# Diophantine¶

## Diophantine equations¶

The word “Diophantine” comes with the name Diophantus, a mathematician lived in the great city of Alexandria sometime around 250 AD. Often referred to as the “father of Algebra”, Diophantus in his famous work “Arithmetica” presented 150 problems that marked the early beginnings of number theory, the field of study about integers and their properties. Diophantine equations play a central and an important part in number theory.

We call a “Diophantine equation” to an equation of the form, $$f(x_1, x_2, \ldots x_n) = 0$$ where $$n \geq 2$$ and $$x_1, x_2, \ldots x_n$$ are integer variables. If we can find $$n$$ integers $$a_1, a_2, \ldots a_n$$ such that $$x_1 = a_1, x_2 = a_2, \ldots x_n = a_n$$ satisfies the above equation, we say that the equation is solvable. You can read more about Diophantine equations in [1] and [2].

Currently, following five types of Diophantine equations can be solved using diophantine() and other helper functions of the Diophantine module.

• Linear Diophantine equations: $$a_1x_1 + a_2x_2 + \ldots + a_nx_n = b$$.
• General binary quadratic equation: $$ax^2 + bxy + cy^2 + dx + ey + f = 0$$
• Homogeneous ternary quadratic equation: $$ax^2 + by^2 + cz^2 + dxy + eyz + fzx = 0$$
• Extended Pythagorean equation: $$a_{1}x_{1}^2 + a_{2}x_{2}^2 + \ldots + a_{n}x_{n}^2 = a_{n+1}x_{n+1}^2$$
• General sum of squares: $$x_{1}^2 + x_{2}^2 + \ldots + x_{n}^2 = k$$

## Module structure¶

This module contains diophantine() and helper functions that are needed to solve certain Diophantine equations. It’s structured in the following manner.

When an equation is given to diophantine(), it factors the equation(if possible) and solves the equation given by each factor by calling diop_solve() separately. Then all the results are combined using merge_solution().

diop_solve() internally uses classify_diop() to find the type of the equation(and some other details) given to it and then calls the appropriate solver function based on the type returned. For example, if classify_diop() returned “linear” as the type of the equation, then diop_solve() calls diop_linear() to solve the equation.

Each of the functions, diop_linear(), diop_quadratic(), diop_ternary_quadratic(), diop_general_pythagorean() and diop_general_sum_of_squares() solves a specific type of equations and the type can be easily guessed by it’s name.

Apart from these functions, there are a considerable number of other functions in the “Diophantine Module” and all of them are listed under User functions and Internal functions.

## Tutorial¶

First, let’s import the highest API of the Diophantine module.

>>> from sympy.solvers.diophantine import diophantine


Before we start solving the equations, we need to define the variables.

>>> from sympy import symbols
>>> x, y, z = symbols("x, y, z", integer=True)


Let’s start by solving the easiest type of Diophantine equations, i.e. linear Diophantine equations. Let’s solve $$2x + 3y = 5$$. Note that although we write the equation in the above form, when we input the equation to any of the functions in Diophantine module, it needs to be in the form $$eq = 0$$.

>>> diophantine(2*x + 3*y - 5)
set([(3*t_0 - 5, -2*t_0 + 5)])


Note that stepping one more level below the highest API, we can solve the very same equation by calling diop_solve().

>>> from sympy.solvers.diophantine import diop_solve
>>> diop_solve(2*x + 3*y - 5)
(3*t_0 - 5, -2*t_0 + 5)


Note that it returns a tuple rather than a set. diophantine() always return a set of tuples. But diop_solve() may return a single tuple or a set of tuples depending on the type of the equation given.

We can also solve this equation by calling diop_linear(), which is what diop_solve() calls internally.

>>> from sympy.solvers.diophantine import diop_linear
>>> diop_linear(2*x + 3*y - 5)
(3*t_0 - 5, -2*t_0 + 5)


If the given equation has no solutions then the outputs will look like below.

>>> diophantine(2*x + 4*y - 3)
set()
>>> diop_solve(2*x + 4*y - 3)
(None, None)
>>> diop_linear(2*x + 4*y - 3)
(None, None)


Note that except for the highest level API, in case of no solutions, a tuple of $$None$$ are returned. Size of the tuple is the same as the number of variables. Also, one can specifically set the parameter to be used in the solutions by passing a customized parameter. Consider the following example:

>>> m = symbols("m", integer=True)
>>> diop_solve(2*x + 3*y - 5, m)
(3*m_0 - 5, -2*m_0 + 5)


For linear Diophantine equations, the customized parameter is the prefix used for each free variable in the solution. Consider the following example:

>>> diop_solve(2*x + 3*y - 5*z + 7, m)
(m_0, -9*m_0 - 5*m_1 - 14, -5*m_0 - 3*m_1 - 7)


In the solution above, m_0 and m_1 are independent free variables.

Please note that for the moment, users can set the parameter only for linear Diophantine equations and binary quadratic equations.

Let’s try solving a binary quadratic equation which is an equation with two variables and has a degree of two. Before trying to solve these equations, an idea about various cases associated with the equation would help a lot. Please refer [3] and [4] for detailed analysis of different cases and the nature of the solutions. Let us define $$\Delta = b^2 - 4ac$$ w.r.t. the binary quadratic $$ax^2 + bxy + cy^2 + dx + ey + f = 0$$.

When $$\Delta < 0$$, there are either no solutions or only a finite number of solutions.

>>> diophantine(x**2 - 4*x*y + 8*y**2 - 3*x + 7*y - 5)
set([(2, 1), (5, 1)])


In the above equation $$\Delta = (-4)^2 - 4*1*8 = -16$$ and hence only a finite number of solutions exist.

When $$\Delta = 0$$ we might have either no solutions or parameterized solutions.

>>> diophantine(3*x**2 - 6*x*y + 3*y**2 - 3*x + 7*y - 5)
set()
>>> diophantine(x**2 - 4*x*y + 4*y**2 - 3*x + 7*y - 5)
set([(-2*t**2 - 7*t + 10, -t**2 - 3*t + 5)])
>>> diophantine(x**2 + 2*x*y + y**2 - 3*x - 3*y)
set([(t_0, -t_0), (t_0, -t_0 + 3)])


The most interesting case is when $$\Delta > 0$$ and it is not a perfect square. In this case, the equation has either no solutions or an infinte number of solutions. Consider the below cases where $$\Delta = 8$$.

>>> diophantine(x**2 - 4*x*y + 2*y**2 - 3*x + 7*y - 5)
set()
>>> from sympy import sqrt
>>> n = symbols("n", integer=True)
>>> s = diophantine(x**2 -  2*y**2 - 2*x - 4*y, n)
>>> x_1, y_1 = s.pop()
>>> x_2, y_2 = s.pop()
>>> x_n = -(-2*sqrt(2) + 3)**n/2 + sqrt(2)*(-2*sqrt(2) + 3)**n/2 - sqrt(2)*(2*sqrt(2) + 3)**n/2 - (2*sqrt(2) + 3)**n/2 + 1
>>> x_1 == x_n or x_2 == x_n
True
>>> y_n = -sqrt(2)*(-2*sqrt(2) + 3)**n/4 + (-2*sqrt(2) + 3)**n/2 + sqrt(2)*(2*sqrt(2) + 3)**n/4 + (2*sqrt(2) + 3)**n/2 - 1
>>> y_1 == y_n or y_2 == y_n
True


Here $$n$$ is an integer. Although x_n and y_n may not look like integers, substituting in specific values for n (and simplifying) shows that they are. For example consider the following example where we set n equal to 9.

>>> from sympy import simplify
>>> simplify(x_n.subs({n: 9}))
-9369318


Any binary quadratic of the form $$ax^2 + bxy + cy^2 + dx + ey + f = 0$$ can be transformed to an equivalent form $$X^2 - DY^2 = N$$.

>>> from sympy.solvers.diophantine import find_DN, diop_DN, transformation_to_DN
>>> find_DN(x**2 - 3*x*y + y**2 - 7*x + 5*y - 3)
(5, 920)


So, the above equation is equivalent to the equation $$X^2 - 5Y^2 = 920$$ after a linear transformation. If we want to find the linear transformation, we can use transformation_to_DN()

>>> A, B = transformation_to_DN(x**2 - 3*x*y + y**2 - 7*x + 5*y - 3)


Here A is a 2 X 2 matrix and B is a 2 X 1 matrix such that the transformation

$\begin{split}\begin{bmatrix} X\\Y \end{bmatrix} = A \begin{bmatrix} x\\y \end{bmatrix} + B\end{split}$

gives the equation $$X^2 -5Y^2 = 920$$. Values of $$A$$ and $$B$$ are as belows.

