Finding Solutions to Congruence

Finding Solutions to Congruences

A congruence is a mathematical statement that expresses a relationship between two numbers with respect to a modulus. The general form of a congruence is:

which reads as "a is congruent to b modulo m", meaning that when $a$ and $b$ are divided by $m$, they leave the same remainder. In other words, $m$ divides $a - b$. Solving congruences involves finding values of $a$ that satisfy the given congruence.

1. Linear Congruences

A linear congruence is an equation of the form:

where $a$, $b$, and $m$ are given integers, and $x$ is the unknown we need to solve for. To solve a linear congruence, we want to find all values of $x$ that satisfy this equation modulo $m$.

Steps to Solve a Linear Congruence $ax \equiv b \pmod{m}$:

1. Check if a solution exists: A solution to the linear congruence $ax \equiv b \pmod{m}$ exists if and only if the greatest common divisor $\gcd(a, m)$ divides $b$. That is, the equation has a solution if and only if:

   $$\gcd(a, m) \mid b$$

   If this condition is not satisfied, then the congruence has no solution.

2. Simplify the equation (if possible): If $\gcd(a, m) = d$, then we can divide the entire equation by $d$ to simplify it. After dividing, the modulus will be reduced by dividing by $d$, and the new congruence will be:

   $$\frac{a}{d} x \equiv \frac{b}{d} \pmod{\frac{m}{d}}$$

3. Use the Extended Euclidean Algorithm: If $\gcd(a, m) = 1$, the linear congruence $ax \equiv b \pmod{m}$ can be solved using the Extended Euclidean Algorithm. This method will help find the multiplicative inverse of $a$ modulo $m$, denoted by $a^{-1}$, such that:

   $$a \cdot a^{-1} \equiv 1 \pmod{m}$$

   Once the inverse is found, multiply both sides of the congruence $ax \equiv b \pmod{m}$ by $a^{-1}$ to solve for $x$:

   $$x \equiv a^{-1} \cdot b \pmod{m}$$

4. General Solution: If $\gcd(a, m) = d > 1$, after dividing by $d$, we find the solution to the reduced congruence. The general solution will be given by:

   $$x = x_0 + k \cdot \frac{m}{d}$$

   where $x_0$ is a particular solution, and $k$ is any integer.

2. Example 1: Solving a Linear Congruence

Solve the congruence $3x \equiv 12 \pmod{15}$.

1. Check if a solution exists: First, compute $\gcd(3, 15)$:

   $$\gcd(3, 15) = 3$$

   Since $3 \mid 12$, a solution exists.

2. Simplify the equation: Divide the entire congruence by $\gcd(3, 15) = 3$:

   $$\frac{3}{3} x \equiv \frac{12}{3} \pmod{\frac{15}{3}}$$

   This simplifies to:

   $$x \equiv 4 \pmod{5}$$

3. Solve the simplified congruence: The simplified congruence is $x \equiv 4 \pmod{5}$, which has the solution $x = 4 + 5k$ for any integer $k$.

4. General solution: The general solution to the original congruence is:

   $$x = 4 + 5k \quad \text{for any integer} \, k$$

   This means $x = 4, 9, 14, 19, \dots$.

3. Solving Systems of Congruences: Chinese Remainder Theorem (CRT)

The Chinese Remainder Theorem (CRT) is a method used to solve a system of simultaneous linear congruences, where each congruence has the same modulus or different moduli. The CRT guarantees that if the moduli are pairwise coprime (i.e., their pairwise GCDs are 1), then there is a unique solution modulo the product of the moduli.

General Form:

Solve the system of congruences:

where $m_1, m_2, \dots, m_n$ are pairwise coprime.

Steps to Solve Using the Chinese Remainder Theorem:

1. Find the product of all moduli: Let $M = m_1 \times m_2 \times \dots \times m_n$.

2. Compute partial products: For each $i$, calculate $M_i = \frac{M}{m_i}$, which is the product of all moduli except $m_i$.

3. Find the modular inverses: For each $i$, find the inverse of $M_i$ modulo $m_i$, denoted by $y_i$, such that:

   $$M_i \cdot y_i \equiv 1 \pmod{m_i}$$

4. Form the solution: The solution to the system of congruences is:

   $$x = \sum_{i=1}^{n} a_i \cdot M_i \cdot y_i \pmod{M}$$

   where $M$ is the product of all the moduli.

4. Example 2: Solving a System Using the Chinese Remainder Theorem

Solve the system of congruences:

1. Find the product of the moduli:

   $$M = 3 \times 5 = 15$$

2. Compute the partial products:

   • $M_1 = \frac{15}{3} = 5$
   • $M_2 = \frac{15}{5} = 3$

3. Find the modular inverses:

   • Find $y_1$ such that $5 \cdot y_1 \equiv 1 \pmod{3}$:

     $$5 \equiv 2 \pmod{3} \quad \text{so} \quad 2 \cdot y_1 \equiv 1 \pmod{3}$$

     The inverse of 2 modulo 3 is 2, so $y_1 = 2$.

   • Find $y_2$ such that $3 \cdot y_2 \equiv 1 \pmod{5}$:

     $$3 \cdot 2 = 6 \equiv 1 \pmod{5} \quad \text{so} \quad y_2 = 2$$

4. Form the solution: The solution is:

   $$x = (2 \cdot 5 \cdot 2) + (3 \cdot 3 \cdot 2) \pmod{15}$$ $$x = (20) + (18) = 38 \pmod{15}$$ $$x \equiv 8 \pmod{15}$$

Thus, the solution to the system is $x \equiv 8 \pmod{15}$.

5. Conclusion

• To solve a linear congruence $ax \equiv b \pmod{m}$, check if $\gcd(a, m)$ divides $b$. If it does, simplify the equation and use the Extended Euclidean Algorithm to solve for $x$.
• The Chinese Remainder Theorem provides a method for solving systems of simultaneous congruences when the moduli are pairwise coprime.
• Solving congruences is important in number theory and has applications in cryptography, computer science, and coding theory.