1 /* Written by Lenka Fibikova <fibikova@exp-math.uni-essen.de>
2 * and Bodo Moeller for the OpenSSL project. */
3 /* ====================================================================
4 * Copyright (c) 1998-2000 The OpenSSL Project. All rights reserved.
5 *
6 * Redistribution and use in source and binary forms, with or without
7 * modification, are permitted provided that the following conditions
8 * are met:
9 *
10 * 1. Redistributions of source code must retain the above copyright
11 * notice, this list of conditions and the following disclaimer.
12 *
13 * 2. Redistributions in binary form must reproduce the above copyright
14 * notice, this list of conditions and the following disclaimer in
15 * the documentation and/or other materials provided with the
16 * distribution.
17 *
18 * 3. All advertising materials mentioning features or use of this
19 * software must display the following acknowledgment:
20 * "This product includes software developed by the OpenSSL Project
21 * for use in the OpenSSL Toolkit. (http://www.openssl.org/)"
22 *
23 * 4. The names "OpenSSL Toolkit" and "OpenSSL Project" must not be used to
24 * endorse or promote products derived from this software without
25 * prior written permission. For written permission, please contact
26 * openssl-core@openssl.org.
27 *
28 * 5. Products derived from this software may not be called "OpenSSL"
29 * nor may "OpenSSL" appear in their names without prior written
30 * permission of the OpenSSL Project.
31 *
32 * 6. Redistributions of any form whatsoever must retain the following
33 * acknowledgment:
34 * "This product includes software developed by the OpenSSL Project
35 * for use in the OpenSSL Toolkit (http://www.openssl.org/)"
36 *
37 * THIS SOFTWARE IS PROVIDED BY THE OpenSSL PROJECT ``AS IS'' AND ANY
38 * EXPRESSED OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
39 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR
40 * PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE OpenSSL PROJECT OR
41 * ITS CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL,
42 * SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT
43 * NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES;
44 * LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
45 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT,
46 * STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE)
47 * ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED
48 * OF THE POSSIBILITY OF SUCH DAMAGE.
49 * ====================================================================
50 *
51 * This product includes cryptographic software written by Eric Young
52 * (eay@cryptsoft.com). This product includes software written by Tim
53 * Hudson (tjh@cryptsoft.com). */
54
55 #include <openssl/bn.h>
56
57 #include <openssl/err.h>
58
59 #include "internal.h"
60
61
BN_mod_sqrt(BIGNUM * in,const BIGNUM * a,const BIGNUM * p,BN_CTX * ctx)62 BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) {
63 /* Compute a square root of |a| mod |p| using the Tonelli/Shanks algorithm
64 * (cf. Henri Cohen, "A Course in Algebraic Computational Number Theory",
65 * algorithm 1.5.1). |p| is assumed to be a prime. */
66
67 BIGNUM *ret = in;
68 int err = 1;
69 int r;
70 BIGNUM *A, *b, *q, *t, *x, *y;
71 int e, i, j;
72
73 if (!BN_is_odd(p) || BN_abs_is_word(p, 1)) {
74 if (BN_abs_is_word(p, 2)) {
75 if (ret == NULL) {
76 ret = BN_new();
77 }
78 if (ret == NULL) {
79 goto end;
80 }
81 if (!BN_set_word(ret, BN_is_bit_set(a, 0))) {
82 if (ret != in) {
83 BN_free(ret);
84 }
85 return NULL;
86 }
87 return ret;
88 }
89
90 OPENSSL_PUT_ERROR(BN, BN_R_P_IS_NOT_PRIME);
91 return (NULL);
92 }
93
94 if (BN_is_zero(a) || BN_is_one(a)) {
95 if (ret == NULL) {
96 ret = BN_new();
97 }
98 if (ret == NULL) {
99 goto end;
100 }
101 if (!BN_set_word(ret, BN_is_one(a))) {
102 if (ret != in) {
103 BN_free(ret);
104 }
105 return NULL;
106 }
107 return ret;
108 }
109
110 BN_CTX_start(ctx);
111 A = BN_CTX_get(ctx);
112 b = BN_CTX_get(ctx);
113 q = BN_CTX_get(ctx);
114 t = BN_CTX_get(ctx);
115 x = BN_CTX_get(ctx);
116 y = BN_CTX_get(ctx);
117 if (y == NULL) {
118 goto end;
119 }
120
121 if (ret == NULL) {
122 ret = BN_new();
123 }
124 if (ret == NULL) {
125 goto end;
126 }
127
128 /* A = a mod p */
129 if (!BN_nnmod(A, a, p, ctx)) {
130 goto end;
131 }
132
133 /* now write |p| - 1 as 2^e*q where q is odd */
134 e = 1;
135 while (!BN_is_bit_set(p, e)) {
136 e++;
137 }
138 /* we'll set q later (if needed) */
139
140 if (e == 1) {
141 /* The easy case: (|p|-1)/2 is odd, so 2 has an inverse
142 * modulo (|p|-1)/2, and square roots can be computed
143 * directly by modular exponentiation.
