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Magazine Chapter 9: Problem 20
Using the Generalized Quadratic Reciprocity Law, determine whether the congruence \(x^{2}=231\) (mod 1105 ) is solvable.
Short Answer
Expert verified
The congruence is not solvable.
Step by step solution
01
Identify the Problem
We want to determine if the congruence \(x^2 \equiv 231 \pmod{1105}\) is solvable by checking if 231 is a quadratic residue modulo 1105.
02
Prime Factorization of Modulus
First, identify the prime factorization of the modulus, 1105. We find that \(1105 = 5 \times 13 \times 17\). We need to analyze the solvability for each prime factor.
03
Check Quadratic Residue for Each Prime Factor
We will check if 231 is a quadratic residue modulo 5, 13, and 17 separately. Use the Legendre symbol \((231/p)\) where \(p\) is each prime factor of 1105.
04
Quadratic Residue Modulo 5
Calculate \(231 \equiv 1 \pmod{5}\). Thus, the problem reduces to finding if 1 is a quadratic residue mod 5. Since 1 is a perfect square, \(231\) is a quadratic residue modulo 5.
05
Quadratic Residue Modulo 13
Calculate \(231 \equiv 10 \pmod{13}\). Now, use the Legendre symbol: \((10/13) = (5/13)(2/13)\). Check if 5 and 2 are quadratic residues modulo 13 using reciprocity laws. Both are not quadratic residues, meaning \((10/13) = -1\). Thus, 231 is not a quadratic residue modulo 13.
06
Quadratic Residue Modulo 17
Calculate \(231 \equiv 10 \pmod{17}\). Use similar steps as modulo 13: \((10/17) = (2/17)(5/17)\). Since 5 is a quadratic residue but 2 is not a quadratic residue modulo 17, \((10/17) = -1\). Thus, 231 is not a quadratic residue modulo 17.
07
Conclusion from Results
Since 231 is not a quadratic residue modulo 13 and 17, the congruence \(x^2 \equiv 231 \pmod{1105}\) is not solvable.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Legendre Symbol
The Legendre symbol \(\left( \frac{a}{p} \right)\) is a mathematical tool used to determine if an integer \('a'\) is a quadratic residue modulo a prime \(p\). It provides a simple way to check the solvability of quadratic congruences.
- If \(\left( \frac{a}{p} \right) = 1\), then \(a\) is a quadratic residue modulo \(p\). This means there exists some integer \(x\) such that \((x^2 \equiv a \pmod{p})\).
- If \(\left( \frac{a}{p} \right) = -1\), then \(a\) is not a quadratic residue modulo \(p\), indicating no such \(x\) exists.
- If \(a\) is divisible by \(p\), then \(\left( \frac{a}{p} \right) = 0\).
What is a Quadratic Residue?
A quadratic residue is an integer \(n\) such that there exists another integer \(x\) where \(x^2 \equiv n \pmod{m}\). Simply put, \(n\) is a quadratic residue modulo \(m\) if the congruence can be solved to yield a perfect square.
- Consider modulo 5: The quadratic residues are 0, 1, and 4 because these are the squares of integers 0, 1, 2, and 3 modulo 5.
- When engaging with non-prime moduli, like 1105, ensuring all the components are quadratic residues under their prime factors is key to broader conclusions.
The Role of Prime Factorization
Prime factorization breaks down a number into a product of prime numbers. This concept is particularly valuable in modular arithmetic to tackle congruence problems involving large numbers.
- For example, since 1105 can be expressed as \(5 \times 13 \times 17\), we analyze individual residues modulo 5, 13, and 17 to draw conclusions about mod 1105.
- This step provides a roadmap and simplifies complex operations by leveraging the properties of smaller prime numbers.
Congruence Solvability Simplified
Congruence solvability refers to the ability to find solutions to equations like \(x^2 \equiv n \pmod{m}\). Several factors influence whether such solutions exist, among them are the concepts of quadratic residues and the Legendre symbol.
- We assess the congruence \(x^2 \equiv 231 \pmod{1105}\) by checking residue status across all prime factors of 1105.
- Finding that 231 is not a quadratic residue modulo either 13 or 17 results in the conclusion that the overall congruence is unsolvable.
