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# Linear Diophantine Equations

A Diophantine equation is a polynomial equation, usually in two or more unknowns, such that only the integral solutions are required. An Integral solution is a solution such that all the unknown variables take only integer values.

Given three integers a, b, c representing a linear equation of the form : ax + by = c. Determine if the equation has a solution such that x and y are both integral values.

Examples:

```Input : a = 3, b = 6, c = 9
Output: Possible
Explanation : The Equation turns out to be,
3x + 6y = 9 one integral solution would be
x = 1 , y = 1

Input : a = 3, b = 6, c = 8
Output : Not Possible
Explanation : o integral values of x and y
exists that can satisfy the equation 3x + 6y = 8

Input : a = 2, b = 5, c = 1
Output : Possible
Explanation : Various integral solutions
possible are, (-2,1) , (3,-1) etc.```

Solution:

For linear Diophantine equation equations, integral solutions exist if and only if, the GCD of coefficients of the two variables divides the constant term perfectly. In other words, the integral solution exists if, GCD(a ,b) divides c.
Thus the algorithm to determine if an equation has integral solution is pretty straightforward.

• Find GCD of a and b
• Check if c % GCD(a ,b) ==0
• If yes then print Possible
• Else print Not Possible

Below is the implementation of the above approach.

## C++

 `// C++ program to check for solutions of diophantine``// equations``#include ``using` `namespace` `std;` `//utility function to find the GCD of two numbers``int` `gcd(``int` `a, ``int` `b)``{``    ``return` `(a%b == 0)? ``abs``(b) : gcd(b,a%b);``}` `// This function checks if integral solutions are``// possible``bool` `isPossible(``int` `a, ``int` `b, ``int` `c)``{``    ``return` `(c%gcd(a,b) == 0);``}` `//driver function``int` `main()``{``    ``// First example``    ``int` `a = 3, b = 6, c = 9;``    ``isPossible(a, b, c)? cout << ``"Possible\n"` `:``                        ``cout << ``"Not Possible\n"``;` `    ``// Second example``    ``a = 3, b = 6, c = 8;``    ``isPossible(a, b, c)? cout << ``"Possible\n"` `:``                       ``cout << ``"Not Possible\n"``;` `    ``// Third example``    ``a = 2, b = 5, c = 1;``    ``isPossible(a, b, c)? cout << ``"Possible\n"` `:``                       ``cout << ``"Not Possible\n"``;` `    ``return` `0;``}`

## Java

 `// Java program to check for solutions of``// diophantine equations``import` `java.io.*;` `class` `GFG {``    ` `    ``// Utility function to find the GCD``    ``// of two numbers``    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{``        ``return` `(a % b == ``0``) ?``                ``Math.abs(b) : gcd(b,a%b);``    ``}``    ` `    ``// This function checks if integral``    ``// solutions are possible``    ``static` `boolean` `isPossible(``int` `a,``                            ``int` `b, ``int` `c)``    ``{``        ``return` `(c % gcd(a, b) == ``0``);``    ``}``    ` `    ``// Driver function``    ``public` `static` `void` `main (String[] args)``    ``{``        ``// First example``        ``int` `a = ``3``, b = ``6``, c = ``9``;``        ``if``(isPossible(a, b, c))``            ``System.out.println( ``"Possible"` `);``        ``else``            ``System.out.println( ``"Not Possible"``);``    ` `        ``// Second example``        ``a = ``3``; b = ``6``; c = ``8``;``        ``if``(isPossible(a, b, c))``            ``System.out.println( ``"Possible"``) ;``        ``else``            ``System.out.println( ``"Not Possible"``);``    ` `        ``// Third example``        ``a = ``2``; b = ``5``; c = ``1``;``        ``if``(isPossible(a, b, c))``            ``System.out.println( ``"Possible"` `);``        ``else``            ``System.out.println( ``"Not Possible"``);``    ``}``}` `// This code is contributed by anuj_67.`

## Python3

 `# Python 3 program to check for solutions``# of diophantine equations``from` `math ``import` `gcd` `# This function checks if integral``# solutions are possible``def` `isPossible(a, b, c):``    ``return` `(c ``%` `gcd(a, b) ``=``=` `0``)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# First example``    ``a ``=` `3``    ``b ``=` `6``    ``c ``=` `9``    ``if` `(isPossible(a, b, c)):``        ``print``(``"Possible"``)``    ``else``:``        ``print``(``"Not Possible"``)` `    ``# Second example``    ``a ``=` `3``    ``b ``=` `6``    ``c ``=` `8``    ``if` `(isPossible(a, b, c)):``        ``print``(``"Possible"``)``    ``else``:``        ``print``(``"Not Possible"``)` `    ``# Third example``    ``a ``=` `2``    ``b ``=` `5``    ``c ``=` `1``    ``if` `(isPossible(a, b, c)):``        ``print``(``"Possible"``)``    ``else``:``        ``print``(``"Not Possible"``)``        ` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// C# program to check for``// solutions of diophantine``// equations``using` `System;` `class` `GFG``{``    ` `    ``// Utility function to find``    ``// the GCD of two numbers``    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{``        ``return` `(a % b == 0) ?``                ``Math.Abs(b) :``               ``gcd(b, a % b);``    ``}``    ` `    ``// This function checks``    ``// if integral solutions``    ``// are possible``    ``static` `bool` `isPossible(``int` `a,``                           ``int` `b,``                           ``int` `c)``    ``{``        ``return` `(c % gcd(a, b) == 0);``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main ()``    ``{``        ``// First example``        ``int` `a = 3, b = 6, c = 9;``        ``if``(isPossible(a, b, c))``            ``Console.WriteLine(``"Possible"``);``        ``else``            ``Console.WriteLine(``"Not Possible"``);``    ` `        ``// Second example``        ``a = 3; b = 6; c = 8;``        ``if``(isPossible(a, b, c))``            ``Console.WriteLine(``"Possible"``) ;``        ``else``            ``Console.WriteLine(``"Not Possible"``);``    ` `        ``// Third example``        ``a = 2; b = 5; c = 1;``        ``if``(isPossible(a, b, c))``            ``Console.WriteLine(``"Possible"``);``        ``else``            ``Console.WriteLine(``"Not Possible"``);``    ``}``}` `// This code is contributed by anuj_67.`

## PHP

 ``

## Javascript

 ``

Output :

