# Linear Diophantine Equation

# Problem

Given the value of integers $A, B$ and $C$ find a pair of integers $(x,y)$ such that it satisfies the equation $Ax + By = C$.

For example, if $A = 2, B = 3$ and $C = 7$, then possible solution of $(x,y)$ for equation $2x + 3y = 7$ would be $(2,1)$ or $(5,-1)$.

The problem above is a type of Diophantine problem. In the Diophantine problem, only integer solutions to an equation are required. Since $Ax + By = C$ is a linear equation, this problem is a **Linear Diophantine Problem** where we have to find a solution for a **Linear Diophantine Equation**.

For now, let us assume that $A$ and $B$ are non-zero integers.

# Existence of Solution

Before we jump in to find the solution for the equation, we need to determine whether it even has a solution. For example, is there any solution for $2x + 2y = 3$? On the left side we have $2x + 2y$ which is even no matter what integer value of $(x,y)$ is used and on the right side, we have $3$ which is odd. This equation is impossible to satisfy using integer values.

So how do we determine if the equation has a solution? Suppose $g = gcd(A,B)$. Then $Ax + By$ is a multiple of $g$. In order to have a valid solution, since left side of the equation is divisible by $g$, the right side too must be divisible by $g$. Therefore, if $g \not| \ C$, then there is no solution.

# Simplifying the Equation

Since both side of equation is divisible by $g$, i.e, $g \ | \ { \ (Ax + By), C \ }$, we can safely divide both side by $g$ resulting in a equivalent equation.

Let $a = \frac{A}{g}$, $b = \frac{B}{g}$ and $c = \frac{C}{g}$. Then,

$$(Ax + By = C) \ \equiv \ (ax + by = c)$$

After simplification, $gcd(a,b)$ is either $1$ or $-1$. If it is $-1$, then we need multiply $-1$ with $a,b$ and $c$ so that $gcd(a,b)$ becomes $1$ and the equation remains unchanged. Why did we make the $gcd(a,b)$ positive? You will find the reason below.

# Using Extended Euclidean Algorithm

Recall that in a previous post “Extended Euclidean Algorithm“, we learned how to solve the Bezout’s Identity $Ax + By = gcd(A, B)$. Can we apply that here in any way?

Yes. Using `ext_gcd()`

function, we can find Bezout’s coefficient for $ax + by = gcd(a,b)$. But we need to find solution for $ax + by = c$. Note that $gcd(a,b) = 1$, so when we use `ext_gcd()`

we find a solution for $ax + by = 1$. Let this solution be $(x_1,y_1)$. We can extend this solution to solve our original problem.

Since we already have a solution where $ax_1 + by_1 = 1$, multiplying both sides with $c$ gives us $a(x_1c) + b(y_1c) = c$. So our result is $(x,y) = (x_1c, y_1c)$. This is why we had to make sure that $gcd(a,b)$ was $1$ and not $-1$. Otherwise, multiplying $c$ would have resulted $ax + by = -c$ instead.

# Summary of Solution

Here is a quick summary of what I described above. We can find solution for Linear Diophantine Equation $Ax + By = C$ in 3 steps:

**No Solution**: First check if solution exists for given equation. Let $g = gcd(A,B)$. If $g \not| \ C$ then no solution exists.**Simplify Equation**: Let $a = \frac{A}{g}, b = \frac{B}{g}$ and $c = \frac{C}{g}$. Then finding solution for $Ax + By = C$ is same as finding solution for $ax + by = c$. In simplified equation, make sure $GCD(a,b)$ is $1$. If not, multiply $-1$ with $a,b,c$.**Extended Euclidean Algorithm**: Use`ext_gcd()`

to find solution $(x_1,y_1)$ for $ax + by = 1$. Then multiply the solution with $c$ to get solution for $ax + by = c$, where $x = x_1 \times c, y = y_1 \times c$.

Let us try few examples.

## Example 1: $2x + 3y = 7$

**Step $1$:** $g = GCD(2,3) = 1$. Since $1$ divides $7$, solution exists.

**Step $2$:** Since $g$ is already $1$ there is nothing to simplify.

**Step $3$:** Using `ext_gcd()`

we get $(x,y) = (-1,1)$. But this is for $ax + by = 1$. We need to multiply $7$. So our solution is $(-7,7)$.

$2 \times -7 + 3 \times 7 = -14 + 21 = 7$. The solution is correct.

