/* $OpenBSD: bn_mod_sqrt.c,v 1.4 2025/05/10 05:54:38 tb Exp $ */ /* * Copyright (c) 2022 Theo Buehler <tb@openbsd.org> * * Permission to use, copy, modify, and distribute this software for any * purpose with or without fee is hereby granted, provided that the above * copyright notice and this permission notice appear in all copies. * * THE SOFTWARE IS PROVIDED "AS IS" AND THE AUTHOR DISCLAIMS ALL WARRANTIES * WITH REGARD TO THIS SOFTWARE INCLUDING ALL IMPLIED WARRANTIES OF * MERCHANTABILITY AND FITNESS. IN NO EVENT SHALL THE AUTHOR BE LIABLE FOR * ANY SPECIAL, DIRECT, INDIRECT, OR CONSEQUENTIAL DAMAGES OR ANY DAMAGES * WHATSOEVER RESULTING FROM LOSS OF USE, DATA OR PROFITS, WHETHER IN AN * ACTION OF CONTRACT, NEGLIGENCE OR OTHER TORTIOUS ACTION, ARISING OUT OF * OR IN CONNECTION WITH THE USE OR PERFORMANCE OF THIS SOFTWARE. */ #include "bn_local.h" #include "err_local.h" /* * Tonelli-Shanks according to H. Cohen "A Course in Computational Algebraic * Number Theory", Section 1.5.1, Springer GTM volume 138, Berlin, 1996. * * Under the assumption that p is prime and a is a quadratic residue, we know: * * a^[(p-1)/2] = 1 (mod p). (*) * * To find a square root of a (mod p), we handle three cases of increasing * complexity. In the first two cases, we can compute a square root using an * explicit formula, thus avoiding the probabilistic nature of Tonelli-Shanks. * * 1. p = 3 (mod 4). * * Set n = (p+1)/4. Then 2n = 1 + (p-1)/2 and (*) shows that x = a^n (mod p) * is a square root of a: x^2 = a^(2n) = a * a^[(p-1)/2] = a (mod p). * * 2. p = 5 (mod 8). * * This uses a simplification due to Atkin. By Theorem 1.4.7 and 1.4.9, the * Kronecker symbol (2/p) evaluates to (-1)^[(p^2-1)/8]. From p = 5 (mod 8) * we get (p^2-1)/8 = 1 (mod 2), so (2/p) = -1, and thus * * 2^[(p-1)/2] = -1 (mod p). (**) * * Set b = (2a)^[(p-5)/8]. With (p-1)/2 = 2 + (p-5)/2, (*) and (**) show * * i = 2 a b^2 is a square root of -1 (mod p). * * Indeed, i^2 = 2^2 a^2 b^4 = 2^[(p-1)/2] a^[(p-1)/2] = -1 (mod p). Because * of (i-1)^2 = -2i (mod p) and i (-i) = 1 (mod p), a square root of a is * * x = a b (i-1) * * as x^2 = a^2 b^2 (-2i) = a (2 a b^2) (-i) = a (mod p). * * 3. p = 1 (mod 8). * * This is the Tonelli-Shanks algorithm. For a prime p, the multiplicative * group of GF(p) is cyclic of order p - 1 = 2^s q, with odd q. Denote its * 2-Sylow subgroup by S. It is cyclic of order 2^s. The squares in S have * order dividing 2^(s-1). They are the even powers of any generator z of S. * If a is a quadratic residue, 1 = a^[(p-1)/2] = (a^q)^[2^(s-1)], so b = a^q * is a square in S. Therefore there is an integer k such that b z^(2k) = 1. * Set x = a^[(q+1)/2] z^k, and find x^2 = a (mod p). * * The problem is thus reduced to finding a generator z of the 2-Sylow * subgroup S of GF(p)* and finding k. An iterative constructions avoids * the need for an explicit k, a generator is found by a randomized search. * * While we do not actually know that p is a prime number, we can still apply * the formulas in cases 1 and 2 and verify that we have indeed found a square * root of