Elements of $\ZZ/n\ZZ$ ¶

An element of the integers modulo $n$ .

There are three types of integer_mod classes, depending on the size of the modulus.

IntegerMod_int stores its value in a int_fast32_t (typically an int); this is used if the modulus is less than $\sqrt{2^{31}-1}$ .
IntegerMod_int64 stores its value in a int_fast64_t (typically a long long); this is used if the modulus is less than $2^{31}-1$ .
IntegerMod_gmp stores its value in a mpz_t; this can be used for an arbitrarily large modulus.

All extend IntegerMod_abstract.

For efficiency reasons, it stores the modulus (in all three forms, if possible) in a common (cdef) class NativeIntStruct rather than in the parent.

AUTHORS:

Robert Bradshaw: most of the work
Didier Deshommes: bit shifting
William Stein: editing and polishing; new arith architecture
Robert Bradshaw: implement native is_square and square_root
William Stein: sqrt
Maarten Derickx: moved the valuation code from the global valuation function to here

TESTS:

sage: R = Integers(101^3)
sage: a = R(824362); b = R(205942)
sage: a * b
851127

sage: type(IntegerModRing(2^31-1).an_element())
<type 'sage.rings.finite_rings.integer_mod.IntegerMod_int64'>
sage: type(IntegerModRing(2^31).an_element())
<type 'sage.rings.finite_rings.integer_mod.IntegerMod_gmp'>

class sage.rings.finite_rings.integer_mod.Int_to_IntegerMod¶

Bases: sage.rings.finite_rings.integer_mod.IntegerMod_hom

EXAMPLES:

We make sure it works for every type.

sage: from sage.rings.finite_rings.integer_mod import Int_to_IntegerMod
sage: Rs = [Integers(2**k) for k in range(1,50,10)]
sage: [type(R(0)) for R in Rs]
[<type 'sage.rings.finite_rings.integer_mod.IntegerMod_int'>, <type 'sage.rings.finite_rings.integer_mod.IntegerMod_int'>, <type 'sage.rings.finite_rings.integer_mod.IntegerMod_int64'>, <type 'sage.rings.finite_rings.integer_mod.IntegerMod_gmp'>, <type 'sage.rings.finite_rings.integer_mod.IntegerMod_gmp'>]
sage: fs = [Int_to_IntegerMod(R) for R in Rs]
sage: [f(-1) for f in fs]
[1, 2047, 2097151, 2147483647, 2199023255551]

sage.rings.finite_rings.integer_mod.IntegerMod(parent, value)¶

Create an integer modulo $n$ with the given parent.

This is mainly for internal use.

class sage.rings.finite_rings.integer_mod.IntegerMod_abstract¶

Bases: sage.rings.finite_rings.element_base.FiniteRingElement

EXAMPLES:

sage: a = Mod(10,30^10); a
10
sage: loads(a.dumps()) == a
True

additive_order()¶

Returns the additive order of self.

This is the same as self.order().

EXAMPLES:

sage: Integers(20)(2).additive_order()
10
sage: Integers(20)(7).additive_order()
20
sage: Integers(90308402384902)(2).additive_order()
45154201192451

centerlift(*args, **kwds)¶: Deprecated: Use lift_centered() instead. See trac ticket #15804 for details.

charpoly(var='x')¶

Returns the characteristic polynomial of this element.

EXAMPLES:

sage: k = GF(3)
sage: a = k.gen()
sage: a.charpoly('x')
x + 2
sage: a + 2
0

AUTHORS:

Craig Citro

crt(other)¶

Use the Chinese Remainder Theorem to find an element of the integers modulo the product of the moduli that reduces to self and to other. The modulus of other must be coprime to the modulus of self.

EXAMPLES:

sage: a = mod(3,5)
sage: b = mod(2,7)
sage: a.crt(b)
23

sage: a = mod(37,10^8)
sage: b = mod(9,3^8)
sage: a.crt(b)
125900000037

sage: b = mod(0,1)
sage: a.crt(b) == a
True
sage: a.crt(b).modulus()
100000000

TESTS:

sage: mod(0,1).crt(mod(4,2^127))
4
sage: mod(4,2^127).crt(mod(0,1))
4
sage: mod(4,2^30).crt(mod(0,1))
4
sage: mod(0,1).crt(mod(4,2^30))
4
sage: mod(0,1).crt(mod(4,2^15))
4
sage: mod(4,2^15).crt(mod(0,1))
4

AUTHORS:

Robert Bradshaw

generalised_log()¶

Return integers $[n_1, \ldots, n_d]$ such that

..math:

\prod_{i=1}^d x_i^{n_i} = \text{self},

where $x_1, \dots, x_d$ are the generators of the unit group returned by self.parent().unit_gens().

EXAMPLES:

sage: m = Mod(3, 1568)
sage: v = m.generalised_log(); v
[1, 3, 1]
sage: prod([Zmod(1568).unit_gens()[i] ** v[i] for i in [0..2]])
3

See also

The method log().

