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Chapter 5: Problem 9

Using Wilson's theorem, prove that for any odd prime \(p\), $$ 1^{2} \cdot 3^{2} \cdot 5^{2} \cdots(p-2)^{2} \equiv(-1)^{(p+1) / 2}(\bmod p) $$ [Hint: Because \(k \equiv-(p-k)(\bmod p)\), it follows that $$ \left.2 \cdot 4 \cdot 6 \cdots(p-1) \equiv(-1)^{(p-1) / 2} 1 \cdot 3 \cdot 5 \cdots(p-2)(\bmod p) .\right] $$

Short Answer

Expert verified
The proof uses the sequence manipulation and Wilson's theorem to show that the product of squares of odd numbers is congruent to \((-1)^{(p+1)/2}\) modulo \(p\).

Step by step solution

01

Understanding Wilson's Theorem

Wilson's Theorem states that for any prime number \(p\), \[(p-1)! \equiv -1 \pmod{p}\]This theorem helps relate factorials and modular arithmetic for primes. It will be instrumental in proving the given congruence.
02

Initial Setup of Equation

We begin by considering the product \(1 \cdot 3 \cdot 5 \cdots (p-2)\), the odd integers up to \(p-2\). If we square this entire product, we get:\[\left(1 \cdot 3 \cdot 5 \cdots (p-2)\right)^2 = 1^2 \cdot 3^2 \cdot 5^2 \cdots (p-2)^2\]Our goal is to find this expression modulo \(p\), making use of Wilson's theorem.
03

Use Given Hint in the Problem

According to the hint provided, \(k \equiv -(p-k) \pmod{p}\). Using this:\[2 \cdot 4 \cdot 6 \cdots (p-1) \equiv (-1)^{(p-1)/2} \cdot 1 \cdot 3 \cdot 5 \cdots (p-2) \pmod{p}\]This is derived from rearranging even numbers as the negative of corresponding odd numbers modulo \(p\).
04

Applying Wilson's Theorem Again

Now consider the factorial \((p-1)!\) using Wilson's theorem:\[(p-1)! \equiv -1 \pmod{p}\]Express \((p-1)!\) in terms of the sequences of even and odd numbers:\[(2 \cdot 4 \cdot 6 \cdots (p-1)) \cdot (1 \cdot 3 \cdot 5 \cdots (p-2)) \equiv (p-1)! \equiv -1 \pmod{p}\]Using the earlier hint, substitute the result for \(2 \cdot 4 \cdot 6 \cdots (p-1)\).
05

Connecting the Results

By substituting for \(2 \cdot 4 \cdot 6 \cdots (p-1)\) from the hint, we find:\[(-1)^{(p-1)/2} \cdot (1 \cdot 3 \cdot 5 \cdots (p-2))^2 \equiv -1 \pmod{p}\]Since \((p-1)! = (2 \cdot 4 \cdot 6 \cdots (p-1)) \cdot (1 \cdot 3 \cdot 5 \cdots (p-2))\), substitute back into the congruence from the theorem.
06

