def carmichael_lambda(n):
"""Returns the smallest positive integer m such that a**m = 1 (mod n) for
all integers a coprime to n.
Parameters
* n: int or list (n > 1)
The modulus. This value can be an int or a factorization.
Returns:
* m: int
Raises:
* ValueError: If n <= 0 or n is not an integer or factorization.
Examples:
>>> carmichael_lambda(100)
20
>>> carmichael_lambda([(2, 2), (5, 2)])
20
>>> carmichael_lambda(113)
112
>>> carmichael_lambda(0)
Traceback (most recent call last):
...
ValueError: carmichael_lambda: Must have n > 0.
Details:
The Carmichael lambda function gives the exponent of the multiplicative
group of integers modulo n. For each n, lambda(n) divides euler_phi(n).
This method uses the standard definition of the Carmichael lambda
function. In particular, lambda(n) is the least common multiple of the
values of lambda(p**e) for each prime power in the factorization of n.
See "Fundamental Number Theory with Applications" by Mollin for details.
"""
if isinstance(n, (int, long)):
if n == 1:
return 1
elif n <= 0:
raise ValueError("carmichael_lambda: Must have n > 0.")
n_factorization = factor.factor(n)
elif isinstance(n, list) and n[0][0] > 0:
n_factorization = n
else:
raise ValueError("carmichael_lambda: Input must be a positive integer or a factorization.")
terms = []
for (p, e) in n_factorization:
if p != 2:
term = p**(e - 1)*(p - 1)
else:
if e <= 2:
term = 1 << (e - 1)
else:
term = 1 << (e - 2)
terms.append(term)
value = lcm_list(terms)
return value
def tau(n):
"""Returns the number of divisors of n.
Parameters
* n: int (n > 0)
The value of n can be an int or a factorization.
Returns:
* s: int
Raises:
* ValueError: If n <= 0 or if n is not an int or factorization.
Examples:
>>> tau(9)
3
>>> tau([(2, 2), (5, 2)])
9
>>> tau(100)
9
>>> tau('cat')
Traceback (most recent call last):
...
ValueError: tau: Input must be a positive integer or a factorization.
>>> tau(-1)
Traceback (most recent call last):
...
ValueError: tau: Must have n > 0.
Details:
The divisor tau function is a multiplicative function satisfying
tau(p**k) = 1 + k for prime powers p**k. For positive integers n, this
method computes the factorization of n, and then uses this product
definition. If instead a factorization of n is given, then the product
formula is used directly.
"""
if isinstance(n, (int, long)):
if n == 1:
return 1
elif n <= 0:
raise ValueError("tau: Must have n > 0.")
n_factorization = factor.factor(n)
elif isinstance(n, list) and n[0][0] > 0:
n_factorization = n
else:
raise ValueError("tau: Input must be a positive integer or a factorization.")
prod = 1
for (p, e) in n_factorization:
prod *= (e + 1)
return prod
def sqrts_mod_n(a, n, n_factorization=None):
"""Returns the solutions x to the congruence x**2 = a (mod n).
Given integers a and n, this function returns all solutions to the
congruence x**2 = a (mod n).
Input:
* a: int
A square modulo n.
* n: int
The modulus.
* n_factorization: list (default=None)
The factorization of n.
Returns:
* roots: list
A sorted list of all square roots of a modulo n.
Examples:
>>> sqrts_mod_n(3, 11**3*23**3)
[16152263, 28026114, 150110933, 161984784]
>>> sqrts_mod_n(49, 53**3*61**4)
[7, 721465236980, 1339862033577, 2061327270550]
>>> sqrts_mod_n(-1, 5**3*7**2)
[]
"""
if n_factorization is None:
n_factorization = factor.factor(n)
congruences = []
moduli = []
for (p, k) in n_factorization:
roots = _sqrts_mod_pk(a, p, k)
pk = p**k
pairs = [(r, pk) for r in roots]
congruences.append(pairs)
moduli.append(pk)
if len(congruences) == 1:
values = [r for (r, _) in congruences[0]]
else:
preconditioning_data = crt_preconditioning_data(moduli)
values = []
for system in itertools.product(*congruences):
values.append(chinese_preconditioned(system, preconditioning_data))
values.sort()
return values
def sqrts_mod_n(a, n, n_factorization=None):
"""Returns the solutions x to the congruence x**2 = a (mod n).
