Study started since 2011年10月17日 素性测试算法，素性测定(11Y11) primality testing

14.19.Elliptic Curve Primality Testing

Elliptic curve

http://calistamusic.dreab.com/p-Elliptic_curve_primality_testing
http://www.peach.dreab.com/p-Elliptic_curve_primality_proving

Elliptic curve primality testing techniques are among the quickest and most widely used methods in primality proving. It is an idea forwarded by Shafi Goldwasser and Joe Kilian in 1986 and turned into an algorithm by A.O.L. Atkin the same year. The algorithm was altered and improved by several collaborators subsequently, and notably by Atkin and Francois Morain, in 1993. The concept of using elliptic curves in factorization had been developed by H.W. Lenstra in 1985, and the implications for its use in primality testing (and proving) followed quickly.

The elliptic curve test, proves primality (or compositeness) with a quickly verifiable certificate.
Elliptic curve primality proving provides an alternative to (among others) the Pocklington primality test, which can be difficult to implement in practice. Interestingly, the elliptic curve primality tests are based on criteria which is analogous to the Pocklington criterion, on which that test is based, where the group is replaced by , and E is a properly chosen elliptic curve. We will now state a proposition on which to base our test, which is analogous to the Pocklington criterion, and gives rise to the Goldwasser-Kilian-Atkin form of the elliptic curve primality test.

Contents

14.19.1.Proposition

Let N be a positive integer, and E be the set which is defined by the equation y² = x³ + ax + b(mod N). Consider E over, use the usual addition law on E, and write O for the neutral element on E.

Let m be an integer. If there is a prime q which divides m, and is greater than (N^{1 / 4} + 1)² and there exists a point P on E such that
(1) mP = 0

(2) (m/q)P is defined and not equal to 0

Then N is prime.

14.19.2.Proof

If N is composite, then there exists a prime that divides N. Define Ep as the elliptic curve defined by the same equation as E but evaluated modulo p rather than modulo N. Define mp as the order of the group Ep. By Hasse's theorem on elliptic curves we know

and thus gcd(q,m_p) = 1 and there exists an integer u with the property that

Let Pp be the point P evaluated modulo p. Thus, on Ep we have

by (1), as mPp is calculated using the same method as m_P, except modulo p rather than modulo N (and ).
This contradicts (2), because if (m/q)P is defined and not equal to 0 (mod N), then the same method calculated modulo p instead of modulo N will yield

14.19.3.Goldwasser–Kilian algorithm

From this proposition an algorithm can be constructed to prove an integer, N, is prime. This is done as follows:

Choose three integers at random, a, x, y and set

Now P = (x,y) is a point on E, where we have that E is defined by y² = x³ + ax + b. Next we need an algorithm to count the number of points on E. Applied to E, this algorithm (Koblitz and others suggest Schoof's algorithm) produces a number m which is the number of points on curve E over FN, provided N is prime. Next we have a criterion for deciding whether our curve E is acceptable.
If we can write m in the form m = kq where is a small integer and q a probable prime (it has passed some previous probabilistic primality test, for example), then we do not discard E. If it is not possible to write m in this form, we discard our curve and randomly select another triple (a, x, y) to start over.
Assuming we find a curve which passes the criterion, proceed to calculate mP and kP. If at any stage in the calculation we encounter an undefined expression (from calculating the multiples of P or in our point counting algorithm), it gives us a non-trivial factor of N.
If it is clear that N is not prime, because if N were prime then E would have order m, and any element of E would become 0 on multiplication by m. If kP = 0 then we have hit a dead-end and must start again with a different triple.
Now if mP = 0 and kP \neq 0 then our previous proposition tells us that N is prime. However there is one possible problem, which is the primality of q. This must be verified, using the same algorithm. So we have described a down-run procedure, where the primality of N can be proven through the primality of q and indeed smaller 'probable primes' until we have reached certain primes and are finished.

14.19.4.Problems with the algorithm

Atkin and Morain state "the problem with GK is that Schoof's algorithm seems almost impossible to implement. It is very slow and cumbersome to count all of the points on E using Schoof's algorithm, which is the preferred algorithm for the Goldwasser–Kilian algorithm. However, the original algorithm by Schoof is not effcicient enough to provide the number of points in short time. These comments have to be seen in the historical context, before the improvements by Elkies and Atkin to Schoof's method.

A second problem Koblitz notes is the difficulty of finding the curve E whose number of points is of the form kq, as above. There is no known theorem which guarantees we can find a suitable E in polynomially many attempts. The distribution of primes on the Hasse interval , which contains m, is not the same as the distribution of primes in the group orders, counting curves with multiplity. However, this is not a significant problem in practice.

