# 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.

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.

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

>>> x, y, z, t, p, q = symbols("x, y, z, t, p, q", integer=True)
>>> t1, t2, t3, t4, t5 = symbols("t1:6", 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)
{(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 diofant.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 diofant.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, m_0 + 5*m_1 - 14, 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. 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)
{(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)
{(-2*t**2 - 7*t + 10, -t**2 - 3*t + 5)}
>>> diophantine(x**2 + 2*x*y + y**2 - 3*x - 3*y)
{(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()
>>> 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.

>>> 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 diofant.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 has no solution.

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)
{(0, 0, 0)}
>>> diophantine(3*x**2 + 4*y**2 - 5*z**2 + 4*x*y - 7*y*z + 7*z*x)
{(-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 in 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 diofant.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 diofant.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 diofant.abc import e, f
>>> diophantine(9*a**2 + 16*b**2 + c**2 + 49*d**2 + 4*e**2 - 25*f**2)
{(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. Either you can call diop_general_pythagorean() or use the high level API. For the general sum of squares, this is also true, but one advantage of calling diop_general_sum_of_squares() is that you can control how many solutions are returned.

>>> from diofant.solvers.diophantine import diop_general_sum_of_squares
>>> eq = a**2 + b**2 + c**2 + d**2 - 18
>>> diophantine(eq)
{(0, 0, 3, 3), (0, 1, 1, 4), (1, 2, 2, 3)}
>>> diop_general_sum_of_squares(eq, 2)
{(0, 0, 3, 3), (1, 2, 2, 3)}


The sum_of_squares() routine will providean iterator that returns solutions and one may control whether the solutions contain zeros or not (and the solutions not containing zeros are returned first):

>>> from diofant.solvers.diophantine import sum_of_squares
>>> sos = sum_of_squares(18, 4, zeros=True)
>>> next(sos)
(1, 2, 2, 3)
>>> next(sos)
(0, 0, 3, 3)


Simple Eqyptian fractions can be found with the Diophantine module, too. For example, here are the ways that one might represent 1/2 as a sum of two unit fractions:

>>> diophantine(Eq(1/x + 1/y, Rational(1, 2)))
{(-2, 1), (1, -2), (3, 6), (4, 4), (6, 3)}


To get a more thorough understanding of the Diophantine module, please refer to the following blog.

https://thilinaatsympy.wordpress.com/

## User Functions¶

This function is imported into the global namespace with from diofant import *:

### diophantine¶

diofant.solvers.diophantine.diophantine(eq, param=Symbol('t', integer=True), syms=None)[source]

Simplify the solution procedure of diophantine equation eq by converting it into a product of terms which should equal zero.

$$(x + y)(x - y) = 0$$ 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. The elements of the tuple are the solutions for each variable in the the equation and are arranged according to the alphabetic ordering of the variables. e.g. For an equation with two variables, $$a$$ and $$b$$, the first element of the tuple is the solution for $$a$$ and the second for $$b$$.

Parameters: eq (Relational or Expr) – an equation (to be solved) t (Symbol, optional) – the parameter to be used in the solution. syms (list of Symbol’s, optional) – which determines the order of the elements in the returned tuple.

Examples

>>> diophantine(x**2 - y**2)
{(t_0, -t_0), (t_0, t_0)}

>>> diophantine(x*(2*x + 3*y - z))
{(0, n1, n2), (t_0, t_1, 2*t_0 + 3*t_1)}
>>> diophantine(x**2 + 3*x*y + 4*x)
{(0, n1), (3*t_0 - 4, -t_0)}


And this function is imported with from diofant.solvers.diophantine import *:

### classify_diop¶

diofant.solvers.diophantine.classify_diop(eq, _dict=True)[source]

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

Parameters: eq (Expr) – an expression, which is assumed to be zero.

Examples

>>> from diofant.abc import w
>>> 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**2: 1, x*y: -1}, 'binary_quadratic')

Returns: 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). The type is one of the following – binary_quadratic cubic_thue general_pythagorean general_sum_of_even_powers general_sum_of_squares homogeneous_general_quadratic homogeneous_ternary_quadratic homogeneous_ternary_quadratic_normal inhomogeneous_general_quadratic inhomogeneous_ternary_quadratic linear univariate

## Internal Functions¶

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

### diop_solve¶

diofant.solvers.diophantine.diop_solve(eq, param=Symbol('t', integer=True))[source]

Solves the diophantine equation eq.

Unlike diophantine(), factoring of eq is not attempted. Uses classify_diop() to determine the type of the equation and calls the appropriate solver function.

Parameters: eq (Expr) – an expression, which is assumed to be zero. t (Symbol, optional) – a parameter, to be used in the solution.

