Problem 5
Question
(p-1) / 2] !^{2} \equiv(-1)^{(p+1) / 2}(\bmod p)\(, for any prime \)p>2 .$ (HINT: Use Wilson's theorem.)
Step-by-Step Solution
Verified Answer
The expression \((\frac{p-1}{2})!^2 \equiv (-1)^{(p+1)/2} \mod p\) holds for primes \( p>2 \). This is shown using Wilson's theorem.
1Step 1: Understand Wilson's Theorem
Wilson's theorem states that for any prime number \( p \), \((p-1)! \equiv -1 \mod p\). This means the factorial of \( p-1 \) is congruent to \(-1\) modulo \( p \). We will use this theorem to help solve the problem.
2Step 2: Apply the Theorem to (p-1)
First, express the factorial of \( \frac{p-1}{2} \) as \( \left(\frac{p-1}{2}\right)! \). In Wilson's theorem, we can consider elements from 1 to \( p-1 \). The middle element in this range divides it into two equal halves because \( p \) is an odd prime.
3Step 3: Relate with Factored Wilson's Theorem
The expression \((p-1)!\) incorporates terms \( 1, 2, \ldots, \frac{p-1}{2}, \frac{p+1}{2}, \ldots, p-1 \). These terms can be grouped into two sets: \( 1 \times 2 \times \ldots \times \frac{p-1}{2} \) and \( \frac{p+1}{2} \times \ldots \times (p-1) \). By Wilson's theorem, multiplying these halves gives:\[ (p-1)! = \left(\left(\frac{p-1}{2}\right)!\right)^2 (-1)^{\frac{p+1}{2}} \equiv -1 \mod p \]
4Step 4: Equate Components
Solving the previous equivalency, by equating both sides of Wilson's theorem, we verify the expression we need to prove:\[ \left(\left(\frac{p-1}{2}\right)!\right)^2 \equiv (-1)^{\frac{p+1}{2}} \mod p \] This shows the given statement from the problem is consistent with Wilson's theorem.
Key Concepts
Prime NumbersModular ArithmeticFactorialsCongruences
Prime Numbers
A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. This fundamental property makes prime numbers the building blocks of arithmetic. Each prime number is unique based on its primes present in sequences.
Prime numbers include examples like:
They are especially important in number theory. Applications of prime numbers include encryption algorithms, where their unique properties secure information.
Prime numbers include examples like:
- 2 (the only even prime number)
- 3
- 5
- 7
- 11
They are especially important in number theory. Applications of prime numbers include encryption algorithms, where their unique properties secure information.
Modular Arithmetic
Modular arithmetic, sometimes referred to as "clock arithmetic," is a system of arithmetic for integers, where numbers "wrap around" upon reaching a certain value known as the modulus. In simple terms, it's like starting over from zero after you reach a certain number, much like how clocks reset to midnight after 12 hours.
In the context of congruences, we express it using notation such as:
Modular arithmetic helps solve problems involving cycles and remainders efficiently. It is crucial for the understanding of congruences and is applied in cryptography, computer science, and coding theory.
In the context of congruences, we express it using notation such as:
- \( a \equiv b \pmod{n} \),
Modular arithmetic helps solve problems involving cycles and remainders efficiently. It is crucial for the understanding of congruences and is applied in cryptography, computer science, and coding theory.
Factorials
The factorial of a non-negative integer \( n \), represented as \( n! \), is the product of all positive integers less than or equal to \( n \).
For example:
In Wilson's Theorem, we see factorials like \((p-1)!\) which involve prime numbers, underscoring the interaction between primes and products of consecutive numbers. Factorials provide a foundation for understanding higher mathematics through combinatorics and algebra.
For example:
- \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
- \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
In Wilson's Theorem, we see factorials like \((p-1)!\) which involve prime numbers, underscoring the interaction between primes and products of consecutive numbers. Factorials provide a foundation for understanding higher mathematics through combinatorics and algebra.
Congruences
Congruences form the backbone of modular arithmetic, making comparisons of integers after division by a modulus.
The expression \( a \equiv b \pmod{n} \) interprets as "\( a \) is congruent to \( b \) modulo \( n \)", implying \((a-b)\) is divisible by \(n\).
Congruences allow us to:
The expression \( a \equiv b \pmod{n} \) interprets as "\( a \) is congruent to \( b \) modulo \( n \)", implying \((a-b)\) is divisible by \(n\).
Congruences allow us to:
- Simplify complex calculations by focusing on remainders.
- Analyze periodic properties of sequences.
- Establish equivalencies among numbers in extended arithmetic operations.
Other exercises in this chapter
Problem 5
Suppose \(m\) has a primitive root, and let \(n\) be relatively prime to \(\phi(m) .\) (Suppose \(n>0 .\) Prove that if \(a\) is relatively prime to \(m\), then
View solution Problem 5
Prove: if \(a \equiv b(\bmod p)\), then \(\left(\frac{a}{p}\right)=\left(\frac{b}{p}\right)\). In particular, \(\left(\frac{a+k p}{p}\right)=\left(\frac{a}{p}\r
View solution Problem 5
Prove the following for all integers \(a, b, c\) and all positive integers \(m\) and \(n\). If \(a \equiv b(\bmod m\) ) and \(a \equiv b(\bmod n)\) where \(\ope
View solution Problem 5
Prove the following for all integers \(a, b, c, d\) and all positive integers \(m\) and \(n .\) If \(a b \equiv 0(\bmod p)\), where \(p\) is a prime, then \(a \
View solution