Coverage for sympy/solvers/diophantine.py: 89%

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from __future__ import print_function, division

from sympy import (Poly, igcd, divisors, sign, symbols, S, Integer, Wild, Symbol, factorint,

Add, Mul, solve, ceiling, floor, sqrt, sympify, Subs, ilcm, Matrix, factor_list, perfect_power,

isprime, nextprime, integer_nthroot)

from sympy.core.function import _mexpand

from sympy.simplify.simplify import rad_rationalize

from sympy.utilities import default_sort_key, numbered_symbols

from sympy.core.numbers import igcdex

from sympy.ntheory.residue_ntheory import sqrt_mod

from sympy.core.compatibility import range

from sympy.core.relational import Eq

from sympy.solvers.solvers import check_assumptions

__all__ = ['diophantine', 'diop_solve', 'classify_diop', 'diop_linear', 'base_solution_linear',

'diop_quadratic', 'diop_DN', 'cornacchia', 'diop_bf_DN', 'transformation_to_DN', 'find_DN',

'diop_ternary_quadratic', 'square_factor', 'descent', 'diop_general_pythagorean',

'diop_general_sum_of_squares', 'partition', 'sum_of_three_squares', 'sum_of_four_squares']

def diophantine(eq, param=symbols("t", integer=True)):

"""

Simplify the solution procedure of diophantine equation ``eq`` by

converting it into a product of terms which should equal zero.

For example, when solving, `x^2 - y^2 = 0` this is treated as

`(x + y)(x - y) = 0` and `x+y = 0` and `x-y = 0` are solved independently

and combined. Each term is solved by calling ``diop_solve()``.

Output of ``diophantine()`` is a set of tuples. Each tuple represents a

solution of the input equation. In a tuple, solution for each variable is

listed according to the alphabetic order of input variables. i.e. if we have

an equation with two variables `a` and `b`, first element of the tuple will

give the solution for `a` and the second element will give the solution for

`b`.

Usage

=====

``diophantine(eq, t)``: Solve the diophantine equation ``eq``.

``t`` is the parameter to be used by ``diop_solve()``.

Details

=======

``eq`` should be an expression which is assumed to be zero.

``t`` is the parameter to be used in the solution.

Examples

========

>>> from sympy.solvers.diophantine import diophantine

>>> from sympy.abc import x, y, z

>>> diophantine(x**2 - y**2)

set([(-t, -t), (t, -t)])

#>>> diophantine(x*(2*x + 3*y - z))

#set([(0, n1, n2), (3*t - z, -2*t + z, z)])

#>>> diophantine(x**2 + 3*x*y + 4*x)

#set([(0, n1), (3*t - 4, -t)])

See Also

========

diop_solve()

"""

if isinstance(eq, Eq):

eq = eq.lhs - eq.rhs

eq = Poly(eq).as_expr()

if not eq.is_polynomial() or eq.is_number:

raise TypeError("Equation input format not supported")

var = list(eq.expand(force=True).free_symbols)

var.sort(key=default_sort_key)

terms = factor_list(eq)[1]

sols = set([])

for term in terms:

base = term[0]

var_t, jnk, eq_type = classify_diop(base)

solution = diop_solve(base, param)

if eq_type in ["linear", "homogeneous_ternary_quadratic", "general_pythagorean"]:

if merge_solution(var, var_t, solution) != ():

sols.add(merge_solution(var, var_t, solution))

elif eq_type in ["binary_quadratic", "general_sum_of_squares", "univariate"]:

for sol in solution:

if merge_solution(var, var_t, sol) != ():

sols.add(merge_solution(var, var_t, sol))

return sols

def merge_solution(var, var_t, solution):

"""

This is used to construct the full solution from the solutions of sub

equations.

For example when solving the equation `(x - y)(x^2 + y^2 - z^2) = 0`,

solutions for each of the equations `x-y = 0` and `x^2 + y^2 - z^2` are

found independently. Solutions for `x - y = 0` are `(x, y) = (t, t)`. But

we should introduce a value for z when we output the solution for the

original equation. This function converts `(t, t)` into `(t, t, n_{1})`

where `n_{1}` is an integer parameter.

"""

l = []

if None in solution:

return ()

solution = iter(solution)

params = numbered_symbols("n", Integer=True, start=1)

for v in var:

if v in var_t:

l.append(next(solution))

else:

l.append(next(params))

for val, symb in zip(l, var):

if check_assumptions(val, **symb.assumptions0) is False:

return tuple()

return tuple(l)

def diop_solve(eq, param=symbols("t", integer=True)):

"""

Solves the diophantine equation ``eq``.

Similar to ``diophantine()`` but doesn't try to factor ``eq`` as latter

does. Uses ``classify_diop()`` to determine the type of the eqaution and

calls the appropriate solver function.

Usage

=====

``diop_solve(eq, t)``: Solve diophantine equation, ``eq`` using ``t``

as a parameter if needed.

Details

=======

``eq`` should be an expression which is assumed to be zero.

``t`` is a parameter to be used in the solution.

Examples

========

>>> from sympy.solvers.diophantine import diop_solve

>>> from sympy.abc import x, y, z, w

>>> diop_solve(2*x + 3*y - 5)

(3*t - 5, -2*t + 5)

>>> diop_solve(4*x + 3*y -4*z + 5)

(3*t + 4*z - 5, -4*t - 4*z + 5, z)

>>> diop_solve(x + 3*y - 4*z + w -6)

(t, -t - 3*y + 4*z + 6, y, z)

>>> diop_solve(x**2 + y**2 - 5)

set([(-2, -1), (-2, 1), (2, -1), (2, 1)])

See Also

========

diophantine()

"""

var, coeff, eq_type = classify_diop(eq)

if eq_type == "linear":

return _diop_linear(var, coeff, param)

elif eq_type == "binary_quadratic":

return _diop_quadratic(var, coeff, param)

elif eq_type == "homogeneous_ternary_quadratic":

x_0, y_0, z_0 = _diop_ternary_quadratic(var, coeff)

return _parametrize_ternary_quadratic((x_0, y_0, z_0), var, coeff)

elif eq_type == "general_pythagorean":

return _diop_general_pythagorean(var, coeff, param)

elif eq_type == "univariate":

l = solve(eq)

s = set([])

for soln in l:

if isinstance(soln, Integer):

s.add((soln,))

return s

elif eq_type == "general_sum_of_squares":

return _diop_general_sum_of_squares(var, coeff)

def classify_diop(eq):

"""

Helper routine used by diop_solve() to find the type of the ``eq`` etc.

Returns a tuple containing the type of the diophantine equation along with

the variables(free symbols) and their coefficients. Variables are returned

as a list and coefficients are returned as a dict with the key being the

respective term and the constant term is keyed to Integer(1). Type is an

element in the set {"linear", "binary_quadratic", "general_pythagorean",

"homogeneous_ternary_quadratic", "univariate", "general_sum_of_squares"}

Usage

=====

``classify_diop(eq)``: Return variables, coefficients and type of the

``eq``.

Details

=======

``eq`` should be an expression which is assumed to be zero.

Examples

========

>>> from sympy.solvers.diophantine import classify_diop

>>> from sympy.abc import x, y, z, w, t

>>> classify_diop(4*x + 6*y - 4)

([x, y], {1: -4, x: 4, y: 6}, 'linear')

>>> classify_diop(x + 3*y -4*z + 5)

([x, y, z], {1: 5, x: 1, y: 3, z: -4}, 'linear')

>>> classify_diop(x**2 + y**2 - x*y + x + 5)

([x, y], {1: 5, x: 1, x**2: 1, y: 0, y**2: 1, x*y: -1}, 'binary_quadratic')

"""

eq = eq.expand(force=True)

var = list(eq.free_symbols)

var.sort(key=default_sort_key)

coeff = {}

diop_type = None

coeff = dict([reversed(t.as_independent(*var)) for t in eq.args])

for v in coeff:

if not isinstance(coeff[v], Integer):

raise TypeError("Coefficients should be Integers")

if len(var) == 1:

diop_type = "univariate"

elif Poly(eq).total_degree() == 1:

diop_type = "linear"

elif Poly(eq).total_degree() == 2 and len(var) == 2:

diop_type = "binary_quadratic"

x, y = var[:2]

if isinstance(eq, Mul):

coeff = {x**2: 0, x*y: eq.args[0], y**2: 0, x: 0, y: 0, Integer(1): 0}

else:

for term in [x**2, y**2, x*y, x, y, Integer(1)]:

if term not in coeff.keys():

coeff[term] = Integer(0)

elif Poly(eq).total_degree() == 2 and len(var) == 3 and Integer(1) not in coeff.keys():

for v in var:

if v in coeff.keys():

diop_type = "inhomogeneous_ternary_quadratic"

break

else:

diop_type = "homogeneous_ternary_quadratic"

x, y, z = var[:3]

for term in [x**2, y**2, z**2, x*y, y*z, x*z]:

if term not in coeff.keys():

coeff[term] = Integer(0)

elif Poly(eq).degree() == 2 and len(var) >= 3:

for v in var:

if v in coeff.keys():

diop_type = "inhomogeneous_general_quadratic"

break

else:

if Integer(1) in coeff.keys():

constant_term = True

else:

constant_term = False

non_square_degree_2_terms = False

for v in var:

for u in var:

if u != v and u*v in coeff.keys():

non_square_degree_2_terms = True

break

if non_square_degree_2_terms:

break

if constant_term and non_square_degree_2_terms:

diop_type = "inhomogeneous_general_quadratic"

elif constant_term and not non_square_degree_2_terms:

for v in var:

if coeff[v**2] != 1:

break

else:

diop_type = "general_sum_of_squares"

elif not constant_term and non_square_degree_2_terms:

diop_type = "homogeneous_general_quadratic"

else:

coeff_sign_sum = 0

for v in var:

if not isinstance(sqrt(abs(Integer(coeff[v**2]))), Integer):

break

coeff_sign_sum = coeff_sign_sum + sign(coeff[v**2])

else:

if abs(coeff_sign_sum) == len(var) - 2 and not constant_term:

diop_type = "general_pythagorean"

elif Poly(eq).total_degree() == 3 and len(var) == 2:

x, y = var[:2]

diop_type = "cubic_thue"

for term in [x**3, x**2*y, x*y**2, y**3, Integer(1)]:

if term not in coeff.keys():

coeff[term] == Integer(0)

if diop_type is not None:

return var, coeff, diop_type

else:

raise NotImplementedError("Still not implemented")

def diop_linear(eq, param=symbols("t", integer=True)):

"""

Solves linear diophantine equations.