p. Similarly, in case 3, we can try to find a quadratic non-residue, * which will fail for example if p is a square. The iterative construction * may or may not find a candidate square root which we can then validate. */ /* * Handle the cases where p is 2, p isn't odd or p is one. Since BN_mod_sqrt() * can run on untrusted data, a primality check is too expensive. Also treat * the obvious cases where a is 0 or 1. */ static int bn_mod_sqrt_trivial_cases(int *done, BIGNUM *out_sqrt, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) { *done = 1; if (BN_abs_is_word(p, 2)) return BN_set_word(out_sqrt, BN_is_odd(a)); if (!BN_is_odd(p) || BN_abs_is_word(p, 1)) { BNerror(BN_R_P_IS_NOT_PRIME); return 0; } if (BN_is_zero(a) || BN_is_one(a)) return BN_set_word(out_sqrt, BN_is_one(a)); *done = 0; return 1; } /* * Case 1. We know that (a/p) = 1 and that p = 3 (mod 4). */ static int bn_mod_sqrt_p_is_3_mod_4(BIGNUM *out_sqrt, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) { BIGNUM *n; int ret = 0; BN_CTX_start(ctx); if ((n = BN_CTX_get(ctx)) == NULL) goto err; /* Calculate n = (|p| + 1) / 4. */ if (!BN_uadd(n, p, BN_value_one())) goto err; if (!BN_rshift(n, n, 2)) goto err; /* By case 1 above, out_sqrt = a^n is a square root of a (mod p). */ if (!BN_mod_exp_ct(out_sqrt, a, n, p, ctx)) goto err; ret = 1; err: BN_CTX_end(ctx); return ret; } /* * Case 2. We know that (a/p) = 1 and that p = 5 (mod 8). */ static int bn_mod_sqrt_p_is_5_mod_8(BIGNUM *out_sqrt, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) { BIGNUM *b, *i, *n, *tmp; int ret = 0; BN_CTX_start(ctx); if ((b = BN_CTX_get(ctx)) == NULL) goto err; if ((i = BN_CTX_get(ctx)) == NULL) goto err; if ((n = BN_CTX_get(ctx)) == NULL) goto err; if ((tmp = BN_CTX_get(ctx)) == NULL) goto err; /* Calculate n = (|p| - 5) / 8. Since p = 5 (mod 8), simply shift. */ if (!BN_rshift(n, p, 3)) goto err; BN_set_negative(n, 0); /* Compute tmp = 2a (mod p) for later use. */ if (!BN_mod_lshift1(tmp, a, p, ctx)) goto err; /* Calculate b = (2a)^n (mod p). */ if (!BN_mod_exp_ct(b, tmp, n, p, ctx)) goto err; /* Calculate i = 2 a b^2 (mod p). */ if (!BN_mod_sqr(i, b, p, ctx)) goto err; if (!BN_mod_mul(i, tmp, i, p, ctx)) goto err; /* A square root is out_sqrt = a b (i-1) (mod p). */ if (!BN_sub_word(i, 1)) goto err; if (!BN_mod_mul(out_sqrt, a, b, p, ctx)) goto err; if (!BN_mod_mul(out_sqrt, out_sqrt, i, p, ctx)) goto err; ret = 1; err: BN_CTX_end(ctx); return ret; } /* * Case 3. We know that (a/p) = 1 and that p = 1 (mod 8). */ /* * Simple helper. To find a generator of the 2-Sylow subgroup of GF(p)*, we * need to find a quadratic non-residue of p, i.e., n such that (n/p) = -1. */ static int bn_mod_sqrt_n_is_non_residue(int *is_non_residue, const BIGNUM *n, const BIGNUM *p, BN_CTX *ctx) { switch (BN_kronecker(n, p, ctx)) { case -1: *is_non_residue = 1; return 1; case 1: *is_non_residue = 0; return 1; case 0: /* n divides p, so ... */ BNerror(BN_R_P_IS_NOT_PRIME); return 0; default: return 0; } } /* * The following is the only non-deterministic part preparing Tonelli-Shanks. * * If we find n