Given integers a and n, this function returns all solutions to the
congruence x**2 = a (mod n).
Input:
* a: int
A square modulo n.
* n: int
The modulus.
* n_factorization: list (default=None)
The factorization of n.
Returns:
* roots: list
A sorted list of all square roots of a modulo n.
Examples:
>>> sqrts_mod_n(3, 11**3*23**3)
[16152263, 28026114, 150110933, 161984784]
>>> sqrts_mod_n(49, 53**3*61**4)
[7, 721465236980, 1339862033577, 2061327270550]
>>> sqrts_mod_n(-1, 5**3*7**2)
[]
"""
if n_factorization is None:
n_factorization = factor.factor(n)
congruences = []
moduli = []
for (p, k) in n_factorization:
roots = _sqrts_mod_pk(a, p, k)
pk = p**k
pairs = [(r, pk) for r in roots]
congruences.append(pairs)
moduli.append(pk)
if len(congruences) == 1:
values = [r for (r, _) in congruences[0]]
else:
preconditioning_data = crt_preconditioning_data(moduli)
values = []
for system in itertools.product(*congruences):
values.append(chinese_preconditioned(system, preconditioning_data))
values.sort()
return values
def test_factor(self):
factorizations = {
5**77 - 1: [(2, 2), (19531, 1), (12207031L, 1), (527093491L, 1),
(8090594434231L, 1), (162715052426691233701L, 1)],
5**93 + 1: [(2, 1), (3, 2), (7, 1), (1303, 1), (21207101L, 1),
(258065887L, 1), (28086211607L, 1), (75005167927L, 1),
(53345671490722200466369L, 1)],
1303**6 * 11213**4: [(1303, 6), (11213, 4)],
2**91 - 1: [(127, 1), (911, 1), (8191, 1), (112901153L, 1),
(23140471537L, 1)],
10**22 + 1: [(89, 1), (101, 1), (1052788969L, 1), (1056689261L, 1)]
}
for (n, n_fac) in factorizations.items():
self.assertEqual(n_fac, factor(n))
def test_factor(self):
factorizations = {
5**77 - 1: [
(2, 2),
(19531, 1),
(12207031L, 1),
(527093491L, 1),
(8090594434231L, 1),
(162715052426691233701L, 1)
],
5**93 + 1: [
(2, 1),
(3, 2),
(7, 1),
(1303, 1),
(21207101L, 1),
(258065887L, 1),
(28086211607L, 1),
(75005167927L, 1),
(53345671490722200466369L, 1)
],
1303**6*11213**4: [
(1303, 6),
(11213, 4)
],
2**91 - 1: [
(127, 1),
(911, 1),
(8191, 1),
(112901153L, 1),
(23140471537L, 1)
],
10**22 + 1: [
(89, 1),
(101, 1),
(1052788969L, 1),
(1056689261L, 1)
]
}
for (n, n_fac) in factorizations.items():
self.assertEqual(n_fac, factor(n))
def moebius(n=None, factorization=None):
r"""Returns the value of the Moebius function ``mu(n)``.
Parameters
----------
n : int, optional
factorization : list, optional
Factorization of `n` given as a list of (prime, exponent) pairs.
Returns
-------
mu : int
The value moebius(n).
Raises
------
ValueError
If ``n <= 0``.
See Also
--------
moebius_sum, moebius_list
Notes
-----
The Moebius mu function is a multiplicative function satisfying
.. math::
mu(p_1 p_2 \cdots p_k) = (-1)^k
and ``mu(p**e) = 0`` for ``e >= 2``. For positive integers `n`, this
method computes the factorization of `n`, and then uses this product
definition. If instead a factorization of `n` is given, then the
product formula is used directly. See section 2.2 of [1] for details,
and see section 3.9 for information related to the growth rate.
References
----------
.. [1] T.M. Apostol, "Introduction to Analytic Number Theory",
Springer-Verlag, New York, 1976.