A linear diophantine equation is an equation of the form `a_{1}x_{1} +

a_{2}x_{2} + .. + a_{n}x_{n} = 0` where `a_{1}, a_{2}, ..a_{n}` are

integer constants and `x_{1}, x_{2}, ..x_{n}` are integer variables.

Usage

=====

``diop_linear(eq)``: Returns a tuple containing solutions to the

diophantine equation ``eq``. Values in the tuple is arranged in the same

order as the sorted variables.

Details

=======

``eq`` is a linear diophantine equation which is assumed to be zero.

``param`` is the parameter to be used in the solution.

Examples

========

>>> from sympy.solvers.diophantine import diop_linear

>>> from sympy.abc import x, y, z, t

>>> from sympy import Integer

>>> diop_linear(2*x - 3*y - 5) #solves equation 2*x - 3*y -5 = 0

(-3*t - 5, -2*t - 5)

Here x = -3*t - 5 and y = -2*t - 5

>>> diop_linear(2*x - 3*y - 4*z -3)

(-3*t - 4*z - 3, -2*t - 4*z - 3, z)

See Also

========

diop_quadratic(), diop_ternary_quadratic(), diop_general_pythagorean(),

diop_general_sum_of_squares()

"""

var, coeff, diop_type = classify_diop(eq)

if diop_type == "linear":

return _diop_linear(var, coeff, param)

def _diop_linear(var, coeff, param):

x, y = var[:2]

a = coeff[x]

b = coeff[y]

if len(var) == len(coeff):

c = 0

else:

c = -coeff[Integer(1)]

if len(var) == 2:

sol_x, sol_y = base_solution_linear(c, a, b, param)

return (sol_x, sol_y)

elif len(var) > 2:

X = []

Y = []

for v in var[2:]:

sol_x, sol_y = base_solution_linear(-coeff[v], a, b)

X.append(sol_x*v)

Y.append(sol_y*v)

sol_x, sol_y = base_solution_linear(c, a, b, param)

X.append(sol_x)

Y.append(sol_y)

l = []

if None not in X and None not in Y:

l.append(Add(*X))

l.append(Add(*Y))

for v in var[2:]:

l.append(v)

else:

for v in var:

l.append(None)

return tuple(l)

def base_solution_linear(c, a, b, t=None):

"""

Return the base solution for a linear diophantine equation with two

variables.

Used by ``diop_linear()`` to find the base solution of a linear

Diophantine equation. If ``t`` is given then the parametrized solution is

returned.

Usage

=====

``base_solution_linear(c, a, b, t)``: ``a``, ``b``, ``c`` are coefficients

in `ax + by = c` and ``t`` is the parameter to be used in the solution.

Examples

========

>>> from sympy.solvers.diophantine import base_solution_linear

>>> from sympy.abc import t

>>> base_solution_linear(5, 2, 3) # equation 2*x + 3*y = 5

(-5, 5)

>>> base_solution_linear(0, 5, 7) # equation 5*x + 7*y = 0

(0, 0)

>>> base_solution_linear(5, 2, 3, t) # equation 2*x + 3*y = 5

(3*t - 5, -2*t + 5)

>>> base_solution_linear(0, 5, 7, t) # equation 5*x + 7*y = 0

(7*t, -5*t)

"""

d = igcd(a, igcd(b, c))

a = a // d

b = b // d

c = c // d

if c == 0:

if t != None:

return (b*t , -a*t)

else:

return (S.Zero, S.Zero)

else:

x0, y0, d = extended_euclid(int(abs(a)), int(abs(b)))

x0 = x0 * sign(a)

y0 = y0 * sign(b)

if divisible(c, d):

if t != None:

return (c*x0 + b*t, c*y0 - a*t)

else:

return (Integer(c*x0), Integer(c*y0))

else:

return (None, None)

def extended_euclid(a, b):

"""

For given ``a``, ``b`` returns a tuple containing integers `x`, `y` and `d`

such that `ax + by = d`. Here `d = gcd(a, b)`.

Usage

=====

``extended_euclid(a, b)``: returns `x`, `y` and `\gcd(a, b)`.

Details

=======

``a`` Any instance of Integer.

``b`` Any instance of Integer.

Examples

========

>>> from sympy.solvers.diophantine import extended_euclid

>>> extended_euclid(4, 6)

(-1, 1, 2)

>>> extended_euclid(3, 5)

(2, -1, 1)

"""

if b == 0:

return (1, 0, a)

x0, y0, d = extended_euclid(b, a%b)

x, y = y0, x0 - (a//b) * y0

return x, y, d

def divisible(a, b):

"""

Returns `True` if ``a`` is divisible by ``b`` and `False` otherwise.

"""

return igcd(int(a), int(b)) == abs(int(b))

def diop_quadratic(eq, param=symbols("t", integer=True)):

"""

Solves quadratic diophantine equations.

i.e. equations of the form `Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0`. Returns a

set containing the tuples `(x, y)` which contains the solutions. If there

are no solutions then `(None, None)` is returned.

Usage

=====

``diop_quadratic(eq, param)``: ``eq`` is a quadratic binary diophantine

equation. ``param`` is used to indicate the parameter to be used in the

solution.

Details

=======

``eq`` should be an expression which is assumed to be zero.

``param`` is a parameter to be used in the solution.

Examples

========

>>> from sympy.abc import x, y, t

>>> from sympy.solvers.diophantine import diop_quadratic

>>> diop_quadratic(x**2 + y**2 + 2*x + 2*y + 2, t)

set([(-1, -1)])

References

==========

.. [1] Methods to solve Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0,[online],

Available: http://www.alpertron.com.ar/METHODS.HTM

.. [2] Solving the equation ax^2+ bxy + cy^2 + dx + ey + f= 0, [online],

Available: http://www.jpr2718.org/ax2p.pdf

See Also

========

diop_linear(), diop_ternary_quadratic(), diop_general_sum_of_squares(),

diop_general_pythagorean()

"""

var, coeff, diop_type = classify_diop(eq)

if diop_type == "binary_quadratic":

return _diop_quadratic(var, coeff, param)

def _diop_quadratic(var, coeff, t):

x, y = var[:2]

for term in [x**2, y**2, x*y, x, y, Integer(1)]:

if term not in coeff.keys():

coeff[term] = Integer(0)

A = coeff[x**2]

B = coeff[x*y]

C = coeff[y**2]

D = coeff[x]

E = coeff[y]

F = coeff[Integer(1)]

d = igcd(A, igcd(B, igcd(C, igcd(D, igcd(E, F)))))

A = A // d

B = B // d

C = C // d

D = D // d

E = E // d

F = F // d

# (1) Linear case: A = B = C = 0 ==> considered under linear diophantine equations

# (2) Simple-Hyperbolic case:A = C = 0, B != 0

# In this case equation can be converted to (Bx + E)(By + D) = DE - BF

# We consider two cases; DE - BF = 0 and DE - BF != 0

# More details, http://www.alpertron.com.ar/METHODS.HTM#SHyperb

l = set([])

if A == 0 and C == 0 and B != 0:

if D*E - B*F == 0:

if divisible(int(E), int(B)):

l.add((-E/B, t))

if divisible(int(D), int(B)):

l.add((t, -D/B))

else:

div = divisors(D*E - B*F)

div = div + [-term for term in div]

for d in div:

if divisible(int(d - E), int(B)):

x0 = (d - E) // B

if divisible(int(D*E - B*F), int(d)):

if divisible(int((D*E - B*F)// d - D), int(B)):

y0 = ((D*E - B*F) // d - D) // B

l.add((x0, y0))

# (3) Parabolic case: B**2 - 4*A*C = 0

# There are two subcases to be considered in this case.

# sqrt(c)D - sqrt(a)E = 0 and sqrt(c)D - sqrt(a)E != 0

# More Details, http://www.alpertron.com.ar/METHODS.HTM#Parabol

elif B**2 - 4*A*C == 0:

if A == 0:

s = _diop_quadratic([y, x], coeff, t)

for soln in s:

l.add((soln[1], soln[0]))

else:

g = igcd(A, C)

g = abs(g) * sign(A)

a = A // g

b = B // g

c = C // g

e = sign(B/A)

if e*sqrt(c)*D - sqrt(a)*E == 0:

z = symbols("z", real=True)

roots = solve(sqrt(a)*g*z**2 + D*z + sqrt(a)*F)

for root in roots:

if isinstance(root, Integer):

l.add((diop_solve(sqrt(a)*x + e*sqrt(c)*y - root)[0], diop_solve(sqrt(a)*x + e*sqrt(c)*y - root)[1]))

elif isinstance(e*sqrt(c)*D - sqrt(a)*E, Integer):

solve_x = lambda u: e*sqrt(c)*g*(sqrt(a)*E - e*sqrt(c)*D)*t**2 - (E + 2*e*sqrt(c)*g*u)*t\

- (e*sqrt(c)*g*u**2 + E*u + e*sqrt(c)*F) // (e*sqrt(c)*D - sqrt(a)*E)

solve_y = lambda u: sqrt(a)*g*(e*sqrt(c)*D - sqrt(a)*E)*t**2 + (D + 2*sqrt(a)*g*u)*t \

+ (sqrt(a)*g*u**2 + D*u + sqrt(a)*F) // (e*sqrt(c)*D - sqrt(a)*E)

for z0 in range(0, abs(e*sqrt(c)*D - sqrt(a)*E)):

if divisible(sqrt(a)*g*z0**2 + D*z0 + sqrt(a)*F, e*sqrt(c)*D - sqrt(a)*E):

l.add((solve_x(z0), solve_y(z0)))

# (4) Method used when B**2 - 4*A*C is a square, is descibed in p. 6 of the below paper

# by John P. Robertson.

# http://www.jpr2718.org/ax2p.pdf

elif isinstance(sqrt(B**2 - 4*A*C), Integer):

if A != 0:

r = sqrt(B**2 - 4*A*C)

u, v = symbols("u, v", integer=True)

eq = _mexpand(4*A*r*u*v + 4*A*D*(B*v + r*u + r*v - B*u) + 2*A*4*A*E*(u - v) + 4*A*r*4*A*F)

sol = diop_solve(eq, t)

sol = list(sol)

for solution in sol:

s0 = solution[0]

t0 = solution[1]

x_0 = S(B*t0 + r*s0 + r*t0 - B*s0)/(4*A*r)

y_0 = S(s0 - t0)/(2*r)

if isinstance(s0, Symbol) or isinstance(t0, Symbol):

if check_param(x_0, y_0, 4*A*r, t) != (None, None):

l.add((check_param(x_0, y_0, 4*A*r, t)[0], check_param(x_0, y_0, 4*A*r, t)[1]))

elif divisible(B*t0 + r*s0 + r*t0 - B*s0, 4*A*r):

if divisible(s0 - t0, 2*r):

if is_solution_quad(var, coeff, x_0, y_0):

l.add((x_0, y_0))

else:

