Number Theory and Congruence
1. a) If a,b,c are integers such that ac≡bc  (mod m),m>0 is a fixed integer and d=          gcd⁡( c,m) then prove that a≡b (mod m/d)


To prove that if \( a, b, c \) are integers such that \( ac \equiv bc \pmod{m} \) and \( d = \gcd(c, m) \), then \( a \equiv b \pmod{\frac{m}{d}} \), we can proceed as follows:

### Proof

1. **Start with the given congruence**:
   \[
   ac \equiv bc \pmod{m}
   \]
   This implies that \( m \) divides \( ac - bc \), or equivalently:
   \[
   m \mid c(a - b)
   \]
   which means there exists some integer \( k \) such that:
   \[
   c(a - b) = km
   \]

2. **Express the gcd condition**:
   We are given that \( d = \gcd(c, m) \). By definition, this means that \( d \) divides both \( c \) and \( m \). Hence, we can write \( c = d \cdot c' \) and \( m = d \cdot m' \) for some integers \( c' \) and \( m' \) where \( \gcd(c', m') = 1 \).

3. **Substitute in terms of \( d \), \( c' \), and \( m' \)**:
   Substitute \( c = d \cdot c' \) and \( m = d \cdot m' \) into the equation \( c(a - b) = km \):
   \[
   (d \cdot c')(a - b) = k(d \cdot m')
   \]
   Dividing both sides by \( d \) (since \( d \neq 0 \)):
   \[
   c'(a - b) = k m'
   \]

4. **Use the fact that \( \gcd(c', m') = 1 \)**:
   Since \( c' \) and \( m' \) are coprime, \( c' \) has no common factors with \( m' \). Therefore, for \( c'(a - b) = k m' \) to hold, \( m' \) must divide \( a - b \). This implies:
   \[
   a - b \equiv 0 \pmod{m'}
   \]
   or equivalently,
   \[
   a \equiv b \pmod{\frac{m}{d}}
   \]
   as \( m' = \frac{m}{d} \).

### Conclusion
We have shown that if \( ac \equiv bc \pmod{m} \) and \( d = \gcd(c, m) \), then it follows that \( a \equiv b \pmod{\frac{m}{d}} \).

Let me know if you'd like more details or have any questions about the steps!

---

Here are 5 follow-up questions to explore this topic further:

1. Can we generalize this result for cases where \( d = 1 \) or \( d = m \)?
2. How does this proof change if \( m \) is not prime?
3. What if \( a \equiv b \pmod{\frac{m}{d}} \) does not imply \( ac \equiv bc \pmod{m} \)? Why?
4. Could you prove similar results if \( c \) and \( m \) are coprime?
5. How can this theorem be applied to solve congruence equations in modular arithmetic?

**Tip:** In modular arithmetic, always check if numbers are coprime to simplify divisibility arguments and factorize easily.

If a,b,c are integers such that ac≡bc (mod m), m>0 is a fixed integer and d= gcd(c,m) then prove that a≡b (mod m/d)

Explore a step-by-step proof showing that if ac ≡ bc (mod m) and d = gcd(c, m), then a ≡ b (mod m/d). This solution covers key concepts in number theory, modular arithmetic, and the properties of gcd, suitable for advanced high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation