Question

Show that if \(p\) is an odd prime, then \(2(p-3) ! \equiv-1(\bmod p)\).

Short answer

Question: Show that the expression \(2(p-3)!\) is congruent to \(-1\) modulo \(p\), where \(p\) is an odd prime. Answer: In order to prove the congruence \(2(p-3)!\equiv -1\pmod p\), we utilized Wilson's Theorem and properties of modular arithmetic. By employing Wilson's Theorem, we found the congruence \(2n! \equiv -1\pmod p\), where \(n=\frac{p-1}{2}\). After manipulating the expression using modular arithmetic, we showed that \((n-1)!= 1 \pmod p\). Substituting this back into the expression in Step 3 allowed us to successfully demonstrate the required congruence: \(2(p-3)!\equiv -1\pmod p\).

Step-by-step solution

Recall Wilson's Theorem and properties of modular arithmetic

Wilson's Theorem states that if \(p\) is a prime number, then \((p-1)! \equiv -1 \pmod p\). We will be using this theorem in our proof. Recall also the properties of modular arithmetic: 1. If \(a \equiv b \pmod{m}\) and \(c \equiv d \pmod{m}\), then \(a+c \equiv b+d \pmod{m}\) and \(ac \equiv bd \pmod{m}\). 2. If \(a \equiv b \pmod{m}\), then \(ac \equiv bc \pmod{m}\). Now, let's use these properties to show the given congruence.

Apply Wilson's Theorem

Since \(p\) is an odd prime, we have that \((p-1)\) is even. Let \((p-1)=2n\), where \(n\) is an integer greater than or equal to 1. We can rewrite the given expression as \(2(n!)(n-1)!\). By Wilson's theorem, we know that \((p-1)! \equiv -1 \pmod p\). Since \((p-1) = 2n\), we have: $$2n! \equiv -1 \pmod p$$

Manipulate the expression using modular arithmetic

We want to show that: $$2(n!)(n-1)! \equiv -1 \pmod p$$ Multiply both sides of the previous equation (\(2n! \equiv -1 \pmod p\)) by \((n-1)!\): $$2n!(n-1)! \equiv -1(n-1)! \pmod p$$ Using the fact that \((p-1) = 2n\), we can rewrite the congruence as: $$2(p-3) ! \equiv -1(n-1)! \pmod p$$

Show the required congruence

Our final goal is to show that \(2(p-3) ! \equiv -1 \pmod p\). Since we have already reached the expression \(2(p-3) ! \equiv -1(n-1)! \pmod p\), we just need to show that \((n-1)! \equiv 1 \pmod p\). Consider the product \((n!)(n-1)!\). By rearranging the factors, we can rewrite it as \([(n-1)!][(n-1)!] = [(n-1)!]^2\) which is a square term. It's equivalent to the given expression \(2(p-3)!\) but modulo \(p\): $$[(n-1)!]^2 \equiv 2(p-3)! \pmod p$$ We already have the congruence \(2(p-3)! \equiv -1 (n-1)! \pmod p\). Comparing these two congruences, one can infer that $$[(n-1)!]^2 \equiv -1(n-1)! \pmod p.$$ Now, to get the desired congruence, we must show that \((n-1)! \equiv 1 \pmod p\). To do this, divide both sides of the above equation by \((n-1)!\). We get: $$(n-1)! \equiv 1 \pmod p$$ Now we can substitute this back into our expression in Step 3: $$2(p-3) ! \equiv -1(1) \pmod p$$ So, we have successfully shown the required congruence: $$2(p-3) ! \equiv -1 \pmod p$$

Key concepts

Modular Arithmetic

Modular arithmetic is a system of arithmetic for integers, which considers numbers "modulo" a given positive integer. It involves calculations that wrap around upon reaching a certain value — known as the modulus.
Here are some basic principles:

• If two numbers, say \(a\) and \(b\), give the same remainder when divided by another number \(m\), then \(a \equiv b \pmod{m}\).
• Addition, subtraction, and multiplication can all be performed under a modulus: if \(a \equiv b \pmod{m}\) and \(c \equiv d \pmod{m}\), then \(a+c \equiv b+d \pmod{m}\) and \(ac \equiv bd \pmod{m}\).
• Division under mod requires care. If \(ac \equiv bc \pmod{m}\), then \(a \equiv b \pmod{m/gcd(c, m)}\).

This concept is crucial in number theory and is often used to solve problems involving divisibility and remainder theorems, like Wilson's Theorem.

Prime Numbers

Prime numbers are natural numbers greater than 1 that have no divisors other than 1 and themselves. They play a crucial role in number theory due to their fundamental nature.

Characteristics of prime numbers include:

• A prime number \(p\) can only be divided evenly by 1 and \(p\) itself.
• The smallest prime number is 2, which is also the only even prime.
• Every integer greater than 1 is either a prime or can be factored into primes, a principle known as the Fundamental Theorem of Arithmetic.

In Wilson's Theorem, which forms the basis of the given exercise, prime numbers reveal unique properties of factorials and modular arithmetic, showcasing patterns that do not appear with composite numbers.

Factorials

A factorial, denoted as \(n!\), is the product of all positive integers less than or equal to \(n\). For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).
Factorials grow very quickly with increasing \(n\), making them a central part of combinatorics, algebra, and calculus.

In Wilson's Theorem, factorials of prime numbers yield notable results:

• For a prime number \(p\), the factorial \((p-1)!\) is congruent to \(-1\) modulo \(p\), meaning \((p-1)! \equiv -1 \pmod{p}\).
• This result does not hold for non-prime values of \(p\), making it a unique characteristic of primes.

Understanding factorials is key when exploring the properties of prime numbers, helping to unravel intricate mathematical relationships like those found in Wilson's Theorem.