144 * We have
145 * 2 * (|p|+1)/4 == 1 (mod (|p|-1)/2),
146 * so we can use exponent (|p|+1)/4, i.e. (|p|-3)/4 + 1.
147 */
148 if (!BN_rshift(q, p, 2)) {
149 goto end;
150 }
151 q->neg = 0;
152 if (!BN_add_word(q, 1) ||
153 !BN_mod_exp_mont(ret, A, q, p, ctx, NULL)) {
154 goto end;
155 }
156 err = 0;
157 goto vrfy;
158 }
159
160 if (e == 2) {
161 /* |p| == 5 (mod 8)
162 *
163 * In this case 2 is always a non-square since
164 * Legendre(2,p) = (-1)^((p^2-1)/8) for any odd prime.
165 * So if a really is a square, then 2*a is a non-square.
166 * Thus for
167 * b := (2*a)^((|p|-5)/8),
168 * i := (2*a)*b^2
169 * we have
170 * i^2 = (2*a)^((1 + (|p|-5)/4)*2)
171 * = (2*a)^((p-1)/2)
172 * = -1;
173 * so if we set
174 * x := a*b*(i-1),
175 * then
176 * x^2 = a^2 * b^2 * (i^2 - 2*i + 1)
177 * = a^2 * b^2 * (-2*i)
178 * = a*(-i)*(2*a*b^2)
179 * = a*(-i)*i
180 * = a.
181 *
182 * (This is due to A.O.L. Atkin,
183 * <URL:
184 *http://listserv.nodak.edu/scripts/wa.exe?A2=ind9211&L=nmbrthry&O=T&P=562>,
185 * November 1992.)
186 */
187
188 /* t := 2*a */
189 if (!BN_mod_lshift1_quick(t, A, p)) {
190 goto end;
191 }
192
193 /* b := (2*a)^((|p|-5)/8) */
194 if (!BN_rshift(q, p, 3)) {
195 goto end;
196 }
197 q->neg = 0;
198 if (!BN_mod_exp_mont(b, t, q, p, ctx, NULL)) {
199 goto end;
200 }
201
202 /* y := b^2 */
203 if (!BN_mod_sqr(y, b, p, ctx)) {
204 goto end;
205 }
206
207 /* t := (2*a)*b^2 - 1*/
208 if (!BN_mod_mul(t, t, y, p, ctx) ||
209 !BN_sub_word(t, 1)) {
210 goto end;
211 }
212
213 /* x = a*b*t */
214 if (!BN_mod_mul(x, A, b, p, ctx) ||
215 !BN_mod_mul(x, x, t, p, ctx)) {
216 goto end;
217 }
218
219 if (!BN_copy(ret, x)) {
220 goto end;
221 }
222 err = 0;
223 goto vrfy;
224 }
225
226 /* e > 2, so we really have to use the Tonelli/Shanks algorithm.
227 * First, find some y that is not a square. */
228 if (!BN_copy(q, p)) {
229 goto end; /* use 'q' as temp */
230 }
231 q->neg = 0;
232 i = 2;
233 do {
234 /* For efficiency, try small numbers first;
235 * if this fails, try random numbers.
