# Symbolic Equations and Inequalities¶

Sage can solve symbolic equations and inequalities. For example, we derive the quadratic formula as follows:

sage: a,b,c = var('a,b,c')
sage: qe = (a*x^2 + b*x + c == 0)
sage: qe
a*x^2 + b*x + c == 0
sage: print solve(qe, x)
[
x == -1/2*(b + sqrt(b^2 - 4*a*c))/a,
x == -1/2*(b - sqrt(b^2 - 4*a*c))/a
]


## The operator, left hand side, and right hand side¶

Operators:

sage: eqn = x^3 + 2/3 >= x - pi
sage: eqn.operator()
<built-in function ge>
sage: (x^3 + 2/3 < x - pi).operator()
<built-in function lt>
sage: (x^3 + 2/3 == x - pi).operator()
<built-in function eq>


Left hand side:

sage: eqn = x^3 + 2/3 >= x - pi
sage: eqn.lhs()
x^3 + 2/3
sage: eqn.left()
x^3 + 2/3
sage: eqn.left_hand_side()
x^3 + 2/3


Right hand side:

sage: (x + sqrt(2) >= sqrt(3) + 5/2).right()
sqrt(3) + 5/2
sage: (x + sqrt(2) >= sqrt(3) + 5/2).rhs()
sqrt(3) + 5/2
sage: (x + sqrt(2) >= sqrt(3) + 5/2).right_hand_side()
sqrt(3) + 5/2


## Arithmetic¶

sage: var('a,b')
(a, b)
sage: m = 144 == -10 * a + b
sage: n = 136 == 10 * a + b
sage: m + n
280 == 2*b
sage: int(-144) + m
0 == -10*a + b - 144


Subtract two symbolic equations:

sage: var('a,b')
(a, b)
sage: m = 144 == 20 * a + b
sage: n = 136 == 10 * a + b
sage: m - n
8 == 10*a
sage: int(144) - m
0 == -20*a - b + 144


Multiply two symbolic equations:

sage: x = var('x')
sage: m = x == 5*x + 1
sage: n = sin(x) == sin(x+2*pi)
sage: m * n
x*sin(x) == (5*x + 1)*sin(2*pi + x)
sage: m = 2*x == 3*x^2 - 5
sage: int(-1) * m
-2*x == -3*x^2 + 5


Divide two symbolic equations:

sage: x = var('x')
sage: m = x == 5*x + 1
sage: n = sin(x) == sin(x+2*pi)
sage: m/n
x/sin(x) == (5*x + 1)/sin(2*pi + x)
sage: m = x != 5*x + 1
sage: n = sin(x) != sin(x+2*pi)
sage: m/n
x/sin(x) != (5*x + 1)/sin(2*pi + x)


## Substitution¶

Substitution into relations:

sage: x, a = var('x, a')
sage: eq = (x^3 + a == sin(x/a)); eq
x^3 + a == sin(x/a)
sage: eq.substitute(x=5*x)
125*x^3 + a == sin(5*x/a)
sage: eq.substitute(a=1)
x^3 + 1 == sin(x)
sage: eq.substitute(a=x)
x^3 + x == sin(1)
sage: eq.substitute(a=x, x=1)
x + 1 == sin(1/x)
sage: eq.substitute({a:x, x:1})
x + 1 == sin(1/x)


## Solving¶

We can solve equations:

sage: x = var('x')
sage: S = solve(x^3 - 1 == 0, x)
sage: S
[x == 1/2*I*sqrt(3) - 1/2, x == -1/2*I*sqrt(3) - 1/2, x == 1]
sage: S[0]
x == 1/2*I*sqrt(3) - 1/2
sage: S[0].right()
1/2*I*sqrt(3) - 1/2
sage: S = solve(x^3 - 1 == 0, x, solution_dict=True)
sage: S
[{x: 1/2*I*sqrt(3) - 1/2}, {x: -1/2*I*sqrt(3) - 1/2}, {x: 1}]
sage: z = 5
sage: solve(z^2 == sqrt(3),z)
Traceback (most recent call last):
...
TypeError: 5 is not a valid variable.