Warning

The output is given relative to the set of generators obtained by passing algorithm='sage' to the method unit_gens() of the parent (which is the default). Specifying algorithm='pari' usually yields a different set of generators that is incompatible with this method.

is_nilpotent()¶

Return True if self is nilpotent, i.e., some power of self is zero.

EXAMPLES:

sage: a = Integers(90384098234^3)
sage: factor(a.order())
2^3 * 191^3 * 236607587^3
sage: b = a(2*191)
sage: b.is_nilpotent()
False
sage: b = a(2*191*236607587)
sage: b.is_nilpotent()
True

ALGORITHM: Let $m \geq \log_2(n)$ , where $n$ is the modulus. Then $x \in \ZZ/n\ZZ$ is nilpotent if and only if $x^m = 0$ .

PROOF: This is clear if you reduce to the prime power case, which you can do via the Chinese Remainder Theorem.

We could alternatively factor $n$ and check to see if the prime divisors of $n$ all divide $x$ . This is asymptotically slower :-).

is_one()¶

is_primitive_root()¶

Determines whether this element generates the group of units modulo n.

This is only possible if the group of units is cyclic, which occurs if n is 2, 4, a power of an odd prime or twice a power of an odd prime.

EXAMPLES:

sage: mod(1,2).is_primitive_root()
True
sage: mod(3,4).is_primitive_root()
True
sage: mod(2,7).is_primitive_root()
False
sage: mod(3,98).is_primitive_root()
True
sage: mod(11,1009^2).is_primitive_root()
True

TESTS:

sage: for p in prime_range(3,12):
....:     for k in range(1,4):
....:         for even in [1,2]:
....:             n = even*p^k
....:             phin = euler_phi(n)
....:             for _ in range(6):
....:                 a = Zmod(n).random_element()
....:                 if not a.is_unit(): continue
....:                 if a.is_primitive_root().__xor__(a.multiplicative_order()==phin):
....:                     print("mod(%s,%s) incorrect" % (a, n))

is_square()¶

EXAMPLES:

sage: Mod(3,17).is_square()
False
sage: Mod(9,17).is_square()
True
sage: Mod(9,17*19^2).is_square()
True
sage: Mod(-1,17^30).is_square()
True
sage: Mod(1/9, next_prime(2^40)).is_square()
True
sage: Mod(1/25, next_prime(2^90)).is_square()
True

TESTS:

sage: Mod(1/25, 2^8).is_square()
True
sage: Mod(1/25, 2^40).is_square()
True

sage: for p,q,r in cartesian_product_iterator([[3,5],[11,13],[17,19]]): # long time
....:     for ep,eq,er in cartesian_product_iterator([[0,1,2,3],[0,1,2,3],[0,1,2,3]]):
....:         for e2 in [0, 1, 2, 3, 4]:
....:             n = p^ep * q^eq * r^er * 2^e2
....:             for _ in range(2):
....:                 a = Zmod(n).random_element()
....:                 if a.is_square().__xor__(a._pari_().issquare()):
....:                     print(a, n)

ALGORITHM: Calculate the Jacobi symbol $(\mathtt{self}/p)$ at each prime $p$ dividing $n$ . It must be 1 or 0 for each prime, and if it is 0 mod $p$ , where $p^k || n$ , then $ord_p(\mathtt{self})$ must be even or greater than $k$ .

The case $p = 2$ is handled separately.

AUTHORS:

Robert Bradshaw

is_unit()¶

lift_centered()¶

Lift self to a centered congruent integer.

OUTPUT:

The unique integer $i$ such that $-n/2 < i \leq n/2$ and $i = self \mod n$ (where $n$ denotes the modulus).

EXAMPLES:

sage: Mod(0,5).lift_centered()
0
sage: Mod(1,5).lift_centered()
1
sage: Mod(2,5).lift_centered()
2
sage: Mod(3,5).lift_centered()
-2
sage: Mod(4,5).lift_centered()
-1
sage: Mod(50,100).lift_centered()
50
sage: Mod(51,100).lift_centered()
-49
sage: Mod(-1,3^100).lift_centered()
-1

log(b=None)¶

Return an integer $x$ such that $b^x = a$ , where $a$ is self.

INPUT:

self - unit modulo $n$
b - a unit modulo $n$ . If b is not given, R.multiplicative_generator() is used, where R is the parent of self.

OUTPUT: Integer $x$ such that $b^x = a$ , if this exists; a ValueError otherwise.

Note

If the modulus is prime and b is a generator, this calls Pari’s znlog function, which is rather fast. If not, it falls back on the generic discrete log implementation in sage.groups.generic.discrete_log().

EXAMPLES:

sage: r = Integers(125)
sage: b = r.multiplicative_generator()^3
sage: a = b^17
sage: a.log(b)
17
sage: a.log()
51

A bigger example:

sage: FF = FiniteField(2^32+61)
sage: c = FF(4294967356)
sage: x = FF(2)
sage: a = c.log(x)
sage: a
2147483678
sage: x^a
4294967356

Things that can go wrong. E.g., if the base is not a generator for the multiplicative group, or not even a unit.

sage: Mod(3, 7).log(Mod(2, 7))
Traceback (most recent call last):
...
ValueError: No discrete log of 3 found to base 2
sage: a = Mod(16, 100); b = Mod(4,100)
sage: a.log(b)
Traceback (most recent call last):
...
ZeroDivisionError: Inverse does not exist.