such that (n/p) = -1, then n^q (mod p) is a generator of the * 2-Sylow subgroup of GF(p)*. To find such n, first try some small numbers, * then random ones. */ static int bn_mod_sqrt_find_sylow_generator(BIGNUM *out_generator, const BIGNUM *p, const BIGNUM *q, BN_CTX *ctx) { BIGNUM *n, *p_abs; int i, is_non_residue; int ret = 0; BN_CTX_start(ctx); if ((n = BN_CTX_get(ctx)) == NULL) goto err; if ((p_abs = BN_CTX_get(ctx)) == NULL) goto err; for (i = 2; i < 32; i++) { if (!BN_set_word(n, i)) goto err; if (!bn_mod_sqrt_n_is_non_residue(&is_non_residue, n, p, ctx)) goto err; if (is_non_residue) goto found; } if (!bn_copy(p_abs, p)) goto err; BN_set_negative(p_abs, 0); for (i = 0; i < 128; i++) { if (!bn_rand_interval(n, 32, p_abs)) goto err; if (!bn_mod_sqrt_n_is_non_residue(&is_non_residue, n, p, ctx)) goto err; if (is_non_residue) goto found; } /* * The probability to get here is < 2^(-128) for prime p. For squares * it is easy: for p = 1369 = 37^2 this happens in ~3% of runs. */ BNerror(BN_R_TOO_MANY_ITERATIONS); goto err; found: /* * If p is prime, n^q generates the 2-Sylow subgroup S of GF(p)*. */ if (!BN_mod_exp_ct(out_generator, n, q, p, ctx)) goto err; /* Sanity: p is not necessarily prime, so we could have found 0 or 1. */ if (BN_is_zero(out_generator) || BN_is_one(out_generator)) { BNerror(BN_R_P_IS_NOT_PRIME); goto err; } ret = 1; err: BN_CTX_end(ctx); return ret; } /* * Initialization step for Tonelli-Shanks. * * In the end, b = a^q (mod p) and x = a^[(q+1)/2] (mod p). Cohen optimizes this * to minimize taking powers of a. This is a bit confusing and distracting, so * factor this into a separate function. */ static int bn_mod_sqrt_tonelli_shanks_initialize(BIGNUM *b, BIGNUM *x, const BIGNUM *a, const BIGNUM *p, const BIGNUM *q, BN_CTX *ctx) { BIGNUM *k; int ret = 0; BN_CTX_start(ctx); if ((k = BN_CTX_get(ctx)) == NULL) goto err; /* k = (q-1)/2. Since q is odd, we can shift. */ if (!BN_rshift1(k, q)) goto err; /* x = a^[(q-1)/2] (mod p). */ if (!BN_mod_exp_ct(x, a, k, p, ctx)) goto err; /* b = ax^2 = a^q (mod p). */ if (!BN_mod_sqr(b, x, p, ctx)) goto err; if (!BN_mod_mul(b, a, b, p, ctx)) goto err; /* x = ax = a^[(q+1)/2] (mod p). */ if (!BN_mod_mul(x, a, x, p, ctx)) goto err; ret = 1; err: BN_CTX_end(ctx); return ret; } /* * Find smallest exponent m such that b^(2^m) = 1 (mod p). Assuming that a * is a quadratic residue and p is a prime, we know that 1 <= m < r. */ static int bn_mod_sqrt_tonelli_shanks_find_exponent(int *out_exponent, const BIGNUM *b, const BIGNUM *p, int r, BN_CTX *ctx) { BIGNUM *x; int m; int ret = 0; BN_CTX_start(ctx); if ((x = BN_CTX_get(ctx)) == NULL) goto err; /* * If r <= 1, the Tonelli-Shanks iteration should have terminated as * r == 1 implies b == 1. */ if (r <= 1) { BNerror(BN_R_P_IS_NOT_PRIME); goto err; } /* * Sanity check to ensure taking squares actually does something: * If b is 1, the Tonelli-Shanks iteration should have terminated. * If b is 0, something's very wrong, in particular p can't be prime. */ if (BN_is_zero(b) || BN_is_one(b)) { BNerror(BN_R_P_IS_NOT_PRIME); goto err; } if (!bn_copy(x, b)) goto err; for (m = 1; m < r; m++) { if (!BN_mod_sqr(x, x, p, ctx)) goto err; if (BN_is_one(x)) break; } if (m >= r) { /* This means a is not a quadratic residue. As (a/p) = 1, ... */ BNerror(BN_R_P_IS_NOT_PRIME); goto err; } *out_exponent = m; ret = 1; err: BN_CTX_end(ctx); return ret; } /* * The update step. With the minimal m such that b^(2^m) = 1 (mod m), * set t = y^[2^(r-m-1)] (mod p) and update x = xt, y = t^2, b = by. * This preserves the loop invariants a b = x^2, y^[2^(r-1)] = -1 and * b^[2^(r-1)] = 1. */ static int bn_mod_sqrt_tonelli_shanks_update(BIGNUM *b, BIGNUM *x, BIGNUM *y, const BIGNUM *p, int m, int r, BN_CTX *ctx) { BIGNUM *t; int ret = 0; BN_CTX_start(ctx); if ((t = BN_CTX_get(ctx)) == NULL) goto err; /* t = y^[2^(r-m-1)] (mod p). */ if (!BN_set_bit(t, r - m - 1)) goto err; if (!BN_mod_exp_ct(t, y, t, p, ctx)) goto err; /* x = xt (mod p). */ if (!BN_mod_mul(x, x, t, p, ctx)) goto err; /* y = t^2 = y^[2^(r-m)] (mod p). */ if (!BN_mod_sqr(y, t, p, ctx)) goto err; /* b = by (mod p). */ if (!BN_mod_mul(b, b, y, p, ctx)) goto err; ret = 1; err: BN_CTX_end(ctx); return ret; } static int bn_mod_sqrt_p_is_1_mod_8(BIGNUM *out_sqrt, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) { BIGNUM *b, *q, *x, *y; int e, m, r; int ret = 0; BN_CTX_start(ctx); if ((b = BN_CTX_get(ctx)) == NULL) goto err; if ((q = BN_CTX_get(ctx)) == NULL) goto err; if ((x = BN_CTX_get(ctx)) == NULL) goto err; if ((y = BN_CTX_get(ctx)) == NULL) goto err; /* * Factor p - 1 = 2^e q with odd q. Since p = 1 (mod 8), we know e >= 3. */ e = 1; while (!BN_is_bit_set(p, e)) e++; if (!BN_rshift(q, p, e)) goto err; if (!bn_mod_sqrt_find_sylow_generator(y, p, q, ctx)) goto err; /* * Set b = a^q (mod p) and x = a^[(q+1)/2] (mod p). */ if (!bn_mod_sqrt_tonelli_shanks_initialize(b, x, a, p, q, ctx)) goto err; /* * The Tonelli-Shanks iteration. Starting with r = e, the following loop * invariants hold at the start of the loop. * * a b = x^2 (mod p) * y^[2^(r-1)] = -1 (mod p) * b^[2^(r-1)] = 1 (mod p) * * In particular, if b = 1 (mod p), x is a square root of a. * * Since p - 1 = 2^e q, we have 2^(e-1) q = (p - 1) / 2, so in the first * iteration this follows from (a/p) = 1, (n/p) = -1, y = n^q, b = a^q. * * In subsequent iterations, t = y^[2^(r-m-1)], where m is the smallest * m such that b^(2^m) = 1. With x = xt (mod p) and b = bt^2 (mod p) the * first invariant is preserved, the second and third follow from * y = t^2 (mod p) and r = m as well as the choice of m. * * Finally, r is strictly decreasing in each iteration. If p is prime, * let S be the 2-Sylow subgroup of GF(p)*. We can prove the algorithm * stops: Let S_r be the subgroup of S consisting of elements of order * dividing 2^r. Then S_r = <y> and b is in S_(r-1). The S_r form a * descending filtration of S and when r = 1, then b = 1. */ for (r = e; r >= 1; r = m) { /* * Termination condition. If b == 1 then x is a square root. */ if (BN_is_one(b)) goto done; /* Find smallest exponent 1 <= m < r such that b^(2^m) == 1. */ if (!bn_mod_sqrt_tonelli_shanks_find_exponent(&m, b, p, r, ctx)) goto