Examples
--------
>>> moebius(1)
1
>>> moebius(35)
1
>>> moebius(factorization=[(5, 1), (7, 1)])
1
>>> moebius(100)
0
>>> moebius(2)
-1
>>> moebius(-1)
Traceback (most recent call last):
...
ValueError: moebius: Must have n > 0.
"""
if n is None and factorization is None:
raise TypeError("euler_phi: Expected at least one argument.")
elif factorization is None:
if n == 1:
return 1
elif n <= 0:
raise ValueError("moebius: Must have n > 0.")
factorization = factor.factor(n)
else:
if factorization[0][0] == -1:
raise ValueError("moebius: Must have n > 0.")
if factor.is_squarefree(factorization=factorization):
k = len(factorization)
return (-1)**k
else:
return 0
def euler_phi_inverse(n):
"""Returns a sorted list of all positive integers k such that euler_phi(k) = n.
Parameters
* n: int or list (n > 0)
The value of n can be an int or a factorization.
Returns:
* values: list
Raises:
* ValueError: If n <= 0 or n is not an int or factorization.
Examples:
>>> euler_phi_inverse(40)
[41, 55, 75, 82, 88, 100, 110, 132, 150]
>>> euler_phi_inverse(100)
[101, 125, 202, 250]
>>> euler_phi_inverse([(2, 2), (5, 2)])
[101, 125, 202, 250]
>>> euler_phi_inverse(103)
[]
>>> euler_phi_inverse(-1)
Traceback (most recent call last):
...
ValueError: euler_phi_inverse: Must have n > 0.
Details:
This uses the algorithm outlined in "Discovering Mathematics with Magma"
by Bosma and Cannon. For a different method, see the paper "Euler's
Totient Function and its Inverse" by Gupta.
"""
if isinstance(n, (int, long)):
if n == 1:
return [1, 2]
elif n > 1 and n % 2 == 1:
# We know euler_phi(n) is even for n >= 3
return []
elif n <= 0:
raise ValueError("euler_phi_inverse: Must have n > 0.")
n_factorization = factor.factor(n)
elif isinstance(n, list) and n[0][0] > 0:
n_factorization = n
n = factor.factor_back(n_factorization)
else:
raise ValueError("euler_phi_inverse: Input must be a positive integer or a factorization.")
powers_of_two = set([1])
if n % 2 == 0:
for i in xrange(1, n_factorization[0][1] + 1):
powers_of_two.add(2**i)
# Odd primes p that divide n must have the property that p - 1 divides m.
prime_list = []
for d in factor.divisors(factorization=n_factorization)[1:]:
if primality.is_prime(d + 1):
prime_list.append(d + 1)
# Here, we store pairs (a, b) such that a is odd and phi(a) = m/b, with b
# even or 1. Every pair contributes at least one solution. When b = 1, the
# pair contributes two solutions.
pairs = [(1, n)]
for p in reversed(prime_list):
new_pairs = []
for (a, b) in pairs:
if b == 1:
continue
pk = p
d = b//(p - 1)
mmod = b % (p - 1)
while mmod == 0:
if d % 2 == 0 or d == 1:
new_pairs.append((pk*a, d))
pk *= p
mmod = d % p
d = d//p
pairs.extend(new_pairs)
# When b = 2^k, we have the solution 2*b*a.
values = []
for (a, b) in pairs:
if b in powers_of_two:
values.append(2*b*a)
if b == 1:
values.append(a)
values.sort()
return values
def euler_phi(n=None, factorization=None):
r"""Returns the Euler totient of `n`.
For positive integers `n`, this method returns the number of positive
integers less than or equal to `n` which are relatively prime to `n`.
Parameters
----------
n : int, optional
factorization : list, optional
Factorization of `n` given as a list of (prime, exponent) pairs.
Returns
-------
phi : int
The value euler_phi(n)
Raises
------
ValueError
If ``n <= 0``.
See Also
--------
euler_phi_sum, euler_phi_list
Notes
-----
The Euler phi function is a multiplicative function satisfying
.. math::
\phi(p^k) = (p - 1) p^{k - 1}
for prime powers ``p^k``. For positive integers `n`, this method
computes the factorization of `n`, and then uses this product
definition. If instead a factorization of `n` is given, then the
product formula is used directly. See section 2.3 of [1] for details,
and see section 3.7 for information related to the growth rate.
