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)
{(3*t_0  5, 5  2*t_0)}
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, 5  2*t_0)
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, 5  2*t_0)
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, 5  2*m_0)
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. 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)
{(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, 3  t_0)}
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 infinite 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
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 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)
{(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 sympy.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 sympy.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:
>>> from sympy import Eq, S
>>> diophantine(Eq(1/x + 1/y, S(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.
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¶
This functions is imported into the global namespace
with from sympy import *
:
diophantine¶

sympy.solvers.diophantine.
diophantine
(eq, param=t, syms=None, permute=False)[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()
. (Although it is possible to calldiop_solve()
directly, one must be careful to pass an equation in the correct form and to interpret the output correctly;diophantine()
is the publicfacing function to use in general.)Output of
diophantine()
is a set of tuples. The elements of the tuple are the solutions for each variable in 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 \(b\).Usage
diophantine(eq, t, syms)
: Solve the diophantine equationeq
.t
is the optional parameter to be used bydiop_solve()
.syms
is an optional list of symbols which determines the order of the elements in the returned tuple.By default, only the base solution is returned. If
permute
is set to True then permutations of the base solution and/or permutations of the signs of the values will be returned when applicable.>>> from sympy.solvers.diophantine import diophantine >>> from sympy.abc import a, b >>> eq = a**4 + b**4  (2**4 + 3**4) >>> diophantine(eq) {(2, 3)} >>> diophantine(eq, permute=True) {(3, 2), (3, 2), (2, 3), (2, 3), (2, 3), (2, 3), (3, 2), (3, 2)}
Details
eq
should be an expression which is assumed to be zero.t
is the parameter to be used in the solution.Examples
>>> from sympy.abc import x, y, z >>> 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 sympy.solvers.diophantine import *
:
Internal Functions¶
These functions are intended for internal use in the Diophantine module.
diop_solve¶

sympy.solvers.diophantine.
diop_solve
(eq, param=t)[source]¶ Solves the diophantine equation
eq
.Unlike
diophantine()
, factoring ofeq
is not attempted. Usesclassify_diop()
to determine the type of the equation and calls the appropriate solver function.Use of
diophantine()
is recommended over other helper functions.diop_solve()
can return either a set or a tuple depending on the nature of the equation.Usage
diop_solve(eq, t)
: Solve diophantine equation,eq
usingt
as a parameter if needed.Details
eq
should be an expression which is assumed to be zero.t
is a parameter to be used in the solution.Examples
>>> from sympy.solvers.diophantine import diop_solve >>> from sympy.abc import x, y, z, w >>> diop_solve(2*x + 3*y  5) (3*t_0  5, 5  2*t_0) >>> 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) {(2, 1), (2, 1), (1, 2), (1, 2), (1, 2), (1, 2), (2, 1), (2, 1)}
See also
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.
Usage
diop_linear(eq)
: Returns a tuple containing solutions to the diophantine equationeq
. Values in the tuple is arranged in the same order as the sorted variables.Details
eq
is a linear diophantine equation which is assumed to be zero.param
is the parameter to be used in the solution.Examples
>>> from sympy.solvers.diophantine import diop_linear >>> from sympy.abc import x, y, z, t >>> 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, 2*t_0 + 4*t_1 + 3, t_0  3*t_1  3)
base_solution_linear¶

sympy.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. Ift
is given then the parametrized solution is returned.Usage
base_solution_linear(c, a, b, t)
:a
,b
,c
are coefficients in \(ax + by = c\) andt
is the parameter to be used in the solution.Examples
>>> from sympy.solvers.diophantine import base_solution_linear >>> from sympy.abc import t >>> base_solution_linear(5, 2, 3) # equation 2*x + 3*y = 5 (5, 5) >>> base_solution_linear(0, 5, 7) # equation 5*x + 7*y = 0 (0, 0) >>> base_solution_linear(5, 2, 3, t) # equation 2*x + 3*y = 5 (3*t  5, 5  2*t) >>> base_solution_linear(0, 5, 7, t) # equation 5*x + 7*y = 0 (7*t, 5*t)
diop_quadratic¶