>>> A
Matrix([
[1/10, 3/10],
[   0,  1/5]])
>>> B
Matrix([
[  1/5],
[-11/5]])


We can solve an equation of the form $$X^2 - DY^2 = N$$ by passing $$D$$ and $$N$$ to diop_DN()

>>> diop_DN(5, 920)
[]


Unfortunately, our equation does not have solutions.

Now let’s turn to homogeneous ternary quadratic equations. These equations are of the form $$ax^2 + by^2 + cz^2 + dxy + eyz + fzx = 0$$. These type of equations either have infinitely many solutions or no solutions (except the obvious solution (0, 0, 0))

>>> diophantine(3*x**2 + 4*y**2 - 5*z**2 + 4*x*y + 6*y*z + 7*z*x)
set()
>>> diophantine(3*x**2 + 4*y**2 - 5*z**2 + 4*x*y - 7*y*z + 7*z*x)
set([(-16*p**2 + 28*p*q + 20*q**2, 3*p**2 + 38*p*q - 25*q**2, 4*p**2 - 24*p*q + 68*q**2)])


If you are only interested about a base solution rather than the parameterized general solution (to be more precise, one of the general solutions), you can use diop_ternary_quadratic().

>>> from sympy.solvers.diophantine import diop_ternary_quadratic
>>> diop_ternary_quadratic(3*x**2 + 4*y**2 - 5*z**2 + 4*x*y - 7*y*z + 7*z*x)
(-4, 5, 1)


diop_ternary_quadratic() first converts the given equation to an equivalent equation of the form $$w^2 = AX^2 + BY^2$$ and then it uses descent() to solve the latter equation. You can refer to the docs of transformation_to_normal() to find more on this. The equation $$w^2 = AX^2 + BY^2$$ can be solved more easily by using the Aforementioned descent().

>>> from sympy.solvers.diophantine import descent
>>> descent(3, 1) # solves the equation w**2 = 3*Y**2 + Z**2
(1, 0, 1)


Here the solution tuple is in the order (w, Y, Z)

The extended Pythagorean equation, $$a_{1}x_{1}^2 + a_{2}x_{2}^2 + \ldots + a_{n}x_{n}^2 = a_{n+1}x_{n+1}^2$$ and the general sum of squares equation, $$x_{1}^2 + x_{2}^2 + \ldots + x_{n}^2 = k$$ can also be solved using the Diophantine module.

>>> from sympy.abc import a, b, c, d, e, f
>>> diophantine(9*a**2 + 16*b**2 + c**2 + 49*d**2 + 4*e**2 - 25*f**2)
set([(70*t1**2 + 70*t2**2 + 70*t3**2 + 70*t4**2 - 70*t5**2, 105*t1*t5, 420*t2*t5, 60*t3*t5, 210*t4*t5, 42*t1**2 + 42*t2**2 + 42*t3**2 + 42*t4**2 + 42*t5**2)])


function diop_general_pythagorean() can also be called directly to solve the same equation. This is true about the general sum of squares too. Either you can call diop_general_pythagorean() or use the high level API.

>>> diophantine(a**2 + b**2 + c**2 + d**2 + e**2 + f**2 - 112)
set([(8, 4, 4, 4, 0, 0)])


If you want to get a more thorough idea about the the Diophantine module please refer to the following blog.

http://thilinaatsympy.wordpress.com/

## References¶

 [1] Andreescu, Titu. Andrica, Dorin. Cucurezeanu, Ion. An Introduction to Diophantine Equations
 [2] Diophantine Equation, Wolfram Mathworld, [online]. Available: http://mathworld.wolfram.com/DiophantineEquation.html
 [3] Methods to solve Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0,[online], Available: http://www.alpertron.com.ar/METHODS.HTM
 [4] Solving the equation ax^2+ bxy + cy^2 + dx + ey + f= 0, [online], Available: http://www.jpr2718.org/ax2p.pdf

## User Functions¶

These are functions that are imported into the global namespace with from sympy import *. These functions are intended for use by ordinary users of SymPy.

### diophantine()¶

sympy.solvers.diophantine.diophantine(eq, param=t)[source]

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 $$b$$.

diop_solve

Examples

>>> from sympy.solvers.diophantine import diophantine
>>> from sympy.abc import x, y, z
>>> diophantine(x**2 - y**2)
set([(-t_0, -t_0), (t_0, -t_0)])


#>>> 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)])

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.

