ECPP in PARI/GP. Jared Asuncion. 29 January Asuncion, J. ECPP in PARI/GP 29 January / 27
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1 ECPP in PARI/GP Jared Asuncion 29 January 2018 Asuncion, J. ECPP in PARI/GP 29 January / 27
2 Main Theorem Problem Problem Given an integer N, prove that it is prime. Asuncion, J. ECPP in PARI/GP 29 January / 27
3 Main Theorem Main Theorem Proposition Let N > 6 be an integer. If there exists: such that an integer m a prime q an elliptic curve E over Z/NZ and a point P on E m = qs for some s Z q > ( N 1/4 + 1 ) 2 mp = sp then N is prime. Asuncion, J. ECPP in PARI/GP 29 January / 27
4 Main Theorem Naive Example Manual verification? Asuncion, J. ECPP in PARI/GP 29 January / 27
5 Main Theorem Naive Example Manual verification? N = 97;? Asuncion, J. ECPP in PARI/GP 29 January / 27
6 Main Theorem Naive Example Manual verification? N = 97;? m = 93;? Asuncion, J. ECPP in PARI/GP 29 January / 27
7 Main Theorem Naive Example Manual verification? N = 97;? m = 93;? q = 31;? Asuncion, J. ECPP in PARI/GP 29 January / 27
8 Main Theorem Naive Example Manual verification? N = 97;? m = 93;? q = 31;? E = ellinit( [Mod(69, N), Mod(2, N)] );? Asuncion, J. ECPP in PARI/GP 29 January / 27
9 Main Theorem Naive Example Manual verification? N = 97;? m = 93;? q = 31;? E = ellinit( [Mod(69, N), Mod(2, N)] );? P = [12, 91];? Asuncion, J. ECPP in PARI/GP 29 January / 27
10 Main Theorem Naive Example Manual verification? N = 97;? m = 93;? q = 31;? E = ellinit( [Mod(69, N), Mod(2, N)] );? P = [12, 91];? s = m/q;? Asuncion, J. ECPP in PARI/GP 29 January / 27
11 Main Theorem Naive Example Manual verification? N = 97;? m = 93;? q = 31;? E = ellinit( [Mod(69, N), Mod(2, N)] );? P = [12, 91];? s = m/q;? (N^(1/4) + 1)^2 % = ? Asuncion, J. ECPP in PARI/GP 29 January / 27
12 Main Theorem Naive Example Manual verification? N = 97;? m = 93;? q = 31;? E = ellinit( [Mod(69, N), Mod(2, N)] );? P = [12, 91];? s = m/q;? (N^(1/4) + 1)^2 % = ? ellmul(e,p,m) % = [0]? Asuncion, J. ECPP in PARI/GP 29 January / 27
13 Main Theorem Naive Example Manual verification? N = 97;? m = 93;? q = 31;? E = ellinit( [Mod(69, N), Mod(2, N)] );? P = [12, 91];? s = m/q;? (N^(1/4) + 1)^2 % = ? ellmul(e,p,m) % = [0]? ellmul(e,p,s) % = [Mod(23, 97), Mod(46, 97)] Asuncion, J. ECPP in PARI/GP 29 January / 27
14 Main Theorem Naive Example Manual verification? N = 97;? m = 93;? q = 31;? E = ellinit( [Mod(69, N), Mod(2, N)] );? P = [12, 91];? s = m/q;? (N^(1/4) + 1)^2 % = ? ellmul(e,p,m) % = [0]? ellmul(e,p,s) % = [Mod(23, 97), Mod(46, 97)] If q = 31 is prime, then N = 97 is prime. Asuncion, J. ECPP in PARI/GP 29 January / 27
15 Usage ecpp First examples of ecpp? Asuncion, J. ECPP in PARI/GP 29 January / 27
16 Usage ecpp First examples of ecpp? N = 97;? Asuncion, J. ECPP in PARI/GP 29 January / 27
17 Usage ecpp First examples of ecpp? N = 97;? ecpp(n) % = 97? Asuncion, J. ECPP in PARI/GP 29 January / 27
18 Usage ecpp First examples of ecpp? N = 97;? ecpp(n) % = 97? N = 51;? Asuncion, J. ECPP in PARI/GP 29 January / 27
19 Usage ecpp First examples of ecpp? N = 97;? ecpp(n) % = 97? N = 51;? ecpp(n) % = 0? Asuncion, J. ECPP in PARI/GP 29 January / 27
20 Usage ecpp First examples of ecpp? N = 97;? ecpp(n) % = 97? N = 51;? ecpp(n) % = 0? N = ecpp(2^100000);? Asuncion, J. ECPP in PARI/GP 29 January / 27
21 Usage ecpp First examples of ecpp? N = 97;? ecpp(n) % = 97? N = 51;? ecpp(n) % = 0? N = ecpp(2^100000);? ecpp(n) % = 0 Asuncion, J. ECPP in PARI/GP 29 January / 27
22 Usage ecpp Bigger example for ecpp? Asuncion, J. ECPP in PARI/GP 29 January / 27
23 Usage ecpp Bigger example for ecpp? N = randomprime(10^50) % = ? Asuncion, J. ECPP in PARI/GP 29 January / 27
24 Usage ecpp Bigger example for ecpp? N = randomprime(10^50) % = ? cert = ecpp(n) Asuncion, J. ECPP in PARI/GP 29 January / 27
