# Lucas–Lehmer test for Mersenne numbers

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 $N=k 2^n-1$, with $2^n > k$, based on the LLT: see Lucas-Lehmer-Riesel test. There is also a Lucas-Lehmer-Reix test for Fermat numbers $F_n=2^\left\{2^n\right\}+1$, 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: see Lucas–Lehmer test."

In mathematics, the Lucas–Lehmer test is a primality test for Mersenne numbers. The test was originally developed by Edouard 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 and Derrick 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 like trial 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:$s_\left\{p-2\right\}equiv0pmod\left\{M_p\right\}.$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 &larr; 4 var M &larr; 2"p" − 1 repeat p − 2 times: s &larr; ((s &times; 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 &times; 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:

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

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 &times; 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, 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("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&ndash;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, 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 &larr; ((4 &times; 4) − 2) mod 7 = 0

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

On the other hand, M11 = 2047 = 23 &times; 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 &larr; ((4 &times; 4) − 2) mod 2047 = 14
* s &larr; ((14 &times; 14) − 2) mod 2047 = 194
* s &larr; ((194 &times; 194) − 2) mod 2047 = 788
* s &larr; ((788 &times; 788) − 2) mod 2047 = 701
* s &larr; ((701 &times; 701) − 2) mod 2047 = 119
* s &larr; ((119 &times; 119) − 2) mod 2047 = 1877
* s &larr; ((1877 &times; 1877) − 2) mod 2047 = 240
* s &larr; ((240 &times; 240) − 2) mod 2047 = 282
* s &larr; ((282 &times; 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: :Then our theorem is that "M""p" is prime iff $s_\left\{p-2\right\}equiv0pmod\left\{M_p\right\}.$

We begin by noting that $\left\{langle\right\}s_i\left\{ angle\right\}$ is a recurrence relation with a closed-form solution. Define $omega = 2 + sqrt\left\{3\right\}$ 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 $s_\left\{p-2\right\}equiv0pmod\left\{M_p\right\}$ implies that $M_p$ 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&ndash;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 $s_\left\{p-2\right\} equiv 0 pmod\left\{M_p\right\}$. 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 $X = \left\{a + bsqrt\left\{3\right\} | a, b in mathbb\left\{Z\right\}_q\right\}$ with "q"2 elements, where $mathbb\left\{Z\right\}_q$ is the integers mod "q", a finite field. The multiplication operation in $X$ is defined by:

:$\left(a + bsqrt\left\{3\right\}\right)\left(c + dsqrt\left\{3\right\}\right) = \left[\left(ac + 3bd\right) hbox\left\{ mod \right\} q\right] + \left[\left(bc + ad\right) hbox\left\{ mod \right\} q\right] sqrt\left\{3\right\}$.

Since "q" > 2, $omega$ and 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 $q^2-1$ (since 0 has no inverse).

Now, since $M_p equiv 0 pmod q$, and $omega in X$, we have $kM_pomega^\left\{2^\left\{p-2 = 0$ in "X", which by equation (1) gives $omega^\left\{2^\left\{p-1 = -1$. Squaring both sides gives $omega^\left\{2^p\right\} = 1$, showing that $omega$ is invertible with inverse $omega^\left\{2^\left\{p\right\}-1\right\}$ and so lies in "X"*, and moreover has an order dividing $2^p$. In fact the order must equal $2^p$, since $omega^\left\{2^\left\{p-1 eq 1$ and so the order does not divide $2^\left\{p-1\right\}$. 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 $M_p$, we must have $q^2 leq M_p = 2^p-1$, yielding the contradiction $2^p < 2^p-1$. So $M_p$ is prime.

Necessity

In the other direction, we suppose $M_p$ is prime and show $s_\left\{p-2\right\} equiv0pmod\left\{M_p\right\}$. 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 the Legendre symbol properties tell us $\left(3|M_p\right)$ is −1. Euler's criterion then gives us:

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

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

: $2^\left\{\left(M_p-1\right)/2\right\} equiv 1 pmod\left\{M_p\right\}$.

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

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

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

: $a^\left\{M_p\right\} equiv a pmod\left\{M_p\right\}$

for every integer "a" (Fermat's little theorem)

Then, in the group "X"* we have:

:

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

:

where we use the fact that

: $24^\left\{\left(M_p-1\right)/2\right\} = \left(2^\left\{\left(M_p-1\right)/2\right\}\right)^3\left(3^\left\{\left(M_p-1\right)/2\right\}\right) = \left(1\right)^3\left(-1\right) = -1.$

Since $M_p equiv 3 pmod 4$, 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 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.

* Lucas-Lehmer test
* Mersenne's conjecture

References

* Section 4.2.1: The Lucas–Lehmer test, pp.167&ndash;170.

Notes

* [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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