We illustrate finding multiplicities of solutions:

sage: f = (x-1)^5*(x^2+1)
sage: solve(f == 0, x)
[x == -I, x == I, x == 1]
sage: solve(f == 0, x, multiplicities=True)
([x == -I, x == I, x == 1], [1, 1, 5])


We can also solve many inequalities:

sage: solve(1/(x-1)<=8,x)
[[x < 1], [x >= (9/8)]]


We can numerically find roots of equations:

sage: (x == sin(x)).find_root(-2,2)
0.0
sage: (x^5 + 3*x + 2 == 0).find_root(-2,2,x)
-0.6328345202421523
sage: (cos(x) == sin(x)).find_root(10,20)
19.634954084936208


We illustrate some valid error conditions:

sage: (cos(x) != sin(x)).find_root(10,20)
Traceback (most recent call last):
...
ValueError: Symbolic equation must be an equality.
sage: (SR(3)==SR(2)).find_root(-1,1)
Traceback (most recent call last):
...
RuntimeError: no zero in the interval, since constant expression is not 0.


There must be at most one variable:

sage: x, y = var('x,y')
sage: (x == y).find_root(-2,2)
Traceback (most recent call last):
...
NotImplementedError: root finding currently only implemented in 1 dimension.


## Assumptions¶

Forgetting assumptions:

sage: var('x,y')
(x, y)
sage: forget() #Clear assumptions
sage: assume(x>0, y < 2)
sage: assumptions()
[x > 0, y < 2]
sage: (y < 2).forget()
sage: assumptions()
[x > 0]
sage: forget()
sage: assumptions()
[]


## Miscellaneous¶

Conversion to Maxima:

sage: x = var('x')
sage: eq = (x^(3/5) >= pi^2 + e^i)
sage: eq._maxima_init_()
'(_SAGE_VAR_x)^(3/5) >= ((%pi)^(2))+(exp(0+%i*1))'
sage: e1 = x^3 + x == sin(2*x)
sage: z = e1._maxima_()
sage: z.parent() is sage.calculus.calculus.maxima
True
sage: z = e1._maxima_(maxima)
sage: z.parent() is maxima
True
sage: z = maxima(e1)
sage: z.parent() is maxima
True


Conversion to Maple:

sage: x = var('x')
sage: eq = (x == 2)
sage: eq._maple_init_()
'x = 2'


Comparison:

sage: x = var('x')
sage: (x>0) == (x>0)
True
sage: (x>0) == (x>1)
False
sage: (x>0) != (x>1)
True


Variables appearing in the relation:

sage: var('x,y,z,w')
(x, y, z, w)
sage: f =  (x+y+w) == (x^2 - y^2 - z^3);   f
w + x + y == -z^3 + x^2 - y^2
sage: f.variables()
(w, x, y, z)


LaTeX output:

sage: latex(x^(3/5) >= pi)
x^{\frac{3}{5}} \geq \pi


When working with the symbolic complex number $$I$$, notice that comparison do not automatically simplifies even in trivial situations:

sage: I^2 == -1
-1 == -1
sage: I^2 < 0
-1 < 0
sage: (I+1)^4 > 0
-4 > 0


Nevertheless, if you force the comparison, you get the right answer (trac ticket #7160):

sage: bool(I^2 == -1)
True
sage: bool(I^2 < 0)
True
sage: bool((I+1)^4 > 0)
False


## More Examples¶

sage: x,y,a = var('x,y,a')
sage: f = x^2 + y^2 == 1
sage: f.solve(x)
[x == -sqrt(-y^2 + 1), x == sqrt(-y^2 + 1)]

sage: f = x^5 + a
sage: solve(f==0,x)
[x == (-a)^(1/5)*e^(2/5*I*pi), x == (-a)^(1/5)*e^(4/5*I*pi), x == (-a)^(1/5)*e^(-4/5*I*pi), x == (-a)^(1/5)*e^(-2/5*I*pi), x == (-a)^(1/5)]