References
----------
.. [1] T.M. Apostol, "Introduction to Analytic Number Theory",
Springer-Verlag, New York, 1976.
Examples
--------
>>> euler_phi(1)
1
>>> euler_phi(7)
6
>>> euler_phi(100)
40
>>> euler_phi(factorization=[(2, 2), (5, 2)])
40
>>> euler_phi(0)
Traceback (most recent call last):
...
ValueError: euler_phi: Must have n > 0.
"""
if n is None and factorization is None:
raise TypeError("euler_phi: Expected at least one argument.")
elif factorization is None:
if n == 1:
return 1
elif n <= 0:
raise ValueError("euler_phi: Must have n > 0.")
factorization = factor.factor(n)
else:
if factorization[0][0] == -1:
raise ValueError("euler_phi: Must have n > 0.")
prod = 1
for (p, e) in factorization:
prod *= p**(e - 1)*(p - 1)
return prod
def sigma(n, k=1):
"""Returns the sum of the kth powers of the divisors of n.
Parameters
* n: int or list (n > 0)
The value of n can be an int or a factorization.
* k: int (k >= 0) (default=1)
Returns:
* s: int
Raises:
* ValueError: If n <= 0 or k < 0.
Examples:
>>> sigma(9)
13
>>> 1 + 3 + 9
13
>>> sigma([(3, 2)])
13
>>> sigma(10, 2)
130
>>> 1**2 + 2**2 + 5**2 + 10**2
130
>>> sigma(10, 0)
4
>>> sigma(-1)
Traceback (most recent call last):
...
ValueError: sigma: Must have n > 0.
>>> sigma(10, -1)
Traceback (most recent call last):
...
ValueError: sigma: Must have k >= 0.
Details:
The divisor sigma function is a multiplicative function satisfying
sigma(p**e, k) = 1 + p**k + p**(2*k) + ... + p**(e*k) for prime powers
p**e. For positive integers n, this method computes the factorization of
n, and then uses this product definition. If instead a factorization
of n is given, then the product formula is used directly.
"""
if k < 0:
raise ValueError("sigma: Must have k >= 0.")
elif k == 0:
return tau(n)
if isinstance(n, (int, long)):
if n == 1:
return 1
elif n <= 0:
raise ValueError("sigma: Must have n > 0.")
n_factorization = factor.factor(n)
elif isinstance(n, list) and n[0][0] > 0:
n_factorization = n
else:
raise ValueError("sigma: Input must be a positive integer or a factorization.")
prod = 1
for (p, e) in n_factorization:
pk = p**k
prod *= (pk**(e + 1) - 1)//(pk - 1)
return prod
def primitive_root(n, n_factorization=None, phi=None, phi_divisors=None):
"""Returns a primitive root modulo n.
Given a positive integer n of the form n = 2, 4, p**a, or 2*p**a for p an
odd prime and a >= 1, this function returns the least primitive root of n.
Input:
* n: int (n > 1)
* n_factorization: list (default=None)
The prime factorization of n.
* phi: int (default=None)
The value of euler_phi(n).
* phi_divisors: list (default=None)
A list of the prime divisors of euler_phi(n).
Returns:
* g: int
The least primitive root of n.
Raises:
* ValueError: if primitive roots don't exist modulo n.
Examples:
>>> primitive_root(7)
3
>>> primitive_root(11)
2
>>> primitive_root(14)
3
>>> primitive_root(12)
Traceback (most recent call last):
...
ValueError: primitive_root: No primitive root for n.
Details:
An integer a is called a primitive root mod n if a generates the
multiplicative group of units modulo n. Equivalently, a is a primitive
root modulo n if the order of a modulo m is euler_phi(n).
As noted above, primitive roots only exist for moduli of the form n = 2,
4, p**a, 2*p**a, where p > 2 is prime, and a >= 1. See Theorem 6.11 of
"Elementary Number Theory" by Jones and Jones for a proof of this fact.
See also Chapter 10 of "Introduction to Analytic Number Theory" by
Apostol.