sympy.solvers.diophantine.
diop_quadratic
(eq, param=t)[source]¶ Solves quadratic diophantine equations.
i.e. equations of the form \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\). Returns a set containing the tuples \((x, y)\) which contains the solutions. If there are no solutions then \((None, None)\) is returned.
Usage
diop_quadratic(eq, param)
:eq
is a quadratic binary diophantine equation.param
is used to indicate the parameter to be used in the solution.Details
eq
should be an expression which is assumed to be zero.param
is a parameter to be used in the solution.Examples
>>> from sympy.abc import x, y, t >>> from sympy.solvers.diophantine import diop_quadratic >>> diop_quadratic(x**2 + y**2 + 2*x + 2*y + 2, t) {(1, 1)}
References
 R669
Methods to solve Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, [online], Available: http://www.alpertron.com.ar/METHODS.HTM
 R670
Solving the equation ax^2+ bxy + cy^2 + dx + ey + f= 0, [online], Available: http://www.jpr2718.org/ax2p.pdf
diop_DN¶

sympy.solvers.diophantine.
diop_DN
(D, N, t=t)[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 [R671] is used to solve this equation.
Returns one solution tuple, (\(x, y)\) for each class of the solutions. Other solutions of the class can be constructed according to the values of
D
andN
.Usage
diop_DN(D, N, t)
: D and N are integers as in \(x^2  Dy^2 = N\) andt
is the parameter to be used in the solutions.Details
D
andN
correspond to D and N in the equation.t
is the parameter to be used in the solutions.Examples
>>> from sympy.solvers.diophantine import diop_DN >>> diop_DN(13, 4) # Solves equation x**2  13*y**2 = 4 [(3, 1), (393, 109), (36, 10)]
The output can be interpreted as follows: There are three fundamental solutions to the equation \(x^2  13y^2 = 4\) given by (3, 1), (393, 109) and (36, 10). Each tuple is in the form (x, y), i.e. solution (3, 1) means that \(x = 3\) and \(y = 1\).
>>> diop_DN(986, 1) # Solves equation x**2  986*y**2 = 1 [(49299, 1570)]
See also
References
 R671(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
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 \(a = b\), only the solutions with \(x \leq y\) are found. For more details, see the References.
Examples
>>> from sympy.solvers.diophantine import cornacchia >>> cornacchia(2, 3, 35) # equation 2x**2 + 3y**2 = 35 {(2, 3), (4, 1)} >>> cornacchia(1, 1, 25) # equation x**2 + y**2 = 25 {(4, 3)}
References
 R672
Nitaj, “L’algorithme de Cornacchia”
 R673
Solving the diophantine equation ax**2 + by**2 = m by Cornacchia’s method, [online], Available: http://www.numbertheory.org/php/cornacchia.html
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 [R674]. Let \((t, u)\) be the minimal positive solution of the equation \(x^2  Dy^2 = 1\). Then this method requires \(\sqrt{\\frac{\mid N \mid (t \pm 1)}{2D}}\) to be small.
Usage
diop_bf_DN(D, N, t)
:D
andN
are coefficients in \(x^2  Dy^2 = N\) andt
is the parameter to be used in the solutions.Details
D
andN
correspond to D and N in the equation.t
is the parameter to be used in the solutions.Examples
>>> from sympy.solvers.diophantine import diop_bf_DN >>> diop_bf_DN(13, 4) [(3, 1), (3, 1), (36, 10)] >>> diop_bf_DN(986, 1) [(49299, 1570)]
See also
References
 R674(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
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 [R675] for more detailed information on the transformation. This function returns a tuple (A, B) where A is a 2 X 2 matrix and B is a 2 X 1 matrix such that,
Transpose([x y]) = A * Transpose([X Y]) + B
Usage
transformation_to_DN(eq)
: whereeq
is the quadratic to be transformed.Examples
>>> from sympy.abc import x, y >>> from sympy.solvers.diophantine import transformation_to_DN >>> from sympy.solvers.diophantine import classify_diop >>> A, B = transformation_to_DN(x**2  3*x*y  y**2  2*y + 1) >>> A Matrix([ [1/26, 3/26], [ 0, 1/13]]) >>> B Matrix([ [6/13], [4/13]])
A, B returned are such that Transpose((x y)) = A * Transpose((X Y)) + B. Substituting these values for \(x\) and \(y\) and a bit of simplifying work will give an equation of the form \(x^2  Dy^2 = N\).
>>> from sympy.abc import X, Y >>> from sympy import Matrix, simplify >>> 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(zip((x, y), (u, 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.See also
References
 R675(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
transformation_to_normal¶

