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
Factorize the Modulus
First, we need to factorize 1105, the modulus. We find that 1105 = 5 × 13 × 17. Hence, 1105 is the product of three distinct primes.
02
Quadratic Residue Modulo Each Prime Factor
We need to check if the congruence \(x^2 \equiv 231}\) is solvable modulo each of the prime factors of 1105. This means checking for 5, 13, and 17.
03
Check Residue Modulo 5
Calculate \(231 \mod 5\). Here, 231 divided by 5 leaves a remainder of 1, so \(231 \equiv 1 \mod{5}\). Since 1 is a quadratic residue modulo 5 (as 1 is 1²), the congruence is solvable modulo 5.
04
Check Residue Modulo 13
Calculate \(231 \mod 13\). Dividing 231 by 13 gives a remainder of 10, so \(231 \equiv 10 \mod{13}\). We check whether 10 is a quadratic residue modulo 13 by computing the Legendre symbol \((10/13)\).
05
Evaluate \(\left(\frac{10}{13}\right)\) Using Quadratic Reciprocity
By quadratic reciprocity, \(\left(\frac{10}{13}\right) = \left(\frac{2}{13}\right)\left(\frac{5}{13}\right)\). Knowing 13 ≡ 1 (mod 4), \(\left(\frac{2}{13}\right) = (-1)^{3} = -1\), and using Euler's Criterion \(\left(\frac{5}{13}\right) = 5^{(13-1)/2} \equiv 1 \mod{13}\). Thus, \((10/13) = (-1)(1) = -1\), non-residue, so not solvable modulo 13.
06
Conclusion
Since 231 is a quadratic residue modulo 5 but not modulo 13, the congruence \(x^2 ≡ 231 \mod{1105}\) is not solvable due to the Chinese Remainder Theorem and properties of quadratic residues over product of primes.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
quadratic residues
In number theory, a quadratic residue is a number that can be expressed as the square of another number modulo some integer. For instance, if we say "10 is a quadratic residue modulo 13", it means there exists an integer, say \( y \), such that \( y^2 \equiv 10 \mod 13\). Recognizing quadratic residues helps solve quadratic congruences, which are equations of the form \( x^2 \equiv a \mod m\). To determine if a given number is a quadratic residue with respect to a modulus \( p \) (usually a prime), we can use the **Legendre symbol** \( \left( \frac{a}{p} \right) \). This notation yields \( 1 \) if \( a \) is a quadratic residue modulo \( p \), and \( -1 \) if it's not. The process involves checking if \( a\) can be a perfect square under that modulus.
For more complex moduli, those which aren't prime, breaking them down into a product of prime factors allows us to determine the residue status individually for each prime factor.
For more complex moduli, those which aren't prime, breaking them down into a product of prime factors allows us to determine the residue status individually for each prime factor.
congruence solvability
Solving a quadratic congruence involves determining if there is an integer \( x \) that satisfies a congruence of the form \( x^2 \equiv a \mod m \). To evaluate solvability, we often break down the modulus into prime factors, especially when it's a composite number like 1105. This is essential due to the properties of quadratic residues, which can differ depending on the prime factors involved. In practice, we calculate whether \( a \), the number on the right-hand side of the congruence, is a quadratic residue for each prime factor. If \( a \) is not a residue for any of the factors, the congruence is unsolvable for the entire modulus. On the other hand, if \( a \) is a residue for all prime factors, solutions exist, and further analysis can help ascertain the specific solutions.
Chinese Remainder Theorem
The Chinese Remainder Theorem (CRT) is a pivotal tool in number theory, especially for solving problems that involve congruences with moduli that share no common factors (i.e., they are coprime). It tells us how to find a solution to a series of congruences, each with a different prime factor as the modulus.
In the context of quadratic congruences, CRT provides a way to integrate the solvability results from each prime factor modulus back into a solution or, as in this case, to conclude the congruence's overall solvability or unsolvability.
For instance, if a number is a quadratic residue not across all components, the CRT implies the original modulus' congruence equation is unsolvable. This is because CRT ensures there is a one-to-one relationship between solving modulo a composite number and solving modulo its individual prime factors, but only when residue conditions are uniform.
For instance, if a number is a quadratic residue not across all components, the CRT implies the original modulus' congruence equation is unsolvable. This is because CRT ensures there is a one-to-one relationship between solving modulo a composite number and solving modulo its individual prime factors, but only when residue conditions are uniform.
Euler's Criterion
Euler's Criterion is a powerful method for checking whether a number is a quadratic residue modulo a prime. For an odd prime \( p \) and any integer \( a \) coprime to \( p \), Euler's Criterion states that \( a \) is a quadratic residue modulo \( p \) if \( a^{(p-1)/2} \equiv 1 \mod p \), and a non-residue if the equivalence is \( -1 \). This criterion simplifies the use of quadratic reciprocity laws by providing a straightforward calculation. In practice, applying Euler's Criterion requires basic modular arithmetic. It simplifies examining quadratic residuosity, especially when a direct determination might be difficult.
- First, compute \( a^{(p-1)/2} \mod p \).
- Then, check the result: \( 1 \) indicates a residue, while \( -1 \) indicates a non-residue.