236 */
237 if (i < 22) {
238 if (!BN_set_word(y, i)) {
239 goto end;
240 }
241 } else {
242 if (!BN_pseudo_rand(y, BN_num_bits(p), 0, 0)) {
243 goto end;
244 }
245 if (BN_ucmp(y, p) >= 0) {
246 if (!(p->neg ? BN_add : BN_sub)(y, y, p)) {
247 goto end;
248 }
249 }
250 /* now 0 <= y < |p| */
251 if (BN_is_zero(y)) {
252 if (!BN_set_word(y, i)) {
253 goto end;
254 }
255 }
256 }
257
258 r = bn_jacobi(y, q, ctx); /* here 'q' is |p| */
259 if (r < -1) {
260 goto end;
261 }
262 if (r == 0) {
263 /* m divides p */
264 OPENSSL_PUT_ERROR(BN, BN_R_P_IS_NOT_PRIME);
265 goto end;
266 }
267 } while (r == 1 && ++i < 82);
268
269 if (r != -1) {
270 /* Many rounds and still no non-square -- this is more likely
271 * a bug than just bad luck.
272 * Even if p is not prime, we should have found some y
273 * such that r == -1.
274 */
275 OPENSSL_PUT_ERROR(BN, BN_R_TOO_MANY_ITERATIONS);
276 goto end;
277 }
278
279 /* Here's our actual 'q': */
280 if (!BN_rshift(q, q, e)) {
281 goto end;
282 }
283
284 /* Now that we have some non-square, we can find an element
285 * of order 2^e by computing its q'th power. */
286 if (!BN_mod_exp_mont(y, y, q, p, ctx, NULL)) {
287 goto end;
288 }
289 if (BN_is_one(y)) {
290 OPENSSL_PUT_ERROR(BN, BN_R_P_IS_NOT_PRIME);
291 goto end;
292 }
293
294 /* Now we know that (if p is indeed prime) there is an integer
295 * k, 0 <= k < 2^e, such that
296 *
297 * a^q * y^k == 1 (mod p).
298 *
299 * As a^q is a square and y is not, k must be even.
300 * q+1 is even, too, so there is an element
301 *
302 * X := a^((q+1)/2) * y^(k/2),
303 *
304 * and it satisfies
305 *
306 * X^2 = a^q * a * y^k
307 * = a,
308 *
309 * so it is the square root that we are looking for.
310 */
311
312 /* t := (q-1)/2 (note that q is odd) */
313 if (!BN_rshift1(t, q)) {
314 goto end;
315 }
316
317 /* x := a^((q-1)/2) */
318 if (BN_is_zero(t)) /* special case: p = 2^e + 1 */
319 {
320 if (!BN_nnmod(t, A, p, ctx)) {
321 goto end;
322 }
323 if (BN_is_zero(t)) {
324 /* special case: a == 0 (mod p) */
325 BN_zero(ret);
326 err = 0;
327 goto end;
328 } else if (!BN_one(x)) {
329 goto end;
330 }
331 } else {
332 if (!BN_mod_exp_mont(x, A, t, p, ctx, NULL)) {
333 goto end;
334 }
335 if (BN_is_zero(x)) {
336 /* special case: a == 0 (mod p) */
337 BN_zero(ret);
338 err = 0;
339 goto end;
340 }
341 }
342
343 /* b := a*x^2 (= a^q) */
344 if (!BN_mod_sqr(b, x, p, ctx) ||
345 !BN_mod_mul(b, b, A, p, ctx)) {
346 goto end;
347 }
348
349 /* x := a*x (= a^((q+1)/2)) */
350 if (!BN_mod_mul(x, x, A, p, ctx)) {
351 goto end;
352 }
353
354 while (1) {
355 /* Now b is a^q * y^k for some even k (0 <= k < 2^E
356 * where E refers to the original value of e, which we
357 * don't keep in a variable), and x is a^((q+1)/2) * y^(k/2).
358 *
359 * We have a*b = x^2,
360 * y^2^(e-1) = -1,
361 * b^2^(e-1) = 1.