Examples

>>> from diofant.abc import 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, 8*t_0 + 4*t_1 + 5, 7*t_0 + 3*t_1 + 5)
>>> diop_solve(x + 3*y - 4*z + w -6)
(t_0, t_0 + t_1, 6*t_0 + 5*t_1 + 4*t_2 - 6, 5*t_0 + 4*t_1 + 3*t_2 - 6)
>>> diop_solve(x**2 + y**2 - 5)
{(-1, -2), (-1, 2), (1, -2), (1, 2)}


### diop_linear¶

diofant.solvers.diophantine.diop_linear(eq, param=Symbol('t', integer=True))[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.

Parameters: eq (Expr) – is a linear diophantine equation which is assumed to be zero. param (Symbol, optional) – is the parameter to be used in the solution.

Examples

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


### base_solution_linear¶

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

Return the base solution for the linear equation, $$ax + by = c$$.

Used by diop_linear() to find the base solution of a linear Diophantine equation. If t is given then the parametrized solution is returned.

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

>>> 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)


diofant.solvers.diophantine.diop_quadratic(eq, param=Symbol('t', integer=True))[source]

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.

Parameters: eq (Expr) – should be a quadratic bivariate expression which is assumed to be zero. param (Symbol, optional) – is a parameter to be used in the solution.

Examples

>>> diop_quadratic(x**2 + y**2 + 2*x + 2*y + 2, t)
{(-1, -1)}


References

### diop_DN¶

diofant.solvers.diophantine.diop_DN(D, N, t=Symbol('t', integer=True))[source]

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

Mainly concerned with the case $$D > 0, D$$ is not a perfect square, which is the same as the generalized Pell equation. The LMM algorithm is used to solve this equation.

Returns: A tuple of pairs, ($$x, y)$$, for each class of the solutions. Other solutions of the class can be constructed according to the values of D and N. D, N (Integer) – correspond to D and N in the equation. t (Symbol, optional) – is the parameter to be used in the solutions.

Examples

>>> 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)]


References

### cornacchia¶

diofant.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 $$a = b$$, only the solutions with $$x \leq y$$ are found. For more details, see the References.

Examples

>>> cornacchia(2, 3, 35)  # equation 2x**2 + 3y**2 = 35
{(2, 3), (4, 1)}
>>> cornacchia(1, 1, 25)  # equation x**2 + y**2 = 25
{(3, 4)}


References

### diop_bf_DN¶

diofant.solvers.diophantine.diop_bf_DN(D, N, t=Symbol('t', integer=True))[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. 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.

Parameters: D, N (Integer) – correspond to D and N in the equation. t (Symbol, optional) – is the parameter to be used in the solutions.

Examples

>>> diop_bf_DN(13, -4)
[(3, 1), (-3, 1), (36, 10)]
>>> diop_bf_DN(986, 1)
[(49299, 1570)]


References

### transformation_to_DN¶

diofant.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. 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

Parameters: eq (Expr) – the quadratic expression to be transformed.

Examples

>>> 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 diofant.abc import X, Y
>>> 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((x**2 - 3*x*y - y**2 - 2*y + 1).subs({x: u, y: v}))
>>> 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.

References

### find_DN¶

diofant.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().

Parameters: eq (Expr) – is the quadratic expression to be transformed.

Examples

>>> 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().

References

diofant.solvers.diophantine.diop_ternary_quadratic(eq)[source]

Solves the general quadratic ternary form, $$ax^2 + by^2 + cz^2 + fxy + gyz + hxz = 0$$.

Returns: tuple – which is a base solution for the above equation. If there are no solutions, (None, None, None) is returned. eq (Expr) – should be an homogeneous expression of degree two in three variables and it is assumed to be zero.

Examples

>>> 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)


### square_factor¶

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

Returns an integer $$c$$ s.t. $$a = c^2k, \ c,k \in Z$$. Here $$k$$ is square free. $$a$$ can be given as an integer or a dictionary of factors.

Examples

>>> square_factor(24)
2
>>> square_factor(-36*3)
6
>>> square_factor(1)
1
>>> square_factor({3: 2, 2: 1, -1: 1})
3


### descent¶

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

Returns a non-trivial solution, (x, y, z), to $$x^2 = Ay^2 + Bz^2$$ using Lagrange’s descent method with lattice-reduction. $$A$$ and $$B$$ are assumed to be valid for such a solution to exist.

This is faster than the normal Lagrange’s descent algorithm because the Gaussian reduction is used.

Examples

>>> 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

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

### diop_general_pythagorean¶

diofant.solvers.diophantine.diop_general_pythagorean(eq, param=Symbol('m', integer=True))[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.

Parameters: eq (Expr) – is a general pythagorean equation which is assumed to be zero param (Symbol, optional) – is the base parameter used to construct other parameters by subscripting.

Examples

>>> from diofant.abc import 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)


### diop_general_sum_of_squares¶

diofant.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.

