Solving Equations in Modular Arithmetic: A Tricky Example of 109x103y5
When dealing with equations in modular arithmetic, particularly ones involving linear Diophantine equations, finding integer solutions can be a challenging but rewarding task. This article will walk you through an interesting example: the equation 109x103y5, where x and y are integers. We will explore various methods to solve this equation and present the solutions step-by-step.
Understanding the Equation
Given the equation 109x103y5, our goal is to find integer values for x and y. We can start by simplifying this equation using properties of modular arithmetic.
Simplification Using Modulo 103
First, let's simplify the equation using modular arithmetic:
109x ≡ 5 (mod 103)
To solve this, we need to find the multiplicative inverse of 109 modulo 103. The multiplicative inverse of a number a modulo m is a number b such that ab ≡ 1 (mod m).
To find the inverse of 109 modulo 103, we can use the Extended Euclidean Algorithm or simply recognize that:
109 ≡ 6 (mod 103) because 109 - 103 6
Now, we need to find the inverse of 6 modulo 103. We do this by finding an integer k such that:
6k ≡ 1 (mod 103)
Using the Extended Euclidean Algorithm, we find that:
6k 103 - 17k
Therefore, k 18 satisfies the equation:
6 * 18 108 ≡ 1 (mod 103)
Hence, the multiplicative inverse of 6 (and thus 109) modulo 103 is 18. So:
x ≡ 5 * 18 (mod 103)
x ≡ 90 (mod 103)
or
x 90 103k, where k is any integer.
Substitution and Solution Finding
Substituting x 90 103k into the original equation:
109(90 103k)103y 5
109 * 90 109 * 103k * 103y 5
109 * 90 109 * 103k * 103y 5
Since 109 * 103 ≡ 0 (mod 103), we have:
109 * 90 ≡ 5 (mod 103)
109 * 103k * 103y ≡ 0 (mod 103)
Thus, the equation simplifies to:
y ≡ -109k (mod 103)
y ≡ -85 (mod 103)
or
y -85 103n, where n is any integer.
Conclusion and General Solution
Therefore, a particular solution to the equation 109x103y5 is:
x 90, y -85
The general solution can be expressed as:
x 90 103k, y -85 - 109k, where k and n are integers.
Furthermore, since both 109 and 103 are prime numbers, we can use the principle of Euclidean algorithm to find more solutions. Using the Euclidean algorithm:
109 103 * 1 6
103 17 * 6 1
Therefore:
103 - 17 * 109 - 103 ≡ 1 (mod 103)
This gives us:
109 * -17 103 * 18 ≡ 1 (mod 103)
Multiplying both sides by 5:
109 * -85 103 * 90 ≡ 5 (mod 103)
Hence, one particular solution is:
x -85, y 90
The general solution will be:
x -85 103n, y 90 - 109n, where n is any integer.
In conclusion, solving the equation 109x103y5 involves using modular arithmetic and the Euclidean algorithm. The solutions provide insight into the nature of linear Diophantine equations, highlighting the importance of understanding modular arithmetic and prime numbers in solving such equations.