};
/* Combined Miller-Rabin test to the base a, and checking the
- conditions from Pocklington's theorem. */
+ conditions from Pocklington's theorem, nm1dq holds (n-1)/q, with q
+ prime. */
static int
miller_rabin_pocklington(mpz_t n, mpz_t nm1, mpz_t nm1dq, mpz_t a)
{
return is_prime;
}
-/* The algorithm is based on the following special case of
- Pocklington's theorem:
+/* The most basic variant of Pocklingtons theorem:
- Assume that n = 1 + f q, where q is a prime, q > sqrt(n) - 1. If we
- can find an a such that
+ Assume that q^e | (n-1), with q prime. If we can find an a such that
a^{n-1} = 1 (mod n)
- gcd(a^f - 1, n) = 1
+ gcd(a^{(n-1)/q} - 1, n) = 1
- then n is prime.
+ then any prime divisor p of n satisfies p = 1 (mod q^e).
- Proof: Assume that n is composite, with smallest prime factor p <=
- sqrt(n). Since q is prime, and q > sqrt(n) - 1 >= p - 1, q and p-1
- are coprime, so that we can define u = q^{-1} (mod (p-1)). The
- assumption a^{n-1} = 1 (mod n) implies that also a^{n-1} = 1 (mod
- p). Since p is prime, we have a^{(p-1)} = 1 (mod p). Now, r =
- (n-1)/q = (n-1) u (mod (p-1)), and it follows that a^r = a^{(n-1)
- u} = 1 (mod p). Then p is a common factor of a^r - 1 and n. This
- contradicts gcd(a^r - 1, n) = 1, and concludes the proof.
+ Proof (Cohen, 8.3.2): Assume p is a prime factor of n. The central
+ idea of the proof is to consider the order, modulo p, of a. Denote
+ this by d.
- If n is specified as k bits, we need q of size ceil(k/2) + 1 bits
- (or more) to make the theorem apply.
+ a^{n-1} = 1 (mod n) implies a^{n-1} = 1 (mod p), hence d | (n-1).
+ Next, the condition gcd(a^{(n-1)/q} - 1, n) = 1 implies that
+ a^{(n-1)/q} != 1, hence d does not divide (n-1)/q. Since q is
+ prime, this means that q^e | d.
+
+ Finally, we have a^{p-1} = 1 (mod p), hence d | (p-1). So q^e | d |
+ (p-1), which gives the desired result: p = 1 (mod q^e).
+
+
+ * Variant, slightly stronger than Fact 4.59, HAC:
+
+ Assume n = 1 + 2rq, q an odd prime, r <= 2q, and
+
+ a^{n-1} = 1 (mod n)
+ gcd(a^{(n-1)/q} - 1, n) = 1
+
+ Then n is prime.
+
+ Proof: By Pocklington's theorem, any prime factor p satisfies p = 1
+ (mod q). Neither 1 or q+1 are primes, hence p >= 1 + 2q. If n is
+ composite, we have n >= (1+2q)^2. But the assumption r <= 2q
+ implies n <= 1 + 4q^2, a contradiction.
+
+ In bits, the requirement is that #n <= 2 #q, then
+
+ r = (n-1)/2q < 2^{#n - #q} <= 2^#q = 2 2^{#q-1}< 2 q
+
+
+ * Another variant with an extra test (Variant of Fact 4.42, HAC):
+
+ Assume n = 1 + 2rq, n odd, q an odd prime, 8 q^3 >= n
+
+ a^{n-1} = 1 (mod n)
+ gcd(a^{(n-1)/q} - 1, n) = 1
+
+ Also let x = floor(r / 2q), y = r mod 2q,
+
+ If y^2 - 4x is not a square, then n is prime.
+
+ Proof (adapted from Maurer, Journal of Cryptology, 8 (1995)):
+
+ Assume n is composite. There are at most two factors, both odd,
+
+ n = (1+2m_1 q)(1+2m_2 q) = 1 + 4 m_1 m_2 q^2 + 2 (m_1 + m_2) q
+
+ where we can assume m_1 >= m_2. Then the bound n <= 8 q^3 implies m_1
+ m_2 < 2q, restricting (m_1, m_2) to the domain 0 < m_2 <
+ sqrt(2q), 0 < m_1 < 2q / m_2.
