prove each of the given statements, assuming hat a, b, c, d, and n are integers with n>1 and that = c (mod n) and b = d (mod n). a = c² (mod n)
prove each of the given statements, assuming hat a, b, c, d, and n are integers with n>1 and that = c (mod n) and b = d (mod n). a = c² (mod n)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Mathematical Proof Problem
#### Objective:
Prove each of the given statements, assuming that \( a, b, c, d, \) and \( n \) are integers with \( n > 1 \) and that
- \( a \equiv c \pmod{n} \)
- \( b \equiv d \pmod{n} \)
#### Statement to Prove:
\[ a^2 \equiv c^2 \pmod{n} \]
### Explanation:
Given:
- Two congruences: \( a \equiv c \pmod{n} \) and \( b \equiv d \pmod{n} \).
To prove:
- \( a^2 \equiv c^2 \pmod{n} \)
The problem requires using modular arithmetic, specifically working with congruences, to show that the square of equivalent numbers modulo \( n \) are themselves equivalent modulo \( n \).
### Steps for Proof:
1. From the given, \( a \equiv c \pmod{n} \) implies that \( n \) divides \( (a - c) \). That is, there exists an integer \( k \) such that:
\[ a = c + kn \]
2. Squaring both sides:
\[ a^2 = (c + kn)^2 \]
3. Expanding the squared term:
\[ a^2 = c^2 + 2ckn + k^2n^2 \]
4. Notice that both \( 2ckn \) and \( k^2n^2 \) are clearly divisible by \( n \).
5. Thus, \( a^2 \equiv c^2 \pmod{n} \).
This congruence shows that if two numbers are congruent modulo \( n \), their squares will also be congruent modulo \( n \).
The approach is a simple application of algebra and understanding of modular arithmetic, illustrating basic properties of congruences.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F282904df-64d2-44b1-a59e-7b2a86d0de90%2F6af8404c-616b-44b8-b09a-cf96d2d9ad6e%2Fswz76o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Mathematical Proof Problem
#### Objective:
Prove each of the given statements, assuming that \( a, b, c, d, \) and \( n \) are integers with \( n > 1 \) and that
- \( a \equiv c \pmod{n} \)
- \( b \equiv d \pmod{n} \)
#### Statement to Prove:
\[ a^2 \equiv c^2 \pmod{n} \]
### Explanation:
Given:
- Two congruences: \( a \equiv c \pmod{n} \) and \( b \equiv d \pmod{n} \).
To prove:
- \( a^2 \equiv c^2 \pmod{n} \)
The problem requires using modular arithmetic, specifically working with congruences, to show that the square of equivalent numbers modulo \( n \) are themselves equivalent modulo \( n \).
### Steps for Proof:
1. From the given, \( a \equiv c \pmod{n} \) implies that \( n \) divides \( (a - c) \). That is, there exists an integer \( k \) such that:
\[ a = c + kn \]
2. Squaring both sides:
\[ a^2 = (c + kn)^2 \]
3. Expanding the squared term:
\[ a^2 = c^2 + 2ckn + k^2n^2 \]
4. Notice that both \( 2ckn \) and \( k^2n^2 \) are clearly divisible by \( n \).
5. Thus, \( a^2 \equiv c^2 \pmod{n} \).
This congruence shows that if two numbers are congruent modulo \( n \), their squares will also be congruent modulo \( n \).
The approach is a simple application of algebra and understanding of modular arithmetic, illustrating basic properties of congruences.
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