You can also do arithmetic with inequalities, as illustrated below:

sage: var('x y')
(x, y)
sage: f = x + 3 == y - 2
sage: f
x + 3 == y - 2
sage: g = f - 3; g
x == y - 5
sage: h =  x^3 + sqrt(2) == x*y*sin(x)
sage: h
x^3 + sqrt(2) == x*y*sin(x)
sage: h - sqrt(2)
x^3 == x*y*sin(x) - sqrt(2)
sage: h + f
x^3 + x + sqrt(2) + 3 == x*y*sin(x) + y - 2
sage: f = x + 3 < y - 2
sage: g = 2 < x+10
sage: f - g
x + 1 < -x + y - 12
sage: f + g
x + 5 < x + y + 8
sage: f*(-1)
-x - 3 < -y + 2


TESTS:

We test serializing symbolic equations:

sage: eqn = x^3 + 2/3 >= x
x^3 + 2/3 >= x
True


AUTHORS:

• Bobby Moretti: initial version (based on a trick that Robert Bradshaw suggested).
• William Stein: second version
• William Stein (2007-07-16): added arithmetic with symbolic equations
sage.symbolic.relation.solve(f, *args, **kwds)

Algebraically solve an equation or system of equations (over the complex numbers) for given variables. Inequalities and systems of inequalities are also supported.

INPUT:

• f - equation or system of equations (given by a list or tuple)
• *args - variables to solve for.
• solution_dict - bool (default: False); if True or non-zero, return a list of dictionaries containing the solutions. If there are no solutions, return an empty list (rather than a list containing an empty dictionary). Likewise, if there’s only a single solution, return a list containing one dictionary with that solution.

There are a few optional keywords if you are trying to solve a single equation. They may only be used in that context.

• multiplicities - bool (default: False); if True, return corresponding multiplicities. This keyword is incompatible with to_poly_solve=True and does not make any sense when solving inequalities.
• explicit_solutions - bool (default: False); require that all roots be explicit rather than implicit. Not used when solving inequalities.
• to_poly_solve - bool (default: False) or string; use Maxima’s to_poly_solver package to search for more possible solutions, but possibly encounter approximate solutions. This keyword is incompatible with multiplicities=True and is not used when solving inequalities. Setting to_poly_solve to ‘force’ (string) omits Maxima’s solve command (useful when some solutions of trigonometric equations are lost).

EXAMPLES:

sage: x, y = var('x, y')
sage: solve([x+y==6, x-y==4], x, y)
[[x == 5, y == 1]]
sage: solve([x^2+y^2 == 1, y^2 == x^3 + x + 1], x, y)
[[x == -1/2*I*sqrt(3) - 1/2, y == -sqrt(-1/2*I*sqrt(3) + 3/2)],
[x == -1/2*I*sqrt(3) - 1/2, y == sqrt(-1/2*I*sqrt(3) + 3/2)],
[x == 1/2*I*sqrt(3) - 1/2, y == -sqrt(1/2*I*sqrt(3) + 3/2)],
[x == 1/2*I*sqrt(3) - 1/2, y == sqrt(1/2*I*sqrt(3) + 3/2)],
[x == 0, y == -1],
[x == 0, y == 1]]
sage: solve([sqrt(x) + sqrt(y) == 5, x + y == 10], x, y)
[[x == -5/2*I*sqrt(5) + 5, y == 5/2*I*sqrt(5) + 5], [x == 5/2*I*sqrt(5) + 5, y == -5/2*I*sqrt(5) + 5]]
sage: solutions=solve([x^2+y^2 == 1, y^2 == x^3 + x + 1], x, y, solution_dict=True)
sage: for solution in solutions: print solution[x].n(digits=3), ",", solution[y].n(digits=3)
-0.500 - 0.866*I , -1.27 + 0.341*I
-0.500 - 0.866*I , 1.27 - 0.341*I
-0.500 + 0.866*I , -1.27 - 0.341*I
-0.500 + 0.866*I , 1.27 + 0.341*I
0.000 , -1.00
0.000 , 1.00