362 */
363
364 if (BN_is_one(b)) {
365 if (!BN_copy(ret, x)) {
366 goto end;
367 }
368 err = 0;
369 goto vrfy;
370 }
371
372
373 /* find smallest i such that b^(2^i) = 1 */
374 i = 1;
375 if (!BN_mod_sqr(t, b, p, ctx)) {
376 goto end;
377 }
378 while (!BN_is_one(t)) {
379 i++;
380 if (i == e) {
381 OPENSSL_PUT_ERROR(BN, BN_R_NOT_A_SQUARE);
382 goto end;
383 }
384 if (!BN_mod_mul(t, t, t, p, ctx)) {
385 goto end;
386 }
387 }
388
389
390 /* t := y^2^(e - i - 1) */
391 if (!BN_copy(t, y)) {
392 goto end;
393 }
394 for (j = e - i - 1; j > 0; j--) {
395 if (!BN_mod_sqr(t, t, p, ctx)) {
396 goto end;
397 }
398 }
399 if (!BN_mod_mul(y, t, t, p, ctx) ||
400 !BN_mod_mul(x, x, t, p, ctx) ||
401 !BN_mod_mul(b, b, y, p, ctx)) {
402 goto end;
403 }
404 e = i;
405 }
406
407 vrfy:
408 if (!err) {
409 /* verify the result -- the input might have been not a square
410 * (test added in 0.9.8) */
411
412 if (!BN_mod_sqr(x, ret, p, ctx)) {
413 err = 1;
414 }
415
416 if (!err && 0 != BN_cmp(x, A)) {
417 OPENSSL_PUT_ERROR(BN, BN_R_NOT_A_SQUARE);
418 err = 1;
419 }
420 }
421
422 end:
423 if (err) {
424 if (ret != in) {
425 BN_clear_free(ret);
426 }
427 ret = NULL;
428 }
429 BN_CTX_end(ctx);
430 return ret;
431 }
432
BN_sqrt(BIGNUM * out_sqrt,const BIGNUM * in,BN_CTX * ctx)433 int BN_sqrt(BIGNUM *out_sqrt, const BIGNUM *in, BN_CTX *ctx) {
434 BIGNUM *estimate, *tmp, *delta, *last_delta, *tmp2;
435 int ok = 0, last_delta_valid = 0;
436
437 if (in->neg) {
438 OPENSSL_PUT_ERROR(BN, BN_R_NEGATIVE_NUMBER);
439 return 0;
440 }
441 if (BN_is_zero(in)) {
442 BN_zero(out_sqrt);
443 return 1;
444 }
445
446 BN_CTX_start(ctx);
447 if (out_sqrt == in) {
448 estimate = BN_CTX_get(ctx);
449 } else {
450 estimate = out_sqrt;
451 }
452 tmp = BN_CTX_get(ctx);
453 last_delta = BN_CTX_get(ctx);
454 delta = BN_CTX_get(ctx);
455 if (estimate == NULL || tmp == NULL || last_delta == NULL || delta == NULL) {
456 OPENSSL_PUT_ERROR(BN, ERR_R_MALLOC_FAILURE);
457 goto err;
458 }
459
460 /* We estimate that the square root of an n-bit number is 2^{n/2}. */
461 if (!BN_lshift(estimate, BN_value_one(), BN_num_bits(in)/2)) {
462 goto err;
463 }
464
465 /* This is Newton's method for finding a root of the equation |estimate|^2 -
466 * |in| = 0. */
467 for (;;) {
468 /* |estimate| = 1/2 * (|estimate| + |in|/|estimate|) */
469 if (!BN_div(tmp, NULL, in, estimate, ctx) ||
470 !BN_add(tmp, tmp, estimate) ||
471 !BN_rshift1(estimate, tmp) ||
472 /* |tmp| = |estimate|^2 */
473 !BN_sqr(tmp, estimate, ctx) ||
474 /* |delta| = |in| - |tmp| */
475 !BN_sub(delta, in, tmp)) {
476 OPENSSL_PUT_ERROR(BN, ERR_R_BN_LIB);
477 goto err;
478 }
479
480 delta->neg = 0;
481 /* The difference between |in| and |estimate| squared is required to always
482 * decrease. This ensures that the loop always terminates, but I don't have
483 * a proof that it always finds the square root for a given square. */
484 if (last_delta_valid && BN_cmp(delta, last_delta) >= 0) {
485 break;
486 }
487
488 last_delta_valid = 1;
489
490 tmp2 = last_delta;
491 last_delta = delta;
492 delta = tmp2;
493 }
494
495 if (BN_cmp(tmp, in) != 0) {
496 OPENSSL_PUT_ERROR(BN, BN_R_NOT_A_SQUARE);
497 goto err;
498 }
499
500 ok = 1;
501
502 err:
503 if (ok && out_sqrt == in && !BN_copy(out_sqrt, estimate)) {
504 ok = 0;
505 }
506 BN_CTX_end(ctx);
507 return ok;
508 }
509