We check that trac ticket #9205 is fixed:

sage: Mod(5,9).log(Mod(2, 9))
5

We test against a bug (side effect on PARI) fixed in trac ticket #9438:

sage: R.<a, b> = QQ[]
sage: pari(b)
b
sage: GF(7)(5).log()
5
sage: pari(b)
b

AUTHORS:

David Joyner and William Stein (2005-11)
William Stein (2007-01-27): update to use PARI as requested by David Kohel.
Simon King (2010-07-07): fix a side effect on PARI

minimal_polynomial(var='x')¶

Returns the minimal polynomial of this element.

EXAMPLES:: sage: GF(241, ‘a’)(1).minimal_polynomial(var = ‘z’) z + 240

minpoly(var='x')¶

Returns the minimal polynomial of this element.

EXAMPLES:: sage: GF(241, ‘a’)(1).minpoly() x + 240

modulus()¶

EXAMPLES:

sage: Mod(3,17).modulus()
17

multiplicative_order()¶

Returns the multiplicative order of self.

EXAMPLES:

sage: Mod(-1,5).multiplicative_order()
2
sage: Mod(1,5).multiplicative_order()
1
sage: Mod(0,5).multiplicative_order()
Traceback (most recent call last):
...
ArithmeticError: multiplicative order of 0 not defined since it is not a unit modulo 5

norm()¶

Returns the norm of this element, which is itself. (This is here for compatibility with higher order finite fields.)

EXAMPLES:

sage: k = GF(691)
sage: a = k(389)
sage: a.norm()
389

AUTHORS:

Craig Citro

nth_root(n, extend=False, all=False, algorithm=None, cunningham=False)¶

Returns an $n$ th root of self.

INPUT:

n - integer $\geq 1$
extend - bool (default: True); if True, return an nth root in an extension ring, if necessary. Otherwise, raise a ValueError if the root is not in the base ring. Warning: this option is not implemented!
all - bool (default: False); if True, return all $n$ th roots of self, instead of just one.
algorithm - string (default: None); The algorithm for the prime modulus case. CRT and p-adic log techniques are used to reduce to this case. ‘Johnston’ is the only currently supported option.
cunningham - bool (default: False); In some cases, factorization of n is computed. If cunningham is set to True, the factorization of n is computed using trial division for all primes in the so called Cunningham table. Refer to sage.rings.factorint.factor_cunningham for more information. You need to install an optional package to use this method, this can be done with the following command line sage -i cunningham_tables

OUTPUT:

If self has an $n$ th root, returns one (if all is False) or a list of all of them (if all is True). Otherwise, raises a ValueError (if extend is False) or a NotImplementedError (if extend is True).

Warning

The ‘extend’ option is not implemented (yet).

NOTES:

If :
- if all=True:
  - if self=1: all nonzero elements of the parent are returned in a list. Note that this could be very expensive for large parents.
  - otherwise: an empty list is returned
- if all=False:
  - if self=1: self is returned
  - otherwise; a ValueError is raised
If :
- if self is invertible, the $(-n)$ th root of the inverse of self is returned
- otherwise a ValueError is raised or empty list returned.

EXAMPLES:

sage: K = GF(31)
sage: a = K(22)
sage: K(22).nth_root(7)
13
sage: K(25).nth_root(5)
5
sage: K(23).nth_root(3)
29
sage: mod(225,2^5*3^2).nth_root(4, all=True)
[225, 129, 33, 63, 255, 159, 9, 201, 105, 279, 183, 87, 81, 273, 177, 207, 111, 15, 153, 57, 249, 135, 39, 231]
sage: mod(275,2^5*7^4).nth_root(7, all=True)
[58235, 25307, 69211, 36283, 3355, 47259, 14331]
sage: mod(1,8).nth_root(2,all=True)
[1, 7, 5, 3]
sage: mod(4,8).nth_root(2,all=True)
[2, 6]
sage: mod(1,16).nth_root(4,all=True)
[1, 15, 13, 3, 9, 7, 5, 11]
sage: (mod(22,31)^200).nth_root(200)
5
sage: mod(3,6).nth_root(0,all=True)
[]
sage: mod(3,6).nth_root(0)
Traceback (most recent call last):
...
ValueError
sage: mod(1,6).nth_root(0,all=True)
[1, 2, 3, 4, 5]

TESTS:

sage: for p in [1009,2003,10007,100003]:
...       K = GF(p)
...       for r in (p-1).divisors():
...           if r == 1: continue
...           x = K.random_element()
...           y = x^r
...           if y.nth_root(r)**r != y: raise RuntimeError
...           if (y^41).nth_root(41*r)**(41*r) != y^41: raise RuntimeError
...           if (y^307).nth_root(307*r)**(307*r) != y^307: raise RuntimeError

sage: for t in xrange(200):
....:     n = randint(1,2^63)
....:     K = Integers(n)
....:     b = K.random_element()
....:     e = randint(-2^62, 2^63)
....:     try:
....:         a = b.nth_root(e)
....:         if a^e != b:
....:             print(n, b, e, a)
....:             raise NotImplementedError
....:     except ValueError:
....:         pass

We check that trac ticket #13172 is resolved:

sage: mod(-1, 4489).nth_root(2, all=True)
[]

Check that the code path cunningham might be used:

sage: a = Mod(9,11)
sage: a.nth_root(2, False, True, 'Johnston', cunningham = True) # optional - cunningham
[3, 8]

ALGORITHMS:

The default for prime modulus is currently an algorithm described in the following paper:

Johnston, Anna M. A generalized qth root algorithm. Proceedings of the tenth annual ACM-SIAM symposium on Discrete algorithms. Baltimore, 1999: pp 929-930.