sympy.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.
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()
.Usage
find_DN(eq)
: whereeq
is the quadratic to be transformed.Examples
>>> from sympy.abc import x, y >>> from sympy.solvers.diophantine import find_DN >>> find_DN(x**2  3*x*y  y**2  2*y + 1) (13, 884)
Interpretation of the output is that we get \(X^2 13Y^2 = 884\) after transforming \(x^2  3xy  y^2  2y + 1\) using the transformation returned by
transformation_to_DN()
.See also
References
 R676
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
diop_ternary_quadratic¶

sympy.solvers.diophantine.
diop_ternary_quadratic
(eq)[source]¶ Solves the general quadratic ternary form, \(ax^2 + by^2 + cz^2 + fxy + gyz + hxz = 0\).
Returns a tuple \((x, y, z)\) which is a base solution for the above equation. If there are no solutions, \((None, None, None)\) is returned.
Usage
diop_ternary_quadratic(eq)
: Return a tuple containing a basic solution toeq
.Details
eq
should be an homogeneous expression of degree two in three variables and it is assumed to be zero.Examples
>>> from sympy.abc import x, y, z >>> from sympy.solvers.diophantine import diop_ternary_quadratic >>> diop_ternary_quadratic(x**2 + 3*y**2  z**2) (1, 0, 1) >>> diop_ternary_quadratic(4*x**2 + 5*y**2  z**2) (1, 0, 2) >>> diop_ternary_quadratic(45*x**2  7*y**2  8*x*y  z**2) (28, 45, 105) >>> diop_ternary_quadratic(x**2  49*y**2  z**2 + 13*z*y 8*x*y) (9, 1, 5)
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. \(a\) can be given as an integer or a dictionary of factors.
Examples
>>> from sympy.solvers.diophantine import square_factor >>> square_factor(24) 2 >>> square_factor(36*3) 6 >>> square_factor(1) 1 >>> square_factor({3: 2, 2: 1, 1: 1}) # 18 3
See also
descent¶