14.19.5.Atkin–Morain elliptic curve primality test (ECPP)

In a 1993 paper, Atkin and Morain described an algorithm ECPP which avoided the trouble of relying on a cumbersome point counting algorithm (Schoof's). The algorithm still relies on the proposition stated above, but rather than randomly generating elliptic curves and searching for a proper m, their idea was to construct a curve E where the number of points is easy to compute. Complex multiplication is key in the construction of the curve.

Now, given an N for which primality needs to be proven we need to find a suitable m and q, just as in the Goldwasser-Kilian test, that will fulfill the proposition and prove the primality of N. (Of course, a point P and the curve itself, E, must also be found.)
ECPP uses complex multiplication to construct the curve E, doing so in a way that allows for m (the number of points on E) to be easily computed. We will now describe this method:
Utilization of complex multiplication requires a negative discriminant, D, such that D can be written as the product of two elements , or completely equivalently, we can write the equation:

For some a, b. If we can describe N in terms of either of these forms, we can create an elliptic curve E on with complex multiplication (described in detail below), and the number of points is given by:

For N to split into two the two elements, we need that (where denotes the Legendre symbol). This is a necessary condition, and we achieve sufficiency if the class number h(D) of the order in is 1. This happens for only 13 values of D, which are the elements of {−3, −4, −7, −8, −11, −12, −16, −19, −27, −28, −43, −67, −163}

14.19.6.The test

Pick discriminants D in sequence of increasing h(D). For each D check if and whether 4N can be written as:

This part can be verified using Cornacchia's algorithm. Once acceptable D and a have been discovered, calculate m = N + 1 − a. Now if m has a prime factor q of size
q > (N^{1 / 4} + 1)²

14.19.7.Complex multiplication method

For completeness, we will provide an overview of complex multiplication, the way in which an elliptic curve can be created, given our D (which can be written as a product of two elements).

Assume first that and (these cases are much more easily done). It is necessary to calculate the elliptic j-invariants of the h(D) classes of the order of discriminant D as complex numbers. There are several formulas to calculate these.

Next create the monic polynomial H_D(X), which has roots corresponding to the h(D) values. Note, that H_D(X) is the class polynomial. From complex multiplication theory, we know that H_D(X) has integer coefficients, which allows us to estimate these coefficients accurately enough to discover their true values.

Now, if N is prime, CM tells us that H_D(X) splits modulo N into a product of h(D) linear factors, based on the fact that D was chosen so that N splits as the product of two elements. Now if j is one of the h(D) roots modulo N we can define E as:

c is any quadratic nonresidue mod N, and r is either 0 or 1.

Given a root j there are only two possible nonisomorphic choices of E, one for each choice of r. We have the cardinality of these curves as

or

14.19.8.Discussion

Just as with the Goldwasser–Killian test, this one leads to a down-run procedure. Again, the culprit is q. Once we find a q that works, we must check it to be prime, so in fact we are doing the whole test now for q. Then again we may have to perform the test for factors of q. This leads to a nested certificate where at each level we have an elliptic curve E, an m and the prime in doubt, q.

14.19.9.Example of Atkin–Morain ECPP

We construct an example to prove that N = 167 is prime using the Atkin–Morain ECPP test. First proceed through the set of 13 possible discriminants, testing whether the Legendre Symbol (D / N) = 1, and if 4N can be written as 4N = a² + | D | b².

For our example D = −43 is chosen. This is because (D / N) = ( − 43 / 167) = 1 and also, using Cornacchia's algorithm, we know that and thus a = 25 and b = 1.

The next step is to calculate m. This is easily done as m = N + 1 − a which yields m = 167 + 1 − 25 = 143. Next we need to find a probable prime divisor of m, which was referred to as q. It must satisfy the condition that q > (N^{1 / 4} + 1)²

Now in this case, m = 143 = 11*13. So unfortunately we cannot choose 11 or 13 as our q, for it does not satisfy the necessary inequality. We are saved, however, by an analogous proposition to that which we stated before the Goldwasser–Kilian algorithm, which comes from a paper by Morain. It states, that given our m, we look for an s which divides m, s > (N^{1 / 4} + 1)², but is not necessarily prime, and check whether, for each p_i which divides s

for some point P on our yet to be constructed curve.

If s satisfies the inequality, and its prime factors satisfy the above, then N is prime.

So in our case, we choose s = m = 143. Thus our possible p_i's are 11 and 13. First, it is clear that 143 > (167^{1 / 4} + 1)², and so we need only check the values of

(143 / 11)P = 13P and (143 / 13)P = 11P.