_var = var

_var[0], _var[1] = _var[1], _var[0] # Interchange x and y

s = _diop_quadratic(_var, coeff, t)

while len(s) > 0:

sol = s.pop()

l.add((sol[1], sol[0]))

# (5) B**2 - 4*A*C > 0 and B**2 - 4*A*C not a square or B**2 - 4*A*C < 0

else:

P, Q = _transformation_to_DN(var, coeff)

D, N = _find_DN(var, coeff)

solns_pell = diop_DN(D, N)

if D < 0:

for solution in solns_pell:

for X_i in [-solution[0], solution[0]]:

for Y_i in [-solution[1], solution[1]]:

x_i, y_i = (P*Matrix([X_i, Y_i]) + Q)[0], (P*Matrix([X_i, Y_i]) + Q)[1]

if isinstance(x_i, Integer) and isinstance(y_i, Integer):

l.add((x_i, y_i))

else:

# In this case equation can be transformed into a Pell equation

#n = symbols("n", integer=True)

fund_solns = solns_pell

solns_pell = set(fund_solns)

for X, Y in fund_solns:

solns_pell.add((-X, -Y))

a = diop_DN(D, 1)

T = a[0][0]

U = a[0][1]

if (isinstance(P[0], Integer) and isinstance(P[1], Integer) and isinstance(P[2], Integer)

and isinstance(P[3], Integer) and isinstance(Q[0], Integer) and isinstance(Q[1], Integer)):

for sol in solns_pell:

r = sol[0]

s = sol[1]

x_n = S((r + s*sqrt(D))*(T + U*sqrt(D))**t + (r - s*sqrt(D))*(T - U*sqrt(D))**t)/2

y_n = S((r + s*sqrt(D))*(T + U*sqrt(D))**t - (r - s*sqrt(D))*(T - U*sqrt(D))**t)/(2*sqrt(D))

x_n = _mexpand(x_n)

y_n = _mexpand(y_n)

x_n, y_n = (P*Matrix([x_n, y_n]) + Q)[0], (P*Matrix([x_n, y_n]) + Q)[1]

l.add((x_n, y_n))

else:

L = ilcm(S(P[0]).q, ilcm(S(P[1]).q, ilcm(S(P[2]).q,

ilcm(S(P[3]).q, ilcm(S(Q[0]).q, S(Q[1]).q)))))

k = 1

T_k = T

U_k = U

while (T_k - 1) % L != 0 or U_k % L != 0:

T_k, U_k = T_k*T + D*U_k*U, T_k*U + U_k*T

k += 1

for X, Y in solns_pell:

for i in range(k):

Z = P*Matrix([X, Y]) + Q

x, y = Z[0], Z[1]

if isinstance(x, Integer) and isinstance(y, Integer):

Xt = S((X + sqrt(D)*Y)*(T_k + sqrt(D)*U_k)**t +

(X - sqrt(D)*Y)*(T_k - sqrt(D)*U_k)**t)/ 2

Yt = S((X + sqrt(D)*Y)*(T_k + sqrt(D)*U_k)**t -

(X - sqrt(D)*Y)*(T_k - sqrt(D)*U_k)**t)/ (2*sqrt(D))

Zt = P*Matrix([Xt, Yt]) + Q

l.add((Zt[0], Zt[1]))

X, Y = X*T + D*U*Y, X*U + Y*T

return l

def is_solution_quad(var, coeff, u, v):

"""

Check whether `(u, v)` is solution to the quadratic binary diophantine

equation with the variable list ``var`` and coefficient dictionary

``coeff``.

Not intended for use by normal users.

"""

x, y = var[:2]

eq = x**2*coeff[x**2] + x*y*coeff[x*y] + y**2*coeff[y**2] + x*coeff[x] + y*coeff[y] + coeff[Integer(1)]

return _mexpand(Subs(eq, (x, y), (u, v)).doit()) == 0

def diop_DN(D, N, t=symbols("t", integer=True)):

"""

Solves the equation `x^2 - Dy^2 = N`.

Mainly concerned in the case `D > 0, D` is not a perfect square, which is

the same as generalized Pell equation. To solve the generalized Pell

equation this function Uses LMM algorithm. Refer [1]_ for more details on

the algorithm.

Returns one solution for each class of the solutions. Other solutions of

the class can be constructed according to the values of ``D`` and ``N``.

Returns a list containing the solution tuples `(x, y)`.

Usage

=====

``diop_DN(D, N, t)``: D and N are integers as in `x^2 - Dy^2 = N` and

``t`` is the parameter to be used in the solutions.

Details

=======

``D`` and ``N`` correspond to D and N in the equation.

``t`` is the parameter to be used in the solutions.

Examples

========

>>> from sympy.solvers.diophantine import diop_DN

>>> diop_DN(13, -4) # Solves equation x**2 - 13*y**2 = -4

[(3, 1), (393, 109), (36, 10)]

The output can be interpreted as follows: There are three fundamental

solutions to the equation `x^2 - 13y^2 = -4` given by (3, 1), (393, 109)

and (36, 10). Each tuple is in the form (x, y), i. e solution (3, 1) means

that `x = 3` and `y = 1`.

>>> diop_DN(986, 1) # Solves equation x**2 - 986*y**2 = 1

[(49299, 1570)]

See Also

========

find_DN(), diop_bf_DN()

References

==========

.. [1] Solving the generalized Pell equation x**2 - D*y**2 = N, John P.

Robertson, July 31, 2004, Pages 16 - 17. [online], Available:

http://www.jpr2718.org/pell.pdf

"""

if D < 0:

if N == 0:

return [(S.Zero, S.Zero)]

elif N < 0:

return []

elif N > 0:

d = divisors(square_factor(N))

sol = []

for divisor in d:

sols = cornacchia(1, -D, N // divisor**2)

if sols:

for x, y in sols:

sol.append((divisor*x, divisor*y))

return sol

elif D == 0:

if N < 0 or not isinstance(sqrt(N), Integer):

return []

if N == 0:

return [(S.Zero, t)]

if isinstance(sqrt(N), Integer):

return [(sqrt(N), t)]

else: # D > 0

if isinstance(sqrt(D), Integer):

r = sqrt(D)

if N == 0:

return [(r*t, t)]

else:

sol = []

for y in range(floor(sign(N)*(N - 1)/(2*r)) + 1):

if isinstance(sqrt(D*y**2 + N), Integer):

sol.append((sqrt(D*y**2 + N), y))

return sol

else:

if N == 0:

return [(S.Zero, S.Zero)]

elif abs(N) == 1:

pqa = PQa(0, 1, D)

a_0 = floor(sqrt(D))

l = 0

G = []

B = []

for i in pqa:

a = i[2]

G.append(i[5])

B.append(i[4])

if l != 0 and a == 2*a_0:

break

l = l + 1

if l % 2 == 1:

if N == -1:

x = G[l-1]

y = B[l-1]

else:

count = l

while count < 2*l - 1:

i = next(pqa)

G.append(i[5])

B.append(i[4])

count = count + 1

x = G[count]

y = B[count]

else:

if N == 1:

x = G[l-1]

y = B[l-1]

else:

return []

return [(x, y)]

else:

fs = []

sol = []

div = divisors(N)

for d in div:

if divisible(N, d**2):

fs.append(d)

for f in fs:

m = N // f**2

zs = sqrt_mod(D, abs(m), True)

zs = [i for i in zs if i <= abs(m) // 2 ]

if abs(m) != 2:

zs = zs + [-i for i in zs]

if S.Zero in zs:

zs.remove(S.Zero) # Remove duplicate zero

for z in zs:

pqa = PQa(z, abs(m), D)

l = 0

G = []

B = []

for i in pqa:

a = i[2]

G.append(i[5])

B.append(i[4])

if l != 0 and abs(i[1]) == 1:

r = G[l-1]

s = B[l-1]

if r**2 - D*s**2 == m:

sol.append((f*r, f*s))

elif diop_DN(D, -1) != []:

a = diop_DN(D, -1)

sol.append((f*(r*a[0][0] + a[0][1]*s*D), f*(r*a[0][1] + s*a[0][0])))

break

l = l + 1

if l == length(z, abs(m), D):

break

return sol

def cornacchia(a, b, m):

"""

Solves `ax^2 + by^2 = m` where `\gcd(a, b) = 1 = gcd(a, m)` and `a, b > 0`.

Uses the algorithm due to Cornacchia. The method only finds primitive

solutions, i.e. ones with `\gcd(x, y) = 1`. So this method can't be used to

find the solutions of `x^2 + y^2 = 20` since the only solution to former is

`(x,y) = (4, 2)` and it is not primitive. When ` a = b = 1`, only the

solutions with `x \geq y` are found. For more details, see the References.

Examples

========

>>> from sympy.solvers.diophantine import cornacchia

>>> cornacchia(2, 3, 35) # equation 2x**2 + 3y**2 = 35

set([(2, 3), (4, 1)])

>>> cornacchia(1, 1, 25) # equation x**2 + y**2 = 25

set([(4, 3)])

References

===========

.. [1] A. Nitaj, "L'algorithme de Cornacchia"

.. [2] Solving the diophantine equation ax**2 + by**2 = m by Cornacchia's

method, [online], Available:

http://www.numbertheory.org/php/cornacchia.html

"""

sols = set([])

a1 = igcdex(a, m)[0]

v = sqrt_mod(-b*a1, m, True)

if v is None:

return None

if not isinstance(v, list):

v = [v]

for t in v:

if t < m // 2:

continue

u, r = t, m

while True:

u, r = r, u % r

if a*r**2 < m:

break

m1 = m - a*r**2

if m1 % b == 0:

m1 = m1 // b

if isinstance(sqrt(m1), Integer):

s = sqrt(m1)

sols.add((int(r), int(s)))

return sols

def PQa(P_0, Q_0, D):

"""

Returns useful information needed to solve the Pell equation.

There are six sequences of integers defined related to the continued

fraction representation of `\\frac{P + \sqrt{D}}{Q}`, namely {`P_{i}`},

{`Q_{i}`}, {`a_{i}`},{`A_{i}`}, {`B_{i}`}, {`G_{i}`}. ``PQa()`` Returns

these values as a 6-tuple in the same order as mentioned above. Refer [1]_

for more detailed information.