Whenever possible, answers will be symbolic, but with systems of equations, at times approximations will be given, due to the underlying algorithm in Maxima:

sage: sols = solve([x^3==y,y^2==x],[x,y]); sols[-1], sols[0]
([x == 0, y == 0], [x == (0.3090169943749475 + 0.9510565162951535*I), y == (-0.8090169943749475 - 0.5877852522924731*I)])
sage: sols[0][0].rhs().pyobject().parent()
Complex Double Field


If f is only one equation or expression, we use the solve method for symbolic expressions, which defaults to exact answers only:

sage: solve([y^6==y],y)
[y == e^(2/5*I*pi), y == e^(4/5*I*pi), y == e^(-4/5*I*pi), y == e^(-2/5*I*pi), y == 1, y == 0]
sage: solve( [y^6 == y], y)==solve( y^6 == y, y)
True


Here we demonstrate very basic use of the optional keywords for a single expression to be solved:

sage: ((x^2-1)^2).solve(x)
[x == -1, x == 1]
sage: ((x^2-1)^2).solve(x,multiplicities=True)
([x == -1, x == 1], [2, 2])
sage: solve(sin(x)==x,x)
[x == sin(x)]
sage: solve(sin(x)==x,x,explicit_solutions=True)
[]
sage: solve(abs(1-abs(1-x)) == 10, x)
[abs(abs(x - 1) - 1) == 10]
sage: solve(abs(1-abs(1-x)) == 10, x, to_poly_solve=True)
[x == -10, x == 12]


Note

For more details about solving a single equation, see the documentation for the single-expression solve().

sage: from sage.symbolic.expression import Expression
sage: Expression.solve(x^2==1,x)
[x == -1, x == 1]


We must solve with respect to actual variables:

sage: z = 5
sage: solve([8*z + y == 3, -z +7*y == 0],y,z)
Traceback (most recent call last):
...
TypeError: 5 is not a valid variable.


If we ask for dictionaries containing the solutions, we get them:

sage: solve([x^2-1],x,solution_dict=True)
[{x: -1}, {x: 1}]
sage: solve([x^2-4*x+4],x,solution_dict=True)
[{x: 2}]
sage: res = solve([x^2 == y, y == 4],x,y,solution_dict=True)
sage: for soln in res: print "x: %s, y: %s"%(soln[x], soln[y])
x: 2, y: 4
x: -2, y: 4


If there is a parameter in the answer, that will show up as a new variable. In the following example, r1 is a real free variable (because of the r):

sage: solve([x+y == 3, 2*x+2*y == 6],x,y)
[[x == -r1 + 3, y == r1]]


Especially with trigonometric functions, the dummy variable may be implicitly an integer (hence the z):

sage: solve([cos(x)*sin(x) == 1/2, x+y == 0],x,y)
[[x == 1/4*pi + pi*z79, y == -1/4*pi - pi*z79]]


Expressions which are not equations are assumed to be set equal to zero, as with $$x$$ in the following example:

sage: solve([x, y == 2],x,y)
[[x == 0, y == 2]]


If True appears in the list of equations it is ignored, and if False appears in the list then no solutions are returned. E.g., note that the first 3==3 evaluates to True, not to a symbolic equation.

sage: solve([3==3, 1.00000000000000*x^3 == 0], x)
[x == 0]
sage: solve([1.00000000000000*x^3 == 0], x)
[x == 0]


Here, the first equation evaluates to False, so there are no solutions:

sage: solve([1==3, 1.00000000000000*x^3 == 0], x)
[]


Completely symbolic solutions are supported:

sage: var('s,j,b,m,g')
(s, j, b, m, g)
sage: sys = [ m*(1-s) - b*s*j, b*s*j-g*j ];
sage: solve(sys,s,j)
[[s == 1, j == 0], [s == g/b, j == (b - g)*m/(b*g)]]
sage: solve(sys,(s,j))
[[s == 1, j == 0], [s == g/b, j == (b - g)*m/(b*g)]]
sage: solve(sys,[s,j])
[[s == 1, j == 0], [s == g/b, j == (b - g)*m/(b*g)]]