Parameters: eq (Expr) – is an expression which is assumed to be zero. Also, eq should be in the form, $$x_{1}^2 + x_{2}^2 + . . . + x_{n}^2 - k = 0$$. limit (int, optional) – upper limit (the default is 1) for number of solutions returned.

Notes

When $$n = 3$$ if $$k = 4^a(8m + 7)$$ for some $$a, m \in Z$$ then there will be no solutions.

Examples

>>> from diofant.abc import e, f
>>> diop_general_sum_of_squares(a**2 + b**2 + c**2 + d**2 + e**2 - 2345)
{(15, 22, 22, 24, 24)}


References

### diop_general_sum_of_even_powers¶

diofant.solvers.diophantine.diop_general_sum_of_even_powers(eq, limit=1)[source]

Solves the equation $$x_{1}^e + x_{2}^e + . . . + x_{n}^e - k = 0$$ where $$e$$ is an even, integer power.

Returns at most limit number of solutions.

Parameters: eq (Expr) – An expression which is assumed to be zero. Also, eq should be in the form, $$x_{1}^e + x_{2}^e + . . . + x_{n}^e - k = 0$$. limit (Expr, optional) – Limit number of returned solutions. Default is 1.

Examples

>>> diop_general_sum_of_even_powers(a**4 + b**4 - (2**4 + 3**4))
{(2, 3)}


### partition¶

diofant.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 up to $$n$$. 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 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.

Parameters: n (int) – is a positive integer k (int, optional) – is the size of the partition which is also positive integer. The default is None. zeros (boolean, optional) – 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

>>> f = partition(5)
>>> next(f)
(1, 1, 1, 1, 1)
>>> next(f)
(1, 1, 1, 2)
>>> g = partition(5, 3)
>>> next(g)
(1, 1, 3)
>>> next(g)
(1, 2, 2)
>>> g = partition(5, 3, zeros=True)
>>> next(g)
(0, 0, 5)


### sum_of_three_squares¶

diofant.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 if $$n = 4^a(8m + 7)$$ for some $$a, m \in Z$$.

Parameters: n (int) – a non-negative integer.

Examples

>>> sum_of_three_squares(44542)
(18, 37, 207)


References

### sum_of_four_squares¶

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

Parameters: n (int) – is a non-negative integer.

Examples

>>> sum_of_four_squares(3456)
(8, 8, 32, 48)
>>> sum_of_four_squares(1294585930293)
(0, 1234, 2161, 1137796)


References

### power_representation¶

diofant.solvers.diophantine.power_representation(n, p, k, zeros=False)[source]

Returns a generator for finding k-tuples of integers, $$(n_{1}, n_{2}, ... n_{k})$$, such that $$n = n_{1}^p + n_{2}^p + ... n_{k}^p$$.

Parameters: n (int) – a non-negative integer k, p (int) – parameters to control representation n as a sum of k, p-th powers. zeros (boolean, optional) – if True (the default is False), then the solutions will contain zeros.

Examples

Represent 1729 as a sum of two cubes:

>>> f = power_representation(1729, 3, 2)
>>> next(f)
(9, 10)
>>> next(f)
(1, 12)


If the flag $$zeros$$ is True, the solution may contain tuples with zeros; any such solutions will be generated after the solutions without zeros:

>>> list(power_representation(125, 2, 3, zeros=True))
[(5, 6, 8), (3, 4, 10), (0, 5, 10), (0, 2, 11)]


For even $$p$$ the $$permute_sign$$ function can be used to get all signed values:

>>> from diofant.utilities.iterables import permute_signs
>>> list(permute_signs((1, 12)))
[(1, 12), (-1, 12), (1, -12), (-1, -12)]


All possible signed permutations can also be obtained:

>>> from diofant.utilities.iterables import signed_permutations
>>> list(signed_permutations((1, 12)))
[(1, 12), (-1, 12), (1, -12), (-1, -12), (12, 1), (-12, 1),
(12, -1), (-12, -1)]

diofant.solvers.diophantine.sum_of_powers()

alias of power_representation()

### sum_of_squares¶

diofant.solvers.diophantine.sum_of_squares(n, k, zeros=False)[source]

Return a generator that yields the k-tuples of nonnegative values, the squares of which sum to n. If zeros is False (default) then the solution will not contain zeros. The nonnegative elements of a tuple are sorted.

• If k == 1 and n is square, (n,) is returned.
• If k == 2 then n can only be written as a sum of squares if every prime in the factorization of n that has the form 4*k + 3 has an even multiplicity. If n is prime then it can only be written as a sum of two squares if it is in the form 4*k + 1.
• if k == 3 then n can be written as a sum of squares if it does not have the form 4**m*(8*k + 7).
• all integers can be written as the sum of 4 squares.
• if k > 4 then n can be partitioned and each partition can be written as a sum of 4 squares; if n is not evenly divisible by 4 then n can be written as a sum of squares only if the an additional partition can be written as as sum of squares. For example, if k = 6 then n is partitioned into two parts, the first being written as a sum of 4 squares and the second being written as a sum of 2 squares – which can only be done if the contition above for k = 2 can be met, so this will automatically reject certain partitions of n.