Usage

=====

``PQa(P_0, Q_0, D)``: ``P_0``, ``Q_0`` and ``D`` are integers corresponding

to `P_{0}`, `Q_{0}` and `D` in the continued fraction

`\\frac{P_{0} + \sqrt{D}}{Q_{0}}`.

Also it's assumed that `P_{0}^2 == D mod(|Q_{0}|)` and `D` is square free.

Examples

========

>>> from sympy.solvers.diophantine import PQa

>>> pqa = PQa(13, 4, 5) # (13 + sqrt(5))/4

>>> next(pqa) # (P_0, Q_0, a_0, A_0, B_0, G_0)

(13, 4, 3, 3, 1, -1)

>>> next(pqa) # (P_1, Q_1, a_1, A_1, B_1, G_1)

(-1, 1, 1, 4, 1, 3)

References

==========

.. [1] Solving the generalized Pell equation x^2 - Dy^2 = N, John P.

Robertson, July 31, 2004, Pages 4 - 8. http://www.jpr2718.org/pell.pdf

"""

A_i_2 = B_i_1 = 0

A_i_1 = B_i_2 = 1

G_i_2 = -P_0

G_i_1 = Q_0

P_i = P_0

Q_i = Q_0

while(1):

a_i = floor((P_i + sqrt(D))/Q_i)

A_i = a_i*A_i_1 + A_i_2

B_i = a_i*B_i_1 + B_i_2

G_i = a_i*G_i_1 + G_i_2

yield P_i, Q_i, a_i, A_i, B_i, G_i

A_i_1, A_i_2 = A_i, A_i_1

B_i_1, B_i_2 = B_i, B_i_1

G_i_1, G_i_2 = G_i, G_i_1

P_i = a_i*Q_i - P_i

Q_i = (D - P_i**2)/Q_i

def diop_bf_DN(D, N, t=symbols("t", integer=True)):

"""

Uses brute force to solve the equation, `x^2 - Dy^2 = N`.

Mainly concerned with the generalized Pell equation which is the case when

`D > 0, D` is not a perfect square. For more information on the case refer

[1]_. Let `(t, u)` be the minimal positive solution of the equation

`x^2 - Dy^2 = 1`. Then this method requires

`\sqrt{\\frac{\mid N \mid (t \pm 1)}{2D}}` to be small.

Usage

=====

``diop_bf_DN(D, N, t)``: ``D`` and ``N`` are coefficients in

`x^2 - Dy^2 = N` and ``t`` is the parameter to be used in the solutions.

Details

=======

``D`` and ``N`` correspond to D and N in the equation.

``t`` is the parameter to be used in the solutions.

Examples

========

>>> from sympy.solvers.diophantine import diop_bf_DN

>>> diop_bf_DN(13, -4)

[(3, 1), (-3, 1), (36, 10)]

>>> diop_bf_DN(986, 1)

[(49299, 1570)]

See Also

========

diop_DN()

References

==========

.. [1] Solving the generalized Pell equation x**2 - D*y**2 = N, John P.

Robertson, July 31, 2004, Page 15. http://www.jpr2718.org/pell.pdf

"""

sol = []

a = diop_DN(D, 1)

u = a[0][0]

v = a[0][1]

if abs(N) == 1:

return diop_DN(D, N)

elif N > 1:

L1 = 0

L2 = floor(sqrt(S(N*(u - 1))/(2*D))) + 1

elif N < -1:

L1 = ceiling(sqrt(S(-N)/D))

L2 = floor(sqrt(S(-N*(u + 1))/(2*D))) + 1

else:

if D < 0:

return [(S.Zero, S.Zero)]

elif D == 0:

return [(S.Zero, t)]

else:

if isinstance(sqrt(D), Integer):

return [(sqrt(D)*t, t), (-sqrt(D)*t, t)]

else:

return [(S.Zero, S.Zero)]

for y in range(L1, L2):

if isinstance(sqrt(N + D*y**2), Integer):

x = sqrt(N + D*y**2)

sol.append((x, y))

if not equivalent(x, y, -x, y, D, N):

sol.append((-x, y))

return sol

def equivalent(u, v, r, s, D, N):

"""

Returns True if two solutions `(u, v)` and `(r, s)` of `x^2 - Dy^2 = N`

belongs to the same equivalence class and False otherwise.

Two solutions `(u, v)` and `(r, s)` to the above equation fall to the same

equivalence class iff both `(ur - Dvs)` and `(us - vr)` are divisible by

`N`. See reference [1]_. No test is performed to test whether `(u, v)` and

`(r, s)` are actually solutions to the equation. User should take care of

this.

Usage

=====

``equivalent(u, v, r, s, D, N)``: `(u, v)` and `(r, s)` are two solutions

of the equation `x^2 - Dy^2 = N` and all parameters involved are integers.

Examples

========

>>> from sympy.solvers.diophantine import equivalent

>>> equivalent(18, 5, -18, -5, 13, -1)

True

>>> equivalent(3, 1, -18, 393, 109, -4)

False

References

==========

.. [1] Solving the generalized Pell equation x**2 - D*y**2 = N, John P.

Robertson, July 31, 2004, Page 12. http://www.jpr2718.org/pell.pdf

"""

return divisible(u*r - D*v*s, N) and divisible(u*s - v*r, N)

def length(P, Q, D):

"""

Returns the (length of aperiodic part + length of periodic part) of

continued fraction representation of `\\frac{P + \sqrt{D}}{Q}`.

It is important to remember that this does NOT return the length of the

periodic part but the addition of the legths of the two parts as mentioned

above.

Usage

=====

``length(P, Q, D)``: ``P``, ``Q`` and ``D`` are integers corresponding to

the continued fraction `\\frac{P + \sqrt{D}}{Q}`.

Details

=======

``P``, ``D`` and ``Q`` corresponds to P, D and Q in the continued fraction,

`\\frac{P + \sqrt{D}}{Q}`.

Examples

========

>>> from sympy.solvers.diophantine import length

>>> length(-2 , 4, 5) # (-2 + sqrt(5))/4

>>> length(-5, 4, 17) # (-5 + sqrt(17))/4

"""

x = P + sqrt(D)

y = Q

x = sympify(x)

v, res = [], []

q = x/y

if q < 0:

v.append(q)

res.append(floor(q))

q = q - floor(q)

num, den = rad_rationalize(1, q)

q = num / den

while 1:

v.append(q)

a = int(q)

res.append(a)

if q == a:

return len(res)

num, den = rad_rationalize(1,(q - a))

q = num / den

if q in v:

return len(res)

def transformation_to_DN(eq):

"""

This function transforms general quadratic,

`ax^2 + bxy + cy^2 + dx + ey + f = 0`

to more easy to deal with `X^2 - DY^2 = N` form.

This is used to solve the general quadratic equation by transforming it to

the latter form. Refer [1]_ for more detailed information on the

transformation. This function returns a tuple (A, B) where A is a 2 X 2

matrix and B is a 2 X 1 matrix such that,

Transpose([x y]) = A * Transpose([X Y]) + B

Usage

=====

``transformation_to_DN(eq)``: where ``eq`` is the quadratic to be

transformed.

Examples

========

>>> from sympy.abc import x, y

>>> from sympy.solvers.diophantine import transformation_to_DN

>>> from sympy.solvers.diophantine import classify_diop

>>> A, B = transformation_to_DN(x**2 - 3*x*y - y**2 - 2*y + 1)

>>> A

Matrix([

[1/26, 3/26],

[ 0, 1/13]])

>>> B

Matrix([

[-6/13],

[-4/13]])

A, B returned are such that Transpose((x y)) = A * Transpose((X Y)) + B.

Substituting these values for `x` and `y` and a bit of simplifying work

will give an equation of the form `x^2 - Dy^2 = N`.

>>> from sympy.abc import X, Y

>>> from sympy import Matrix, simplify, Subs

>>> u = (A*Matrix([X, Y]) + B)[0] # Transformation for x

>>> u

X/26 + 3*Y/26 - 6/13

>>> v = (A*Matrix([X, Y]) + B)[1] # Transformation for y

>>> v

Y/13 - 4/13

Next we will substitute these formulas for `x` and `y` and do

``simplify()``.

>>> eq = simplify(Subs(x**2 - 3*x*y - y**2 - 2*y + 1, (x, y), (u, v)).doit())

>>> eq

X**2/676 - Y**2/52 + 17/13

By multiplying the denominator appropriately, we can get a Pell equation

in the standard form.

>>> eq * 676

X**2 - 13*Y**2 + 884

If only the final equation is needed, ``find_DN()`` can be used.

See Also

========

find_DN()

References

==========

.. [1] Solving the equation ax^2 + bxy + cy^2 + dx + ey + f = 0,

John P.Robertson, May 8, 2003, Page 7 - 11.

http://www.jpr2718.org/ax2p.pdf

"""

var, coeff, diop_type = classify_diop(eq)

if diop_type == "binary_quadratic":

return _transformation_to_DN(var, coeff)

def _transformation_to_DN(var, coeff):

x, y = var[:2]

a = coeff[x**2]

b = coeff[x*y]

c = coeff[y**2]

d = coeff[x]

e = coeff[y]

f = coeff[Integer(1)]

g = igcd(a, igcd(b, igcd(c, igcd(d, igcd(e, f)))))

a = a // g

b = b // g

c = c // g

d = d // g

e = e // g

f = f // g

X, Y = symbols("X, Y", integer=True)

if b != Integer(0):

B = (S(2*a)/b).p

C = (S(2*a)/b).q

A = (S(a)/B**2).p

T = (S(a)/B**2).q

# eq_1 = A*B*X**2 + B*(c*T - A*C**2)*Y**2 + d*T*X + (B*e*T - d*T*C)*Y + f*T*B

coeff = {X**2: A*B, X*Y: 0, Y**2: B*(c*T - A*C**2), X: d*T, Y: B*e*T - d*T*C, Integer(1): f*T*B}

A_0, B_0 = _transformation_to_DN([X, Y], coeff)

return Matrix(2, 2, [S(1)/B, -S(C)/B, 0, 1])*A_0, Matrix(2, 2, [S(1)/B, -S(C)/B, 0, 1])*B_0

else:

if d != Integer(0):

B = (S(2*a)/d).p

C = (S(2*a)/d).q

A = (S(a)/B**2).p

T = (S(a)/B**2).q

# eq_2 = A*X**2 + c*T*Y**2 + e*T*Y + f*T - A*C**2

coeff = {X**2: A, X*Y: 0, Y**2: c*T, X: 0, Y: e*T, Integer(1): f*T - A*C**2}

A_0, B_0 = _transformation_to_DN([X, Y], coeff)

return Matrix(2, 2, [S(1)/B, 0, 0, 1])*A_0, Matrix(2, 2, [S(1)/B, 0, 0, 1])*B_0 + Matrix([-S(C)/B, 0])

else:

if e != Integer(0):

B = (S(2*c)/e).p

C = (S(2*c)/e).q

A = (S(c)/B**2).p

T = (S(c)/B**2).q

# eq_3 = a*T*X**2 + A*Y**2 + f*T - A*C**2

coeff = {X**2: a*T, X*Y: 0, Y**2: A, X: 0, Y: 0, Integer(1): f*T - A*C**2}

A_0, B_0 = _transformation_to_DN([X, Y], coeff)

return Matrix(2, 2, [1, 0, 0, S(1)/B])*A_0, Matrix(2, 2, [1, 0, 0, S(1)/B])*B_0 + Matrix([0, -S(C)/B])

else:

# TODO: pre-simplification: Not necessary but may simplify

# the equation.

return Matrix(2, 2, [S(1)/a, 0, 0, 1]), Matrix([0, 0])

def find_DN(eq):

"""

This function returns a tuple, `(D, N)` of the simplified form,

`x^2 - Dy^2 = N`, corresponding to the general quadratic,

`ax^2 + bxy + cy^2 + dx + ey + f = 0`.

