2934be9134
The FIPS provider does not have a default OPENSSL_CTX so, where necessary, we need to ensure we can always access an explicit OPENSSL_CTX. We remove functions from the FIPS provider that use the default OPENSSL_CTX, and fixup some places which were using those removed functions. Reviewed-by: Shane Lontis <shane.lontis@oracle.com> (Merged from https://github.com/openssl/openssl/pull/9310)
359 lines
9.3 KiB
C
359 lines
9.3 KiB
C
/*
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* Copyright 2000-2018 The OpenSSL Project Authors. All Rights Reserved.
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*
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* Licensed under the Apache License 2.0 (the "License"). You may not use
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* this file except in compliance with the License. You can obtain a copy
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* in the file LICENSE in the source distribution or at
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* https://www.openssl.org/source/license.html
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*/
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#include "internal/cryptlib.h"
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#include "bn_lcl.h"
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BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx)
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/*
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* Returns 'ret' such that ret^2 == a (mod p), using the Tonelli/Shanks
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* algorithm (cf. Henri Cohen, "A Course in Algebraic Computational Number
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* Theory", algorithm 1.5.1). 'p' must be prime!
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*/
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{
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BIGNUM *ret = in;
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int err = 1;
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int r;
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BIGNUM *A, *b, *q, *t, *x, *y;
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int e, i, j;
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if (!BN_is_odd(p) || BN_abs_is_word(p, 1)) {
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if (BN_abs_is_word(p, 2)) {
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if (ret == NULL)
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ret = BN_new();
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if (ret == NULL)
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goto end;
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if (!BN_set_word(ret, BN_is_bit_set(a, 0))) {
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if (ret != in)
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BN_free(ret);
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return NULL;
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}
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bn_check_top(ret);
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return ret;
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}
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BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
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return NULL;
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}
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if (BN_is_zero(a) || BN_is_one(a)) {
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if (ret == NULL)
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ret = BN_new();
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if (ret == NULL)
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goto end;
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if (!BN_set_word(ret, BN_is_one(a))) {
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if (ret != in)
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BN_free(ret);
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return NULL;
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}
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bn_check_top(ret);
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return ret;
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}
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BN_CTX_start(ctx);
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A = BN_CTX_get(ctx);
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b = BN_CTX_get(ctx);
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q = BN_CTX_get(ctx);
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t = BN_CTX_get(ctx);
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x = BN_CTX_get(ctx);
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y = BN_CTX_get(ctx);
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if (y == NULL)
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goto end;
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if (ret == NULL)
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ret = BN_new();
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if (ret == NULL)
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goto end;
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/* A = a mod p */
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if (!BN_nnmod(A, a, p, ctx))
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goto end;
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/* now write |p| - 1 as 2^e*q where q is odd */
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e = 1;
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while (!BN_is_bit_set(p, e))
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e++;
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/* we'll set q later (if needed) */
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if (e == 1) {
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/*-
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* The easy case: (|p|-1)/2 is odd, so 2 has an inverse
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* modulo (|p|-1)/2, and square roots can be computed
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* directly by modular exponentiation.
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* We have
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* 2 * (|p|+1)/4 == 1 (mod (|p|-1)/2),
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* so we can use exponent (|p|+1)/4, i.e. (|p|-3)/4 + 1.
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*/
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if (!BN_rshift(q, p, 2))
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goto end;
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q->neg = 0;
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if (!BN_add_word(q, 1))
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goto end;
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if (!BN_mod_exp(ret, A, q, p, ctx))
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goto end;
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err = 0;
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goto vrfy;
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}
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if (e == 2) {
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/*-
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* |p| == 5 (mod 8)
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*
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* In this case 2 is always a non-square since
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* Legendre(2,p) = (-1)^((p^2-1)/8) for any odd prime.
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* So if a really is a square, then 2*a is a non-square.
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* Thus for
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* b := (2*a)^((|p|-5)/8),
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* i := (2*a)*b^2
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* we have
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* i^2 = (2*a)^((1 + (|p|-5)/4)*2)
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* = (2*a)^((p-1)/2)
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* = -1;
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* so if we set
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* x := a*b*(i-1),
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* then
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* x^2 = a^2 * b^2 * (i^2 - 2*i + 1)
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* = a^2 * b^2 * (-2*i)
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* = a*(-i)*(2*a*b^2)
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* = a*(-i)*i
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* = a.
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*
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* (This is due to A.O.L. Atkin,
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* Subject: Square Roots and Cognate Matters modulo p=8n+5.
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* URL: https://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind9211&L=NMBRTHRY&P=4026
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* November 1992.)
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*/
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/* t := 2*a */
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if (!BN_mod_lshift1_quick(t, A, p))
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goto end;
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/* b := (2*a)^((|p|-5)/8) */
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if (!BN_rshift(q, p, 3))
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goto end;
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q->neg = 0;
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if (!BN_mod_exp(b, t, q, p, ctx))
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goto end;
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/* y := b^2 */
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if (!BN_mod_sqr(y, b, p, ctx))
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goto end;
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/* t := (2*a)*b^2 - 1 */
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if (!BN_mod_mul(t, t, y, p, ctx))
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goto end;
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if (!BN_sub_word(t, 1))
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goto end;
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/* x = a*b*t */
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if (!BN_mod_mul(x, A, b, p, ctx))
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goto end;
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if (!BN_mod_mul(x, x, t, p, ctx))
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goto end;
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if (!BN_copy(ret, x))
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goto end;
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err = 0;
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goto vrfy;
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}
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/*
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* e > 2, so we really have to use the Tonelli/Shanks algorithm. First,
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* find some y that is not a square.
