# 5.1: Linear Diophantine Equations

- Page ID
- 7314

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Thinking out loud

Mary went to a park and saw vehicles with \(2\) wheels and \(4 \) wheels. She counted the wheels. When she came home she told her mom that the vehicles she had seen had a total of \(28\) wheels. Her mom asked how many vehicles had \(2\) wheels and how many vehicles had \(4\) wheels. What was Mary's response?

Diophantine Equation

A Diophantine equation is a polynomial equation with 2 or more integer unknowns.

A Linear Diophantine equation (LDE) is an equation with 2 or more integer unknowns and the integer unknowns are each to at most degree of 1.

Linear Diophantine equation in two variables takes the form of \(ax+by=c,\) where \(x, y \in \mathbb{Z}\) and a, b, c are integer constants. x and y are unknown variables.

A **Homogeneous** Linear Diophantine equation (HLDE) is \(ax+by=0, x, y \in \mathbb{Z}\). Note that \(x=0\) and \(y=0\) is a solution, called the trivial solution for this equation.

Example \(\PageIndex{1}\):

Example of a homogeneous linear diophantine equation:

\(5x-3y=0, x, y \in \mathbb{Z}\).

In this case \(x= 3\) , \(y=5\) is a solution as is \(x=6\) , \(y=10\).

Hence \(x=3k \) and \( y=5k, k \in \mathbb{Z}\) represent all the solutions.

Check: \(5(3k)-3(5k)=15k-15k = 0.\)

**** NOTE**** In a homogeneous linear diophantine equation, the minute the equation is an addition, one of the variable is required to be negative.

In the case of \(5x+3y=0, x, y \in \mathbb{Z}\), \(x= -3k\) and \(y= 5k, k \in\mathbb{Z}\) are solutions.

THEOREM: Homogeneous Linear Diophantine Equation

Let \(ax+by=0, x, y \in \mathbb{Z}\) be a homogeneous linear Diophantine equation.

If \(\gcd(a, b)=d\), then the complete family of solutions to the above equation is

\(x=\displaystyle \frac{b}{d} k,\) and \(y=-\displaystyle \frac{a}{d} k, k \in\mathbb{Z}\).

Example \(\PageIndex{2}\): Solve the Homogeneous linear Diophantine equation

\(6x+9y=0, x, y \in \mathbb{Z}\).

**Solution:**

Note that GCD of 6 and 9 is 3. Hence the solutions are

\(x= \frac{9k}{3}=3k\) and \(y= \frac{-6k}{3}=-2k\) with \(k \in\mathbb{Z}\).

**Use the following steps to solve a non-homogeneous linear Diophantine equation.**

Solve the linear Diophantine Equations: \(ax+by=c, x, y \in\mathbb{Z}\).

Use the following steps to solve a non-homogeneous linear Diophantine equation.

**Step 1:** Determine the GCD of a and b. Let suppose \(\gcd(a, b)=d\).

**Step 2:** Check that the GCD of a and b is divides c. NOTE: If YES, continue on to step 3. If NO, STOP as there are no solutions.

**Step 3: **Find a particular solution to \(ax+by=c\) by first finding \(x_0,y_0\) such that \(ax+by=d\). Suppose \(x=\frac{c}{d}x_0\) and \(y=\frac{c}{d}y_0\).

**Step 4**: Use a change of variables: Let \( u=x-\frac{c}{d}x_0\) and \(v=y-\frac{c}{d}y_0\), then we will see that \(au+bv=0\) (important to check your result).

**Step 5**: Solve \(au+bv=0\). That is: \(u=-\frac{b}{d}m\) and \(v=\frac{a}{d}m, m \in\mathbb{Z}\).

**Step 6:** Substitute for \(u\) and \(v\). Thus the general solutions are \(x-\frac{c}{d}x_0=-\frac{b}{d}m\) and \(y-\frac{c}{d}y_0=\frac{a}{d}m, m \in\mathbb{Z}\).

