Solving the Diophantine Equation 6x 15y 10z 53: A Step-by-Step Guide

The equation 6x 15y 10z 53 is a classic example of a Diophantine equation, where the goal is to find integer solutions for the variables x, y, and z. This article provides a comprehensive guide to solving such equations step by step.

Step 1: Find a Specific Solution

The challenge often lies in finding a specific set of integer values that satisfy the equation. One common technique is to start with a guess and adjust until the equation holds true. Let's illustrate this with the given equation.

Given equation: 6x 15y 10z 53

By trial and error, we can find a solution:

3(1) 15(1) 10(-2) 3 15 - 20 53 - 53 0

This confirms that x 3, y 1, z -2 is a valid solution.

Step 2: Generalize the Specific Solution

Once a specific solution is found, we can generalize it to find all possible integer solutions. This involves using the concept of homogeneous solutions and linear transformations.

Modulo Operations

First, let's use modulo operations to further simplify the problem:

Modulo 5 yields: x ≡ 3 (mod 5) Modulo 2 yields: y ≡ 1 (mod 2) Modulo 3 yields: z ≡ 2 (mod 3)

These congruences help us express x, y, and z in terms of integers m, n, and p:

x 5m 3 y 2n 1 z 3p 2

Substitution into the Original Equation

Substituting these expressions back into the original equation:

6(5m 3) 15(2n 1) 10(3p 2) 53

Expanding and simplifying:

30m 18 30n 15 30p 20 53

30m 30n 30p 53 53

30m 30n 30p 0

30(m n p) 0

m n p 0

This implies that p -m - n.

General Solution

The general solution can be written as:

x 5m 3

y 2n 1

z -3(m n) 2

Where m and n are integers in (mathbb{Z}).

Alternative Method: Homogeneous Solutions and Linear Combinations

Alternatively, we can use a more generalized approach by finding homogeneous solutions and combining them:

Subtract multiples of one variable to zero out that variable's coefficient. Add multiples of the resulting equations to the original equation to satisfy the constant term.

By adding multiples of (15, -6, 0), we can eliminate one of the variables, and similarly, multiplications can be used to generate integer solutions.

For example:

(15, -1, 0) - (0, 2, -2) (15, -3, 2) (15, -1, 0) - (0, 0, -3) (15, -1, -3)

The general solution can be expressed as:

x 5a - 2, y -2a 2b 1, z -2b 2

Positive Integer Solutions

For a more practical and positive solution set, we can scale the specific solution to ensure all variables are positive:

3(1) 1(1) - 2(2) 3 1 - 4 0

Using the same method, we can derive the general solution:

x 5a - 2, y -2a 2b 1, z -2b 2

Where a and b are non-negative integers.

Conclusion

By using these steps, we can systematically find and generalize the integer solutions for Diophantine equations. The methods outlined provide a robust framework for solving similar equations and can be applied to a wide range of integer problems.

Related Keywords

Keyword 1: Diophantine equations

Keyword 2: integer solutions

Keyword 3: system of equations