Solving the general quadratic is then equivalent to solving the equation

`X^2 - DY^2 = N` and transforming the solutions by using the transformation

matrices returned by ``transformation_to_DN()``.

Usage

=====

``find_DN(eq)``: where ``eq`` is the quadratic to be transformed.

Examples

========

>>> from sympy.abc import x, y

>>> from sympy.solvers.diophantine import find_DN

>>> find_DN(x**2 - 3*x*y - y**2 - 2*y + 1)

(13, -884)

Interpretation of the output is that we get `X^2 -13Y^2 = -884` after

transforming `x^2 - 3xy - y^2 - 2y + 1` using the transformation returned

by ``transformation_to_DN()``.

See Also

========

transformation_to_DN()

References

==========

.. [1] Solving the equation ax^2 + bxy + cy^2 + dx + ey + f = 0,

John P.Robertson, May 8, 2003, Page 7 - 11.

http://www.jpr2718.org/ax2p.pdf

"""

var, coeff, diop_type = classify_diop(eq)

if diop_type == "binary_quadratic":

return _find_DN(var, coeff)

def _find_DN(var, coeff):

x, y = var[:2]

X, Y = symbols("X, Y", integer=True)

A , B = _transformation_to_DN(var, coeff)

u = (A*Matrix([X, Y]) + B)[0]

v = (A*Matrix([X, Y]) + B)[1]

eq = x**2*coeff[x**2] + x*y*coeff[x*y] + y**2*coeff[y**2] + x*coeff[x] + y*coeff[y] + coeff[Integer(1)]

simplified = _mexpand(Subs(eq, (x, y), (u, v)).doit())

coeff = dict([reversed(t.as_independent(*[X, Y])) for t in simplified.args])

for term in [X**2, Y**2, Integer(1)]:

if term not in coeff.keys():

coeff[term] = Integer(0)

return -coeff[Y**2]/coeff[X**2], -coeff[Integer(1)]/coeff[X**2]

def check_param(x, y, a, t):

"""

Check if there is a number modulo ``a`` such that ``x`` and ``y`` are both

integers. If exist, then find a parametric representation for ``x`` and

``y``.

Here ``x`` and ``y`` are functions of ``t``.

"""

k, m, n = symbols("k, m, n", integer=True)

p = Wild("p", exclude=[k])

q = Wild("q", exclude=[k])

ok = False

for i in range(a):

z_x = _mexpand(Subs(x, t, a*k + i).doit()).match(p*k + q)

z_y = _mexpand(Subs(y, t, a*k + i).doit()).match(p*k + q)

if (isinstance(z_x[p], Integer) and isinstance(z_x[q], Integer) and

isinstance(z_y[p], Integer) and isinstance(z_y[q], Integer)):

ok = True

break

if ok == True:

x_param = x.match(p*t + q)

y_param = y.match(p*t + q)

if x_param[p] == 0 or y_param[p] == 0:

if x_param[p] == 0:

l1, junk = Poly(y).clear_denoms()

else:

l1 = 1

if y_param[p] == 0:

l2, junk = Poly(x).clear_denoms()

else:

l2 = 1

return x*ilcm(l1, l2), y*ilcm(l1, l2)

eq = S(m - x_param[q])/x_param[p] - S(n - y_param[q])/y_param[p]

lcm_denom, junk = Poly(eq).clear_denoms()

eq = eq * lcm_denom

return diop_solve(eq, t)[0], diop_solve(eq, t)[1]

else:

return (None, None)

def diop_ternary_quadratic(eq):

"""

Solves the general quadratic ternary form,

`ax^2 + by^2 + cz^2 + fxy + gyz + hxz = 0`.

Returns a tuple `(x, y, z)` which is a base solution for the above

equation. If there are no solutions, `(None, None, None)` is returned.

Usage

=====

``diop_ternary_quadratic(eq)``: Return a tuple containing a basic solution

to ``eq``.

Details

=======

``eq`` should be an homogeneous expression of degree two in three variables

and it is assumed to be zero.

Examples

========

>>> from sympy.abc import x, y, z

>>> from sympy.solvers.diophantine import diop_ternary_quadratic

>>> diop_ternary_quadratic(x**2 + 3*y**2 - z**2)

(1, 0, 1)

>>> diop_ternary_quadratic(4*x**2 + 5*y**2 - z**2)

(1, 0, 2)

>>> diop_ternary_quadratic(45*x**2 - 7*y**2 - 8*x*y - z**2)

(28, 45, 105)

>>> diop_ternary_quadratic(x**2 - 49*y**2 - z**2 + 13*z*y -8*x*y)

(9, 1, 5)

"""

var, coeff, diop_type = classify_diop(eq)

if diop_type == "homogeneous_ternary_quadratic":

return _diop_ternary_quadratic(var, coeff)

def _diop_ternary_quadratic(_var, coeff):

x, y, z = _var[:3]

var = [x]*3

var[0], var[1], var[2] = _var[0], _var[1], _var[2]

# Equations of the form B*x*y + C*z*x + E*y*z = 0 and At least two of the

# coefficients A, B, C are non-zero.

# There are infinitely many solutions for the equation.

# Ex: (0, 0, t), (0, t, 0), (t, 0, 0)

# Equation can be re-written as y*(B*x + E*z) = -C*x*z and we can find rather

# unobviuos solutions. Set y = -C and B*x + E*z = x*z. The latter can be solved by

# using methods for binary quadratic diophantine equations. Let's select the

# solution which minimizes |x| + |z|

if coeff[x**2] == 0 and coeff[y**2] == 0 and coeff[z**2] == 0:

if coeff[x*z] != 0:

sols = diophantine(coeff[x*y]*x + coeff[y*z]*z - x*z)

s = sols.pop()

min_sum = abs(s[0]) + abs(s[1])

for r in sols:

if abs(r[0]) + abs(r[1]) < min_sum:

s = r

min_sum = abs(s[0]) + abs(s[1])

x_0, y_0, z_0 = s[0], -coeff[x*z], s[1]

else:

var[0], var[1] = _var[1], _var[0]

y_0, x_0, z_0 = _diop_ternary_quadratic(var, coeff)

return simplified(x_0, y_0, z_0)

if coeff[x**2] == 0:

# If the coefficient of x is zero change the variables

if coeff[y**2] == 0:

var[0], var[2] = _var[2], _var[0]

z_0, y_0, x_0 = _diop_ternary_quadratic(var, coeff)

else:

var[0], var[1] = _var[1], _var[0]

y_0, x_0, z_0 = _diop_ternary_quadratic(var, coeff)

else:

if coeff[x*y] != 0 or coeff[x*z] != 0:

# Apply the transformation x --> X - (B*y + C*z)/(2*A)

A = coeff[x**2]

B = coeff[x*y]

C = coeff[x*z]

D = coeff[y**2]

E = coeff[y*z]

F = coeff[z**2]

_coeff = dict()

_coeff[x**2] = 4*A**2

_coeff[y**2] = 4*A*D - B**2

_coeff[z**2] = 4*A*F - C**2

_coeff[y*z] = 4*A*E - 2*B*C

_coeff[x*y] = 0

_coeff[x*z] = 0

X_0, y_0, z_0 = _diop_ternary_quadratic(var, _coeff)

if X_0 == None:

return (None, None, None)

l = (S(B*y_0 + C*z_0)/(2*A)).q

x_0, y_0, z_0 = X_0*l - (S(B*y_0 + C*z_0)/(2*A)).p, y_0*l, z_0*l

elif coeff[z*y] != 0:

if coeff[y**2] == 0:

if coeff[z**2] == 0:

# Equations of the form A*x**2 + E*yz = 0.

A = coeff[x**2]

E = coeff[y*z]

b = (S(-E)/A).p

a = (S(-E)/A).q

x_0, y_0, z_0 = b, a, b

else:

# Ax**2 + E*y*z + F*z**2 = 0

var[0], var[2] = _var[2], _var[0]

z_0, y_0, x_0 = _diop_ternary_quadratic(var, coeff)

else:

# A*x**2 + D*y**2 + E*y*z + F*z**2 = 0, C may be zero

var[0], var[1] = _var[1], _var[0]

y_0, x_0, z_0 = _diop_ternary_quadratic(var, coeff)

else:

# Ax**2 + D*y**2 + F*z**2 = 0, C may be zero

x_0, y_0, z_0 = _diop_ternary_quadratic_normal(var, coeff)

return simplified(x_0, y_0, z_0)

def transformation_to_normal(eq):

"""

Returns the transformation Matrix from general ternary quadratic equation

`eq` to normal form.

General form of the ternary quadratic equation is `ax^2 + by^2 cz^2 + dxy +

eyz + fxz`. This function returns a 3X3 transformation Matrix which

transforms the former equation to the form `ax^2 + by^2 + cz^2 = 0`. This

is not used in solving ternary quadratics. Only implemented for the sake

of completeness.