AUTHORS:

David Roe (2010-2-13)

polynomial(var='x')¶

Returns a constant polynomial representing this value.

EXAMPLES:

sage: k = GF(7)
sage: a = k.gen(); a
1
sage: a.polynomial()
1
sage: type(a.polynomial())
<type 'sage.rings.polynomial.polynomial_zmod_flint.Polynomial_zmod_flint'>

rational_reconstruction()¶

Use rational reconstruction to try to find a lift of this element to the rational numbers.

EXAMPLES:

sage: R = IntegerModRing(97)
sage: a = R(2) / R(3)
sage: a
33
sage: a.rational_reconstruction()
2/3

This method is also inherited by prime finite fields elements:

sage: k = GF(97)
sage: a = k(RationalField()('2/3'))
sage: a
33
sage: a.rational_reconstruction()
2/3

sqrt(extend=True, all=False)¶

Returns square root or square roots of self modulo $n$ .

INPUT:

extend - bool (default: True); if True, return a square root in an extension ring, if necessary. Otherwise, raise a ValueError if the square root is not in the base ring.
all - bool (default: False); if True, return {all} square roots of self, instead of just one.

ALGORITHM: Calculates the square roots mod $p$ for each of the primes $p$ dividing the order of the ring, then lifts them $p$ -adically and uses the CRT to find a square root mod $n$ .

See also square_root_mod_prime_power and square_root_mod_prime (in this module) for more algorithmic details.

EXAMPLES:

sage: mod(-1, 17).sqrt()
4
sage: mod(5, 389).sqrt()
86
sage: mod(7, 18).sqrt()
5
sage: a = mod(14, 5^60).sqrt()
sage: a*a
14
sage: mod(15, 389).sqrt(extend=False)
Traceback (most recent call last):
...
ValueError: self must be a square
sage: Mod(1/9, next_prime(2^40)).sqrt()^(-2)
9
sage: Mod(1/25, next_prime(2^90)).sqrt()^(-2)
25

sage: a = Mod(3,5); a
3
sage: x = Mod(-1, 360)
sage: x.sqrt(extend=False)
Traceback (most recent call last):
...
ValueError: self must be a square
sage: y = x.sqrt(); y
sqrt359
sage: y.parent()
Univariate Quotient Polynomial Ring in sqrt359 over Ring of integers modulo 360 with modulus x^2 + 1
sage: y^2
359

We compute all square roots in several cases:

sage: R = Integers(5*2^3*3^2); R
Ring of integers modulo 360
sage: R(40).sqrt(all=True)
[20, 160, 200, 340]
sage: [x for x in R if x^2 == 40]  # Brute force verification
[20, 160, 200, 340]
sage: R(1).sqrt(all=True)
[1, 19, 71, 89, 91, 109, 161, 179, 181, 199, 251, 269, 271, 289, 341, 359]
sage: R(0).sqrt(all=True)
[0, 60, 120, 180, 240, 300]

sage: R = Integers(5*13^3*37); R
Ring of integers modulo 406445
sage: v = R(-1).sqrt(all=True); v
[78853, 111808, 160142, 193097, 213348, 246303, 294637, 327592]
sage: [x^2 for x in v]
[406444, 406444, 406444, 406444, 406444, 406444, 406444, 406444]
sage: v = R(169).sqrt(all=True); min(v), -max(v), len(v)
(13, 13, 104)
sage: all([x^2==169 for x in v])
True

sage: t = FiniteField(next_prime(2^100))(4)
sage: t.sqrt(extend = False, all = True)
[2, 1267650600228229401496703205651]
sage: t = FiniteField(next_prime(2^100))(2)
sage: t.sqrt(extend = False, all = True)
[]

Modulo a power of 2:

sage: R = Integers(2^7); R
Ring of integers modulo 128
sage: a = R(17)
sage: a.sqrt()
23
sage: a.sqrt(all=True)
[23, 41, 87, 105]
sage: [x for x in R if x^2==17]
[23, 41, 87, 105]

square_root(extend=True, all=False)¶

Returns square root or square roots of self modulo $n$ .

INPUT:

extend - bool (default: True); if True, return a square root in an extension ring, if necessary. Otherwise, raise a ValueError if the square root is not in the base ring.
all - bool (default: False); if True, return {all} square roots of self, instead of just one.

ALGORITHM: Calculates the square roots mod $p$ for each of the primes $p$ dividing the order of the ring, then lifts them $p$ -adically and uses the CRT to find a square root mod $n$ .

