Problem 2 - A Book of Abstract Algebra

Question

Solve each of the following pairs of simultaneous congruences: (a) \(10 x \equiv 2(\bmod 12) ; 6 x \equiv 14(\bmod 20)\) (b) \(4 x \equiv 2(\bmod 6) ; 9 x \equiv 3(\bmod 12)\) (c) \(6 x \equiv 2(\bmod 8) ; 10 x \equiv 2(\bmod 12)\)

Step-by-Step Solution

Answer

(a) x = 29, (b) x = 11, (c) x = 23

Step 1: Simplify the First Congruence (Part a)

For the congruence \(10x \equiv 2 \pmod{12}\), simplify by dividing all terms by the greatest common divisor (GCD) of 10, 2, and 12, which is 2. This simplifies to \(5x \equiv 1 \pmod{6}\).

Step 2: Simplify the Second Congruence (Part a)

For \(6x \equiv 14 \pmod{20}\), first solve the equation. Since \( \gcd(6, 20) = 2 \), divide the entire congruence by 2, resulting in \(3x \equiv 7 \pmod{10}\).

Step 3: Solve the First Simplified Congruence (Part a)

Solve \(5x \equiv 1 \pmod{6}\). To find the inverse of 5 modulo 6, note that \(5 \equiv -1 \pmod{6}\) so \(5 \times 5 \equiv 25 \equiv 1 \pmod{6}\). Thus, the solution is \(x \equiv 5 \pmod{6}\).

Step 4: Solve the Second Simplified Congruence (Part a)

For \(3x \equiv 7 \pmod{10}\), try values for \(x\) to satisfy the equation. If \(x = 9\), then \(3 \times 9 = 27 \equiv 7 \pmod{10}\). Therefore, \(x \equiv 9 \pmod{10}\).

Step 5: Apply the Chinese Remainder Theorem (Part a)

To solve \(x \equiv 5 \pmod{6}\) and \(x \equiv 9 \pmod{10}\), note that 6 and 10 are coprime but overlap because of the solution values. Testing \(x = 29\) satisfies both: \(29 \equiv 5 \pmod{6}\) and \(29 \equiv 9 \pmod{10}\).

Step 6: Solve the First Congruence (Part b)

For \(4x \equiv 2 \pmod{6}\), divide through by the GCD, 2, resulting in \(2x \equiv 1 \pmod{3}\). Since 2 is invertible modulo 3, and its inverse is 2, \(x \equiv 2 \pmod{3}\).

Step 7: Solve the Second Congruence (Part b)

Simplify \(9x \equiv 3 \pmod{12}\) by noting \( \gcd(9, 12) = 3\). Dividing by 3 gives \(3x \equiv 1 \pmod{4}\), and testing \(x = 3\), the solution \(x \equiv 3 \pmod{4}\) satisfies it.

Step 8: Apply the Chinese Remainder Theorem (Part b)

To solve \(x \equiv 2 \pmod{3}\) and \(x \equiv 3 \pmod{4}\), use back substitution to find \(x = 11\), which meets both congruences: \(11 \equiv 2 \pmod{3}\) and \(11 \equiv 3 \pmod{4}\).

Step 9: Solve the First Congruence (Part c)

Simplify \(6x \equiv 2 \pmod{8}\) by dividing everything by 2 (the GCD of 6, 2, and 8), to get \(3x \equiv 1 \pmod{4}\). Test and find \(x = 3\) gives: \(3 \times 3 = 9 \equiv 1 \pmod{4}\), so \(x \equiv 3 \pmod{4}\).

Step 10: Solve the Second Congruence (Part c)

For \(10x \equiv 2 \pmod{12}\), simplify by finding the GCD, which is 2, to yield \(5x \equiv 1 \pmod{6}\). Using the reciprocal method, the inverse is 5, giving \(x \equiv 5 \pmod{6}\).

Step 11: Apply the Chinese Remainder Theorem (Part c)

To find a common solution for \(x \equiv 3 \pmod{4}\) and \(x \equiv 5 \pmod{6}\), compute through substitution; trial and error reveals \(x = 23\) satisfies all the conditions: \(23 \equiv 3 \pmod{4}\) and \(23 \equiv 5 \pmod{6}\).

Key Concepts

Chinese Remainder Theorem

The Chinese Remainder Theorem is a classic method used when dealing with simultaneous congruences, particularly when the moduli are pairwise coprime (meaning they share no common factor other than 1). This theorem allows us to find a unique solution modulo the product of the moduli. In essence, if you have two or more congruences:

• \( x \equiv a \pmod{m} \)
• \( x \equiv b \pmod{n} \)

If the greatest common divisor (GCD) of \( m \) and \( n \) is 1, the Chinese Remainder Theorem guarantees a single solution for \( x \) in the form of \( \pmod{mn} \). In practical terms, it involves finding values that satisfy each congruence and then merging these values into a single coherent solution. The process may include trial and error or back-substitution, but it provides an efficient method to stitch together solutions from different moduli into one comprehensive answer.

greatest common divisor (GCD)

The greatest common divisor (GCD) is a fundamental concept in number theory, deeply intertwined with congressive solutions. It defines the largest integer that can divide two or more numbers without leaving a remainder. This concept is crucial when simplifying congruences and solving modular problems.

• For instance, when solving \( 10x \equiv 2 \pmod{12} \), the GCD of 10, 2, and 12 is key.
• The GCD helps in reducing the terms, by dividing them out, simplifying the equation to \( 5x \equiv 1 \pmod{6} \).

Recognizing when you can divide through by the GCD helps avoid unnecessary complexity and pushes you closer to a simpler, more solvable form of the equation. This simplicity often reveals solutions more directly, facilitating a deeper understanding of the problem at hand.

modular arithmetic

Modular arithmetic, sometimes known as 'clock arithmetic', is a system of arithmetic for integers, where numbers wrap around a certain value—the modulus. This means that the numbers reset after reaching this modulus, much like how the numbers on a clock reset after going past 12.

• In modular arithmetic, equivalence is expressed as
  \( a \equiv b \pmod{m} \), meaning when \( a \) is divided by \( m \), it leaves the same remainder as \( b \).
• In the example of \( 6x \equiv 2 \pmod{8} \), you simplify this by realizing that the numbers loop every 8.

Understanding modular arithmetic is essential when working with simultaneous congruences. It provides the framework needed to interpret, simplify, and solve problems that reset upon reaching a particular boundary, allowing for neat and organized solutions.