Inequalities can be also solved:

sage: solve(x^2>8,x)
[[x < -2*sqrt(2)], [x > 2*sqrt(2)]]


We use use_grobner in Maxima if no solution is obtained from Maxima’s to_poly_solve:

sage: x,y=var('x y'); c1(x,y)=(x-5)^2+y^2-16; c2(x,y)=(y-3)^2+x^2-9
sage: solve([c1(x,y),c2(x,y)],[x,y])
[[x == -9/68*sqrt(55) + 135/68, y == -15/68*sqrt(11)*sqrt(5) + 123/68], [x == 9/68*sqrt(55) + 135/68, y == 15/68*sqrt(11)*sqrt(5) + 123/68]]


TESTS:

sage: solve([sin(x)==x,y^2==x],x,y)
[sin(x) == x, y^2 == x]
sage: solve(0==1,x)
Traceback (most recent call last):
...
TypeError:  The first argument must be a symbolic expression or a list of symbolic expressions.


Test if the empty list is returned, too, when (a list of) dictionaries (is) are requested (#8553):

sage: solve([SR(0)==1],x)
[]
sage: solve([SR(0)==1],x,solution_dict=True)
[]
sage: solve([x==1,x==-1],x)
[]
sage: solve([x==1,x==-1],x,solution_dict=True)
[]
sage: solve((x==1,x==-1),x,solution_dict=0)
[]


Relaxed form, suggested by Mike Hansen (#8553):

sage: solve([x^2-1],x,solution_dict=-1)
[{x: -1}, {x: 1}]
sage: solve([x^2-1],x,solution_dict=1)
[{x: -1}, {x: 1}]
sage: solve((x==1,x==-1),x,solution_dict=-1)
[]
sage: solve((x==1,x==-1),x,solution_dict=1)
[]


This inequality holds for any real x (trac #8078):

sage: solve(x^4+2>0,x)
[x < +Infinity]


Test for user friendly input handling trac ticket #13645:

sage: poly.<a,b> = PolynomialRing(RR)
sage: solve([a+b+a*b == 1], a)
Traceback (most recent call last):
...
TypeError: The first argument to solve() should be a symbolic expression or a list of symbolic expressions, cannot handle <type 'bool'>
sage: solve([a, b], (1, a))
Traceback (most recent call last):
...
TypeError: 1 is not a valid variable.
sage: solve([x == 1], (1, a))
Traceback (most recent call last):
...
TypeError: (1, a) are not valid variables.


Test that the original version of a system in the French Sage book now works (trac ticket #14306):

sage: var('y,z')
(y, z)
sage: solve([x^2 * y * z == 18, x * y^3 * z == 24, x * y * z^4 == 6], x, y, z)
[[x == 3, y == 2, z == 1], [x == (1.337215067... - 2.685489874...*I), y == (-1.700434271... + 1.052864325...*I), z == (0.9324722294... - 0.3612416661...*I)], ...]

sage.symbolic.relation.solve_ineq(ineq, vars=None)

Solves inequalities and systems of inequalities using Maxima. Switches between rational inequalities (sage.symbolic.relation.solve_ineq_rational) and fourier elimination (sage.symbolic.relation.solve_ineq_fouried). See the documentation of these functions for more details.

INPUT:

• ineq - one inequality or a list of inequalities

Case1: If ineq is one equality, then it should be rational expression in one varible. This input is passed to sage.symbolic.relation.solve_ineq_univar function.

Case2: If ineq is a list involving one or more inequalities, than the input is passed to sage.symbolic.relation.solve_ineq_fourier function. This function can be used for system of linear inequalities and for some types of nonlinear inequalities. See http://maxima.cvs.sourceforge.net/viewvc/maxima/maxima/share/contrib/fourier_elim/rtest_fourier_elim.mac for a big gallery of problems covered by this algorithm.

• vars - optional parameter with list of variables. This list is used only if fourier elimination is used. If omitted or if rational inequality is solved, then variables are determined automatically.