See also square_root_mod_prime_power and square_root_mod_prime (in this module) for more algorithmic details.

EXAMPLES:

sage: mod(-1, 17).sqrt()
4
sage: mod(5, 389).sqrt()
86
sage: mod(7, 18).sqrt()
5
sage: a = mod(14, 5^60).sqrt()
sage: a*a
14
sage: mod(15, 389).sqrt(extend=False)
Traceback (most recent call last):
...
ValueError: self must be a square
sage: Mod(1/9, next_prime(2^40)).sqrt()^(-2)
9
sage: Mod(1/25, next_prime(2^90)).sqrt()^(-2)
25

sage: a = Mod(3,5); a
3
sage: x = Mod(-1, 360)
sage: x.sqrt(extend=False)
Traceback (most recent call last):
...
ValueError: self must be a square
sage: y = x.sqrt(); y
sqrt359
sage: y.parent()
Univariate Quotient Polynomial Ring in sqrt359 over Ring of integers modulo 360 with modulus x^2 + 1
sage: y^2
359

We compute all square roots in several cases:

sage: R = Integers(5*2^3*3^2); R
Ring of integers modulo 360
sage: R(40).sqrt(all=True)
[20, 160, 200, 340]
sage: [x for x in R if x^2 == 40]  # Brute force verification
[20, 160, 200, 340]
sage: R(1).sqrt(all=True)
[1, 19, 71, 89, 91, 109, 161, 179, 181, 199, 251, 269, 271, 289, 341, 359]
sage: R(0).sqrt(all=True)
[0, 60, 120, 180, 240, 300]

sage: R = Integers(5*13^3*37); R
Ring of integers modulo 406445
sage: v = R(-1).sqrt(all=True); v
[78853, 111808, 160142, 193097, 213348, 246303, 294637, 327592]
sage: [x^2 for x in v]
[406444, 406444, 406444, 406444, 406444, 406444, 406444, 406444]
sage: v = R(169).sqrt(all=True); min(v), -max(v), len(v)
(13, 13, 104)
sage: all([x^2==169 for x in v])
True

sage: t = FiniteField(next_prime(2^100))(4)
sage: t.sqrt(extend = False, all = True)
[2, 1267650600228229401496703205651]
sage: t = FiniteField(next_prime(2^100))(2)
sage: t.sqrt(extend = False, all = True)
[]

Modulo a power of 2:

sage: R = Integers(2^7); R
Ring of integers modulo 128
sage: a = R(17)
sage: a.sqrt()
23
sage: a.sqrt(all=True)
[23, 41, 87, 105]
sage: [x for x in R if x^2==17]
[23, 41, 87, 105]

trace()¶

Returns the trace of this element, which is itself. (This is here for compatibility with higher order finite fields.)

EXAMPLES:

sage: k = GF(691)
sage: a = k(389)
sage: a.trace()
389

AUTHORS:

Craig Citro

valuation(p)¶

The largest power r such that m is in the ideal generated by p^r or infinity if there is not a largest such power. However it is an error to take the valuation with respect to a unit.

Note

This is not a valuation in the mathematical sense. As shown with the examples below.

EXAMPLES:

This example shows that the (a*b).valuation(n) is not always the same as a.valuation(n) + b.valuation(n)

sage: R=ZZ.quo(9)
sage: a=R(3)
sage: b=R(6)
sage: a.valuation(3)
1
sage: a.valuation(3) + b.valuation(3)
2
sage: (a*b).valuation(3)
+Infinity

The valuation with respect to a unit is an error

sage: a.valuation(4)
Traceback (most recent call last):
...
ValueError: Valuation with respect to a unit is not defined.

TESTS:

sage: R=ZZ.quo(12)
sage: a=R(2)
sage: b=R(4)
sage: a.valuation(2)
1
sage: b.valuation(2)
+Infinity
sage: ZZ.quo(1024)(16).valuation(4)
2

class sage.rings.finite_rings.integer_mod.IntegerMod_gmp¶

Bases: sage.rings.finite_rings.integer_mod.IntegerMod_abstract

Elements of $\ZZ/n\ZZ$ for n not small enough to be operated on in word size.

AUTHORS:

Robert Bradshaw (2006-08-24)

gcd(other)¶

Greatest common divisor

Returns the “smallest” generator in $\ZZ / N\ZZ$ of the ideal generated by self and other.

INPUT:

other – an element of the same ring as this one.

EXAMPLES:

sage: mod(2^3*3^2*5, 3^3*2^2*17^8).gcd(mod(2^4*3*17, 3^3*2^2*17^8))
12
sage: mod(0,17^8).gcd(mod(0,17^8))
0

is_one()¶

Returns True if this is $1$ , otherwise False.

EXAMPLES:

sage: mod(1,5^23).is_one()
True
sage: mod(0,5^23).is_one()
False

is_unit()¶

Return True iff this element is a unit.

EXAMPLES:

sage: mod(13, 5^23).is_unit()
True
sage: mod(25, 5^23).is_unit()
False

lift()¶

Lift an integer modulo $n$ to the integers.