But before we can do this, we must construct our curve, and choose a point P. In order to construct the curve, we make use of complex multiplication. In our case we compute the J-invariant

Next we compute and we know our elliptic curve is of the form:

y² = x³ + 3kc²x + 2kc³,

where k is as described previously and c a non-square in . So we can begin with

, which yields

E: y² = x³ + 140x + 149(mod 167)

Now, utilizing the point P = (6,6) on E it can be verified that 143P = .

It is simple to check that 13(6, 6) = (12, 65) and 11P = (140, 147), and so, by Morain's proposition, N is prime.

14.19.10.Complexity and running times

Goldwasser and Kilian's elliptic curve primality proving algorithm terminates in expected polynomial time for at least

of prime inputs.

14.19.11.Conjecture

Let π(x) be the number of primes smaller than x

for sufficiently large x.

If one accepts this conjecture then the Goldwasser–Kilian algorithm terminates in expected polynomial time for every input. Also, if our N is of length k, then the algorithm creates a certificate of size O(k²) that can be verified in O(k⁴).

Now consider another conjecture, which will give us a bound on the total time of the algorithm.

14.19.12.Conjecture 2

Suppose there exist positive constants c₁ and c₂ such that the amount of primes in the interval

is larger than

Then the Goldwasser Kilian algorithm proves the primality of N in an expected time of

For the Atkin–Morain algorithm, the running time stated is

O((logN)^{6 + ε}) for some ε > 0

14.19.13.Primes of special form

For some forms of numbers, it is possible to find 'short-cuts' to a primality proof. This is the case for the Mersenne numbers. In fact, due to their special structure, which allows for easier verification of primality, the largest known prime number is a Mersenne number. There has been a method in use for some time to verify primality of Mersenne numbers, known as the Lucas–Lehmer test. This test does not rely on elliptic curves. However we present a result where numbers of the form N = 2^kn − 1 where , n odd can be proven prime (or composite) using elliptic curves. Of course this will also provide a method for proving primality of Mersenne numbers, which correspond to the case where n = 1. It should be noted that there is a method in place for testing this form of number without elliptic curves (with a limitation on the size of k) known as the Lucas–Lehmer–Riesel test. The following method is drawn from the paper Primality Test for 2^kn − 1 using Elliptic Curves, by Yu Tsumura.

14.19.14.Group structure of E(F_N)

We take E as our elliptic curve, where E is of the form y² = x³ − mx for , , where is prime, and odd.

14.19.15.Theorem 1

#E

14.19.16.Theorem 2

E or

E

Depending on whether or not m is a quadratic residue modulo p.

14.19.17.Theorem 3

Let be prime, E, k, n, m as above. Take Q = (x,y) on E, x a quadratic nonresidue modulo p.

Then the order of Q is divisible by 2^k in the cyclic group .

First we will present the case where n is relatively small with respect to 2^k, and this will require one more theorem.

14.19.18.Theorem 4

Choose a λ > 1. E, k, n, m are specified as above with the added restrictions that

and

p is a prime if and only if there exists a Q = (x,y) which is on E, such that the

gcd(S_i,p) = 1 for i = 1, 2, ...,k − 1 and

where S_i is a sequence with initial value S₀ = x

14.19.19.The algorithm

We provide the following algorithm, which relies mainly on Theorems 3 and 4. To verify the primality of a given number N, perform the following steps:

(1) Chose such that , and find y such that

Take

Then Q' = (x,y) is on E: y² = x³ − mx where

Calculate Q = mQ'. If then N is composite, otherwise proceed to (2).

(2) Set S_i as the sequence with initial value Q. Calculate S_i for i = 1,2,..., k − 1

If gcd(S_i,N) > 1 for an i, where then N is composite. Otherwise, proceed to (3).

(3) If then N is prime. Otherwise, N is composite. This completes the test.

14.19.20.Justification of the algorithm

In (1), and elliptic curve, E is picked, along with a point Q on E, such that the x-coordinate of Q is a quadratic nonresidue. We can say

Thus, if N is prime, Q' has order divisible by 2^k, by Theorem 3,

and therefore the order of Q' is 2^kd d | n.

This means Q = nQ' has order 2^k. Therefore, if (1) concludes that N is composite, it truly is composite. (2) and (3) check if Q has order 2^k. Thus, if (2) or (3) conclude N is composite, it is composite.

Now, if the algorithm concludes that N is prime, then that means S₁ satisfies the condition of Theorem 4, and so N is truly prime.

There is an algorithm as well for when n is large, however for this we refer to the aforementioned article.

14.19.21.References