```Possible
Not Possible
Possible```

Time Complexity: O(min(a,b))

Auxiliary Space: O(1)

How does this work? Let GCD of ‘a’ and ‘b’ be ‘g’. g divides a and b. This implies g also divides (ax + by) (if x and y are integers). This implies gcd also divides ‘c’ using the relation that ax + by = c. Refer this wiki link for more details.

New Approach:-

1. Find GCD of a and b using Euclidean algorithm:

• Divide the larger number by the smaller number and find the remainder.
• Repeat the process with the divisor (smaller number) and the remainder.
• Continue this process until the remainder becomes zero.
• The GCD will be the last non-zero remainder.

2. Check if c is divisible by GCD(a, b).

• If c is divisible by GCD(a, b), then there exist integer solutions.
• Otherwise, there are no integer solutions.

Below is the implementation of the above approach:-

## C++

 `#include ``using` `namespace` `std;` `// Function to find the GCD of two numbers``int` `gcd(``int` `a, ``int` `b)``{``    ``if` `(a == 0) {``        ``return` `b;``    ``}``    ``return` `gcd(b % a, a);``}` `// Function to check if integral solutions are possible``bool` `isPossible(``int` `a, ``int` `b, ``int` `c)``{``    ``int` `gcd_val = gcd(a, b);``    ``return` `(c % gcd_val == 0);``}` `// Driver function``int` `main()``{``    ``int` `a = 3, b = 6, c = 9;``    ``if` `(isPossible(a, b, c)) {``        ``cout << ``"Possible"` `<< endl;``    ``}``    ``else` `{``        ``cout << ``"Not Possible"` `<< endl;``    ``}``    ``return` `0;``}`

## Java

 `import` `java.util.*;` `class` `LinearDiophantineEquations {``    ` `    ``// Function to find the GCD of two numbers``    ``public` `static` `int` `gcd(``int` `a, ``int` `b) {``        ``if` `(a == ``0``) {``            ``return` `b;``        ``}``        ``return` `gcd(b % a, a);``    ``}``    ` `    ``// Function to check if integral solutions are possible``    ``public` `static` `boolean` `isPossible(``int` `a, ``int` `b, ``int` `c) {``        ``int` `gcd = gcd(a, b);``        ``return` `(c % gcd == ``0``);``    ``}``    ` `    ``// Driver function``    ``public` `static` `void` `main(String[] args) {``        ``int` `a = ``3``, b = ``6``, c = ``9``;``        ``if` `(isPossible(a, b, c)) {``            ``System.out.println(``"Possible"``);``        ``} ``else` `{``            ``System.out.println(``"Not Possible"``);``        ``}``    ``}``}`

## Python3

 `class` `LinearDiophantineEquations:``    ` `    ``# Function to find the GCD of two numbers``    ``def` `gcd(``self``, a, b):``        ``if` `a ``=``=` `0``:``            ``return` `b``        ``return` `self``.gcd(b ``%` `a, a)``    ` `    ``# Function to check if integral solutions are possible``    ``def` `isPossible(``self``, a, b, c):``        ``gcd ``=` `self``.gcd(a, b)``        ``return` `c ``%` `gcd ``=``=` `0``    ` `    ``# Driver function``    ``def` `main(``self``):``        ``a, b, c ``=` `3``, ``6``, ``9``        ``if` `self``.isPossible(a, b, c):``            ``print``(``"Possible"``)``        ``else``:``            ``print``(``"Not Possible"``)``        ` `# Create an object of the class and call the main function``ld ``=` `LinearDiophantineEquations()``ld.main()`

## Javascript

 `    ``// Function to find the GCD of two numbers``    ``function` `gcd(a, b) {``    ``if` `(a === 0) {``        ``return` `b;``    ``}``    ``return` `gcd(b % a, a);``}` `// Function to check if integral solutions are possible``function` `isPossible(a, b, c) {``    ``let gcdValue = gcd(a, b);``    ``return` `(c % gcdValue === 0);``}` ` ``// Driver function``let a = 3,``    ``b = 6,``    ``c = 9;``if` `(isPossible(a, b, c)) {``    ``console.log(``"Possible"``);``} ``else` `{``    ``console.log(``"Not Possible"``);``}`

Output:-

`Possible`

Time Complexity:-The time complexity of this program is dominated by the gcd function, which uses the Euclidean algorithm to compute the GCD of two numbers. The time complexity of the Euclidean algorithm is O(log min(a, b)), so the time complexity of the gcd function is O(log min(a, b)). Since the isPossible function calls the gcd function once, its time complexity is also O(log min(a, b)).

Auxiliary Space:-The space complexity of this program is O(1), since we are only using a few integer variables (a, b, c, and gcd) and a few boolean and string variables (isPossible and the output string). None of these variables grow with the input size, so the space complexity is constant.

Therefore, the time complexity of this program is O(log min(a, b)) and the space complexity is O(1).

This article is contributed by Ashutosh Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.