EXAMPLES:

sage: a = Mod(8943, 2^70); type(a)
<type 'sage.rings.finite_rings.integer_mod.IntegerMod_gmp'>
sage: lift(a)
8943
sage: a.lift()
8943

class sage.rings.finite_rings.integer_mod.IntegerMod_hom¶: Bases: sage.categories.morphism.Morphism

class sage.rings.finite_rings.integer_mod.IntegerMod_int¶

Bases: sage.rings.finite_rings.integer_mod.IntegerMod_abstract

Elements of $\ZZ/n\ZZ$ for n small enough to be operated on in 32 bits

AUTHORS:

Robert Bradshaw (2006-08-24)

gcd(other)¶

Greatest common divisor

Returns the “smallest” generator in $\ZZ / N\ZZ$ of the ideal generated by self and other.

INPUT:

other – an element of the same ring as this one.

EXAMPLES:

sage: R = Zmod(60); S = Zmod(72)
sage: a = R(40).gcd(S(30)); a
2
sage: a.parent()
Ring of integers modulo 12
sage: b = R(17).gcd(60); b
1
sage: b.parent()
Ring of integers modulo 60

sage: mod(72*5, 3^3*2^2*17^2).gcd(mod(48*17, 3^3*2^2*17^2))
12
sage: mod(0,1).gcd(mod(0,1))
0

is_one()¶

Returns True if this is $1$ , otherwise False.

EXAMPLES:

sage: mod(6,5).is_one()
True
sage: mod(0,5).is_one()
False

is_unit()¶

Return True iff this element is a unit

EXAMPLES:

sage: a=Mod(23,100)
sage: a.is_unit()
True
sage: a=Mod(24,100)
sage: a.is_unit()
False

lift()¶

Lift an integer modulo $n$ to the integers.

EXAMPLES:

sage: a = Mod(8943, 2^10); type(a)
<type 'sage.rings.finite_rings.integer_mod.IntegerMod_int'>
sage: lift(a)
751
sage: a.lift()
751

sqrt(extend=True, all=False)¶

Returns square root or square roots of self modulo $n$ .

INPUT:

extend - bool (default: True); if True, return a square root in an extension ring, if necessary. Otherwise, raise a ValueError if the square root is not in the base ring.
all - bool (default: False); if True, return {all} square roots of self, instead of just one.

ALGORITHM: Calculates the square roots mod $p$ for each of the primes $p$ dividing the order of the ring, then lifts them $p$ -adically and uses the CRT to find a square root mod $n$ .

See also square_root_mod_prime_power and square_root_mod_prime (in this module) for more algorithmic details.

EXAMPLES:

sage: mod(-1, 17).sqrt()
4
sage: mod(5, 389).sqrt()
86
sage: mod(7, 18).sqrt()
5
sage: a = mod(14, 5^60).sqrt()
sage: a*a
14
sage: mod(15, 389).sqrt(extend=False)
Traceback (most recent call last):
...
ValueError: self must be a square
sage: Mod(1/9, next_prime(2^40)).sqrt()^(-2)
9
sage: Mod(1/25, next_prime(2^90)).sqrt()^(-2)
25

sage: a = Mod(3,5); a
3
sage: x = Mod(-1, 360)
sage: x.sqrt(extend=False)
Traceback (most recent call last):
...
ValueError: self must be a square
sage: y = x.sqrt(); y
sqrt359
sage: y.parent()
Univariate Quotient Polynomial Ring in sqrt359 over Ring of integers modulo 360 with modulus x^2 + 1
sage: y^2
359

We compute all square roots in several cases:

sage: R = Integers(5*2^3*3^2); R
Ring of integers modulo 360
sage: R(40).sqrt(all=True)
[20, 160, 200, 340]
sage: [x for x in R if x^2 == 40]  # Brute force verification
[20, 160, 200, 340]
sage: R(1).sqrt(all=True)
[1, 19, 71, 89, 91, 109, 161, 179, 181, 199, 251, 269, 271, 289, 341, 359]
sage: R(0).sqrt(all=True)
[0, 60, 120, 180, 240, 300]
sage: GF(107)(0).sqrt(all=True)
[0]

sage: R = Integers(5*13^3*37); R
Ring of integers modulo 406445
sage: v = R(-1).sqrt(all=True); v
[78853, 111808, 160142, 193097, 213348, 246303, 294637, 327592]
sage: [x^2 for x in v]
[406444, 406444, 406444, 406444, 406444, 406444, 406444, 406444]
sage: v = R(169).sqrt(all=True); min(v), -max(v), len(v)
(13, 13, 104)
sage: all([x^2==169 for x in v])
True

Modulo a power of 2:

sage: R = Integers(2^7); R
Ring of integers modulo 128
sage: a = R(17)
sage: a.sqrt()
23
sage: a.sqrt(all=True)
[23, 41, 87, 105]
sage: [x for x in R if x^2==17]
[23, 41, 87, 105]

class sage.rings.finite_rings.integer_mod.IntegerMod_int64¶

Bases: sage.rings.finite_rings.integer_mod.IntegerMod_abstract

Elements of $\ZZ/n\ZZ$ for n small enough to be operated on in 64 bits

AUTHORS:

Robert Bradshaw (2006-09-14)

gcd(other)¶

Greatest common divisor

Returns the “smallest” generator in $\ZZ / N\ZZ$ of the ideal generated by self and other.

