diff options
Diffstat (limited to 'crypto/bn/bn_sqrt.c')
| -rw-r--r-- | crypto/bn/bn_sqrt.c | 122 |
1 files changed, 61 insertions, 61 deletions
diff --git a/crypto/bn/bn_sqrt.c b/crypto/bn/bn_sqrt.c index 772c8080bb..232af99a21 100644 --- a/crypto/bn/bn_sqrt.c +++ b/crypto/bn/bn_sqrt.c @@ -132,14 +132,14 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) /* we'll set q later (if needed) */ if (e == 1) { - /*- - * The easy case: (|p|-1)/2 is odd, so 2 has an inverse - * modulo (|p|-1)/2, and square roots can be computed - * directly by modular exponentiation. - * We have - * 2 * (|p|+1)/4 == 1 (mod (|p|-1)/2), - * so we can use exponent (|p|+1)/4, i.e. (|p|-3)/4 + 1. - */ + /*- + * The easy case: (|p|-1)/2 is odd, so 2 has an inverse + * modulo (|p|-1)/2, and square roots can be computed + * directly by modular exponentiation. + * We have + * 2 * (|p|+1)/4 == 1 (mod (|p|-1)/2), + * so we can use exponent (|p|+1)/4, i.e. (|p|-3)/4 + 1. + */ if (!BN_rshift(q, p, 2)) goto end; q->neg = 0; @@ -152,32 +152,32 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) } if (e == 2) { - /*- - * |p| == 5 (mod 8) - * - * In this case 2 is always a non-square since - * Legendre(2,p) = (-1)^((p^2-1)/8) for any odd prime. - * So if a really is a square, then 2*a is a non-square. - * Thus for - * b := (2*a)^((|p|-5)/8), - * i := (2*a)*b^2 - * we have - * i^2 = (2*a)^((1 + (|p|-5)/4)*2) - * = (2*a)^((p-1)/2) - * = -1; - * so if we set - * x := a*b*(i-1), - * then - * x^2 = a^2 * b^2 * (i^2 - 2*i + 1) - * = a^2 * b^2 * (-2*i) - * = a*(-i)*(2*a*b^2) - * = a*(-i)*i - * = a. - * - * (This is due to A.O.L. Atkin, - * <URL: http://listserv.nodak.edu/scripts/wa.exe?A2=ind9211&L=nmbrthry&O=T&P=562>, - * November 1992.) - */ + /*- + * |p| == 5 (mod 8) + * + * In this case 2 is always a non-square since + * Legendre(2,p) = (-1)^((p^2-1)/8) for any odd prime. + * So if a really is a square, then 2*a is a non-square. + * Thus for + * b := (2*a)^((|p|-5)/8), + * i := (2*a)*b^2 + * we have + * i^2 = (2*a)^((1 + (|p|-5)/4)*2) + * = (2*a)^((p-1)/2) + * = -1; + * so if we set + * x := a*b*(i-1), + * then + * x^2 = a^2 * b^2 * (i^2 - 2*i + 1) + * = a^2 * b^2 * (-2*i) + * = a*(-i)*(2*a*b^2) + * = a*(-i)*i + * = a. + * + * (This is due to A.O.L. Atkin, + * <URL: http://listserv.nodak.edu/scripts/wa.exe?A2=ind9211&L=nmbrthry&O=T&P=562>, + * November 1992.) + */ /* t := 2*a */ if (!BN_mod_lshift1_quick(t, A, p)) @@ -277,24 +277,24 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) goto end; } - /*- - * Now we know that (if p is indeed prime) there is an integer - * k, 0 <= k < 2^e, such that - * - * a^q * y^k == 1 (mod p). - * - * As a^q is a square and y is not, k must be even. - * q+1 is even, too, so there is an element - * - * X := a^((q+1)/2) * y^(k/2), - * - * and it satisfies - * - * X^2 = a^q * a * y^k - * = a, - * - * so it is the square root that we are looking for. - */ + /*- + * Now we know that (if p is indeed prime) there is an integer + * k, 0 <= k < 2^e, such that + * + * a^q * y^k == 1 (mod p). + * + * As a^q is a square and y is not, k must be even. + * q+1 is even, too, so there is an element + * + * X := a^((q+1)/2) * y^(k/2), + * + * and it satisfies + * + * X^2 = a^q * a * y^k + * = a, + * + * so it is the square root that we are looking for. + */ /* t := (q-1)/2 (note that q is odd) */ if (!BN_rshift1(t, q)) @@ -333,15 +333,15 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) goto end; while (1) { - /*- - * Now b is a^q * y^k for some even k (0 <= k < 2^E - * where E refers to the original value of e, which we - * don't keep in a variable), and x is a^((q+1)/2) * y^(k/2). - * - * We have a*b = x^2, - * y^2^(e-1) = -1, - * b^2^(e-1) = 1. - */ + /*- + * Now b is a^q * y^k for some even k (0 <= k < 2^E + * where E refers to the original value of e, which we + * don't keep in a variable), and x is a^((q+1)/2) * y^(k/2). + * + * We have a*b = x^2, + * y^2^(e-1) = -1, + * b^2^(e-1) = 1. + */ if (BN_is_one(b)) { if (!BN_copy(ret, x)) |