25 Usage ecpp Bigger example for ecpp? N = randomprime(10^50) % = ? cert = ecpp(n) % = [[ , , , 0, [ , ]], [ , , , , [ , ]], [ , , , 0, [ , ]], [ , , , 0, [ , ]]] Asuncion, J. ECPP in PARI/GP 29 January / 27
26 Usage ecpp Interpreting the certificate? N = ;? cert = ecpp(n);? Asuncion, J. ECPP in PARI/GP 29 January / 27
27 Usage ecpp Interpreting the certificate? N = ;? cert = ecpp(n);???ecpp? Asuncion, J. ECPP in PARI/GP 29 January / 27
28 Usage ecpp Interpreting the certificate? N = ;? cert = ecpp(n);???ecpp? #cert % = 4? Asuncion, J. ECPP in PARI/GP 29 January / 27
29 Usage ecpp Interpreting the certificate? N = ;? cert = ecpp(n);???ecpp? #cert % = 4? N = cert[1][1] % = ? Asuncion, J. ECPP in PARI/GP 29 January / 27
30 Usage ecpp Interpreting the certificate? N = ;? cert = ecpp(n);???ecpp? #cert % = 4? N = cert[1][1] % = ? t = cert[1][2] % = ? Asuncion, J. ECPP in PARI/GP 29 January / 27
31 Usage ecpp Interpreting the certificate? N = ;? cert = ecpp(n);???ecpp? #cert % = 4? N = cert[1][1] % = ? t = cert[1][2] % = ? s = cert[1][3] % = ? Asuncion, J. ECPP in PARI/GP 29 January / 27
32 Usage ecpp Interpreting the certificate? N = ;? cert = ecpp(n);???ecpp? #cert % = 4? N = cert[1][1] % = ? t = cert[1][2] % = ? s = cert[1][3] % = ? a4 = cert[1][4] % = 0? Asuncion, J. ECPP in PARI/GP 29 January / 27
33 Usage ecpp Interpreting the certificate? N = ;? cert = ecpp(n);???ecpp? #cert % = 4? N = cert[1][1] % = ? t = cert[1][2] % = ? s = cert[1][3] % = ? a4 = cert[1][4] % = 0? P = cert[1][5] % = [ , ] Asuncion, J. ECPP in PARI/GP 29 January / 27
34 Usage ecppexport Using ecppexport? N = ;? cert = ecpp(n);? Asuncion, J. ECPP in PARI/GP 29 January / 27
35 Usage ecppexport Using ecppexport? N = ;? cert = ecpp(n);? expo = ecppexport(cert) % = "\n[1]\n N = [+++]? Asuncion, J. ECPP in PARI/GP 29 January / 27
36 Usage ecppexport Using ecppexport? N = ;? cert = ecpp(n);? expo = ecppexport(cert) % = "\n[1]\n N = [+++]? print(expo) [1] N = t = s = a4 = 0 D = -3 m = q = E = [0, ] P = [ [+++] Asuncion, J. ECPP in PARI/GP 29 January / 27
37 Usage ecppexport Exporting to MAGMA/PRIMO? N = ;? cert = ecpp(n);? Asuncion, J. ECPP in PARI/GP 29 January / 27
38 Usage ecppexport Exporting to MAGMA/PRIMO? N = ;? cert = ecpp(n);? magma = ecppexport(cert,1) % = "[* [* [+++]? Asuncion, J. ECPP in PARI/GP 29 January / 27
39 Usage ecppexport Exporting to MAGMA/PRIMO? N = ;? cert = ecpp(n);? magma = ecppexport(cert,1) % = "[* [* [+++]? primo = ecppexport(cert,2) % = "[PRIMO - Primality Certificate]\nFormat=4\nTestCou[+++]? Asuncion, J. ECPP in PARI/GP 29 January / 27
40 Usage ecppexport Exporting to MAGMA/PRIMO? N = ;? cert = ecpp(n);? magma = ecppexport(cert,1) % = "[* [* [+++]? primo = ecppexport(cert,2) % = "[PRIMO - Primality Certificate]\nFormat=4\nTestCou[+++]? write("primo.out", primo); Asuncion, J. ECPP in PARI/GP 29 January / 27
41 Usage ecppisvalid Is the certificate valid?? N = ;? cert = ecpp(n);? Asuncion, J. ECPP in PARI/GP 29 January / 27
42 Usage ecppisvalid Is the certificate valid?? N = ;? cert = ecpp(n);? ecppisvalid(cert) % = 1? Asuncion, J. ECPP in PARI/GP 29 January / 27
43 Usage ecppisvalid Is the certificate valid?? N = ;? cert = ecpp(n);? ecppisvalid(cert) % = 1? cert[4][5] = [ , ];? Asuncion, J. ECPP in PARI/GP 29 January / 27
44 Usage ecppisvalid Is the certificate valid?? N = ;? cert = ecpp(n);? ecppisvalid(cert) % = 1? cert[4][5] = [ , ];? ecppisvalid(cert) % = 0? Asuncion, J. ECPP in PARI/GP 29 January / 27
45 Usage ecppisvalid Is the certificate valid?? N = ;? cert = ecpp(n);? ecppisvalid(cert) % = 1? cert[4][5] = [ , ];? ecppisvalid(cert) % = 0? ecppisvalid(vector(100, i, i)) % = 0 Asuncion, J. ECPP in PARI/GP 29 January / 27