INPUT:

other – an element of the same ring as this one.

EXAMPLES:

sage: mod(2^3*3^2*5, 3^3*2^2*17^5).gcd(mod(2^4*3*17, 3^3*2^2*17^5))
12
sage: mod(0,17^5).gcd(mod(0,17^5))
0

is_one()¶

Returns True if this is $1$ , otherwise False.

EXAMPLES:

sage: (mod(-1,5^10)^2).is_one()
True
sage: mod(0,5^10).is_one()
False

is_unit()¶

Return True iff this element is a unit.

EXAMPLES:

sage: mod(13, 5^10).is_unit()
True
sage: mod(25, 5^10).is_unit()
False

lift()¶

Lift an integer modulo $n$ to the integers.

EXAMPLES:

sage: a = Mod(8943, 2^25); type(a)
<type 'sage.rings.finite_rings.integer_mod.IntegerMod_int64'>
sage: lift(a)
8943
sage: a.lift()
8943

class sage.rings.finite_rings.integer_mod.IntegerMod_to_Integer¶

Bases: sage.categories.map.Map

Map to lift elements to Integer.

EXAMPLES:

sage: ZZ.convert_map_from(GF(2))
Lifting map:
  From: Finite Field of size 2
  To:   Integer Ring

class sage.rings.finite_rings.integer_mod.IntegerMod_to_IntegerMod¶

Bases: sage.rings.finite_rings.integer_mod.IntegerMod_hom

Very fast IntegerMod to IntegerMod homomorphism.

EXAMPLES:

sage: from sage.rings.finite_rings.integer_mod import IntegerMod_to_IntegerMod
sage: Rs = [Integers(3**k) for k in range(1,30,5)]
sage: [type(R(0)) for R in Rs]
[<type 'sage.rings.finite_rings.integer_mod.IntegerMod_int'>, <type 'sage.rings.finite_rings.integer_mod.IntegerMod_int'>, <type 'sage.rings.finite_rings.integer_mod.IntegerMod_int64'>, <type 'sage.rings.finite_rings.integer_mod.IntegerMod_int64'>, <type 'sage.rings.finite_rings.integer_mod.IntegerMod_gmp'>, <type 'sage.rings.finite_rings.integer_mod.IntegerMod_gmp'>]
sage: fs = [IntegerMod_to_IntegerMod(S, R) for R in Rs for S in Rs if S is not R and S.order() > R.order()]
sage: all([f(-1) == f.codomain()(-1) for f in fs])
True
sage: [f(-1) for f in fs]
[2, 2, 2, 2, 2, 728, 728, 728, 728, 177146, 177146, 177146, 43046720, 43046720, 10460353202]

class sage.rings.finite_rings.integer_mod.Integer_to_IntegerMod¶

Bases: sage.rings.finite_rings.integer_mod.IntegerMod_hom

Fast $\ZZ \rightarrow \ZZ/n\ZZ$ morphism.

EXAMPLES:

We make sure it works for every type.

sage: from sage.rings.finite_rings.integer_mod import Integer_to_IntegerMod
sage: Rs = [Integers(10), Integers(10^5), Integers(10^10)]
sage: [type(R(0)) for R in Rs]
[<type 'sage.rings.finite_rings.integer_mod.IntegerMod_int'>, <type 'sage.rings.finite_rings.integer_mod.IntegerMod_int64'>, <type 'sage.rings.finite_rings.integer_mod.IntegerMod_gmp'>]
sage: fs = [Integer_to_IntegerMod(R) for R in Rs]
sage: [f(-1) for f in fs]
[9, 99999, 9999999999]

section()¶

sage.rings.finite_rings.integer_mod.Mod(n, m, parent=None)¶

Return the equivalence class of $n$ modulo $m$ as an element of $\ZZ/m\ZZ$ .

EXAMPLES:

sage: x = Mod(12345678, 32098203845329048)
sage: x
12345678
sage: x^100
1017322209155072

You can also use the lowercase version:

sage: mod(12,5)
2

Illustrates that trac ticket #5971 is fixed. Consider $n$ modulo $m$ when $m = 0$ . Then $\ZZ/0\ZZ$ is isomorphic to $\ZZ$ so $n$ modulo $0$ is equivalent to $n$ for any integer value of $n$ :

sage: Mod(10, 0)
10
sage: a = randint(-100, 100)
sage: Mod(a, 0) == a
True

class sage.rings.finite_rings.integer_mod.NativeIntStruct¶

Bases: object

We store the various forms of the modulus here rather than in the parent for efficiency reasons.

We may also store a cached table of all elements of a given ring in this class.

precompute_table(parent, inverses=True)¶

Function to compute and cache all elements of this class.

If inverses == True, also computes and caches the inverses of the invertible elements.