This method uses the fact that an integer a coprime to n is a primitive
root if and only if a**(euler_phi(n)/p) != 1 (mod n) for each prime p
dividing euler_phi(n). See Lemma 6.4 in Jones and Jones for a proof of
this.
Note also that there is another way to construct primitive roots for
composite moduli. To find a primitive root modulu p**a for p an odd
prime and a >= 2, first compute g, a primitive root modulo p using the
above criteria. Next, compute g1 = g**(p - 1) (mod p**2). If g1 != 1,
then g is a primitive root modulo p**a for every a >= 1. Otherwise, g +
p is. Finally, note that when p is an odd prime, if g is a primitive
root modulo p**a, then either g or g + p**a (whichever is odd) is a
primitive root modulo 2*p**a. See Lemma 1.4.5 of "A Course in
Computational Algebraic Number Theory" by Cohen for more details.
"""
if n_factorization is None:
n_factorization = factor.factor(n)
if n % 4 == 0 and n != 4:
raise ValueError("primitive_root: No primitive root for n.")
if len(n_factorization) > 2:
raise ValueError("primitive_root: No primitive root for n.")
if n % 2 == 1 and len(n_factorization) > 1:
raise ValueError("primitive_root: No primitive root for n.")
if phi is None:
phi = functions.euler_phi(n_factorization)
if phi_divisors is None:
phi_divisors = factor.prime_divisors(phi)
for g in xrange(1, n):
if is_primitive_root(g, n, phi=phi, phi_divisors=phi_divisors):
return g
def quadratic_congruence(coeff_list, n, n_factorization=None):
"""Returns a list of the roots of the quadratic equation modulo n.
Input:
* coeff_list: list
A list of the coefficients of the quadratic.
* n: int
The modulus.
* n_factorization: list (default=None)
The factorization of the modulus n.
Returns:
* roots: list
A sorted list of the roots of the quadratic modulo n.
Examples:
>>> quadratic_congruence([1, 3, -18], 1000)
[3, 378, 619, 994]
>>> quadratic_congruence([1, -31, -12], 36)
[7, 15, 16, 24]
>>> quadratic_congruence([11, 5, 18], 29)
[22, 25]
Details:
The algorithm proceeds by finding roots modulo p**k for each prime power
p**k in the factorization of the modulus n. These roots are then
combined using the Chinese Remainder Theorem. See Chapter 5 of "The
Theory of Numbers - A Text and Source Book of Problems" by Adler and
Coury for detailed information.
"""
if n_factorization is None:
n_factorization = factor.factor(n)
(a, b, c) = coeff_list
discriminant = b * b - 4 * a * c
all_roots = []
if gcd(2 * a, n) == 1:
# This is the easy case where we can complete the square.
# First, we solve the congruence y^2 = b^2 - 4*a*c (mod n) for y.
discriminant_roots = sqrts_mod_n(discriminant, n, n_factorization)
# Next, we solve the congruence 2*a*x = y - b (mod n) for x to obtain
# the solutions to the quadratic. Since gcd(2*a, n) == 1, this is
# simple, and each each value of y leads to one value of x.
inverse = inverse_mod(2 * a, n)
for y in discriminant_roots:
x = (y - b) * inverse % n
all_roots.append(x)
else:
# Here, gcd(4*a, n) != 1, so we can't complete the square as usual.
# Write 4*a = a1*a2, with a2 coprime to n.
a1 = 1
a2 = 4 * a
d = gcd(n, a2)
while d > 1:
a1 *= d
a2 /= d
d = gcd(n, a2)
# We solve the congruence y^2 = b^2 - 4*a*c (mod a1*n) for y.
discriminant_roots = sqrts_mod_n(discriminant, a1 * n)
# For each solution y, we solve 2*a*x = y - b (mod a1*n) for x. Since
# gcd(2*a, n) > 1, each solution y leads to multiple values of x.
for y in discriminant_roots:
roots = linear_congruence(2 * a, y - b, a1 * n)
all_roots.extend(roots)
# Eliminate repeated solutions, and reduce modulo the original modulus.
distinct_roots = {x % n for x in all_roots}
all_roots = list(distinct_roots)
all_roots.sort()
return all_roots
def euler_phi_inverse(n):
"""Returns a sorted list of all positive integers k such that euler_phi(k) = n.