### diop_solve()¶

sympy.solvers.diophantine.diop_solve(eq, param=t)[source]

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.

diophantine

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_0 - 5, -2*t_0 + 5)
>>> diop_solve(4*x + 3*y -4*z + 5)
(t_0, -4*t_1 + 5, t_0 - 3*t_1 + 5)
>>> diop_solve(x + 3*y - 4*z + w -6)
(t_0, t_0 + t_1, -2*t_0 - 3*t_1 - 4*t_2 - 6, -t_0 - 2*t_1 - 3*t_2 - 6)
>>> diop_solve(x**2 + y**2 - 5)
set([(-2, -1), (-2, 1), (2, -1), (2, 1)])


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.

### classify_diop()¶

sympy.solvers.diophantine.classify_diop(eq)[source]

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”}

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')


Usage

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

Details

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

### diop_linear()¶

sympy.solvers.diophantine.diop_linear(eq, param=t)[source]

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.

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_0 - 5, -2*t_0 - 5)


Here x = -3*t_0 - 5 and y = -2*t_0 - 5

>>> diop_linear(2*x - 3*y - 4*z -3)
(t_0, -6*t_0 - 4*t_1 + 3, 5*t_0 + 3*t_1 - 3)


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.

### base_solution_linear()¶

sympy.solvers.diophantine.base_solution_linear(c, a, b, t=None)[source]

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.

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)


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.

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.

References

 [R445] Methods to solve Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0,[online], Available: http://www.alpertron.com.ar/METHODS.HTM
 [R446] Solving the equation ax^2+ bxy + cy^2 + dx + ey + f= 0, [online], Available: http://www.jpr2718.org/ax2p.pdf

Examples

>>> from sympy.abc import x, y, t
>>> diop_quadratic(x**2 + y**2 + 2*x + 2*y + 2, t)
set([(-1, -1)])


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.

### diop_DN()¶

sympy.solvers.diophantine.diop_DN(D, N, t=t)[source]

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 [R447] 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 $$(x, y)$$.

find_DN, diop_bf_DN

References

 [R447] (1, 2) 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

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)]


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.

### cornacchia()¶

sympy.solvers.diophantine.cornacchia(a, b, m)[source]

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.

References

 [R448] Nitaj, “L’algorithme de Cornacchia”
 [R449] Solving the diophantine equation ax**2 + by**2 = m by Cornacchia’s method, [online], Available: http://www.numbertheory.org/php/cornacchia.html

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)])


### diop_bf_DN()¶

sympy.solvers.diophantine.diop_bf_DN(D, N, t=t)[source]

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 [R450]. 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.

diop_DN

References

 [R450] (1, 2) 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

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)]


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.

### transformation_to_DN()¶

sympy.solvers.diophantine.transformation_to_DN(eq)[source]

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 [R451] 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

find_DN

References

 [R451] (1, 2) 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

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.

Usage

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

### find_DN()¶

sympy.solvers.diophantine.find_DN(eq)[source]

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().

transformation_to_DN

References

 [R452] 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

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().

Usage

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

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.

Examples

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


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.

### square_factor()¶

sympy.solvers.diophantine.square_factor(a)[source]

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)
2
>>> square_factor(36)
6
>>> square_factor(1)
1


### descent()¶

sympy.solvers.diophantine.descent(A, B)[source]

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.

References

 [R453] Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin, Mathematics of Computation, Volume 00, Number 0.

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)


### diop_general_pythagorean()¶

sympy.solvers.diophantine.diop_general_pythagorean(eq, param=m)[source]

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.

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)


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.

### diop_general_sum_of_squares()¶

sympy.solvers.diophantine.diop_general_sum_of_squares(eq, limit=1)[source]

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.

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)])


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 [R454] for more details.

Reference

 [R454] Representing an Integer as a sum of three squares, [online], Available: http://www.proofwiki.org/wiki/Integer_as_Sum_of_Three_Squares

### partition()¶

sympy.solvers.diophantine.partition(n, k=None, zeros=False)[source]

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.

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)


Details

partition(n, k): Here n is a positive integer and k is the size of the partition which is also positive integer.

Reference

 [R455] Generating Integer Partitions, [online], Available: http://jeromekelleher.net/partitions.php

### sum_of_three_squares()¶

sympy.solvers.diophantine.sum_of_three_squares(n)[source]

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 [R456] for more details.