Final Deduction

Notice that from the given information:\[\left(1 \cdot 3 \cdot 5 \cdots (p-2)\right)^2 \equiv (-1)^{(p+1)/2} \pmod{p}\]This matches what we need to prove:\[1^2 \cdot 3^2 \cdot 5^2 \cdots (p-2)^2 \equiv (-1)^{(p+1)/2} \pmod{p}\]Thus, the relationship is confirmed via Wilson's theorem and the manipulations of sequences provided.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Modular Arithmetic
Modular arithmetic is a system of arithmetic for integers where numbers "wrap around" after reaching a certain value, the modulus. This concept is akin to clock arithmetic. Imagine a clock that resets anytime the hand passes 12.
In modular arithmetic, we say two integers are congruent modulo a number if they give the same remainder when divided by that number. The notation used for this is:
  • If two integers, say \(a\) and \(b\), are congruent modulo \(m\), then \(a \equiv b \pmod{m}\).
  • This means \(m\) divides their difference: \(a - b = km\), where \(k\) is an integer.
Take the numbers 17 and 5 with modulus 12 as an example. Dividing 17 by 12 gives a remainder of 5. Therefore, 17 is congruent to 5 modulo 12: \(17 \equiv 5 \pmod{12}\).
Modular arithmetic is fundamental to areas like number theory and cryptography, especially when dealing with primes and divisibility, which is showcased in Wilson's theorem.
Factorials
A factorial, denoted by an exclamation mark, such as \(n!\), is the product of all positive integers less than or equal to \(n\). It is calculated as follows:
  • For any positive integer \(n\), \(n! = n \times (n-1) \times (n-2) \times ... \times 1\).
  • For example, \(4! = 4 \times 3 \times 2 \times 1 = 24\).
  • The factorial of zero is defined as 1: \(0! = 1\).
Factorials are crucial when exploring the properties of combinatorial mathematics and sequences. They also play an essential role in the application of Wilson's theorem in problem-solving, especially in identifying patterns in sequences involving primes. The relationship between factorials and primes is deeply explored through Wilson's theorem, where the factorial of one less than a prime number holds a particular significance in modular arithmetic.
Congruences
Congruences form the backbone of modular arithmetic. When a number is divided by another, the remainder signifies congruence. Congruence relations allow us to simplify complex arithmetic problems.
  • If \(a \equiv b \pmod{m}\), this relation can be used to simplify equations and solve integer restrictions.
  • Congruences help in the computing of powers, roots, and inverse within modular arithmetic.
  • They ensure a unique residue class for any integer \(a\), where each \(a\) has a specific remainder \(b\) when divided by \(m\).
Understanding congruences helps us prove the expressions seen in the exercise. Here, substituting even and odd products into congruences showed how Wilson's theorem derives patterns in number sequences, particularly when dealing with primes.
Odd Primes
Odd primes are prime numbers that are not divisible by 2, making them all odd. They appear regularly in number theory because they hold properties distinct from even numbers, especially in simplifying modular operations.
  • The smallest odd prime number is 3, and it helps set patterns for larger odd primes like 5, 7, 11, and so on.
  • In the application of Wilson's theorem, for an odd prime \(p\), the expression \((p-1)! \equiv -1 \pmod{p}\) offers reduction opportunities.
  • They allow special properties in sequences, like the product of all integers up to \(p-2\) squared, which transforms through congruence to solve problems efficiently.
Wilson's theorem utilizes the properties of odd primes to expand on congruences and factorial attributes, cementing their importance in the proof, and treatment of modular arithmetic related exercises.

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Most popular questions from this chapter

(a) When \(n=2 p\), where \(p\) is an odd prime, prove that \(a^{n-1} \equiv a(\bmod n)\) for any integer \(a\). (b) For \(n=195=3 \cdot 5 \cdot 13\), verify that \(a^{n-2} \equiv a(\bmod n)\) for any integer \(a\).

Prove that if \(p\) is an odd prime and \(k\) is an integer satisfying \(1 \leq k \leq p-1\), then the binomial coefficient $$ \left(\begin{array}{c} p-1 \\ k \end{array}\right) \equiv(-1)^{k}(\bmod p) $$

Prove that a perfect square must end in one of the following pairs of digits: \(00,01,04,09\), \(16,21,24,25,29,36,41,44,49,56,61,64,69,76,81,84,89,96\) [Hint: Because \(x^{2} \equiv(50+x)^{2}(\bmod 100)\) and \(x^{2} \equiv(50-x)^{2}\) (mod 100), it suffices to examine the final digits of \(x^{2}\) for the 26 values \(\left.x=0,1,2, \ldots, 25 .\right]\)

If \(\operatorname{gcd}(a, 30)=1\), show that 60 divides \(a^{4}+59\).

Assume that \(p\) and \(q\) are distinct odd primes such that \(p-1 \mid q-1\). If \(\operatorname{gcd}(a, p q)=1\), show that \(a^{q-1} \equiv 1(\bmod p q)\)
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