46 Implementation Details Overview Assume that we have a list L of discriminants we want to try. PHASE I: Find m and q. For each discriminant D under consideration: Write m = N + 1 U where (U, V ) is an integer solution to U 2 + D V 2 = 4N. a involves taking a square root of D modulo N involves doing a half-gcd Try to write m as m = qs where q and s are as in the theorem. involves finding small factors of m good idea to check pseudoprimality of q a m can take at most six other forms as well Asuncion, J. ECPP in PARI/GP 29 January / 27
47 Implementation Details Overview Assume that we have the corresponding (D, m, q) from Phase 1. PHASE II: Find E and P. Find a root j modulo N of the Hilbert class polynomial of Q( D). involves computing a class polynomial (of degree h) involves finding a root of this polynomial modulo N Find an elliptic curve E modulo N with j-invariant j and a point P satisfying the conditions in the theorem. finding the twist involves finding a quadratic non-residue Asuncion, J. ECPP in PARI/GP 29 January / 27
48 Implementation Details Tricks Proposition A fundamental discriminant D is of the form: D = t pi i=0 where p 0 { 8, 4, 1, 8} pi 1 (mod 4) and pi is prime for i > 0. Proposition If U 2 + D V 2 = 4N has integer solutions (U, V ), then ( D ) N = 1 ( ) p = 1 for all p dividing D. N Asuncion, J. ECPP in PARI/GP 29 January / 27
49 Implementation Details Tricks Finding a solution to U 2 + D V 2 = 4N. Globally: Keep the prime factorization of each D on the list L. Only consider D which are B-smooth for some B. Only consider D which are associated to low class numbers. For each N: For each D, test for necessary conditions for the existence of a solution. Find D modulo N by finding the square roots of primes dividing D. Remember the square roots of the primes for future use. Find (and remember) a quadratic nonresidue g. Asuncion, J. ECPP in PARI/GP 29 January / 27
50 Implementation Details Tricks Removing small factors Precompute the product P of all primes less than S. Divide m by its gcd with P. Do this repeatedly until they are coprime. There is a clever way to treat multiple m at once. Asuncion, J. ECPP in PARI/GP 29 January / 27
51 Implementation Details Tricks Removing small factors Precompute the product P of all primes less than S. Divide m by its gcd with P. Do this repeatedly until they are coprime. There is a clever way to treat multiple m at once. Let P = = Asuncion, J. ECPP in PARI/GP 29 January / 27
52 Implementation Details Tricks Removing small factors Precompute the product P of all primes less than S. Divide m by its gcd with P. Do this repeatedly until they are coprime. There is a clever way to treat multiple m at once. Let P = = Let m 1 = 1295, m 2 = 2781, and M = m 1 m 2 = Asuncion, J. ECPP in PARI/GP 29 January / 27
53 Implementation Details Tricks Removing small factors Precompute the product P of all primes less than S. Divide m by its gcd with P. Do this repeatedly until they are coprime. There is a clever way to treat multiple m at once. Let P = = Let m 1 = 1295, m 2 = 2781, and M = m 1 m 2 = P = (mod M) Asuncion, J. ECPP in PARI/GP 29 January / 27
54 Implementation Details Tricks Removing small factors Precompute the product P of all primes less than S. Divide m by its gcd with P. Do this repeatedly until they are coprime. There is a clever way to treat multiple m at once. Let P = = Let m 1 = 1295, m 2 = 2781, and M = m 1 m 2 = P = (mod M) P (mod 1295). P (mod 2781). Asuncion, J. ECPP in PARI/GP 29 January / 27
55 Implementation Details Tricks Removing small factors Precompute the product P of all primes less than S. Divide m by its gcd with P. Do this repeatedly until they are coprime. There is a clever way to treat multiple m at once. Let P = = Let m 1 = 1295, m 2 = 2781, and M = m 1 m 2 = P = (mod M) P (mod 1295). gcd(0, 1295) = P (mod 2781). Asuncion, J. ECPP in PARI/GP 29 January / 27
