 = 23 × 89 is not prime. To show this, we start with s set to 4 and update it 11−2 = 9 times, taking the results mod 2047:

• s ← ((4 × 4) − 2) mod 2047 = 14
• s ← ((14 × 14) − 2) mod 2047 = 194
• s ← ((194 × 194) − 2) mod 2047 = 788
• s ← ((788 × 788) − 2) mod 2047 = 701
• s ← ((701 × 701) − 2) mod 2047 = 119
• s ← ((119 × 119) − 2) mod 2047 = 1877
• s ← ((1877 × 1877) − 2) mod 2047 = 240
• s ← ((240 × 240) − 2) mod 2047 = 282
• s ← ((282 × 282) − 2) mod 2047 = 1736

Because s is not zero, M11=2047 is not prime. Notice that we learn nothing about the factors of 2047, only its Lucas–Lehmer residue, 1736.

Proof of
correctness

Lehmer's original proof of the correctness of this test is complex, so we'll depend upon more recent refinements. Recall the definition:

$s_i= \begin{cases} 4 & \mbox{if }i=0; \ s_{i-1}^2-2 & \mbox{otherwise.} \end{cases}$

Then our theorem is that Mp is prime iff $s_{p-2}\equiv0\pmod{M_p}.$

We begin by noting that ${\langle}s_i{\rangle}$ is a recurrence relation with a closed-form solution. Define $\omega = 2 + \sqrt{3}$ and $\bar{\omega} = 2 - \sqrt{3}$; then we can verify by induction that $s_i = \omega^{2^i} + \bar{\omega}^{2^i}$ for all i:

$s_0 = \omega^{2^0} + \bar{\omega}^{2^0} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4.$
$\begin{array}{lcl}s_n & = & s_{n-1}^2 - 2 \ & = & \left(\omega^{2^{n-1}} + \bar{\omega}^{2^{n-1}}\right)^2 - 2 \ & = & \omega^{2^n} + \bar{\omega}^{2^n} + 2(\omega\bar{\omega})^{2^{n-1}} - 2 \ & = & \omega^{2^n} + \bar{\omega}^{2^n}, \ \end{array}$

where the last step follows from $\omega\bar{\omega} = (2 + \sqrt{3})(2 - \sqrt{3}) = 1$. We will use this in both parts.

Sufficiency

In this direction we wish to show that $s_{p-2}\equiv0\pmod{M_p}$ implies that Mp is prime. We relate a straightforward proof exploiting elementary group theory given by J. W. Bruce[2] as related by Jason Wojciechowski[3].

Suppose $s_{p-2} \equiv 0 \pmod{M_p}$. Then $\omega^{2^{p-2}} + \bar{\omega}^{2^{p-2}} = kM_p$ for some integer k, and:

$\begin{align} \omega^{2^{p-2}} & = kM_p - \bar{\omega}^{2^{p-2}} \ \left(\omega^{2^{p-2}}\right)^2 & = kM_p\omega^{2^{p-2}} - (\omega\bar{\omega})^{2^{p-2}} \ \omega^{2^{p-1}} & = kM_p\omega^{2^{p-2}} - 1.\quad\quad\quad\quad\quad(1) \end{align}$

Now suppose Mp is composite with nontrivial prime factor q > 2 (all Mersenne numbers are odd). Define the set $X = \{a + b\sqrt{3} | a, b \in \mathbb{Z}_q\}$ with q2 elements, where $\mathbb{Z}_q$ is the integers mod q, a finite field. The multiplication operation in X is defined by:

$(a + b\sqrt{3})(c + d\sqrt{3}) = [(ac + 3bd) \hbox{ mod } q] + [(bc + ad) \hbox{ mod } q]\sqrt{3}.$

Since q > 2, ω and $\bar{\omega}$ are in X. Any product of two numbers in X is in X, but it's not a group under multiplication because not every element x has an inverse y such that xy = 1. If we consider only the elements that have inverses, we get a group X* of size at most q2 − 1 (since 0 has no inverse).

Now, since $M_p \equiv 0 \pmod q$, and $\omega \in X$, we have $kM_p\omega^{2^{p-2}} = 0$ in X, which by equation (1) gives $\omega^{2^{p-1}} = -1$. Squaring both sides gives $\omega^{2^p} = 1$, showing that ω is invertible with inverse $\omega^{2^{p}-1}$ and so lies in X*, and moreover has an order dividing 2p. In fact the order must equal 2p, since $\omega^{2^{p-1}} \neq 1$ and so the order does not divide 2p − 1. Since the order of an element is at most the order (size) of the group, we conclude that $2^p \leq q^2 - 1 < q^2$. But since q is a nontrivial prime factor of Mp, we must have $q^2 \leq M_p = 2^p-1$, yielding the contradiction 2p < 2p − 1. So Mp is prime.