## Example 2: $4x + 10y = 8$

**Step $1$:** $g = GCD(4,10) = 2$. Since $2$ divides $8$, solution exists.

**Step $2$:** $a = \frac{4}{2}, b = \frac{10}{2}, c = \frac{8}{2}$. We will find solution of $2x + 5y = 4$.

**Step $3$:** Using `ext_gcd()`

we get $(x,y) = (-2,1)$. But this is for $ax + by = 1$. We need to multiply $4$. So our solution is $(-8,4)$.

$ax + by = 2 \times -8 + 5 \times 4 = -16 + 20 = 4 = c$.

Also, $Ax + By = 4 \times -8 + 10 \times 4 = -32 + 40 = 8 = C$. The solution is correct. Both $ax + by = c$ and $Ax + By = C$ are satisfied.

# Finding More Solutions

We can now find a possible solution for $Ax + By = C$, but what if we want to find more? How many solutions are there? Since the solution for $Ax + By = C$ is derived from Bezout’s Identity, there are infinite solutions.

Suppose we found a solution $(x,y)$ for $Ax + By = C$. Then we can find more solutions using the formula: $( x + k \frac{B}{g}, y – k \frac{A}{g})$, where $k$ is any integer.

# Code

Let us convert our idea into code.

bool linearDiophantine ( int A, int B, int C, int *x, int *y ) { int g = gcd ( A, B ); if ( C % g != 0 ) return false; //No Solution int a = A / g, b = B / g, c = C / g; ext_gcd( a, b, x, y ); //Solve ax + by = 1 if ( g < 0 ) { //Make Sure gcd(a,b) = 1 a *= -1; b *= -1; c *= -1; } *x *= c; *y *= c; //ax + by = c return true; //Solution Exists } int main () { int x, y, A = 2, B = 3, C = 5; bool res = linearDiophantine ( A, B, C, &x, &y ); if ( res == false ) printf ( "No Solution\n" ); else { printf ( "One Possible Solution (%d %d)\n", x, y ); int g = gcd ( A, B ); int k = 1; //Use different value of k to get different solutions printf ( "Another Possible Solution (%d %d)\n", x + k * ( B / g ), y - k * ( A / g ) ); } return 0; }

`linearDiophantine()`

function finds a possible solution for equation $Ax + By = C$. It takes in $5$ parameters. $A,B,C$ defines the coefficients of equation and `*x`

, `*y`

are two pointers that will carry our solution. The function will return $true$ if solution exists and $false$ if not.

In line $2$ we calculate $gcd(A,B)$ and in line $3$ we check if $C$ is divisible by $g$ or not. If not, we return $false$.

Next on line $5$ we define $a,b,c$ for simplified equation. On line $6$ we solve for $ax + by = 1$ using `ext_gcd()`

. Then we check if $g < 0$. If so, we multiply $-1$ with $a,b,c$ to make it positive. Then we multiply $c$ with $x,y$ so that our solution satisfies $ax + by = c$. A solution is found so we return true.

In $\text{main}()$ function, we call $\text{linearDiophantine}()$ using $A=2,B=3,C=5$. In line $22$ we print a possible solution. In line $27$ we print another possible solution using formula for more solutions.

# $A$ and $B$ with Value $0$

Till now we assumed ${A, B}$ have non-zero values. What happens if they have value $0$?

## When Both $A = B = 0$

When both $A$ and $B$ are zero, the value of $Ax + By$ will always be $0$. Therefore, if $C \neq 0$ then there is no solution. Otherwise, any pair of value for $(x,y)$ will act as a solution for the equation.

## When $A$ or $B$ is $0$

Suppose only $A$ is $0$. Then equation $Ax + By = C$ becomes $0x + By = C \ \equiv \ By = C$. Therefore $y = \frac{C}{B}$. If $B$ does not divide $C$ then there is no solution. Else solution will be $(x,y) = (k, \frac{C}{B})$, where $k$ is any intger.

Using same logic, when $B$ is $0$, solution will be $(x,y) = ( \frac{C}{A}, k )$.

# Coding Pitfalls

When we use $gcd(a,b)$ in our code, we mean the result from Euclidean Algorithm, not what we understand mathematically. $gcd(4,-2)$ is $-2$ according to Euclidean Algorithm whereas it is $2$ in common sense.

# Resources

- Wiki - Diophantine Equation - https://en.wikipedia.org/wiki/Diophantine_equation
- forthright48 - Extended Euclidean Algorithm - https://forthright48.com/2015/07/extended-euclidean-algorithm.html