56 Implementation Details Tricks Removing small factors Precompute the product P of all primes less than S. Divide m by its gcd with P. Do this repeatedly until they are coprime. There is a clever way to treat multiple m at once. Let P = = Let m 1 = 1295, m 2 = 2781, and M = m 1 m 2 = P = (mod M) P (mod 1295). gcd(0, 1295) = q = 1. P (mod 2781). Asuncion, J. ECPP in PARI/GP 29 January / 27
57 Implementation Details Tricks Removing small factors Precompute the product P of all primes less than S. Divide m by its gcd with P. Do this repeatedly until they are coprime. There is a clever way to treat multiple m at once. Let P = = Let m 1 = 1295, m 2 = 2781, and M = m 1 m 2 = P = (mod M) P (mod 1295). gcd(0, 1295) = q = 1. P (mod 2781). gcd(2154, 3) = 3. Take 2154/3 = 927. Asuncion, J. ECPP in PARI/GP 29 January / 27
58 Implementation Details Tricks Removing small factors Precompute the product P of all primes less than S. Divide m by its gcd with P. Do this repeatedly until they are coprime. There is a clever way to treat multiple m at once. Let P = = Let m 1 = 1295, m 2 = 2781, and M = m 1 m 2 = P = (mod M) P (mod 1295). gcd(0, 1295) = q = 1. P (mod 2781). gcd(2154, 3) = 3. Take 2154/3 = 927. gcd(927, 3) = 3. Take 927/3 = 309. Asuncion, J. ECPP in PARI/GP 29 January / 27
59 Implementation Details Tricks Removing small factors Precompute the product P of all primes less than S. Divide m by its gcd with P. Do this repeatedly until they are coprime. There is a clever way to treat multiple m at once. Let P = = Let m 1 = 1295, m 2 = 2781, and M = m 1 m 2 = P = (mod M) P (mod 1295). gcd(0, 1295) = q = 1. P (mod 2781). gcd(2154, 3) = 3. Take 2154/3 = 927. gcd(927, 3) = 3. Take 927/3 = 309. gcd(309, 3) = 3. Take 309/3 = 103. Asuncion, J. ECPP in PARI/GP 29 January / 27
60 Implementation Details Tricks Removing small factors Precompute the product P of all primes less than S. Divide m by its gcd with P. Do this repeatedly until they are coprime. There is a clever way to treat multiple m at once. Let P = = Let m 1 = 1295, m 2 = 2781, and M = m 1 m 2 = P = (mod M) P (mod 1295). gcd(0, 1295) = q = 1. P (mod 2781). gcd(2154, 3) = 3. Take 2154/3 = 927. gcd(927, 3) = 3. Take 927/3 = 309. gcd(309, 3) = 3. Take 309/3 = 103. gcd(103, 3) = 1. Asuncion, J. ECPP in PARI/GP 29 January / 27
61 Implementation Details Tricks Removing small factors Precompute the product P of all primes less than S. Divide m by its gcd with P. Do this repeatedly until they are coprime. There is a clever way to treat multiple m at once. Let P = = Let m 1 = 1295, m 2 = 2781, and M = m 1 m 2 = P = (mod M) P (mod 1295). gcd(0, 1295) = q = 1. P (mod 2781). gcd(2154, 3) = 3. Take 2154/3 = 927. gcd(927, 3) = 3. Take 927/3 = 309. gcd(309, 3) = 3. Take 309/3 = 103. gcd(103, 3) = 1. q = 103. Asuncion, J. ECPP in PARI/GP 29 January / 27
62 Implementation Details Tricks Summary Look at next D. Check necessary conditions. If fail, go back to start. Find D modulo N. by finding q (mod N) for all q D or just accessing it from memory if it has been done before Solve for U 2 + D V 2 = 4N. Let m = N + 1 ± U (or more if D = 3, 4). Add the m s to the list to be factored. If list is not large enough, go back to start. If list is large enough, do the batch factoring. For each m, check if the remaining q is pseudoprime. If yes, try to prove q is prime. Asuncion, J. ECPP in PARI/GP 29 January / 27
63 Implementation Details Tricks What to do when the list L runs out? If we are at the top of the recursion tree, then expand the list to include higher class numbers. Otherwise, mark that instance as FAIL and resume the computations one level up. This results in solving one polynomial of huge degree. The rest have small degrees. Other option explored: Expand the list while in the middle of the tree. Asuncion, J. ECPP in PARI/GP 29 January / 27