OUTPUT:

• list – output is list of solutions as a list of simple inequalities output [A,B,C] means (A or B or C) each A, B, C is again a list and if A=[a,b], then A means (a and b).

EXAMPLES:

sage: from sage.symbolic.relation import solve_ineq


Inequalities in one variable. The variable is detected automatically:

sage: solve_ineq(x^2-1>3)
[[x < -2], [x > 2]]

sage: solve_ineq(1/(x-1)<=8)
[[x < 1], [x >= (9/8)]]


System of inequalities with automatically detected inequalities:

sage: y=var('y')
sage: solve_ineq([x-y<0,x+y-3<0],[y,x])
[[x < y, y < -x + 3, x < (3/2)]]
sage: solve_ineq([x-y<0,x+y-3<0],[x,y])
[[x < min(-y + 3, y)]]


Note that although Sage will detect the variables automatically, the order it puts them in may depend on the system, so the following command is only guaranteed to give you one of the above answers:

sage: solve_ineq([x-y<0,x+y-3<0]) # not tested - random
[[x < y, y < -x + 3, x < (3/2)]]


ALGORITHM:

Calls solve_ineq_fourier if inequalities are list and solve_ineq_univar of the inequality is symbolic expression. See the description of these commands for more details related to the set of inequalities which can be solved. The list is empty if there is no solution.

AUTHORS:

• Robert Marik (01-2010)
sage.symbolic.relation.solve_ineq_fourier(ineq, vars=None)

Solves sytem of inequalities using Maxima and fourier elimination

Can be used for system of linear inequalities and for some types of nonlinear inequalities. For examples see the section EXAMPLES below and http://maxima.cvs.sourceforge.net/viewvc/maxima/maxima/share/contrib/fourier_elim/rtest_fourier_elim.mac

INPUT:

• ineq - list with system of inequalities
• vars - optionally list with variables for fourier elimination.

OUTPUT:

• list - output is list of solutions as a list of simple inequalities output [A,B,C] means (A or B or C) each A, B, C is again a list and if A=[a,b], then A means (a and b). The list is empty if there is no solution.

EXAMPLES:

sage: from sage.symbolic.relation import solve_ineq_fourier
sage: y=var('y')
sage: solve_ineq_fourier([x+y<9,x-y>4],[x,y])
[[y + 4 < x, x < -y + 9, y < (5/2)]]
sage: solve_ineq_fourier([x+y<9,x-y>4],[y,x])
[[y < min(x - 4, -x + 9)]]

sage: solve_ineq_fourier([x^2>=0])
[[x < +Infinity]]

sage: solve_ineq_fourier([log(x)>log(y)],[x,y])
[[y < x, 0 < y]]
sage: solve_ineq_fourier([log(x)>log(y)],[y,x])
[[0 < y, y < x, 0 < x]]


Note that different systems will find default variables in different orders, so the following is not tested:

sage: solve_ineq_fourier([log(x)>log(y)]) # not tested - one of the following appears
[[0 < y, y < x, 0 < x]]
[[y < x, 0 < y]]


ALGORITHM:

Calls Maxima command fourier_elim

AUTHORS:

• Robert Marik (01-2010)
sage.symbolic.relation.solve_ineq_univar(ineq)

Function solves rational inequality in one variable.

INPUT:

• ineq - inequality in one variable

OUTPUT:

• list – output is list of solutions as a list of simple inequalities output [A,B,C] means (A or B or C) each A, B, C is again a list and if A=[a,b], then A means (a and b). The list is empty if there is no solution.

EXAMPLES:

sage: from sage.symbolic.relation import solve_ineq_univar
sage: solve_ineq_univar(x-1/x>0)
[[x > -1, x < 0], [x > 1]]

sage: solve_ineq_univar(x^2-1/x>0)
[[x < 0], [x > 1]]

sage: solve_ineq_univar((x^3-1)*x<=0)
[[x >= 0, x <= 1]]


ALGORITHM:

Calls Maxima command solve_rat_ineq

AUTHORS:

• Robert Marik (01-2010)
sage.symbolic.relation.solve_mod(eqns, modulus, solution_dict=False)

Return all solutions to an equation or list of equations modulo the given integer modulus. Each equation must involve only polynomials in 1 or many variables.

