/*  $OpenBSD: bn_mod_sqrt.c,v 1.3 2023/08/03 18:53:55 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 <openssl/err.h>

#include "bn_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);