sympy.solvers.diophantine.
descent
(A, B)[source]¶ Returns a nontrivial solution, (x, y, z), to \(x^2 = Ay^2 + Bz^2\) using Lagrange’s descent method with latticereduction. \(A\) 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
>>> from sympy.solvers.diophantine import descent >>> descent(3, 1) # x**2 = 3*y**2 + z**2 (1, 0, 1)
\((x, y, z) = (1, 0, 1)\) is a solution to the above equation.
>>> descent(41, 113) (16, 3, 1)
References
 R677
Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin, Mathematics of Computation, Volume 00, Number 0.
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.
Usage
diop_general_pythagorean(eq, param)
: whereeq
is a general pythagorean equation which is assumed to be zero andparam
is the base parameter used to construct other parameters by subscripting.Examples
>>> from sympy.solvers.diophantine import diop_general_pythagorean >>> from sympy.abc import a, b, c, d, e >>> diop_general_pythagorean(a**2 + b**2 + c**2  d**2) (m1**2 + m2**2  m3**2, 2*m1*m3, 2*m2*m3, m1**2 + m2**2 + m3**2) >>> diop_general_pythagorean(9*a**2  4*b**2 + 16*c**2 + 25*d**2 + e**2) (10*m1**2 + 10*m2**2 + 10*m3**2  10*m4**2, 15*m1**2 + 15*m2**2 + 15*m3**2 + 15*m4**2, 15*m1*m4, 12*m2*m4, 60*m3*m4)
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.Usage
general_sum_of_squares(eq, limit)
: Hereeq
is an expression which is assumed to be zero. Also,eq
should be in the form, \(x_{1}^2 + x_{2}^2 + . . . + x_{n}^2  k = 0\).Details
When \(n = 3\) if \(k = 4^a(8m + 7)\) for some \(a, m \in Z\) then there will be no solutions. Refer [R678] for more details.
Examples
>>> from sympy.solvers.diophantine import diop_general_sum_of_squares >>> from sympy.abc import a, b, c, d, e, f >>> diop_general_sum_of_squares(a**2 + b**2 + c**2 + d**2 + e**2  2345) {(15, 22, 22, 24, 24)}
Reference
 R678
Representing an integer as a sum of three squares, [online], Available: http://www.proofwiki.org/wiki/Integer_as_Sum_of_Three_Squares
diop_general_sum_of_even_powers¶

sympy.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.Usage
general_sum_of_even_powers(eq, limit)
: Hereeq
is an expression which is assumed to be zero. Also,eq
should be in the form, \(x_{1}^e + x_{2}^e + . . . + x_{n}^e  k = 0\).Examples
>>> from sympy.solvers.diophantine import diop_general_sum_of_even_powers >>> from sympy.abc import a, b >>> diop_general_sum_of_even_powers(a**4 + b**4  (2**4 + 3**4)) {(2, 3)}
See also
power_representation¶

sympy.solvers.diophantine.
power_representation
(n, p, k, zeros=False)[source]¶ Returns a generator for finding ktuples of integers, \((n_{1}, n_{2}, . . . n_{k})\), such that \(n = n_{1}^p + n_{2}^p + . . . n_{k}^p\).
Usage
power_representation(n, p, k, zeros)
: Represent nonnegative numbern
as a sum ofk
p
th powers. Ifzeros
is true, then the solutions is allowed to contain zeros.Examples
>>> from sympy.solvers.diophantine import power_representation
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 sympy.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 sympy.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)]
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 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 sizek
are returned. If thezero
parameter is set to True then a suitable number of zeros are added at the end of every partition of size less thank
.zero
parameter is considered only ifk
is not None. When the partitions are over, the last \(next()\) call throws theStopIteration
exception, so this function should always be used inside a try  except block.Details
partition(n, k)
: Heren
is a positive integer andk
is the size of the partition which is also positive integer.Examples
>>> from sympy.solvers.diophantine import partition >>> f = partition(5) >>> next(f) (1, 1, 1, 1, 1) >>> next(f) (1, 1, 1, 2) >>> g = partition(5, 3) >>> next(g) (1, 1, 3) >>> next(g) (1, 2, 2) >>> g = partition(5, 3, zeros=True) >>> next(g) (0, 0, 5)
sum_of_three_squares¶

sympy.solvers.diophantine.
sum_of_three_squares
(n)[source]¶ Returns a 3tuple \((a, b, c)\) 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 \(a, m \in Z\). See [R679] for more details.
Usage
sum_of_three_squares(n)
: Heren
is a nonnegative integer.Examples
>>> from sympy.solvers.diophantine import sum_of_three_squares >>> sum_of_three_squares(44542) (18, 37, 207)
See also
References
 R679(1,2)
Representing a number as a sum of three squares, [online], Available: http://schorn.ch/lagrange.html
sum_of_four_squares¶