+
+ We have the bound
+
+ m_1 + m_2 < 2q / m_2 + m_2 <= 2q + 1 (maximum value for m_2 = 1)
+
+ And the case m_1 = 2q, m_2 = 1 can be excluded, because it gives n
+ > 8q^3. So in fact, m_1 + m_2 < 2q.
+
+ Next, write r = (n-1)/2q = 2 m_1 m_2 q + m_1 + m_2.
+
+ If follows that m_1 + m_2 = y and m_1 m_2 = x. m_1 and m_2 are
+ thus the roots of the equation
+
+ m^2 - y m + x = 0
+
+ which has integer roots iff y^2 - 4 x is the square of an integer.
+
+ In bits, the requirement is that #n <= 3 #q, then
+
+ n < 2^#n <= 2^{3 #q} = 8 2^{3 (#q-1)} < 8 q^3
*/
/* Generate a prime number p of size bits with 2 p0q dividing (p-1).
- p0 must be of size >= ceil(bits/2) + 1. The extra factor q can be
- omitted. If top_bits_set is one, the top most two bits are one,
- suitable for RSA primes. */
+ p0 must be of size >= ceil(bits/3). The extra factor q can be
+ omitted. If top_bits_set is one, the topmost two bits are set to
+ one, suitable for RSA primes. Also returns r = (p-1)/q. */
void
_nettle_generate_pocklington_prime (mpz_t p, mpz_t r,
unsigned bits, int top_bits_set,
const mpz_t q,
const mpz_t p0q)
{
- mpz_t r_min, r_range, pm1,a;
-
- assert (2*mpz_sizeinbase (p0, 2) > bits + 1);
+ mpz_t r_min, r_range, pm1, a, e;
+ int need_square_test;
+ unsigned p0_bits;
+ mpz_t x, y, p04;
+
+ p0_bits = mpz_sizeinbase (p0, 2);
+
+ assert (bits <= 3*p0_bits);
+ assert (bits > p0_bits);
+
+ need_square_test = (bits > 2 * p0_bits);
mpz_init (r_min);
mpz_init (r_range);
mpz_init (pm1);
mpz_init (a);
+ if (need_square_test)
+ {
+ mpz_init (x);
+ mpz_init (y);
+ mpz_init (p04);
+ mpz_mul_2exp (p04, p0, 2);
+ }
+
+ if (q)
+ mpz_init (e);
+
if (top_bits_set)
{
/* i = floor (2^{bits-3} / p0q), then 3I + 3 <= r <= 4I, with I
mpz_fdiv_q (r_range, r_range, p0q);
mpz_add_ui (r_min, r_range, 1);
}
+
for (;;)
{
uint8_t buf[1];
if (q)
{
- mpz_t e;
- int is_prime;
-
- mpz_init (e);
-
mpz_mul (e, r, q);
- is_prime = miller_rabin_pocklington(p, pm1, e, a);
- mpz_clear (e);
-
- if (is_prime)
- break;
+ if (!miller_rabin_pocklington(p, pm1, e, a))
+ continue;
+
+ if (need_square_test)
+ {
+ /* Our e corresponds to 2r in the theorem */
+ mpz_tdiv_qr (x, y, e, p04);
+ goto square_test;
+ }
}
- else if (miller_rabin_pocklington(p, pm1, r, a))
- break;
+ else
+ {
+ if (!miller_rabin_pocklington(p, pm1, r, a))
+ continue;
+ if (need_square_test)
+ {
+ mpz_tdiv_qr (x, y, r, p04);
+ square_test:
+ /* We have r' = 2r, x = floor (r/2q) = floor(r'/2q),
+ and y' = r' - x 4q = 2 (r - x 2q) = 2y.
+
+ Then y^2 - 4x is a square iff y'^2 - 16 x is a
+ square. */
+
+ mpz_mul (y, y, y);
+ mpz_submul_ui (y, x, 16);
+ if (mpz_perfect_square_p (y))
+ continue;
+ }
+ }
+
+ /* If we passed all the tests, we have found a prime. */
+ break;
}
mpz_clear (r_min);
mpz_clear (r_range);
mpz_clear (pm1);
mpz_clear (a);
+
+ if (need_square_test)
+ {
+ mpz_clear (x);
+ mpz_clear (y);
+ mpz_clear (p04);
+ }
+ if (q)
+ mpz_clear (e);
}
/* Generate random prime of a given size. Maurer's algorithm (Alg.
projects / thirdparty / nettle.git / commitdiff