Parameters
* n: int or list (n > 0)
The value of n can be an int or a factorization.
Returns:
* values: list
Raises:
* ValueError: If n <= 0 or n is not an int or factorization.
Examples:
>>> euler_phi_inverse(40)
[41, 55, 75, 82, 88, 100, 110, 132, 150]
>>> euler_phi_inverse(100)
[101, 125, 202, 250]
>>> euler_phi_inverse([(2, 2), (5, 2)])
[101, 125, 202, 250]
>>> euler_phi_inverse(103)
[]
>>> euler_phi_inverse(-1)
Traceback (most recent call last):
...
ValueError: euler_phi_inverse: Must have n > 0.
Details:
This uses the algorithm outlined in "Discovering Mathematics with Magma"
by Bosma and Cannon. For a different method, see the paper "Euler's
Totient Function and its Inverse" by Gupta.
"""
if isinstance(n, (int, long)):
if n == 1:
return [1, 2]
elif n > 1 and n % 2 == 1:
# We know euler_phi(n) is even for n >= 3
return []
elif n <= 0:
raise ValueError("euler_phi_inverse: Must have n > 0.")
n_factorization = factor.factor(n)
elif isinstance(n, list) and n[0][0] > 0:
n_factorization = n
n = factor.factor_back(n_factorization)
else:
raise ValueError(
"euler_phi_inverse: Input must be a positive integer or a factorization."
)
powers_of_two = set([1])
if n % 2 == 0:
for i in xrange(1, n_factorization[0][1] + 1):
powers_of_two.add(2**i)
# Odd primes p that divide n must have the property that p - 1 divides m.
prime_list = []
for d in factor.divisors(factorization=n_factorization)[1:]:
if primality.is_prime(d + 1):
prime_list.append(d + 1)
# Here, we store pairs (a, b) such that a is odd and phi(a) = m/b, with b
# even or 1. Every pair contributes at least one solution. When b = 1, the
# pair contributes two solutions.
pairs = [(1, n)]
for p in reversed(prime_list):
new_pairs = []
for (a, b) in pairs:
if b == 1:
continue
pk = p
d = b // (p - 1)
mmod = b % (p - 1)
while mmod == 0:
if d % 2 == 0 or d == 1:
new_pairs.append((pk * a, d))
pk *= p
mmod = d % p
d = d // p
pairs.extend(new_pairs)
# When b = 2^k, we have the solution 2*b*a.
values = []
for (a, b) in pairs:
if b in powers_of_two:
values.append(2 * b * a)
if b == 1:
values.append(a)
values.sort()
return values
def primitive_root(n, n_factorization=None, phi=None, phi_divisors=None):
"""Returns a primitive root modulo n.
Given a positive integer n of the form n = 2, 4, p**a, or 2*p**a for p an
odd prime and a >= 1, this function returns the least primitive root of n.
Input:
* n: int (n > 1)
* n_factorization: list (default=None)
The prime factorization of n.
* phi: int (default=None)
The value of euler_phi(n).
* phi_divisors: list (default=None)
A list of the prime divisors of euler_phi(n).
Returns:
* g: int
The least primitive root of n.
Raises:
* ValueError: if primitive roots don't exist modulo n.
Examples:
>>> primitive_root(7)
3
>>> primitive_root(11)
2
>>> primitive_root(14)
3
>>> primitive_root(12)
Traceback (most recent call last):
...
ValueError: primitive_root: No primitive root for n.
Details:
An integer a is called a primitive root mod n if a generates the
multiplicative group of units modulo n. Equivalently, a is a primitive
root modulo n if the order of a modulo m is euler_phi(n).
As noted above, primitive roots only exist for moduli of the form n = 2,
4, p**a, 2*p**a, where p > 2 is prime, and a >= 1. See Theorem 6.11 of
"Elementary Number Theory" by Jones and Jones for a proof of this fact.
See also Chapter 10 of "Introduction to Analytic Number Theory" by
Apostol.