sympy.solvers.diophantine.
sum_of_four_squares
(n)[source]¶ Returns a 4tuple \((a, b, c, d)\) such that \(a^2 + b^2 + c^2 + d^2 = n\).
Here \(a, b, c, d \geq 0\).
Usage
sum_of_four_squares(n)
: Heren
is a nonnegative integer.Examples
>>> from sympy.solvers.diophantine import sum_of_four_squares >>> sum_of_four_squares(3456) (8, 8, 32, 48) >>> sum_of_four_squares(1294585930293) (0, 1234, 2161, 1137796)
See also
References
 R680
Representing a number as a sum of four squares, [online], Available: http://schorn.ch/lagrange.html
sum_of_powers¶

sympy.solvers.diophantine.
sum_of_powers
(n, p, k, zeros=False)[source]¶ Returns a generator for finding ktuples of integers, \((n_{1}, n_{2}, . . . n_{k})\), such that \(n = n_{1}^p + n_{2}^p + . . . n_{k}^p\).
Usage
power_representation(n, p, k, zeros)
: Represent nonnegative numbern
as a sum ofk
p
th powers. Ifzeros
is true, then the solutions is allowed to contain zeros.Examples
>>> from sympy.solvers.diophantine import power_representation
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 sympy.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 sympy.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)]
sum_of_squares¶

sympy.solvers.diophantine.
sum_of_squares
(n, k, zeros=False)[source]¶ Return a generator that yields the ktuples 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 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 condition above for k = 2 can be met, so this will automatically reject certain partitions of n.
Examples
>>> from sympy.solvers.diophantine import sum_of_squares >>> 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¶

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¶
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 6tuple in the same order as mentioned above. Refer [R681] for more detailed information.Usage
PQa(P_0, Q_0, D)
:P_0
,Q_0
andD
are integers corresponding to \(P_{0}\), \(Q_{0}\) and \(D\) in the continued fraction \(\\frac{P_{0} + \sqrt{D}}{Q_{0}}\). Also it’s assumed that \(P_{0}^2 == D mod(Q_{0})\) and \(D\) is square free.Examples
>>> from sympy.solvers.diophantine import PQa >>> pqa = PQa(13, 4, 5) # (13 + sqrt(5))/4 >>> next(pqa) # (P_0, Q_0, a_0, A_0, B_0, G_0) (13, 4, 3, 3, 1, 1) >>> next(pqa) # (P_1, Q_1, a_1, A_1, B_1, G_1) (1, 1, 1, 4, 1, 3)
References
 R681(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
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 [R682]. No test is performed to test whether \((u, v)\) and \((r, s)\) are actually solutions to the equation. User should take care of this.
Usage
equivalent(u, v, r, s, D, N)
: \((u, v)\) and \((r, s)\) are two solutions of the equation \(x^2  Dy^2 = N\) and all parameters involved are integers.Examples
>>> from sympy.solvers.diophantine import equivalent >>> equivalent(18, 5, 18, 5, 13, 1) True >>> equivalent(3, 1, 18, 393, 109, 4) False
References
 R682(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
parametrize_ternary_quadratic¶

sympy.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
>>> from sympy import Tuple, ordered >>> from sympy.abc import x, y, z >>> from sympy.solvers.diophantine import parametrize_ternary_quadratic
The parametrized solution may be returned with three parameters:
>>> parametrize_ternary_quadratic(2*x**2 + y**2  2*z**2) (p**2  2*q**2, 2*p**2 + 4*p*q  4*p*r  4*q**2, p**2  4*p*q + 2*q**2  4*q*r)
There might also be only two parameters:
>>> parametrize_ternary_quadratic(4*x**2 + 2*y**2  3*z**2) (2*p**2  3*q**2, 4*p**2 + 12*p*q  6*q**2, 4*p**2  8*p*q + 6*q**2)
Notes
Consider
p
andq
in the previous 2parameter solution and observe that more than one solution can be represented by a given pair of parameters. If \(p\) andq
are not coprime, this is trivially true since the common factor will also be a common factor of the solution values. But it may also be true even whenp
andq
are coprime:>>> sol = Tuple(*_) >>> p, q = ordered(sol.free_symbols) >>> sol.subs([(p, 3), (q, 2)]) (6, 12, 12) >>> sol.subs([(q, 1), (p, 1)]) (1, 2, 2) >>> sol.subs([(q, 0), (p, 1)]) (2, 4, 4) >>> sol.subs([(q, 1), (p, 0)]) (3, 6, 6)
Except for sign and a common factor, these are equivalent to the solution of (1, 2, 2).
References
 R683
The algorithmic resolution of Diophantine equations, Nigel P. Smart, London Mathematical Society Student Texts 41, Cambridge University Press, Cambridge, 1998.
diop_ternary_quadratic_normal¶