By default the solutions are returned as $$n$$-tuples, where $$n$$ is the number of variables appearing anywhere in the given equations. The variables are in alphabetical order.

INPUT:

• eqns - equation or list of equations
• modulus - an integer
• solution_dict - bool (default: False); if True or non-zero, return a list of dictionaries containing the solutions. If there are no solutions, return an empty list (rather than a list containing an empty dictionary). Likewise, if there’s only a single solution, return a list containing one dictionary with that solution.

EXAMPLES:

sage: var('x,y')
(x, y)
sage: solve_mod([x^2 + 2 == x, x^2 + y == y^2], 14)
[(4, 2), (4, 6), (4, 9), (4, 13)]
sage: solve_mod([x^2 == 1, 4*x  == 11], 15)
[(14,)]


Fermat’s equation modulo 3 with exponent 5:

sage: var('x,y,z')
(x, y, z)
sage: solve_mod([x^5 + y^5 == z^5], 3)
[(0, 0, 0), (0, 1, 1), (0, 2, 2), (1, 0, 1), (1, 1, 2), (1, 2, 0), (2, 0, 2), (2, 1, 0), (2, 2, 1)]


We can solve with respect to a bigger modulus if it consists only of small prime factors:

sage: [d] = solve_mod([5*x + y == 3, 2*x - 3*y == 9], 3*5*7*11*19*23*29, solution_dict = True)
sage: d[x]
12915279
sage: d[y]
8610183


For cases where there are relatively few solutions and the prime factors are small, this can be efficient even if the modulus itself is large:

sage: sorted(solve_mod([x^2 == 41], 10^20))
[(4538602480526452429,), (11445932736758703821,), (38554067263241296179,),
(45461397519473547571,), (54538602480526452429,), (61445932736758703821,),
(88554067263241296179,), (95461397519473547571,)]


We solve a simple equation modulo 2:

sage: x,y = var('x,y')
sage: solve_mod([x == y], 2)
[(0, 0), (1, 1)]


Warning

The current implementation splits the modulus into prime powers, then naively enumerates all possible solutions (starting modulo primes and then working up through prime powers), and finally combines the solution using the Chinese Remainder Theorem. The interface is good, but the algorithm is very inefficient if the modulus has some larger prime factors! Sage does have the ability to do something much faster in certain cases at least by using Groebner basis, linear algebra techniques, etc. But for a lot of toy problems this function as is might be useful. At least it establishes an interface.

TESTS:

Make sure that we short-circuit in at least some cases:

sage: solve_mod([2*x==1], 2*next_prime(10^50))
[]


Try multi-equation cases:

sage: x, y, z = var("x y z")
sage: solve_mod([2*x^2 + x*y, -x*y+2*y^2+x-2*y, -2*x^2+2*x*y-y^2-x-y], 12)
[(0, 0), (4, 4), (0, 3), (4, 7)]
sage: eqs = [-y^2+z^2, -x^2+y^2-3*z^2-z-1, -y*z-z^2-x-y+2, -x^2-12*z^2-y+z]
sage: solve_mod(eqs, 11)
[(8, 5, 6)]


Confirm that modulus 1 now behaves as it should:

sage: x, y = var("x y")
sage: solve_mod([x==1], 1)
[(0,)]
sage: solve_mod([2*x^2+x*y, -x*y+2*y^2+x-2*y, -2*x^2+2*x*y-y^2-x-y], 1)
[(0, 0)]

sage.symbolic.relation.string_to_list_of_solutions(s)

Used internally by the symbolic solve command to convert the output of Maxima’s solve command to a list of solutions in Sage’s symbolic package.

