- Lucas–Lehmer test for Mersenne numbers
:"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 , with , based on the LLT: see
Lucas-Lehmer-Riesel test . There is also a Lucas-Lehmer-Reix test for Fermat numbers , with seed = 5, based on the LLT: see the "External Links". There is also a generalized Lucas–Lehmer test for testing the primality of any number, not based on the LLT: seeLucas–Lehmer test ."In
mathematics , the Lucas–Lehmer test is aprimality test for Mersenne numbers. The test was originally developed byEdouard Lucas in 1856 [http://primes.utm.edu/notes/by_year.html] [http://primes.utm.edu/curios/page.php?number_id=135] , and subsequently improved by Lucas in 1878 andDerrick Henry Lehmer in the 1930s.The test
The Lucas-Lehmer test works as follows. Let "M""p" = 2"p" − 1 be the Mersenne number to test with "p" an
odd prime (because "p" is exponentially smaller than "M""p", we can use a simple algorithm liketrial division for establishing its primality). Define a sequence {"s" "i"} for all "i" ≥ 0 by:
The first few terms of this sequence are 4, 14, 194, 37634, ... OEIS|id=A003010.Then "M""p" is prime
iff :The number "s""p" − 2 mod "M""p" is called the Lucas–Lehmer residue of "p". (Some authors equivalently set "s"1 = 4 and test "s""p"−1 mod "M""p"). In pseudocode, the test might be written:"// Determine if M"p" = 2"p" − 1 is prime Lucas-Lehmer(p) var s ← 4 var M ← 2"p" − 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 inmodular exponentiation .Time complexity
In the algorithm as written above, there are two expensive operations during each iteration: the multiplication
s × s
, and themod M
operation. Themod M
operation can be made particularly efficient on standard binary computers by observing the following simple property::.
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 2"n"−1 without using division. For example:
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 2"n"−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("p"2) 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("p"3). The most efficient known multiplication method, theSchönhage-Strassen algorithm based on theFast Fourier transform , requires O("p" log "p" log log "p") time to square a "p"-bit number, reducing the complexity to O("p"2 log "p" log log "p") or Õ("p"2). [W. N. Colquitt, L. Welsh, Jr. A New Mersenne Prime. "Mathematics of Computation", Vol.56, No.194, pp.867–870. April 1991. "The use of the FFT speeds up the asymptotic time for the Lucas-Lehmer test for M"p" from O("p"3) to O("p"2 log "p" log log "p") bit operations."]By comparison, the most efficient randomized primality test for general integers, the
Miller-Rabin primality test , takes O("k" "p"2 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, theAKS 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 = 1736Because "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: :Then our theorem is that "M""p" is prime
iff We begin by noting that is a
recurrence relation with a closed-form solution. Define and ; then we can verify by induction that for all "i":::where the last step follows from . We will use this in both parts.
Sufficiency
In this direction we wish to show that implies that is prime. We relate a straightforward proof exploiting elementary
group theory given by J. W. Bruce [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.] as related by Jason Wojciechowski [Jason Wojciechowski. Mersenne Primes, An Introduction and Overview. January 2003. http://wonka.hampshire.edu/~jason/math/smithnum/project.ps] .Suppose . Then for some integer "k", and::
Now suppose "M""p" is composite with nontrivial prime factor "q" > 2 (all Mersenne numbers are odd). Define the set with "q"2 elements, where is the integers mod "q", a
finite field . The multiplication operation in is defined by::.
Since "q" > 2, and are in . 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 (since 0 has no inverse).
Now, since , and , we have in "X", which by equation (1) gives . Squaring both sides gives , showing that is invertible with inverse and so lies in "X"*, and moreover has an order dividing . In fact the order must equal , since and so the order does not divide . Since the order of an element is at most the order (size) of the group, we conclude that . But since "q" is a nontrivial prime factor of , we must have , yielding the contradiction . So is prime.
Necessity
In the other direction, we suppose is prime and show . We rely on a simplification of a proof by Öystein J. R.Ödseth. [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] First, notice that 3 is a
quadratic non-residue mod "M""p", since 2 "p" − 1 for odd "p" > 1 only takes on the value 7 mod 12, and so theLegendre symbol properties tell us is −1.Euler's criterion then gives us:: .
On the other hand, 2 is a
quadratic residue mod , since and so . Euler's criterion again gives:: .
Next, define , and define "X"* similarly as before as the multiplicative group of . We will use the following lemmas:
:
(from
Proofs of Fermat's little theorem#Proof_using_the_binomial_theorem ):
for every integer "a" (
Fermat's little theorem )Then, in the group "X"* we have:
:
We chose such that . Consequently, we can use this to compute in the group "X"*:
:
where we use the fact that
:
Since , all that remains is to multiply both sides of this equation by and use :
:
Since "s""p"−2 is an integer and is zero in "X"*, it is also zero mod "M""p".
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. [GIMPS Home Page. Frequently Asked Questions: General Questions: What are Mersenne primes? How are they useful? http://www.mersenne.org/faq.htm#what] 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 fastPépin's test for anyFermat number ) can be tried on a large search space of numbers with the required form before reaching computational limits.See also
*
Lucas-Lehmer test
*Mersenne's conjecture References
* Section 4.2.1: The Lucas–Lehmer test, pp.167–170.
Notes
External links
* [http://mathworld.wolfram.com/Lucas-LehmerTest.html MathWorld: Lucas–Lehmer test]
* [http://www.mersenne.org GIMPS (The Great Internet Mersenne Prime Search)]
* [http://www.jt-actuary.com/lucas-le.htm A proof of Lucas–Lehmer test (for Mersenne numbers)]
* [http://arxiv.org/abs/0705.3664 A proof of Lucas-Lehmer-Reix test (for Fermat numbers)]
* [http://www.mersennewiki.org/index.php/Lucas-Lehmer_Test Lucas–Lehmer test] at MersenneWiki
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