Pocklington primality test

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In mathematics, the Pocklington–Lehmer primality test is a primality test devised by Henry Cabourn Pocklington[1] and Derrick Henry Lehmer.[2] The test uses a partial factorization of to prove that an integer is prime.

It produces a primality certificate to be found with less effort than the Lucas primality test, which requires the full factorization of .

Pocklington criterion

The basic version of the test relies on the Pocklington Theorem (Pocklington criterion) which is formulated as follows:

Let be an integer, and suppose there exist numbers a and p such that

 

 

 

 

(1)

p is prime, and

 

 

 

 

(2)

 

 

 

 

(3)

Then N is prime.[3]

Note: Equation (1) is simply a Fermat primality test. If we find any value of a, not divisible by N, such that equation (1) is false, we may immediately conclude that N is not prime. (This divisibility condition is not explicitly stated because it is implied by equation (3).) For example, let . With , we find that . This is enough to prove that N is not prime.

Proof of this theorem

Suppose N is not prime. This means there must be a prime q, where that divides N.

Since , , and since p is prime, .

Thus there must exist an integer u, a multiplicative inverse of p modulo q−1, with the property that

 

 

 

 

(4)

and therefore, by Fermat's little theorem

 

 

 

 

(5)

This implies

,     by (1) since
,
,     by (5)

This shows that q divides the in (3), and therefore this ; a contradiction.[3]

Given N, if p and a can be found which satisfy the conditions of the theorem, then N is prime. Moreover, the pair (p, a) constitute a primality certificate which can be quickly verified to satisfy the conditions of the theorem, confirming N as prime.

The main difficulty is finding a value of p which satisfies (2). First, it is usually difficult to find a large prime factor of a large number. Second, for many primes N, such a p does not exist. For example, has no suitable p because , and , which violates the inequality in (2).

Given p, finding a is not nearly as difficult.[4] If N is prime, then by Fermat's little theorem, any a in the interval will satisfy (1) (however, the cases and are trivial and will not satisfy (3)). This a will satisfy (3) as long as ord(a) does not divide . Thus a randomly chosen a in the interval has a good chance of working. If a is a generator mod N, its order is and so the method is guaranteed to work for this choice.

Generalized Pocklington method

A generalized version of Pocklington's theorem is more widely applicable because it does not require finding a single large prime factor of ; also, it allows a different value of a to be used for each known prime factor of .[5]:Corollary 1

Corollary: Factor N − 1 as N − 1 = AB, where A and B are relatively prime, , the prime factorization of A is known, but the factorization of B is not necessarily known.

If for each prime factor p of A there exists an integer so that

, and

 

 

 

 

(6)

,

 

 

 

 

(7)

then N is prime.

Proof of Corollary: Let p be a prime dividing A and let be the maximum power of p dividing A. Let q be a prime factor of N. For the from the corollary set . This means and because of also .

This means that the order of is

Thus, . The same observation holds for each prime power factor of A, which implies .

Specifically, this means

If N were composite, it would necessarily have a prime factor which is less than or equal to . It has been shown that there is no such factor, which proves that N is prime.

The test

The Pocklington–Lehmer primality test follows directly from this corollary. To use this corollary, first find enough factors of N − 1 so the product of those factors exceeds . Call this product A. Then let B = (N − 1)/A be the remaining, unfactored portion of N − 1. It does not matter whether B is prime. We merely need to verify that no prime that divides A also divides B, that is, that A and B are relatively prime. Then, for every prime factor p of A, find an which fulfills conditions (6) and (7) of the corollary. If such s can be found, the Corollary implies that N is prime.

According to Koblitz, = 2 often works.[3]

Example

Determine whether

is prime.

First, search for small prime factors of . We quickly find that

.

We must determine whether and meet the conditions of the Corollary. , so . Therefore, we have factored enough of to apply the Corollary. We must also verify that .

It does not matter whether B is prime (in fact, it is not).

Finally, for each prime factor p of A, use trial and error to find an ap that satisfies (6) and (7).

For , try . Raising to this high power can be done efficiently using binary exponentiation:

.

So, satisfies (6) but not (7). As we are allowed a different ap for each p, try instead:

.

So satisfies both (6) and (7).

For , the second prime factor of A, try :

.
.

satisfies both (6) and (7).

This completes the proof that is prime. The certificate of primality for would consist of the two pairs (2, 5) and (3, 2).

We have chosen small numbers for this example, but in practice when we start factoring A we may get factors that are themselves so large their primality is not obvious. We cannot prove N is prime without proving that the factors of A are prime as well. In such a case we use the same test recursively on the large factors of A, until all of the primes are below a reasonable threshold.

In our example, we can say with certainty that 2 and 3 are prime, and thus we have proved our result. The primality certificate is the list of pairs, which can be quickly checked in the corollary.

If our example had included large prime factors, the certificate would be more complicated. It would first consist of our initial round of aps which correspond to the 'prime' factors of A; Next, for each factor of A where primality was uncertain, we would have more ap, and so on for factors of these factors until we reach factors of which primality is certain. This can continue for many layers if the initial prime is large, but the important point is that a certificate can be produced, containing at each level the prime to be tested, and the corresponding aps, which can easily be verified.

Extensions and variants

The 1975 paper by Brillhart, Lehmer, and Selfridge[5] gives a proof for what is shown above as the "generalized Pocklington theorem" as Theorem 4 on page 623. Additional theorems are shown which allow less factoring. This includes their Theorem 3 (a strengthening of an 1878 theorem of Proth):

Let where p is an odd prime such that . If there exists an a for which , but , then N is prime.

If N is large, it is often difficult to factor enough of to apply the above corollary. Theorem 5 of the Brillhart, Lehmer, and Selfridge paper allows a primality proof when the factored part has reached only . Many additional such theorems are presented that allow one to prove the primality of N based on the partial factorization of and .

References

  1. ^ Pocklington, Henry C. (1914–1916). "The determination of the prime or composite nature of large numbers by Fermat's theorem". Proceedings of the Cambridge Philosophical Society. 18: 29–30.
  2. ^ D. H. Lehmer (1927). "Tests for primality by the converse of Fermat's theorem". Bull. Amer. Math. Soc. 33 (3): 327–340. doi:10.1090/s0002-9904-1927-04368-3.
  3. ^ a b c Koblitz, Neal (1994). A Course in Number Theory and Cryptography. Graduate Texts in Mathematics. 144 (2nd ed.). Springer. ISBN 0-387-94293-9.
  4. ^ Roberto Avanzi, Henri Cohen, Christophe Doche, Gerhard Frey, Tanja Lange, Kim Nguyen, Frederik Vercauteren (2005). Handbook of Elliptic and Hyperelliptic Curve Cryptography. Boca Raton: Chapman & Hall/CRC.
  5. ^ a b Brillhart, John; Lehmer, D. H.; Selfridge, J. L. (April 1975). "New Primality Criteria and Factorizations of 2m ± 1" (PDF). Mathematics of Computation. 29 (130): 620–647. doi:10.1090/S0025-5718-1975-0384673-1. JSTOR 2005583.

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