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# 5.1: Linear Diophantine Equations

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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 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 $$\PageIndex{3}$$: Die hard Jug Problem

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: $$gcd(20, 16) = 4.$$ 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)}

The following problem can be found in puzzle books.

Example $$\PageIndex{6}$$

Solution

Let x be the number of dollars Mrs Brown should have received and y be the number of cents she should have received.

Then 2(100x + y) = 100y + x - 5

Note double original amount without spending a nickel.

200x + 2y = 100y + x - 5

199x - 98y = -5.

5 = - 199x + 98y

Step 1: gcd(199,98) = 1. Since 1 | 5, we can continue.

Step 2: A solution is 51=-199(-33)(5) + (98)(-67)(5)

5 = -199(-165) + 98(-335).

Hence x = -165 and y = -335 is a solution to 5 = 98y - 199x.

Step 3: Let u = x + 165 and v = y + 335.

Consider that -199u + 98v = -199(x + 165) + 98(y + 335)

= -199x + 98y - [(199)(165) + (98)(335)]

Thus -199u + 98v = -199x + 98y - 5 = 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(199, 98) = 1.

Thus, u = 98k and v = 199k, k ∈ ℤ.

Step 5: Replace u and v.

x + 165 = 98k and y + 335 = 199k, k ∈ ℤ.

Hence x = -165 + 98k and y = -335 + 199k.

Step 6: Both x & y ≥ 0 and both x, y < 100

-165 + 98k ≥ 0, so k ≥ 1.68

-335 + 199k ≥ 0, so k ≥ 1.68

-165 + 98k < 100, 98k < 265, ∴ k < 2.70

-335 + 199k < 100, 199k < 435, ∴ k < 2.18

Since, 1.68 ≤ k < 2.18 and k ∈ ℤ, k = 2.

Thus, x = 98(2) - 165 = 31 and y = -335 + 199(2) = 63.

Thus, the cheque was for $31.63. To verify, the teller gave Mrs. Brown$63.31, she then spent 5 cents, leaving her with \$63.26 which is twice the check amount $$(2)(\31.63)=\63.26$$.✔

PRACTICAL USES

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