3 Generation of RSA keypairs
5 Copyright (C) 2002 Niels Möller
7 This file is part of GNU Nettle.
9 GNU Nettle is free software: you can redistribute it and/or
10 modify it under the terms of either:
12 * the GNU Lesser General Public License as published by the Free
13 Software Foundation; either version 3 of the License, or (at your
14 option) any later version.
18 * the GNU General Public License as published by the Free
19 Software Foundation; either version 2 of the License, or (at your
20 option) any later version.
22 or both in parallel, as here.
24 GNU Nettle is distributed in the hope that it will be useful,
25 but WITHOUT ANY WARRANTY; without even the implied warranty of
26 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
27 General Public License for more details.
29 You should have received copies of the GNU General Public License and
30 the GNU Lesser General Public License along with this program. If
31 not, see http://www.gnu.org/licenses/.
54 rsa_generate_keypair(struct rsa_public_key *pub,
55 struct rsa_private_key *key,
56 void *random_ctx, nettle_random_func *random,
57 void *progress_ctx, nettle_progress_func *progress,
68 /* We should choose e randomly. Is the size reasonable? */
69 if ((e_size < 16) || (e_size >= n_size) )
74 /* We have a fixed e. Check that it makes sense */
77 if (!mpz_tstbit(pub->e, 0))
81 if (mpz_cmp_ui(pub->e, 3) < 0)
84 /* And size less than n */
85 if (mpz_sizeinbase(pub->e, 2) >= n_size)
89 if (n_size < RSA_MINIMUM_N_BITS)
92 mpz_init(p1); mpz_init(q1); mpz_init(phi); mpz_init(tmp);
97 /* Generate p, such that gcd(p-1, e) = 1 */
100 nettle_random_prime(key->p, (n_size+1)/2, 1,
102 progress_ctx, progress);
104 mpz_sub_ui(p1, key->p, 1);
106 /* If e was given, we must chose p such that p-1 has no factors in
111 mpz_gcd(tmp, pub->e, p1);
113 if (mpz_cmp_ui(tmp, 1) == 0)
115 else if (progress) progress(progress_ctx, 'c');
119 progress(progress_ctx, '\n');
121 /* Generate q, such that gcd(q-1, e) = 1 */
124 nettle_random_prime(key->q, n_size/2, 1,
126 progress_ctx, progress);
129 if (mpz_cmp (key->q, key->p) == 0)
132 mpz_sub_ui(q1, key->q, 1);
134 /* If e was given, we must chose q such that q-1 has no factors in
139 mpz_gcd(tmp, pub->e, q1);
141 if (mpz_cmp_ui(tmp, 1) == 0)
143 else if (progress) progress(progress_ctx, 'c');
146 /* Now we have the primes. Is the product of the right size? */
147 mpz_mul(pub->n, key->p, key->q);
149 assert (mpz_sizeinbase(pub->n, 2) == n_size);
152 progress(progress_ctx, '\n');
154 /* c = q^{-1} (mod p) */
155 if (mpz_invert(key->c, key->q, key->p))
156 /* This should succeed everytime. But if it doesn't,
159 else if (progress) progress(progress_ctx, '?');
162 mpz_mul(phi, p1, q1);
164 /* If we didn't have a given e, generate one now. */
170 nettle_mpz_random_size(pub->e,
174 /* Make sure it's odd and that the most significant bit is
176 mpz_setbit(pub->e, 0);
177 mpz_setbit(pub->e, e_size - 1);
179 /* Needs gmp-3, or inverse might be negative. */
180 if (mpz_invert(key->d, pub->e, phi))
183 if (progress) progress(progress_ctx, 'e');
186 if (retried && progress)
187 progress(progress_ctx, '\n');
191 /* Must always succeed, as we already that e
192 * doesn't have any common factor with p-1 or q-1. */
193 int res = mpz_invert(key->d, pub->e, phi);
197 /* Done! Almost, we must compute the auxillary private values. */
199 mpz_fdiv_r(key->a, key->d, p1);
202 mpz_fdiv_r(key->b, key->d, q1);
204 /* c was computed earlier */
206 pub->size = key->size = (n_size + 7) / 8;
207 assert(pub->size >= RSA_MINIMUM_N_OCTETS);
209 mpz_clear(p1); mpz_clear(q1); mpz_clear(phi); mpz_clear(tmp);