Examples

>>> list(sum_of_squares(25, 2))
[(3, 4)]
>>> list(sum_of_squares(25, 2, True))
[(3, 4), (0, 5)]
>>> list(sum_of_squares(25, 4))
[(1, 2, 2, 4)]


### merge_solution¶

diofant.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¶

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

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

### PQa¶

diofant.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.

Parameters: P_0, Q_0, D (Integer) – 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

>>> 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

### equivalent¶

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

Parameters: u, v, r, s, D, N (Integer)

Examples

>>> equivalent(18, 5, -18, -5, 13, -1)
True
>>> equivalent(3, 1, -18, 393, 109, -4)
False


References

diofant.solvers.diophantine.parametrize_ternary_quadratic(eq)[source]

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

>>> 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

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

diofant.solvers.diophantine.diop_ternary_quadratic_normal(eq)[source]

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

>>> 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)


### ldescent¶

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

Return a non-trivial solution to $$w^2 = Ax^2 + By^2$$ using Lagrange’s method; return None if there is no such solution.

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

>>> 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

• The algorithmic resolution of Diophantine equations, Nigel P. Smart, London Mathematical Society Student Texts 41, Cambridge University Press, Cambridge, 1998.
• Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin, Mathematics of Computation, Volume 00, Number 0, http://eprints.nottingham.ac.uk/60/1/kvxefz87.pdf

### gaussian_reduce¶

diofant.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.

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

References

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

### holzer¶

diofant.solvers.diophantine.holzer(x, y, z, a, b, c)[source]

Simplify the solution $$(x, y, z)$$ of the equation $$ax^2 + by^2 = cz^2$$ with $$a, b, c > 0$$ and $$z^2 \geq \mid ab \mid$$ to a new reduced solution $$(x', y', z')$$ such that $$z'^2 \leq \mid ab \mid$$.

The algorithm is an interpretation of Mordell’s reduction as described on page 8 of Cremona and Rusin’s paper and the work of Mordell.

References

• Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin, Mathematics of Computation, Volume 00, Number 0.
• Diophantine Equations, L. J. Mordell, page 48.

### prime_as_sum_of_two_squares¶

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

Represent a prime $$p$$ as a unique sum of two squares; this can only be done if the prime is congruent to 1 mod 4.

Examples

>>> prime_as_sum_of_two_squares(7)  # can't be done
>>> prime_as_sum_of_two_squares(5)
(1, 2)


References

### square_factor¶

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

Returns an integer $$c$$ s.t. $$a = c^2k, \ c,k \in Z$$. Here $$k$$ is square free. $$a$$ can be given as an integer or a dictionary of factors.

Examples

>>> square_factor(24)
2
>>> square_factor(-36*3)
6
>>> square_factor(1)
1
>>> square_factor({3: 2, 2: 1, -1: 1})
3


### sqf_normal¶

diofant.solvers.diophantine.sqf_normal(a, b, c, steps=False)[source]

Return $$a', b', c'$$, the coefficients of the square-free normal form of $$ax^2 + by^2 + cz^2 = 0$$, where $$a', b', c'$$ are pairwise prime. If $$steps$$ is True then also return three tuples: $$sq$$, $$sqf$$, and $$(a', b', c')$$ where $$sq$$ contains the square factors of $$a$$, $$b$$ and $$c$$ after removing the $$gcd(a, b, c)$$; $$sqf$$ contains the values of $$a$$, $$b$$ and $$c$$ after removing both the $$gcd(a, b, c)$$ and the square factors.

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

>>> sqf_normal(2 * 3**2 * 5, 2 * 5 * 11, 2 * 7**2 * 11)
(11, 1, 5)
>>> sqf_normal(2 * 3**2 * 5, 2 * 5 * 11, 2 * 7**2 * 11, True)
((3, 1, 7), (5, 55, 11), (11, 1, 5))


References

### reconstruct¶

diofant.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 the square-free normal form of the equation, $$a'*x^2 + b'*y^2 + c'*z^2$$, where $$a'$$, $$b'$$ and $$c'$$ are square free and $$gcd(a', b', c') == 1$$.

### transformation_to_normal¶

diofant.solvers.diophantine.transformation_to_normal(eq)[source]

Returns the transformation Matrix that converts a general ternary quadratic equation $$eq$$ ($$ax^2 + by^2 + cz^2 + dxy + eyz + fxz$$) to a form without cross terms: $$ax^2 + by^2 + cz^2 = 0$$. This is not used in solving ternary quadratics; it is only implemented for the sake of completeness.