Necessity

In the other direction, we suppose Mp is prime and show $s_{p-2} \equiv0\pmod{M_p}$. We rely on a simplification of a proof by Öystein J. R. Ödseth.[4] First, notice that 3 is a quadratic non-residue mod Mp, since 2 p − 1 for odd p > 1 only takes on the value 7 mod 12, and so the Legendre symbol properties tell us (3 | Mp) is −1. Euler's criterion then gives us:

$3^{(M_p-1)/2} \equiv -1 \pmod{M_p}.\,$

On the other hand, 2 is a quadratic residue mod Mp, since $2^p \equiv 1 \pmod{M_p}$ and so $2 \equiv 2^{p+1} = \left(2^{(p+1)/2}\right)^2 \pmod{M_p}$. Euler's criterion again gives:

$2^{(M_p-1)/2} \equiv 1 \pmod{M_p}.\,$

Next, define $\sigma = 2\sqrt{3}$, and define X* similarly as before as the multiplicative group of $\{a + b\sqrt{3} | a, b \in \mathbb{Z}_{M_p}\}$. We will use the following lemmas:

$(x+y)^{M_p} \equiv x^{M_p} + y^{M_p} \pmod{M_p}$

(from Proofs of Fermat's little theorem#Proof_using_the_binomial_theorem)

$a^{M_p} \equiv a \pmod{M_p}$

for every integer a (Fermat's little theorem)

Then, in the group X* we have:

$\begin{align} (6+\sigma)^{M_p} & = 6^{M_p} + (2^{M_p})(\sqrt{3}^{M_p}) \ & = 6 + 2(3^{(M_p-1)/2})\sqrt{3} \ & = 6 + 2(-1)\sqrt{3} \ & = 6 - \sigma. \end{align}$

We chose σ such that ω = (6 + σ)2 / 24. Consequently, we can use this to compute $\omega^{(M_p+1)/2}$ in the group X*:

$\begin{align} \omega^{(M_p+1)/2} & = (6 + \sigma)^{M_p+1}/24^{(M_p+1)/2} \ & = (6 + \sigma)^{M_p}(6 + \sigma)/(24 \times 24^{(M_p-1)/2}) \ & = (6 - \sigma)(6 + \sigma)/(-24) \ & = -1. \end{align}$

where we use the fact that

$24^{(M_p-1)/2} = (2^{(M_p-1)/2})^3(3^{(M_p-1)/2}) = (1)^3(-1) = -1.$

Since $M_p \equiv 3 \pmod 4$, all that remains is to multiply both sides of this equation by $\bar{\omega}^{(M_p+1)/4}$ and use $\omega\bar{\omega}=1$:

$\begin{align} \omega^{(M_p+1)/2}\bar{\omega}^{(M_p+1)/4} & = -\bar{\omega}^{(M_p+1)/4} \ \omega^{(M_p+1)/4} + \bar{\omega}^{(M_p+1)/4} & = 0 \ \omega^{(2^p-1+1)/4} + \bar{\omega}^{(2^p-1+1)/4} & = 0 \ \omega^{2^{p-2}} + \bar{\omega}^{2^{p-2}} & = 0 \ s_{p-2} & = 0. \end{align}$

Since sp−2 is an integer and is zero in X*, it is also zero mod Mp.

Applications

The Lucas-Lehmer test is the primality test used by the Great Internet Mersenne Prime Search to locate large primes, and has been successful in locating many of the largest primes known to date.[5] They consider it valuable for finding very large primes because Mersenne numbers are considered somewhat more likely to be prime than randomly chosen odd integers of the same size. Additionally, the test is considered valuable because it can provably test a very large number for primality within affordable time and (in contrast to the equivalently fast Pépin's test for any Fermat number) can be tried on a large search space of numbers with the required form before reaching computational limits.

References

1. ^ Colquitt, W. N.; Welsh, L., Jr. (1991), "A New Mersenne Prime", Mathematics of Computation 56 (194): 867–870, "The use of the FFT speeds up the asymptotic time for the Lucas-Lehmer test for Mp from O(p3) to O(p2 log p log log p) bit operations."
2. ^ J. W. Bruce. A Really Trivial Proof of the Lucas-Lehmer Test. The American Mathematical Monthly, Vol.100, No.4, pp.370–371. April 1993.
3. ^ Jason Wojciechowski. Mersenne Primes, An Introduction and Overview. January 2003. http://wonka.hampshire.edu/~jason/math/smithnum/project.ps
4. ^ Öystein J. R. Ödseth. A note on primality tests for N = h · 2n − 1. Department of Mathematics, University of Bergen. http://www.uib.no/People/nmaoy/papers/luc.pdf
5. ^ GIMPS Home Page. Frequently Asked Questions: General Questions: What are Mersenne primes? How are they useful? http://www.mersenne.org/faq.htm#what

Further
reading

• Crandall, Richard; Pomerance, Carl (2001), Prime Numbers: A Computational Perspective (1st ed.), Berlin: Springer, ISBN 0387947779   Section 4.2.1: The Lucas–Lehmer test, pp.167–170.

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