"""

var, coeff, diop_type = classify_diop(eq)

if diop_type == "homogeneous_ternary_quadratic":

return _transformation_to_normal(var, coeff)

def _transformation_to_normal(var, coeff):

_var = [var[0]]*3

_var[1], _var[2] = var[1], var[2]

x, y, z = var[:3]

if coeff[x**2] == 0:

# If the coefficient of x is zero change the variables

if coeff[y**2] == 0:

_var[0], _var[2] = var[2], var[0]

T = _transformation_to_normal(_var, coeff)

T.row_swap(0, 2)

T.col_swap(0, 2)

return T

else:

_var[0], _var[1] = var[1], var[0]

T = _transformation_to_normal(_var, coeff)

T.row_swap(0, 1)

T.col_swap(0, 1)

return T

else:

# Apply the transformation x --> X - (B*Y + C*Z)/(2*A)

if coeff[x*y] != 0 or coeff[x*z] != 0:

A = coeff[x**2]

B = coeff[x*y]

C = coeff[x*z]

D = coeff[y**2]

E = coeff[y*z]

F = coeff[z**2]

_coeff = dict()

_coeff[x**2] = 4*A**2

_coeff[y**2] = 4*A*D - B**2

_coeff[z**2] = 4*A*F - C**2

_coeff[y*z] = 4*A*E - 2*B*C

_coeff[x*y] = 0

_coeff[x*z] = 0

T_0 = _transformation_to_normal(_var, _coeff)

return Matrix(3, 3, [1, S(-B)/(2*A), S(-C)/(2*A), 0, 1, 0, 0, 0, 1]) * T_0

elif coeff[y*z] != 0:

if coeff[y**2] == 0:

if coeff[z**2] == 0:

# Equations of the form A*x**2 + E*yz = 0.

# Apply transformation y -> Y + Z ans z -> Y - Z

return Matrix(3, 3, [1, 0, 0, 0, 1, 1, 0, 1, -1])

else:

# Ax**2 + E*y*z + F*z**2 = 0

_var[0], _var[2] = var[2], var[0]

T = _transformtion_to_normal(_var, coeff)

T.row_swap(0, 2)

T.col_swap(0, 2)

return T

else:

# A*x**2 + D*y**2 + E*y*z + F*z**2 = 0, F may be zero

_var[0], _var[1] = var[1], var[0]

T = _transformation_to_normal(_var, coeff)

T.row_swap(0, 1)

T.col_swap(0, 1)

return T

else:

return Matrix(3, 3, [1, 0, 0, 0, 1, 0, 0, 0, 1])

def simplified(x, y, z):

"""

Simplify the solution `(x, y, z)`.

"""

if x == None or y == None or z == None:

return (x, y, z)

g = igcd(x, igcd(y, z))

return x // g, y // g, z // g

def parametrize_ternary_quadratic(eq):

"""

Returns the parametrized general solution for the ternary quadratic

equation ``eq`` which has the form

`ax^2 + by^2 + cz^2 + fxy + gyz + hxz = 0`.

Examples

========

>>> from sympy.abc import x, y, z

>>> from sympy.solvers.diophantine import parametrize_ternary_quadratic

>>> parametrize_ternary_quadratic(x**2 + y**2 - z**2)

(2*p*q, p**2 - q**2, p**2 + q**2)

Here `p` and `q` are two co-prime integers.

>>> parametrize_ternary_quadratic(3*x**2 + 2*y**2 - z**2 - 2*x*y + 5*y*z - 7*y*z)

(2*p**2 - 2*p*q - q**2, 2*p**2 + 2*p*q - q**2, 2*p**2 - 2*p*q + 3*q**2)

>>> parametrize_ternary_quadratic(124*x**2 - 30*y**2 - 7729*z**2)

(-1410*p**2 - 363263*q**2, 2700*p**2 + 30916*p*q - 695610*q**2, -60*p**2 + 5400*p*q + 15458*q**2)

References

==========

.. [1] The algorithmic resolution of Diophantine equations, Nigel P. Smart,

London Mathematical Society Student Texts 41, Cambridge University

Press, Cambridge, 1998.

"""

var, coeff, diop_type = classify_diop(eq)

if diop_type == "homogeneous_ternary_quadratic":

x_0, y_0, z_0 = _diop_ternary_quadratic(var, coeff)

return _parametrize_ternary_quadratic((x_0, y_0, z_0), var, coeff)

def _parametrize_ternary_quadratic(solution, _var, coeff):

x, y, z = _var[:3]

x_0, y_0, z_0 = solution[:3]

v = [x]*3

v[0], v[1], v[2] = _var[0], _var[1], _var[2]

if x_0 == None:

return (None, None, None)

if x_0 == 0:

if y_0 == 0:

v[0], v[2] = v[2], v[0]

z_p, y_p, x_p = _parametrize_ternary_quadratic((z_0, y_0, x_0), v, coeff)

return x_p, y_p, z_p

else:

v[0], v[1] = v[1], v[0]

y_p, x_p, z_p = _parametrize_ternary_quadratic((y_0, x_0, z_0), v, coeff)

return x_p, y_p, z_p

x, y, z = v[:3]

r, p, q = symbols("r, p, q", integer=True)

eq = x**2*coeff[x**2] + y**2*coeff[y**2] + z**2*coeff[z**2] + x*y*coeff[x*y] + y*z*coeff[y*z] + z*x*coeff[z*x]

eq_1 = Subs(eq, (x, y, z), (r*x_0, r*y_0 + p, r*z_0 + q)).doit()

eq_1 = _mexpand(eq_1)

A, B = eq_1.as_independent(r, as_Add=True)

x = A*x_0

y = (A*y_0 - _mexpand(B/r*p))

z = (A*z_0 - _mexpand(B/r*q))

return x, y, z

def diop_ternary_quadratic_normal(eq):

"""

Solves the quadratic ternary diophantine equation,

`ax^2 + by^2 + cz^2 = 0`.

Here the coefficients `a`, `b`, and `c` should be non zero. Otherwise the

equation will be a quadratic binary or univariate equation. If solvable,

returns a tuple `(x, y, z)` that satisifes the given equation. If the

equation does not have integer solutions, `(None, None, None)` is returned.

Usage

=====

``diop_ternary_quadratic_normal(eq)``: where ``eq`` is an equation of the form

`ax^2 + by^2 + cz^2 = 0`.

Examples

========

>>> from sympy.abc import x, y, z

>>> from sympy.solvers.diophantine import diop_ternary_quadratic_normal

>>> diop_ternary_quadratic_normal(x**2 + 3*y**2 - z**2)

(1, 0, 1)

>>> diop_ternary_quadratic_normal(4*x**2 + 5*y**2 - z**2)

(1, 0, 2)

>>> diop_ternary_quadratic_normal(34*x**2 - 3*y**2 - 301*z**2)

(4, 9, 1)

"""

var, coeff, diop_type = classify_diop(eq)

if diop_type == "homogeneous_ternary_quadratic":

return _diop_ternary_quadratic_normal(var, coeff)

def _diop_ternary_quadratic_normal(var, coeff):

x, y, z = var[:3]

a = coeff[x**2]

b = coeff[y**2]

c = coeff[z**2]

if a*b*c == 0:

raise ValueError("Try factoring out you equation or using diophantine()")

g = igcd(a, igcd(b, c))

a = a // g

b = b // g

c = c // g

a_0 = square_factor(a)

b_0 = square_factor(b)

c_0 = square_factor(c)

a_1 = a // a_0**2

b_1 = b // b_0**2

c_1 = c // c_0**2

a_2, b_2, c_2 = pairwise_prime(a_1, b_1, c_1)

A = -a_2*c_2

B = -b_2*c_2

# If following two conditions are satisified then there are no solutions

if A < 0 and B < 0:

return (None, None, None)

if (sqrt_mod(-b_2*c_2, a_2) == None or sqrt_mod(-c_2*a_2, b_2) == None or

sqrt_mod(-a_2*b_2, c_2) == None):

return (None, None, None)

z_0, x_0, y_0 = descent(A, B)

if divisible(z_0, c_2) == True:

z_0 = z_0 // abs(c_2)

else:

x_0 = x_0*(S(z_0)/c_2).q

y_0 = y_0*(S(z_0)/c_2).q

z_0 = (S(z_0)/c_2).p

x_0, y_0, z_0 = simplified(x_0, y_0, z_0)

# Holzer reduction

if sign(a) == sign(b):

x_0, y_0, z_0 = holzer(x_0, y_0, z_0, abs(a_2), abs(b_2), abs(c_2))

elif sign(a) == sign(c):

x_0, z_0, y_0 = holzer(x_0, z_0, y_0, abs(a_2), abs(c_2), abs(b_2))

else:

y_0, z_0, x_0 = holzer(y_0, z_0, x_0, abs(b_2), abs(c_2), abs(a_2))

x_0 = reconstruct(b_1, c_1, x_0)

y_0 = reconstruct(a_1, c_1, y_0)

z_0 = reconstruct(a_1, b_1, z_0)

l = ilcm(a_0, ilcm(b_0, c_0))

x_0 = abs(x_0*l//a_0)

y_0 = abs(y_0*l//b_0)

z_0 = abs(z_0*l//c_0)

return simplified(x_0, y_0, z_0)

def square_factor(a):

"""

Returns an integer `c` s.t. `a = c^2k, \ c,k \in Z`. Here `k` is square

free.

Examples

========

>>> from sympy.solvers.diophantine import square_factor

>>> square_factor(24)

>>> square_factor(36)

>>> square_factor(1)

"""

f = factorint(abs(a))

c = 1

for p, e in f.items():

c = c * p**(e//2)

return c

def pairwise_prime(a, b, c):

"""

Transform `ax^2 + by^2 + cz^2 = 0` into an equivalent equation

`a'x^2 + b'y^2 + c'z^2 = 0` where `a', b', c'` are pairwise relatively

prime.

Returns a tuple containing `a', b', c'`. `\gcd(a, b, c)` should equal `1`

for this to work. The solutions for `ax^2 + by^2 + cz^2 = 0` can be

recovered from the solutions of `a'x^2 + b'y^2 + c'z^2 = 0`.

Examples

========

>>> from sympy.solvers.diophantine import pairwise_prime

>>> pairwise_prime(6, 15, 10)

(5, 2, 3)

See Also

========

make_prime(), reocnstruct()

"""

a, b, c = make_prime(a, b, c)

b, c, a = make_prime(b, c, a)

c, a, b = make_prime(c, a, b)

return a, b, c

def make_prime(a, b, c):

"""

Transform the equation `ax^2 + by^2 + cz^2 = 0` to an equivalent equation

`a'x^2 + b'y^2 + c'z^2 = 0` with `\gcd(a', b') = 1`.