EXAMPLES:

We derive the (monic) quadratic formula:

sage: var('x,a,b')
(x, a, b)
sage: solve(x^2 + a*x + b == 0, x)
[x == -1/2*a - 1/2*sqrt(a^2 - 4*b), x == -1/2*a + 1/2*sqrt(a^2 - 4*b)]


Behind the scenes when the above is evaluated the function string_to_list_of_solutions() is called with input the string $$s$$ below:

sage: s = '[x=-(sqrt(a^2-4*b)+a)/2,x=(sqrt(a^2-4*b)-a)/2]'
sage: sage.symbolic.relation.string_to_list_of_solutions(s)
[x == -1/2*a - 1/2*sqrt(a^2 - 4*b), x == -1/2*a + 1/2*sqrt(a^2 - 4*b)]

sage.symbolic.relation.test_relation_maxima(relation)

Return True if this (in)equality is definitely true. Return False if it is false or the algorithm for testing (in)equality is inconclusive.

EXAMPLES:

sage: from sage.symbolic.relation import test_relation_maxima
sage: k = var('k')
sage: pol = 1/(k-1) - 1/k -1/k/(k-1);
sage: test_relation_maxima(pol == 0)
True
sage: f = sin(x)^2 + cos(x)^2 - 1
sage: test_relation_maxima(f == 0)
True
sage: test_relation_maxima( x == x )
True
sage: test_relation_maxima( x != x )
False
sage: test_relation_maxima( x > x )
False
sage: test_relation_maxima( x^2 > x )
False
sage: test_relation_maxima( x + 2 > x )
True
sage: test_relation_maxima( x - 2 > x )
False


Here are some examples involving assumptions:

sage: x, y, z = var('x, y, z')
sage: assume(x>=y,y>=z,z>=x)
sage: test_relation_maxima(x==z)
True
sage: test_relation_maxima(z<x)
False
sage: test_relation_maxima(z>y)
False
sage: test_relation_maxima(y==z)
True
sage: forget()
sage: assume(x>=1,x<=1)
sage: test_relation_maxima(x==1)
True
sage: test_relation_maxima(x>1)
False
sage: test_relation_maxima(x>=1)
True
sage: test_relation_maxima(x!=1)
False
sage: forget()
sage: assume(x>0)
sage: test_relation_maxima(x==0)
False
sage: test_relation_maxima(x>-1)
True
sage: test_relation_maxima(x!=0)
True
sage: test_relation_maxima(x!=1)
False
sage: forget()


TESTS:

Ensure that canonicalize_radical() and simplify_log are not used inappropriately, trac ticket #17389. Either one would simplify f to zero below:

sage: x,y = SR.var('x,y')
sage: assume(y, 'complex')
sage: f = log(x*y) - (log(x) + log(y))
sage: f(x=-1, y=i)
-2*I*pi
sage: test_relation_maxima(f == 0)
False
sage: forget()


Ensure that the sqrt(x^2) -> abs(x) simplification is not performed when testing equality:

sage: assume(x, 'complex')
sage: f = sqrt(x^2) - abs(x)
sage: test_relation_maxima(f == 0)
False
sage: forget()


If assumptions are made, simplify_rectform() is used:

sage: assume(x, 'real')
sage: f1 = ( e^(I*x) - e^(-I*x) ) / ( I*e^(I*x) + I*e^(-I*x) )
sage: f2 = sin(x)/cos(x)
sage: test_relation_maxima(f1 - f2 == 0)
True
sage: forget()


But not if x itself is complex:

sage: assume(x, 'complex')
sage: f1 = ( e^(I*x) - e^(-I*x) ) / ( I*e^(I*x) + I*e^(-I*x) )
sage: f2 = sin(x)/cos(x)
sage: test_relation_maxima(f1 - f2 == 0)
False
sage: forget()


If assumptions are made, then simplify_factorial() is used:

sage: n,k = SR.var('n,k')
sage: assume(n, 'integer')
sage: assume(k, 'integer')
sage: f1 = factorial(n+1)/factorial(n)
sage: f2 = n + 1
sage: test_relation_maxima(f1 - f2 == 0)
True
sage: forget()