Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

5.1: Linear Diophantine Equations

  • Page ID
    7314
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    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.