EXAMPLES:

This is used by the sage.rings.finite_rings.integer_mod_ring.IntegerModRing_generic constructor:

sage: from sage.rings.finite_rings.integer_mod_ring import IntegerModRing_generic
sage: R = IntegerModRing_generic(39, cache=False)
sage: R(5)^-1
8
sage: R(5)^-1 is R(8)
False
sage: R = IntegerModRing_generic(39, cache=True)  # indirect doctest
sage: R(5)^-1 is R(8)
True

Check that the inverse of 0 modulo 1 works, see trac ticket #13639:

sage: R = IntegerModRing_generic(1, cache=True)  # indirect doctest
sage: R(0)^-1 is R(0)
True

sage.rings.finite_rings.integer_mod.is_IntegerMod(x)¶

Return True if and only if x is an integer modulo $n$ .

EXAMPLES:

sage: from sage.rings.finite_rings.integer_mod import is_IntegerMod
sage: is_IntegerMod(5)
False
sage: is_IntegerMod(Mod(5,10))
True

sage.rings.finite_rings.integer_mod.lucas(k, P, Q=1, n=None)¶

Return $[V_k(P, Q) \mod n, Q^{\lfloor k/2 \rfloor} \mod n]$ where $V_k$ is the Lucas function defined by the recursive relation

$V_k(P, Q) = P V_{k-1}(P, Q) - Q V_{k-2}(P, Q)$

with $V_0 = 2, V_1 = P$ .

INPUT:

k – integer; index to compute
P, Q – integers or modular integers; initial values
n – integer (optional); modulus to use if P is not a modular integer

REFERENCES:

[IEEEP1363]

IEEE P1363 / D13 (Draft Version 13). Standard Specifications for Public Key Cryptography Annex A (Informative). Number-Theoretic Background. Section A.2.4

AUTHORS:

Somindu Chaya Ramanna, Shashank Singh and Srinivas Vivek Venkatesh (2011-09-15, ECC2011 summer school)
Robert Bradshaw

TESTS:

sage: from sage.rings.finite_rings.integer_mod import lucas
sage: p = randint(0,100000)
sage: q = randint(0,100000)
sage: n = randint(0,100)
sage: all([lucas(k,p,q,n)[0] == Mod(lucas_number2(k,p,q),n)
...        for k in Integers(20)])
True
sage: from sage.rings.finite_rings.integer_mod import lucas
sage: p = randint(0,100000)
sage: q = randint(0,100000)
sage: n = randint(0,100)
sage: k = randint(0,100)
sage: lucas(k,p,q,n) == [Mod(lucas_number2(k,p,q),n),Mod(q^(int(k/2)),n)]
True

EXAMPLES:

sage: [lucas(k,4,5,11)[0] for k in range(30)]
[2, 4, 6, 4, 8, 1, 8, 5, 2, 5, 10, 4, 10, 9, 8, 9, 7, 5, 7, 3, 10, 3, 6, 9, 6, 1, 7, 1, 2, 3]

sage: lucas(20,4,5,11)
[10, 1]

sage.rings.finite_rings.integer_mod.lucas_q1(mm, P)¶

Return $V_k(P, 1)$ where $V_k$ is the Lucas function defined by the recursive relation

$V_k(P, Q) = PV_{k-1}(P, Q) - QV_{k-2}(P, Q)$

with $V_0 = 2, V_1(P_Q) = P$ .

REFERENCES:

[Pos88]

H. Postl. ‘Fast evaluation of Dickson Polynomials’ Contrib. to General Algebra, Vol. 6 (1988) pp. 223-225

AUTHORS:

Robert Bradshaw

TESTS:

sage: from sage.rings.finite_rings.integer_mod import lucas_q1
sage: all([lucas_q1(k, a) == BinaryRecurrenceSequence(a, -1, 2, a)(k)
....:      for a in Integers(23)
....:      for k in range(13)])
True

sage.rings.finite_rings.integer_mod.makeNativeIntStruct(z)¶: Function to convert a Sage Integer into class NativeIntStruct.

Note

This function is only used for the unpickle override below.

sage.rings.finite_rings.integer_mod.mod(n, m, parent=None)¶

Return the equivalence class of $n$ modulo $m$ as an element of $\ZZ/m\ZZ$ .

EXAMPLES:

sage: x = Mod(12345678, 32098203845329048)
sage: x
12345678
sage: x^100
1017322209155072

You can also use the lowercase version:

sage: mod(12,5)
2

Illustrates that trac ticket #5971 is fixed. Consider $n$ modulo $m$ when $m = 0$ . Then $\ZZ/0\ZZ$ is isomorphic to $\ZZ$ so $n$ modulo $0$ is equivalent to $n$ for any integer value of $n$ :

sage: Mod(10, 0)
10
sage: a = randint(-100, 100)
sage: Mod(a, 0) == a
True

sage.rings.finite_rings.integer_mod.slow_lucas(k, P, Q=1)¶

Lucas function defined using the standard definition, for consistency testing. This is deprecated in trac ticket #11802. Use BinaryRecurrenceSequence(P, -Q, 2, P)(k) instead.

Elements of $\ZZ/n\ZZ$ ¶

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Elements of $\ZZ/n\ZZ$ ¶