1 /* crypto/bn/bn_sqrt.c */
3 * Written by Lenka Fibikova <fibikova@exp-math.uni-essen.de> and Bodo
4 * Moeller for the OpenSSL project.
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54 * This product includes cryptographic software written by Eric Young
55 * (eay@cryptsoft.com). This product includes software written by Tim
56 * Hudson (tjh@cryptsoft.com).
60 #include "internal/cryptlib.h"
63 BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx)
65 * Returns 'ret' such that ret^2 == a (mod p), using the Tonelli/Shanks
66 * algorithm (cf. Henri Cohen, "A Course in Algebraic Computational Number
67 * Theory", algorithm 1.5.1). 'p' must be prime!
73 BIGNUM *A, *b, *q, *t, *x, *y;
76 if (!BN_is_odd(p) || BN_abs_is_word(p, 1)) {
77 if (BN_abs_is_word(p, 2)) {
82 if (!BN_set_word(ret, BN_is_bit_set(a, 0))) {
91 BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
95 if (BN_is_zero(a) || BN_is_one(a)) {
100 if (!BN_set_word(ret, BN_is_one(a))) {
125 if (!BN_nnmod(A, a, p, ctx))
128 /* now write |p| - 1 as 2^e*q where q is odd */
130 while (!BN_is_bit_set(p, e))
132 /* we'll set q later (if needed) */
136 * The easy case: (|p|-1)/2 is odd, so 2 has an inverse
137 * modulo (|p|-1)/2, and square roots can be computed
138 * directly by modular exponentiation.
140 * 2 * (|p|+1)/4 == 1 (mod (|p|-1)/2),
141 * so we can use exponent (|p|+1)/4, i.e. (|p|-3)/4 + 1.
143 if (!BN_rshift(q, p, 2))
146 if (!BN_add_word(q, 1))
148 if (!BN_mod_exp(ret, A, q, p, ctx))
158 * In this case 2 is always a non-square since
159 * Legendre(2,p) = (-1)^((p^2-1)/8) for any odd prime.
160 * So if a really is a square, then 2*a is a non-square.
162 * b := (2*a)^((|p|-5)/8),
165 * i^2 = (2*a)^((1 + (|p|-5)/4)*2)
171 * x^2 = a^2 * b^2 * (i^2 - 2*i + 1)
172 * = a^2 * b^2 * (-2*i)
177 * (This is due to A.O.L. Atkin,
178 * <URL: http://listserv.nodak.edu/scripts/wa.exe?A2=ind9211&L=nmbrthry&O=T&P=562>,
183 if (!BN_mod_lshift1_quick(t, A, p))
186 /* b := (2*a)^((|p|-5)/8) */
187 if (!BN_rshift(q, p, 3))
190 if (!BN_mod_exp(b, t, q, p, ctx))
194 if (!BN_mod_sqr(y, b, p, ctx))
197 /* t := (2*a)*b^2 - 1 */
198 if (!BN_mod_mul(t, t, y, p, ctx))
200 if (!BN_sub_word(t, 1))
204 if (!BN_mod_mul(x, A, b, p, ctx))
206 if (!BN_mod_mul(x, x, t, p, ctx))
209 if (!BN_copy(ret, x))
216 * e > 2, so we really have to use the Tonelli/Shanks algorithm. First,
217 * find some y that is not a square.
220 goto end; /* use 'q' as temp */
225 * For efficiency, try small numbers first; if this fails, try random
229 if (!BN_set_word(y, i))
232 if (!BN_pseudo_rand(y, BN_num_bits(p), 0, 0))
234 if (BN_ucmp(y, p) >= 0) {
235 if (!(p->neg ? BN_add : BN_sub) (y, y, p))
238 /* now 0 <= y < |p| */
240 if (!BN_set_word(y, i))
244 r = BN_kronecker(y, q, ctx); /* here 'q' is |p| */
249 BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
253 while (r == 1 && ++i < 82);
257 * Many rounds and still no non-square -- this is more likely a bug
258 * than just bad luck. Even if p is not prime, we should have found
259 * some y such that r == -1.
261 BNerr(BN_F_BN_MOD_SQRT, BN_R_TOO_MANY_ITERATIONS);
265 /* Here's our actual 'q': */
266 if (!BN_rshift(q, q, e))
270 * Now that we have some non-square, we can find an element of order 2^e
271 * by computing its q'th power.
273 if (!BN_mod_exp(y, y, q, p, ctx))
276 BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
281 * Now we know that (if p is indeed prime) there is an integer
282 * k, 0 <= k < 2^e, such that
284 * a^q * y^k == 1 (mod p).
286 * As a^q is a square and y is not, k must be even.
287 * q+1 is even, too, so there is an element
289 * X := a^((q+1)/2) * y^(k/2),
293 * X^2 = a^q * a * y^k
296 * so it is the square root that we are looking for.
299 /* t := (q-1)/2 (note that q is odd) */
300 if (!BN_rshift1(t, q))
303 /* x := a^((q-1)/2) */
304 if (BN_is_zero(t)) { /* special case: p = 2^e + 1 */
305 if (!BN_nnmod(t, A, p, ctx))
308 /* special case: a == 0 (mod p) */
312 } else if (!BN_one(x))
315 if (!BN_mod_exp(x, A, t, p, ctx))
318 /* special case: a == 0 (mod p) */
325 /* b := a*x^2 (= a^q) */
326 if (!BN_mod_sqr(b, x, p, ctx))
328 if (!BN_mod_mul(b, b, A, p, ctx))
331 /* x := a*x (= a^((q+1)/2)) */
332 if (!BN_mod_mul(x, x, A, p, ctx))
337 * Now b is a^q * y^k for some even k (0 <= k < 2^E
338 * where E refers to the original value of e, which we
339 * don't keep in a variable), and x is a^((q+1)/2) * y^(k/2).
347 if (!BN_copy(ret, x))
353 /* find smallest i such that b^(2^i) = 1 */
355 if (!BN_mod_sqr(t, b, p, ctx))
357 while (!BN_is_one(t)) {
360 BNerr(BN_F_BN_MOD_SQRT, BN_R_NOT_A_SQUARE);
363 if (!BN_mod_mul(t, t, t, p, ctx))
367 /* t := y^2^(e - i - 1) */
370 for (j = e - i - 1; j > 0; j--) {
371 if (!BN_mod_sqr(t, t, p, ctx))
374 if (!BN_mod_mul(y, t, t, p, ctx))
376 if (!BN_mod_mul(x, x, t, p, ctx))
378 if (!BN_mod_mul(b, b, y, p, ctx))
386 * verify the result -- the input might have been not a square (test
390 if (!BN_mod_sqr(x, ret, p, ctx))
393 if (!err && 0 != BN_cmp(x, A)) {
394 BNerr(BN_F_BN_MOD_SQRT, BN_R_NOT_A_SQUARE);