Example \(\PageIndex{3}\):

Solve the linear Diophantine Equations: \(5x+3y=4, x, y \in\mathbb{Z}\).

**Solution:**

**Step 1:** Determine the GCD of 5 and 3 (a and b). Since \(5(2)+3(-3)=1\), \(\gcd(5, 3)=1.\)

**Step 2:** Since \(1\mid 4\), we will continue on to Step 3.

**Step 3: **Find a particular solution to \(5x+3y=4,x,y \in\mathbb{Z}\).

Since \(5(5)+3(-7)=4, x=5\) and \(y=-7\) is a particular solution.

**Step 4:** Let \(u=x-5\) and \(v=y+7.\) Note: The opposite integer of Step 4, so if it's positive in step 4 it will be negative in step 5 and vice versa.

Then \(5u+3v= 5(x-5)+3(y+7)\)

\( = 5x-25+3y+21\)

\( =5x+3y-4\)

\( = 4-4\) (because the equation is \(5x+3y=4\))

\( =0.\)

**Step 5:** Solve 5u+3v=0

The general solutions are \(u=-3m\) and \(v=5m, m \in\mathbb{Z}\).

**Step 6: **\(x-5=-3m\) and \(y+7=5m, m \in\mathbb{Z}\).

Hence the general solutions are \(x=-3m+5, y=5m-7, m \in\mathbb{Z}\).

Example \(\PageIndex{4}\):

Solve the linear Diophantine Equations: \(2x+4y=21, x,y \in\mathbb{Z}\).

**Solution:**

Since \(\gcd(2, 4)=2\) and \(2\) does not divide \(21\), \( 2x+4y=21\) has no solution.

Example \(\PageIndex{5}\)

Solve the linear Diophantine Equation \( 20x+16y=500, x,y \in \mathbb{Z_+}\).

**Solution**

Both \(x, y ≥ 0.

500 = 20(x) + 16(y).\)

**Step 1:** \(gcd(20, 16) = 4. |0 Since 4 | 500, we expect a solution.

**Step 2:** A solution is 4125=20(1)(125)+16(-1)(125).

500= 20(125)+16(-125)

Hence, x = 125 and y = -125 is a solution to 500 = 20x + 16y.

**Step 3:** Let u = x - 125 and v = y + 125.

Consider that 20u + 16v =20x - (20)(125) + 16y +(16)(125)

=20x +16y -[(20)(125) -(16)(125)]

=20x + 16y -500.

Thus, 20u + 16v = 0.

**Step 4:** In general, the solution to ax + by = 0 is x=bdk and y=-adk, kZ \ {0}, d=gcd(a,b). Recall, gcd(20, 16) = 4.

Thus u = 16k/4 = 4k and v = -20k/4 = -5k, k ∈ ℤ.

**Step 5:** Replace u and v.

Consider 4k = x - 125 and -5k = y + 125.

Hence, x = 4k + 125 and y = -5k - 125.

**Step 6: **Both x and y ≥ 0. x ≤ 25 and y ≤ 31 since total is 500.

4k + 125 ≥ 0, k ≥ -125/4, ∴ k ≥ -31.25.

4k + 125 ≤ 25, 4k ≤ -100, ∴ k ≤ -25.

Thus, the possible solutions are:

Let k = -25 then x = 25, y = 0.

Let k = -26 then x = 21, y = 5.

Let k = -27 then x = 17, y = 10.

Let k = -28 then x = 13, y = 15.

Let k = -29 then x = 9, y = 20.

Let k = -30 then x = 5, y = 25.

Let k = -31 then x = 1, y = 30.

Thus the options of \((x,y\) that satisfy the given equation are:

{ (25,0), (21,5), (17,10), (13, 15), (9, 20), (5, 25), (1,30)}

PRACTICAL USES

- Cryptography
- Designing different combinations of a variety of elements.