sympy.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 satisfies the given equation. If the equation does not have integer solutions, \((None, None, None)\) is returned.
Usage
diop_ternary_quadratic_normal(eq)
: whereeq
is an equation of the form \(ax^2 + by^2 + cz^2 = 0\).Examples
>>> from sympy.abc import x, y, z >>> from sympy.solvers.diophantine import diop_ternary_quadratic_normal >>> diop_ternary_quadratic_normal(x**2 + 3*y**2  z**2) (1, 0, 1) >>> diop_ternary_quadratic_normal(4*x**2 + 5*y**2  z**2) (1, 0, 2) >>> diop_ternary_quadratic_normal(34*x**2  3*y**2  301*z**2) (4, 9, 1)
ldescent¶

sympy.solvers.diophantine.
ldescent
(A, B)[source]¶ Return a nontrivial 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
>>> from sympy.solvers.diophantine import ldescent >>> ldescent(1, 1) # w^2 = x^2 + y^2 (1, 1, 0) >>> ldescent(4, 7) # w^2 = 4x^2  7y^2 (2, 1, 0)
This means that \(x = 1, y = 0\) and \(w = 2\) is a solution to the equation \(w^2 = 4x^2  7y^2\)
>>> ldescent(5, 1) # w^2 = 5x^2  y^2 (2, 1, 1)
References
 R684
The algorithmic resolution of Diophantine equations, Nigel P. Smart, London Mathematical Society Student Texts 41, Cambridge University Press, Cambridge, 1998.
 R685
Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin, [online], Available: http://eprints.nottingham.ac.uk/60/1/kvxefz87.pdf
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 \(x^2 + az^2\) is minimal.
Details
Here
w
is a solution of the congruence \(x^2 \equiv a \ (mod \ b)\)References
 R686
Gaussian lattice Reduction [online]. Available: http://home.ie.cuhk.edu.hk/~wkshum/wordpress/?p=404
 R687
Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin, Mathematics of Computation, Volume 00, Number 0.
holzer¶

sympy.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 [R688] and the work of Mordell in reference [R689].
References
prime_as_sum_of_two_squares¶

sympy.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
>>> from sympy.solvers.diophantine import prime_as_sum_of_two_squares >>> prime_as_sum_of_two_squares(7) # can't be done >>> prime_as_sum_of_two_squares(5) (1, 2)
Reference
 R690
Representing a number as a sum of four squares, [online], Available: http://schorn.ch/lagrange.html
See also
sqf_normal¶

sympy.solvers.diophantine.
sqf_normal
(a, b, c, steps=False)[source]¶ Return \(a', b', c'\), the coefficients of the squarefree 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
>>> from sympy.solvers.diophantine import sqf_normal >>> 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))
See also
References
 R691
Legendre’s Theorem, Legrange’s Descent, http://public.csusm.edu/aitken_html/notes/legendre.pdf
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 the squarefree normal form of the equation, \(a'*x^2 + b'*y^2 + c'*z^2\), where \(a'\), \(b'\) and \(c'\) are square free and \(gcd(a', b', c') == 1\).