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*/
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if (!BN_copy(q, p))
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goto end; /* use 'q' as temp */
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q->neg = 0;
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i = 2;
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do {
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/*
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* For efficiency, try small numbers first; if this fails, try random
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* numbers.
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*/
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if (i < 22) {
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if (!BN_set_word(y, i))
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goto end;
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} else {
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if (!BN_priv_rand_ex(y, BN_num_bits(p), 0, 0, ctx))
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goto end;
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if (BN_ucmp(y, p) >= 0) {
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if (!(p->neg ? BN_add : BN_sub) (y, y, p))
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goto end;
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}
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/* now 0 <= y < |p| */
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if (BN_is_zero(y))
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if (!BN_set_word(y, i))
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goto end;
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}
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r = BN_kronecker(y, q, ctx); /* here 'q' is |p| */
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if (r < -1)
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goto end;
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if (r == 0) {
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/* m divides p */
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BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
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goto end;
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}
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}
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while (r == 1 && ++i < 82);
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if (r != -1) {
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/*
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* Many rounds and still no non-square -- this is more likely a bug
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* than just bad luck. Even if p is not prime, we should have found
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* some y such that r == -1.
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*/
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BNerr(BN_F_BN_MOD_SQRT, BN_R_TOO_MANY_ITERATIONS);
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goto end;
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}
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/* Here's our actual 'q': */
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if (!BN_rshift(q, q, e))
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goto end;
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/*
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* Now that we have some non-square, we can find an element of order 2^e
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* by computing its q'th power.
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*/
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if (!BN_mod_exp(y, y, q, p, ctx))
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goto end;
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if (BN_is_one(y)) {
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BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
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goto end;
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}
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/*-
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* Now we know that (if p is indeed prime) there is an integer
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* k, 0 <= k < 2^e, such that
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*
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* a^q * y^k == 1 (mod p).
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*
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* As a^q is a square and y is not, k must be even.
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* q+1 is even, too, so there is an element
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*
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* X := a^((q+1)/2) * y^(k/2),
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*
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* and it satisfies
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*
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* X^2 = a^q * a * y^k
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* = a,
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*
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* so it is the square root that we are looking for.
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*/
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/* t := (q-1)/2 (note that q is odd) */
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if (!BN_rshift1(t, q))
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goto end;
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/* x := a^((q-1)/2) */
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if (BN_is_zero(t)) { /* special case: p = 2^e + 1 */
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if (!BN_nnmod(t, A, p, ctx))
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goto end;
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if (BN_is_zero(t)) {
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/* special case: a == 0 (mod p) */
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BN_zero(ret);
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err = 0;
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goto end;
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} else if (!BN_one(x))
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goto end;
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} else {
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if (!BN_mod_exp(x, A, t, p, ctx))
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goto end;
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if (BN_is_zero(x)) {
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/* special case: a == 0 (mod p) */
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BN_zero(ret);
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err = 0;
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goto end;
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}
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}
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/* b := a*x^2 (= a^q) */
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if (!BN_mod_sqr(b, x, p, ctx))
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goto end;
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if (!BN_mod_mul(b, b, A, p, ctx))
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goto end;
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/* x := a*x (= a^((q+1)/2)) */
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if (!BN_mod_mul(x, x, A, p, ctx))
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goto end;
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while (1) {
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/*-
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* Now b is a^q * y^k for some even k (0 <= k < 2^E
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* where E refers to the original value of e, which we
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* don't keep in a variable), and x is a^((q+1)/2) * y^(k/2).
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*
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* We have a*b = x^2,
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* y^2^(e-1) = -1,
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* b^2^(e-1) = 1.
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*/
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if (BN_is_one(b)) {
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if (!BN_copy(ret, x))
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goto end;
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err = 0;
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goto vrfy;
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}
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/* find smallest i such that b^(2^i) = 1 */
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i = 1;
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if (!BN_mod_sqr(t, b, p, ctx))
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goto end;
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while (!BN_is_one(t)) {
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i++;
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if (i == e) {
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BNerr(BN_F_BN_MOD_SQRT, BN_R_NOT_A_SQUARE);
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goto end;
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}
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if (!BN_mod_mul(t, t, t, p, ctx))
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goto end;
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}
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/* t := y^2^(e - i - 1) */
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if (!BN_copy(t, y))
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goto end;
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for (j = e - i - 1; j > 0; j--) {
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if (!BN_mod_sqr(t, t, p, ctx))
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goto end;
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}
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if (!BN_mod_mul(y, t, t, p, ctx))
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goto end;
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if (!BN_mod_mul(x, x, t, p, ctx))
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goto end;
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if (!BN_mod_mul(b, b, y, p, ctx))
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goto end;
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e = i;
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}
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vrfy:
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if (!err) {
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/*
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* verify the result -- the input might have been not a square (test
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* added in 0.9.8)
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*/
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if (!BN_mod_sqr(x, ret, p, ctx))
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err = 1;
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if (!err && 0 != BN_cmp(x, A)) {
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BNerr(BN_F_BN_MOD_SQRT, BN_R_NOT_A_SQUARE);
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err = 1;
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}
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}
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end:
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if (err) {
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if (ret != in)
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BN_clear_free(ret);
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ret = NULL;
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}
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BN_CTX_end(ctx);
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bn_check_top(ret);
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return ret;
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}
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