Returns a tuple `(a', b', c')` which satisfies above conditions. Note that

in the returned tuple `\gcd(a', c')` and `\gcd(b', c')` can take any value.

Examples

========

>>> from sympy.solvers.diophantine import make_prime

>>> make_prime(4, 2, 7)

(2, 1, 14)

See Also

========

pairwaise_prime(), reconstruct()

"""

g = igcd(a, b)

if g != 1:

f = factorint(g)

for p, e in f.items():

a = a // p**e

b = b // p**e

if e % 2 == 1:

c = p*c

return a, b, c

def reconstruct(a, b, z):

"""

Reconstruct the `z` value of an equivalent solution of `ax^2 + by^2 + cz^2`

from the `z` value of a solution of a transformed version of the above

equation.

"""

g = igcd(a, b)

if g != 1:

f = factorint(g)

for p, e in f.items():

if e %2 == 0:

z = z*p**(e//2)

else:

z = z*p**((e//2)+1)

return z

def ldescent(A, B):

"""

Uses Lagrange's method to find a non trivial solution to

`w^2 = Ax^2 + By^2`.

Here, `A \\neq 0` and `B \\neq 0` and `A` and `B` are square free. Output a

tuple `(w_0, x_0, y_0)` which is a solution to the above equation.

Examples

========

>>> from sympy.solvers.diophantine import ldescent

>>> ldescent(1, 1) # w^2 = x^2 + y^2

(1, 1, 0)

>>> ldescent(4, -7) # w^2 = 4x^2 - 7y^2

(2, -1, 0)

This means that `x = -1, y = 0` and `w = 2` is a solution to the equation

`w^2 = 4x^2 - 7y^2`

>>> ldescent(5, -1) # w^2 = 5x^2 - y^2

(2, 1, -1)

References

==========

.. [1] The algorithmic resolution of Diophantine equations, Nigel P. Smart,

London Mathematical Society Student Texts 41, Cambridge University

Press, Cambridge, 1998.

.. [2] Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin,

Mathematics of Computation, Volume 00, Number 0.

"""

if abs(A) > abs(B):

w, y, x = ldescent(B, A)

return w, x, y

if A == 1:

return (S.One, S.One, 0)

if B == 1:

return (S.One, 0, S.One)

r = sqrt_mod(A, B)

Q = (r**2 - A) // B

if Q == 0:

B_0 = 1

d = 0

else:

div = divisors(Q)

B_0 = None

for i in div:

if isinstance(sqrt(abs(Q) // i), Integer):

B_0, d = sign(Q)*i, sqrt(abs(Q) // i)

break

if B_0 != None:

W, X, Y = ldescent(A, B_0)

return simplified((-A*X + r*W), (r*X - W), Y*(B_0*d))

# In this module Descent will always be called with inputs which have solutions.

def descent(A, B):

"""

Lagrange's `descent()` with lattice-reduction to find solutions to

`x^2 = Ay^2 + Bz^2`.

Here `A` and `B` should be square free and pairwise prime. Always should be

called with suitable ``A`` and ``B`` so that the above equation has

solutions.

This is more faster than the normal Lagrange's descent algorithm because

the gaussian reduction is used.

Examples

========

>>> from sympy.solvers.diophantine import descent

>>> descent(3, 1) # x**2 = 3*y**2 + z**2

(1, 0, 1)

`(x, y, z) = (1, 0, 1)` is a solution to the above equation.

>>> descent(41, -113)

(-16, -3, 1)

References

==========

.. [1] Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin,

Mathematics of Computation, Volume 00, Number 0.

"""

if abs(A) > abs(B):

x, y, z = descent(B, A)

return x, z, y

if B == 1:

return (1, 0, 1)

if A == 1:

return (1, 1, 0)

if B == -1:

return (None, None, None)

if B == -A:

return (0, 1, 1)

if B == A:

x, z, y = descent(-1, A)

return (A*y, z, x)

w = sqrt_mod(A, B)

x_0, z_0 = gaussian_reduce(w, A, B)

t = (x_0**2 - A*z_0**2) // B

t_2 = square_factor(t)

t_1 = t // t_2**2

x_1, z_1, y_1 = descent(A, t_1)

return simplified(x_0*x_1 + A*z_0*z_1, z_0*x_1 + x_0*z_1, t_1*t_2*y_1)

def gaussian_reduce(w, a, b):

"""

Returns a reduced solution `(x, z)` to the congruence

`X^2 - aZ^2 \equiv 0 \ (mod \ b)` so that `x^2 + |a|z^2` is minimal.

Details

=======

Here ``w`` is a solution of the congruence `x^2 \equiv a \ (mod \ b)`

References

==========

.. [1] Gaussian lattice Reduction [online]. Available:

http://home.ie.cuhk.edu.hk/~wkshum/wordpress/?p=404

.. [2] Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin,

Mathematics of Computation, Volume 00, Number 0.

"""

u = (0, 1)

v = (1, 0)

if dot(u, v, w, a, b) < 0:

v = (-v[0], -v[1])

if norm(u, w, a, b) < norm(v, w, a, b):

u, v = v, u

while norm(u, w, a, b) > norm(v, w, a, b):

k = dot(u, v, w, a, b) // dot(v, v, w, a, b)

u, v = v, (u[0]- k*v[0], u[1]- k*v[1])

u, v = v, u

if dot(u, v, w, a, b) < dot(v, v, w, a, b)/2 or norm((u[0]-v[0], u[1]-v[1]), w, a, b) > norm(v, w, a, b):

c = v

else:

c = (u[0] - v[0], u[1] - v[1])

return c[0]*w + b*c[1], c[0]

def dot(u, v, w, a, b):

"""

Returns a special dot product of the vectors `u = (u_{1}, u_{2})` and

`v = (v_{1}, v_{2})` which is defined in order to reduce solution of

the congruence equation `X^2 - aZ^2 \equiv 0 \ (mod \ b)`.

"""

u_1, u_2 = u[:2]

v_1, v_2 = v[:2]

return (w*u_1 + b*u_2)*(w*v_1 + b*v_2) + abs(a)*u_1*v_1

def norm(u, w, a, b):

"""

Returns the norm of the vector `u = (u_{1}, u_{2})` under the dot product

defined by `u \cdot v = (wu_{1} + bu_{2})(w*v_{1} + bv_{2}) + |a|*u_{1}*v_{1}`

where `u = (u_{1}, u_{2})` and `v = (v_{1}, v_{2})`.

"""

u_1, u_2 = u[:2]

return sqrt(dot((u_1, u_2), (u_1, u_2), w, a, b))

def holzer(x_0, y_0, z_0, a, b, c):

"""

Simplify the solution `(x_{0}, y_{0}, z_{0})` of the equation

`ax^2 + by^2 = cz^2` with `a, b, c > 0` and `z_{0}^2 \geq \mid ab \mid` to

a new reduced solution `(x, y, z)` such that `z^2 \leq \mid ab \mid`.

"""

while z_0 > sqrt(a*b):

if c % 2 == 0:

k = c // 2

u_0, v_0 = base_solution_linear(k, y_0, -x_0)

else:

k = 2*c

u_0, v_0 = base_solution_linear(c, y_0, -x_0)

w = -(a*u_0*x_0 + b*v_0*y_0) // (c*z_0)

if c % 2 == 1:

if w % 2 != (a*u_0 + b*v_0) % 2:

w = w + 1

x = (x_0*(a*u_0**2 + b*v_0**2 + c*w**2) - 2*u_0*(a*u_0*x_0 + b*v_0*y_0 + c*w*z_0)) // k

y = (y_0*(a*u_0**2 + b*v_0**2 + c*w**2) - 2*v_0*(a*u_0*x_0 + b*v_0*y_0 + c*w*z_0)) // k

z = (z_0*(a*u_0**2 + b*v_0**2 + c*w**2) - 2*w*(a*u_0*x_0 + b*v_0*y_0 + c*w*z_0)) // k

x_0, y_0, z_0 = x, y, z

return x_0, y_0, z_0

def diop_general_pythagorean(eq, param=symbols("m", integer=True)):

"""

Solves the general pythagorean equation,

`a_{1}^2x_{1}^2 + a_{2}^2x_{2}^2 + . . . + a_{n}^2x_{n}^2 - a_{n + 1}^2x_{n + 1}^2 = 0`.

Returns a tuple which contains a parametrized solution to the equation,

sorted in the same order as the input variables.

Usage

=====

``diop_general_pythagorean(eq, param)``: where ``eq`` is a general

pythagorean equation which is assumed to be zero and ``param`` is the base

parameter used to construct other parameters by subscripting.

Examples

========

>>> from sympy.solvers.diophantine import diop_general_pythagorean

>>> from sympy.abc import a, b, c, d, e

>>> diop_general_pythagorean(a**2 + b**2 + c**2 - d**2)

(m1**2 + m2**2 - m3**2, 2*m1*m3, 2*m2*m3, m1**2 + m2**2 + m3**2)

>>> diop_general_pythagorean(9*a**2 - 4*b**2 + 16*c**2 + 25*d**2 + e**2)

(10*m1**2 + 10*m2**2 + 10*m3**2 - 10*m4**2, 15*m1**2 + 15*m2**2 + 15*m3**2 + 15*m4**2, 15*m1*m4, 12*m2*m4, 60*m3*m4)

"""

var, coeff, diop_type = classify_diop(eq)

if diop_type == "general_pythagorean":

return _diop_general_pythagorean(var, coeff, param)

def _diop_general_pythagorean(var, coeff, t):

if sign(coeff[var[0]**2]) + sign(coeff[var[1]**2]) + sign(coeff[var[2]**2]) < 0:

for key in coeff.keys():

coeff[key] = coeff[key] * -1

n = len(var)

index = 0

for i, v in enumerate(var):

if sign(coeff[v**2]) == -1:

index = i

m = symbols(str(t) + "1:" + str(n), integer=True)

l = []

ith = 0

for m_i in m:

ith = ith + m_i**2

l.append(ith - 2*m[n - 2]**2)

for i in range(n - 2):

l.append(2*m[i]*m[n-2])

sol = l[:index] + [ith] + l[index:]

lcm = 1

for i, v in enumerate(var):

if i == index or (index > 0 and i == 0) or (index == 0 and i == 1):

lcm = ilcm(lcm, sqrt(abs(coeff[v**2])))

else:

lcm = ilcm(lcm, sqrt(coeff[v**2]) if sqrt(coeff[v**2]) % 2 else sqrt(coeff[v**2]) // 2)

for i, v in enumerate(var):

sol[i] = (lcm*sol[i]) / sqrt(abs(coeff[v**2]))

return tuple(sol)

def diop_general_sum_of_squares(eq, limit=1):

"""

Solves the equation `x_{1}^2 + x_{2}^2 + . . . + x_{n}^2 - k = 0`.