64 Implementation Details Tricks Assume that we have the corresponding (D, m, q) from Phase 1. PHASE II: Find E and P. Find a root j modulo N of the Hilbert class polynomial of Q( D). involves computing a class polynomial (of degree h) involves finding a root of this polynomial modulo N Find an elliptic curve E modulo N with j-invariant j and a point P satisfying the conditions in the theorem. finding the twist involves finding a quadratic non-residue Asuncion, J. ECPP in PARI/GP 29 January / 27
65 Implementation Details Tricks Finding a root j of the Hilbert class polynomial. Remember and reuse previous computations. Prioritize discriminants D which give polynomials of lower degree (during the first phase). Use class polynomials obtained from other invariants. Trick to find one root of a totally split polynomial f mod p: Basic trick: gcd ( f, X (p 1)/2 1 ) or gcd ( f, X (p 1)/2 + 1 ) modulo p. This splits the roots into two buckets. New trick: gcd ( f, X (p 1)/n ζ k n ) modulo p for all k. This splits the roots into k buckets. (where k (p 1)). Asuncion, J. ECPP in PARI/GP 29 January / 27
66 Implementation Details Tricks Finding an elliptic curve E and a point P. Use the quadratic nonresidue g to find twists of elliptic curves. Asuncion, J. ECPP in PARI/GP 29 January / 27
67 Runtime Total Average runtime for a sample size of 100 primes, single thread bits 2000 bits 3000 bits 4000 bits Magma ecpp-dj Pari 2.10 isprime Primo Pari 2.10 ecpp Asuncion, J. ECPP in PARI/GP 29 January / 27
68 Runtime Phases Runtime for each phase, single thread: Primo 4.21 Pari ecpp 1000 bits 2000 bits 3000 bits 4000 bits I II * I II (*) is the average time spent for precomputations Asuncion, J. ECPP in PARI/GP 29 January / 27
69 Runtime Per step Average runtime for each step of Pari ecpp: D (mod N) 1000 bits 2000 bits 3000 bits 4000 bits factoring smalls pseudoprimality class polynomial root mod N find E and P Phase I operations are done more frequently as N becomes higher. It is harder to express m = qs as m N becomes bigger. There is a big spike at 4000 bits for computing the roots. This is because the degree can go higher than 40. Asuncion, J. ECPP in PARI/GP 29 January / 27
70 Conclusion Further Use intermediate fields of H K /K to obtain lower degree polynomials and find their roots instead. We will seldom need to find roots of a polynomial of huge degree. Because of this, we can afford a longer list L. We go up the recursion tree less frequently because we have a higher chance to succeed. Parallel computation. Phase II is done. (Pari/GP Atelier 2018) Phase I is trickier. Asuncion, J. ECPP in PARI/GP 29 January / 27
71 Conclusion New libpari commands Fp sqrt i find the square root of x given a quadratic nonresidue mod p cornacchia2 sqrt finds the solution to U 2 + D V 2 = 4N given D (mod N) disc best modinv find the invariant to use with polclass such that the height of the output is low FpX oneroot split finds one root of a totally split polynomial over F p [X ] removing small factors by batch Asuncion, J. ECPP in PARI/GP 29 January / 27
72 Conclusion Thank you for listening! Asuncion, J. ECPP in PARI/GP 29 January / 27
73 Conclusion References A. Atkin and F. Morain, Elliptic curves and primality proving, Mathematics of Computation 61 (1993), no. 203, J. Franke, T. Kleinjung, F. Morain, and T. Wirth, Proving the primality of very large numbers with fastecpp, pp , Springer Berlin Heidelberg, Berlin, Heidelberg, F. Morain, Implementing the asymptotically fast version of the elliptic curve primality proving algorithm. Asuncion, J. ECPP in PARI/GP 29 January / 27
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