This method uses the fact that an integer a coprime to n is a primitive
root if and only if a**(euler_phi(n)/p) != 1 (mod n) for each prime p
dividing euler_phi(n). See Lemma 6.4 in Jones and Jones for a proof of
this.
Note also that there is another way to construct primitive roots for
composite moduli. To find a primitive root modulu p**a for p an odd
prime and a >= 2, first compute g, a primitive root modulo p using the
above criteria. Next, compute g1 = g**(p - 1) (mod p**2). If g1 != 1,
then g is a primitive root modulo p**a for every a >= 1. Otherwise, g +
p is. Finally, note that when p is an odd prime, if g is a primitive
root modulo p**a, then either g or g + p**a (whichever is odd) is a
primitive root modulo 2*p**a. See Lemma 1.4.5 of "A Course in
Computational Algebraic Number Theory" by Cohen for more details.
"""
if n_factorization is None:
n_factorization = factor.factor(n)
if n % 4 == 0 and n != 4:
raise ValueError("primitive_root: No primitive root for n.")
if len(n_factorization) > 2:
raise ValueError("primitive_root: No primitive root for n.")
if n % 2 == 1 and len(n_factorization) > 1:
raise ValueError("primitive_root: No primitive root for n.")
if phi is None:
phi = functions.euler_phi(n_factorization)
if phi_divisors is None:
phi_divisors = factor.prime_divisors(phi)
for g in xrange(1, n):
if is_primitive_root(g, n, phi=phi, phi_divisors=phi_divisors):
return g
def quadratic_congruence(coeff_list, n, n_factorization=None):
"""Returns a list of the roots of the quadratic equation modulo n.
Input:
* coeff_list: list
A list of the coefficients of the quadratic.
* n: int
The modulus.
* n_factorization: list (default=None)
The factorization of the modulus n.
Returns:
* roots: list
A sorted list of the roots of the quadratic modulo n.
Examples:
>>> quadratic_congruence([1, 3, -18], 1000)
[3, 378, 619, 994]
>>> quadratic_congruence([1, -31, -12], 36)
[7, 15, 16, 24]
>>> quadratic_congruence([11, 5, 18], 29)
[22, 25]
Details:
The algorithm proceeds by finding roots modulo p**k for each prime power
p**k in the factorization of the modulus n. These roots are then
combined using the Chinese Remainder Theorem. See Chapter 5 of "The
Theory of Numbers - A Text and Source Book of Problems" by Adler and
Coury for detailed information.
"""
if n_factorization is None:
n_factorization = factor.factor(n)
(a, b, c) = coeff_list
discriminant = b*b - 4*a*c
all_roots = []
if gcd(2*a, n) == 1:
# This is the easy case where we can complete the square.
# First, we solve the congruence y^2 = b^2 - 4*a*c (mod n) for y.
discriminant_roots = sqrts_mod_n(discriminant, n, n_factorization)
# Next, we solve the congruence 2*a*x = y - b (mod n) for x to obtain
# the solutions to the quadratic. Since gcd(2*a, n) == 1, this is
# simple, and each each value of y leads to one value of x.
inverse = inverse_mod(2*a, n)
for y in discriminant_roots:
x = (y - b)*inverse % n
all_roots.append(x)
else:
# Here, gcd(4*a, n) != 1, so we can't complete the square as usual.
# Write 4*a = a1*a2, with a2 coprime to n.
a1 = 1
a2 = 4*a
d = gcd(n, a2)
while d > 1:
a1 *= d
a2 /= d
d = gcd(n, a2)
# We solve the congruence y^2 = b^2 - 4*a*c (mod a1*n) for y.
discriminant_roots = sqrts_mod_n(discriminant, a1*n)
# For each solution y, we solve 2*a*x = y - b (mod a1*n) for x. Since
# gcd(2*a, n) > 1, each solution y leads to multiple values of x.
for y in discriminant_roots:
roots = linear_congruence(2*a, y - b, a1*n)
all_roots.extend(roots)
# Eliminate repeated solutions, and reduce modulo the original modulus.
distinct_roots = {x % n for x in all_roots}
all_roots = list(distinct_roots)
all_roots.sort()
return all_roots