Returns at most ``limit`` number of solutions. Currently there is no way to

set ``limit`` using higher level API's like ``diophantine()`` or

``diop_solve()`` but that will be fixed soon.

Usage

=====

``general_sum_of_squares(eq, limit)`` : Here ``eq`` is an expression which

is assumed to be zero. Also, ``eq`` should be in the form,

`x_{1}^2 + x_{2}^2 + . . . + x_{n}^2 - k = 0`. At most ``limit`` number of

solutions are returned.

Details

=======

When `n = 3` if `k = 4^a(8m + 7)` for some `a, m \in Z` then there will be

no solutions. Refer [1]_ for more details.

Examples

========

>>> from sympy.solvers.diophantine import diop_general_sum_of_squares

>>> from sympy.abc import a, b, c, d, e, f

>>> diop_general_sum_of_squares(a**2 + b**2 + c**2 + d**2 + e**2 - 2345)

set([(0, 48, 5, 4, 0)])

Reference

=========

.. [1] Representing an Integer as a sum of three squares, [online],

Available:

http://www.proofwiki.org/wiki/Integer_as_Sum_of_Three_Squares

"""

var, coeff, diop_type = classify_diop(eq)

if diop_type == "general_sum_of_squares":

return _diop_general_sum_of_squares(var, coeff, limit)

def _diop_general_sum_of_squares(var, coeff, limit=1):

n = len(var)

k = -int(coeff[Integer(1)])

s = set([])

if k < 0:

return set([])

if n == 3:

s.add(sum_of_three_squares(k))

elif n == 4:

s.add(sum_of_four_squares(k))

else:

m = n // 4

f = partition(k, m, True)

for j in range(limit):

soln = []

try:

l = next(f)

except StopIteration:

break

for n_i in l:

a, b, c, d = sum_of_four_squares(n_i)

soln = soln + [a, b, c, d]

soln = soln + [0] * (n % 4)

s.add(tuple(soln))

return s

## Functions below this comment can be more suitably grouped under an Additive number theory module

## rather than the Diophantine equation module.

def partition(n, k=None, zeros=False):

"""

Returns a generator that can be used to generate partitions of an integer

`n`.

A partition of `n` is a set of positive integers which add upto `n`. For

example, partitions of 3 are 3 , 1 + 2, 1 + 1+ 1. A partition is returned

as a tuple. If ``k`` equals None, then all possible partitions are returned

irrespective of their size, otherwise only the partitions of size ``k`` are

returned. If there are no partions of `n` with size `k` then an empty tuple

is returned. If the ``zero`` parameter is set to True then a suitable

number of zeros are added at the end of every partition of size less than

``k``.

``zero`` parameter is considered only if ``k`` is not None. When the

partitions are over, the last `next()` call throws the ``StopIteration``

exception, so this function should always be used inside a try - except

block.

Details

=======

``partition(n, k)``: Here ``n`` is a positive integer and ``k`` is the size

of the partition which is also positive integer.

Examples

========

>>> from sympy.solvers.diophantine import partition

>>> f = partition(5)

>>> next(f)

(1, 1, 1, 1, 1)

>>> next(f)

(1, 1, 1, 2)

>>> g = partition(5, 3)

>>> next(g)

(3, 1, 1)

>>> next(g)

(2, 2, 1)

Reference

=========

.. [1] Generating Integer Partitions, [online],

Available: http://jeromekelleher.net/partitions.php

"""

if n < 1:

yield tuple()

if k is not None:

if k < 1:

yield tuple()

elif k > n:

if zeros:

for i in range(1, n):

for t in partition(n, i):

yield (t,) + (0,) * (k - i)

else:

yield tuple()

else:

a = [1 for i in range(k)]

a[0] = n - k + 1

yield tuple(a)

i = 1

while a[0] >= n // k + 1:

j = 0

while j < i and j + 1 < k:

a[j] = a[j] - 1

a[j + 1] = a[j + 1] + 1

yield tuple(a)

j = j + 1

i = i + 1

if zeros:

for m in range(1, k):

for a in partition(n, m):

yield tuple(a) + (0,) * (k - m)

else:

a = [0 for i in range(n + 1)]

l = 1

y = n - 1

while l != 0:

x = a[l - 1] + 1

l -= 1

while 2*x <= y:

a[l] = x

y -= x

l += 1

m = l + 1

while x <= y:

a[l] = x

a[m] = y

yield tuple(a[:l + 2])

x += 1

y -= 1

a[l] = x + y

y = x + y - 1

yield tuple(a[:l + 1])

def prime_as_sum_of_two_squares(p):

"""

Represent a prime `p` which is congruent to 1 mod 4, as a sum of two

squares.

Examples

========

>>> from sympy.solvers.diophantine import prime_as_sum_of_two_squares

>>> prime_as_sum_of_two_squares(5)

(2, 1)

Reference

=========

.. [1] Representing a number as a sum of four squares, [online],

Available: http://www.schorn.ch/howto.html

"""

if p % 8 == 5:

b = 2

else:

b = 3

while pow(b, (p - 1) // 2, p) == 1:

b = nextprime(b)

b = pow(b, (p - 1) // 4, p)

a = p

while b**2 > p:

a, b = b, a % b

return (b, a % b)

def sum_of_three_squares(n):

"""

Returns a 3-tuple `(a, b, c)` such that `a^2 + b^2 + c^2 = n` and

`a, b, c \geq 0`.

Returns (None, None, None) if `n = 4^a(8m + 7)` for some `a, m \in Z`. See

[1]_ for more details.

Usage

=====

``sum_of_three_squares(n)``: Here ``n`` is a non-negative integer.

Examples

========

>>> from sympy.solvers.diophantine import sum_of_three_squares

>>> sum_of_three_squares(44542)

(207, 37, 18)

References

==========

.. [1] Representing a number as a sum of three squares, [online],

Available: http://www.schorn.ch/howto.html

"""

special = {1:(1, 0, 0), 2:(1, 1, 0), 3:(1, 1, 1), 10: (1, 3, 0), 34: (3, 3, 4), 58:(3, 7, 0),

85:(6, 7, 0), 130:(3, 11, 0), 214:(3, 6, 13), 226:(8, 9, 9), 370:(8, 9, 15),

526:(6, 7, 21), 706:(15, 15, 16), 730:(1, 27, 0), 1414:(6, 17, 33), 1906:(13, 21, 36),

2986: (21, 32, 39), 9634: (56, 57, 57)}

v = 0

if n == 0:

return (0, 0, 0)

while n % 4 == 0:

v = v + 1

n = n // 4

if n % 8 == 7:

return (None, None, None)

if n in special.keys():

x, y, z = special[n]

return (2**v*x, 2**v*y, 2**v*z)

l = int(sqrt(n))

if n == l**2:

return (2**v*l, 0, 0)

x = None

if n % 8 == 3:

l = l if l % 2 else l - 1

for i in range(l, -1, -2):

if isprime((n - i**2) // 2):

x = i

break

y, z = prime_as_sum_of_two_squares((n - x**2) // 2)

return (2**v*x, 2**v*(y + z), 2**v*abs(y - z))

if n % 8 == 2 or n % 8 == 6:

l = l if l % 2 else l - 1

else:

l = l - 1 if l % 2 else l

for i in range(l, -1, -2):

if isprime(n - i**2):

x = i

break

y, z = prime_as_sum_of_two_squares(n - x**2)

return (2**v*x, 2**v*y, 2**v*z)

def sum_of_four_squares(n):

"""

Returns a 4-tuple `(a, b, c, d)` such that `a^2 + b^2 + c^2 + d^2 = n`.

Here `a, b, c, d \geq 0`.

Usage

=====

``sum_of_four_squares(n)``: Here ``n`` is a non-negative integer.

Examples

========

>>> from sympy.solvers.diophantine import sum_of_four_squares

>>> sum_of_four_squares(3456)

(8, 48, 32, 8)

>>> sum_of_four_squares(1294585930293)

(0, 1137796, 2161, 1234)

References

==========

.. [1] Representing a number as a sum of four squares, [online],

Available: http://www.schorn.ch/howto.html

"""

if n == 0:

return (0, 0, 0, 0)

v = 0

while n % 4 == 0:

v = v + 1

n = n // 4

if n % 8 == 7:

d = 2

n = n - 4

elif n % 8 == 6 or n % 8 == 2:

d = 1

n = n - 1

else:

d = 0

x, y, z = sum_of_three_squares(n)

return (2**v*d, 2**v*x, 2**v*y, 2**v*z)

def power_representation(n, p, k, zeros=False):

"""

Returns a generator for finding k-tuples `(n_{1}, n_{2}, . . . n_{k})` such

that `n = n_{1}^p + n_{2}^p + . . . n_{k}^p`.

Here `n` is a non-negative integer. StopIteration exception is raised after

all the solutions are generated, so should always be used within a try-

catch block.

Usage

=====

``power_representation(n, p, k, zeros)``: Represent number ``n`` as a sum

of ``k``, ``p``th powers. If ``zeros`` is true, then the solutions will

contain zeros.

Examples

========

>>> from sympy.solvers.diophantine import power_representation

>>> f = power_representation(1729, 3, 2) # Represent 1729 as a sum of two cubes

>>> next(f)

(12, 1)

>>> next(f)

(10, 9)

"""

if p < 1 or k < 1 or n < 1:

raise ValueError("Expected: n > 0 and k >= 1 and p >= 1")

if k == 1:

if perfect_power(n):

yield (perfect_power(n)[0],)

else:

yield tuple()

elif p == 1:

for t in partition(n, k, zeros):

yield t

else:

l = []

a = integer_nthroot(n, p)[0]

for t in pow_rep_recursive(a, k, n, [], p):

yield t

if zeros:

for i in range(2, k):

for t in pow_rep_recursive(a, i, n, [], p):

yield t + (0,) * (k - i)

def pow_rep_recursive(n_i, k, n_remaining, terms, p):

if k == 0 and n_remaining == 0:

yield tuple(terms)

else:

if n_i >= 1 and k > 0 and n_remaining >= 0:

if n_i**p <= n_remaining:

for t in pow_rep_recursive(n_i, k - 1, n_remaining - n_i**p, terms + [n_i], p):

yield t

for t in pow_rep_recursive(n_i - 1, k, n_remaining, terms, p):

yield t

Coverage for sympy/solvers/diophantine.py : 89%

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