Chapter 9: Problem 6
Let \(p\) be an odd prime and \(\operatorname{gcd}(a, p)=1 .\) Establish that the quadratic congruence \(a x^{2}+b x+c=0(\bmod p)\) is solvable if and only if \(b^{2}-4 a c\) is either zero or a quadratic residue of \(p\).
Short Answer
Expert verified
The quadratic congruence is solvable if \( b^2 - 4ac \) is zero or a quadratic residue modulo \( p \).
Step by step solution
01
Understand the Problem
The problem involves checking the solvability of a quadratic congruence modulo an odd prime \( p \). We need to show that this congruence is solvable if and only if the discriminant \( b^2 - 4ac \) is either zero or a quadratic residue modulo \( p \).
02
Recall the Quadratic Congruence Formula
The quadratic congruence can be solved using the formula similar to the quadratic formula: \( x \equiv \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \pmod{p} \). For this formula to give integer solutions, \( b^2 - 4ac \) must have a square root modulo \( p \).
03
Check the Discriminant
For the quadratic congruence \( ax^2 + bx + c \equiv 0 \pmod{p} \) to have solutions, the discriminant \( b^2 - 4ac \) must satisfy certain conditions. Specifically, \( b^2 - 4ac \equiv 0 \pmod{p} \) or the Legendre symbol \( \left(\frac{b^2 - 4ac}{p}\right) = 1 \) must hold. The latter condition indicates that \( b^2 - 4ac \) is a quadratic residue modulo \( p \).
04
Establish Solvability Conditions
By number theory, an equation modulo \( p \) is solvable if its discriminant is a quadratic residue. Therefore, if \( b^2 - 4ac \equiv 0 \pmod{p} \) or \( \left(\frac{b^2 - 4ac}{p}\right) = 1 \), the quadratic congruence has solutions. Conversely, if neither condition is met, the discriminant is not a quadratic residue, and the congruence is not solvable.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Residue
In number theory, a quadratic residue is an important concept when dealing with congruences and modular arithmetic, especially quadratic congruences. Simply put, given a modulo number, say an odd prime \( p \), an integer \( a \) is known to be a quadratic residue modulo \( p \) if there exists an integer \( x \) such that \( x^2 \equiv a \pmod{p} \). This tells us that \( a \) can be expressed as the square of some integer modulo \( p \).
- If such an \( x \) exists, \( a \) is a quadratic residue modulo \( p \).
- If no such \( x \) can be found, \( a \) is a non-residue.
Legendre Symbol
The Legendre symbol \( \left(\frac{a}{p}\right) \) is a powerful tool in number theory that helps us quickly ascertain the status of an integer \( a \) as a quadratic residue modulo a prime \( p \). It is defined as follows:
The Legendre symbol simplifies the process of checking quadratic residues, offering a binary approach to confirm solvability of congruences.
- \( \left(\frac{a}{p}\right) = 1 \) if \( a \) is a quadratic residue modulo \( p \) and \( a ot\equiv 0 \pmod{p} \).
- \( \left(\frac{a}{p}\right) = -1 \) if \( a \) is a non-residue modulo \( p \).
- \( \left(\frac{a}{p}\right) = 0 \) if \( a \equiv 0 \pmod{p} \).
The Legendre symbol simplifies the process of checking quadratic residues, offering a binary approach to confirm solvability of congruences.
Discriminant in Number Theory
In number theory, the discriminant of a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( b^2 - 4ac \). This discriminant plays a pivotal role in determining the nature of roots of the equation, not just over real numbers but also in the context of modular arithmetic. In quadratic congruences, the discriminant determines:
- Whether the congruence has real integer solutions modulo \( p \).
- If these solutions are unique or not.
- Zero, implying a perfect square. The equation becomes a perfect square trinomial having a double root.
- Also serving as a key in identifying if its square root exists under modular arithmetic, implying solvability.