err; /* * With t = y^[2^(r-m-1)], update x = xt, y = t^2, b = by. */ if (!bn_mod_sqrt_tonelli_shanks_update(b, x, y, p, m, r, ctx)) goto err; /* * Sanity check to make sure we don't loop indefinitely. * bn_mod_sqrt_tonelli_shanks_find_exponent() ensures m < r. */ if (r <= m) goto err; } /* * If p is prime, we should not get here. */ BNerror(BN_R_NOT_A_SQUARE); goto err; done: if (!bn_copy(out_sqrt, x)) goto err; ret = 1; err: BN_CTX_end(ctx); return ret; } /* * Choose the smaller of sqrt and |p| - sqrt. */ static int bn_mod_sqrt_normalize(BIGNUM *sqrt, const BIGNUM *p, BN_CTX *ctx) { BIGNUM *x; int ret = 0; BN_CTX_start(ctx); if ((x = BN_CTX_get(ctx)) == NULL) goto err; if (!BN_lshift1(x, sqrt)) goto err; if (BN_ucmp(x, p) > 0) { if (!BN_usub(sqrt, p, sqrt)) goto err; } ret = 1; err: BN_CTX_end(ctx); return ret; } /* * Verify that a = (sqrt_a)^2 (mod p). Requires that a is reduced (mod p). */ static int bn_mod_sqrt_verify(const BIGNUM *a, const BIGNUM *sqrt_a, const BIGNUM *p, BN_CTX *ctx) { BIGNUM *x; int ret = 0; BN_CTX_start(ctx); if ((x = BN_CTX_get(ctx)) == NULL) goto err; if (!BN_mod_sqr(x, sqrt_a, p, ctx)) goto err; if (BN_cmp(x, a) != 0) { BNerror(BN_R_NOT_A_SQUARE); goto err; } ret = 1; err: BN_CTX_end(ctx); return ret; } static int bn_mod_sqrt_internal(BIGNUM *out_sqrt, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) { BIGNUM *a_mod_p, *sqrt; BN_ULONG lsw; int done; int kronecker; int ret = 0; BN_CTX_start(ctx); if ((a_mod_p = BN_CTX_get(ctx)) == NULL) goto err; if ((sqrt = BN_CTX_get(ctx)) == NULL) goto err; if (!BN_nnmod(a_mod_p, a, p, ctx)) goto err; if (!bn_mod_sqrt_trivial_cases(&done, sqrt, a_mod_p, p, ctx)) goto err; if (done) goto verify; /* * Make sure that the Kronecker symbol (a/p) == 1. In case p is prime * this is equivalent to a having a square root (mod p). The cost of * BN_kronecker() is O(log^2(n)). This is small compared to the cost * O(log^4(n)) of Tonelli-Shanks. */ if ((kronecker = BN_kronecker(a_mod_p, p, ctx)) == -2) goto err; if (kronecker <= 0) { /* This error is only accurate if p is known to be a prime. */ BNerror(BN_R_NOT_A_SQUARE); goto err; } lsw = BN_lsw(p); if (lsw % 4 == 3) { if (!bn_mod_sqrt_p_is_3_mod_4(sqrt, a_mod_p, p, ctx)) goto err; } else if (lsw % 8 == 5) { if (!bn_mod_sqrt_p_is_5_mod_8(sqrt, a_mod_p, p, ctx)) goto err; } else if (lsw % 8 == 1) { if (!bn_mod_sqrt_p_is_1_mod_8(sqrt, a_mod_p, p, ctx)) goto err; } else { /* Impossible to hit since the trivial cases ensure p is odd. */ BNerror(BN_R_P_IS_NOT_PRIME); goto err; } if (!bn_mod_sqrt_normalize(sqrt, p, ctx)) goto err; verify: if (!bn_mod_sqrt_verify(a_mod_p, sqrt, p, ctx)) goto err; if (!bn_copy(out_sqrt, sqrt)) goto err; ret = 1; err: BN_CTX_end(ctx); return ret; } BIGNUM * BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) { BIGNUM *out_sqrt; if ((out_sqrt = in) == NULL) out_sqrt = BN_new(); if (out_sqrt == NULL) goto err; if (!bn_mod_sqrt_internal(out_sqrt, a, p, ctx)) goto err; return out_sqrt; err: if (out_sqrt != in) BN_free(out_sqrt); return NULL; } LCRYPTO_ALIAS(BN_mod_sqrt);