References

 [R456] (1, 2) Representing a number as a sum of three squares, [online], Available: http://www.schorn.ch/howto.html

Examples

>>> from sympy.solvers.diophantine import sum_of_three_squares
>>> sum_of_three_squares(44542)
(207, 37, 18)


Usage

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

### sum_of_four_squares()¶

sympy.solvers.diophantine.sum_of_four_squares(n)[source]

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

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

References

 [R457] Representing a number as a sum of four squares, [online], Available: http://www.schorn.ch/howto.html

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)


Usage

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

## Internal Functions¶

These functions are intended for the internal use in Diophantine module.

### merge_solution¶

sympy.solvers.diophantine.merge_solution(var, var_t, solution)[source]

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.

### divisible¶

sympy.solvers.diophantine.divisible(a, b)[source]

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

### extended_euclid¶

sympy.solvers.diophantine.extended_euclid(a, b)[source]

For given a, b returns a tuple containing integers $$x$$, $$y$$ and $$d$$ such that $$ax + by = d$$. Here $$d = gcd(a, b)$$.

Examples

>>> from sympy.solvers.diophantine import extended_euclid
>>> extended_euclid(4, 6)
(-1, 1, 2)
>>> extended_euclid(3, 5)
(2, -1, 1)


Usage

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

Details

a Any instance of Integer. b Any instance of Integer.

### PQa¶

sympy.solvers.diophantine.PQa(P_0, Q_0, D)[source]

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 [R458] for more detailed information.

References

 [R458] (1, 2) 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

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)


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.

### equivalent¶

sympy.solvers.diophantine.equivalent(u, v, r, s, D, N)[source]

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 [R459]. No test is performed to test whether $$(u, v)$$ and $$(r, s)$$ are actually solutions to the equation. User should take care of this.

References

 [R459] (1, 2) 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

Examples

>>> from sympy.solvers.diophantine import equivalent
>>> equivalent(18, 5, -18, -5, 13, -1)
True
>>> equivalent(3, 1, -18, 393, 109, -4)
False


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.

### simplified¶

sympy.solvers.diophantine.simplified(x, y, z)[source]

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

Returns the parametrized general solution for the ternary quadratic equation eq which has the form $$ax^2 + by^2 + cz^2 + fxy + gyz + hxz = 0$$.

References

 [R460] The algorithmic resolution of Diophantine equations, Nigel P. Smart, London Mathematical Society Student Texts 41, Cambridge University Press, Cambridge, 1998.

Examples

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


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.

Examples

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


Usage

diop_ternary_quadratic_normal(eq): where eq is an equation of the form $$ax^2 + by^2 + cz^2 = 0$$.

### ldescent¶

sympy.solvers.diophantine.ldescent(A, B)[source]

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.

References

 [R461] The algorithmic resolution of Diophantine equations, Nigel P. Smart, London Mathematical Society Student Texts 41, Cambridge University Press, Cambridge, 1998.
 [R462] Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin, Mathematics of Computation, Volume 00, Number 0.

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)


### gaussian_reduce¶

sympy.solvers.diophantine.gaussian_reduce(w, a, b)[source]

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.

References

 [R463] Gaussian lattice Reduction [online]. Available: http://home.ie.cuhk.edu.hk/~wkshum/wordpress/?p=404
 [R464] Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin, Mathematics of Computation, Volume 00, Number 0.

Details

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

### holzer¶

sympy.solvers.diophantine.holzer(x_0, y_0, z_0, a, b, c)[source]

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 $$z^2 \leq \mid ab \mid$$.

### prime_as_sum_of_two_squares¶

sympy.solvers.diophantine.prime_as_sum_of_two_squares(p)[source]

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

 [R465] Representing a number as a sum of four squares, [online], Available: http://www.schorn.ch/howto.html

### pairwise_prime¶

sympy.solvers.diophantine.pairwise_prime(a, b, c)[source]

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 $$a'x^2 + b'y^2 + c'z^2 = 0$$.

make_prime, reocnstruct

Examples

>>> from sympy.solvers.diophantine import pairwise_prime
>>> pairwise_prime(6, 15, 10)
(5, 2, 3)


### make_prime¶

sympy.solvers.diophantine.make_prime(a, b, c)[source]

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.

pairwaise_prime, reconstruct

Examples

>>> from sympy.solvers.diophantine import make_prime
>>> make_prime(4, 2, 7)
(2, 1, 14)


### reconstruct¶

sympy.solvers.diophantine.reconstruct(a, b, z)[source]

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.