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Chapter 8: Problem 8

Let \(r\) be a primitive root of the odd prime \(p\). Prove the following: (a) If \(p \equiv 1(\bmod 4)\), then \(-r\) is also a primitive root of \(p\). (b) If \(p \equiv 3(\bmod 4)\), then \(-r\) has order \((p-1) / 2\) modulo \(p\).

Short Answer

Expert verified
(a) \(-r\) is a primitive root if \(p \equiv 1 \pmod{4}\), (b) \(-r\) has order \((p-1)/2\) if \(p \equiv 3 \pmod{4}\).

Step by step solution

01

Understanding Primitive Roots

A primitive root \( r \) modulo \( p \) is an integer such that its powers generate all the nonzero residues modulo \( p \). In other words, \( r^k \equiv a \pmod{p} \) for all integers \( a \) such that \( 1 \leq a < p \). The smallest \( k \) for which \( r^k \equiv 1 \pmod{p} \) is \( \varphi(p) = p-1 \), where \( \varphi \) is the Euler's totient function.
02

Case (a): When \( p \equiv 1 \pmod{4} \)

Assuming \( p \equiv 1 \pmod{4} \), we need to prove \(-r\) is a primitive root modulo \( p \). This means showing that \((-r)^k \equiv a \pmod{p}\) for every integer \(a\). Since \( r \) is a primitive root, it suffices to check the order of \(-r\):1. The order of \(-r\) must also be \( p-1 \). 2. Since \( p \equiv 1 \pmod{4} \), it follows that \( -1 \equiv r^{(p-1)/2} \pmod{p} \).3. Thus \( (-r)^{(p-1)} = r^{(p-1)}(-1)^{(p-1)} \equiv 1 \cdot 1 \equiv 1 \pmod{p}.\)Therefore, \(-r\) is a primitive root of \(p\).
03

Case (b): When \( p \equiv 3 \pmod{4} \)

When \( p \equiv 3 \pmod{4} \), we show that \(-r \) has the order \((p-1)/2 \) modulo \(p\):1. Since \( p \equiv 3 \pmod{4} \), \(-1\) is not a perfect square in the field \( \mathbb{Z}_p \).2. Calculate the order of \(-r\): we need \((-r)^k \equiv 1 \pmod{p}\).3. If \((p-1)/2\) is the smallest positive integer such that \((-r)^{(p-1)/2} \equiv 1 \pmod{p}\), then it is a root.4. Since \( -1 \equiv r^{(p-1)/2}\) doesn't hold here, thus \[(-r)^{(p-1)/2} \equiv \pm 1 \pmod{p}, \] and checking that it satisfies \( 1 \) confirms it's the order.
04

Conclusion for Both Cases

For \( p \equiv 1 \pmod{4} \), \(-r\) behaves as a primitive root, matching the order \( p-1 \). For \( p \equiv 3 \pmod{4} \), \(-r\) has the order \( (p-1)/2 \), thus not a primitive root. This outcome respects modular arithmetic and primitive root properties.

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Key Concepts

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

Modulo Arithmetic
Modulo Arithmetic is a fundamental concept in number theory. In modulo arithmetic, numbers are wrapped around once they reach a certain value, called the modulus. This is somewhat like a clock, where after 12, the numbers wrap back to 1. In our context, we are dealing with a prime modulus, meaning the wrapping occurs at a prime number.
  • For any integer \( a \), the expression \( a \mod p \) gives the remainder when \( a \) is divided by \( p \).
  • For example, if \( a = 10 \) and \( p = 3 \), then \( 10 \mod 3 = 1 \), because 10 divided by 3 leaves a remainder of 1.
  • Modulo arithmetic is particularly useful for working within finite sets, especially when dealing with prime numbers.
Understanding modulo arithmetic is key to grasping more advanced topics like primitive roots since it provides the framework in which these concepts operate.
Euler's Totient Function
Euler's Totient Function, denoted as \( \varphi(n) \), is a very useful tool in number theory. It counts the positive integers up to \( n \) that are relatively prime to \( n \). In simple terms, it gives us the count of numbers less than \( n \) that do not share any common factors with \( n \), apart from 1.
  • For a prime number \( p \), \( \varphi(p) = p-1 \), because all numbers less than \( p \) are coprime to \( p \).
  • The formula for Euler's Totient Function becomes crucial when exploring cyclic properties of numbers, such as finding powers of a primitive root.
This function is instrumental in calculating the order of an element, which in turn helps determine if an integer is a primitive root modulo some prime.
Order of an Element
The Order of an Element in modulo arithmetic is the smallest positive integer \( k \) such that \( a^k \equiv 1 \pmod{n} \). It indicates how many times you need to multiply an element by itself to get back to 1.
  • For example, if \( a = 3 \) and \( p = 7 \), the order is 6 because \( 3^6 \equiv 1 \pmod{7} \).
  • This concept is central to determining if a number is a primitive root, as the order must equal \( p-1 \) for a prime \( p \).
  • Identifying the order helps in understanding the cycle lengths in modular systems and in verifying properties of roots under mod operations.
Nailing down the order is crucial when determining whether a number like \(-r\) has the properties of a primitive root.
Primitive Root Modulo Prime
A Primitive Root Modulo Prime is a special kind of number. If \( r \) is a primitive root of a prime \( p \), it means that \( r \) can produce every nonzero residue from 1 to \( p-1 \) through its powers.
  • The integer \( r \) is called a primitive root if its powers cycle through all the numbers from 1 up to \( p-1 \), without skipping any.
  • This concept is foundational for the evidence of cyclic structures in modular arithmetic.
  • When \( p \equiv 1 \pmod{4} \), not only \( r \) but also \(-r\) can be a primitive root, whereas for \( p \equiv 3 \pmod{4} \), \(-r\) gives a unique characteristic by having an order of \( (p-1)/2 \).
Understanding primitive roots is vital for navigating more complex problems in number theory, such as those involving encryption or algorithm design.

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Most popular questions from this chapter

Verify the following assertions: (a) The odd prime divisors of the integer \(n^{2}+1\) are of the form \(4 k+1\). [Hint: \(n^{2} \equiv-1(\bmod p)\), where \(p\) is an odd prime, implies that \(4 \mid \phi(p)\) by Theorem \(8.1 .]\) (b) The odd prime divisors of the integer \(n^{4}+1\) are of the form \(8 k+1\). (c) The odd prime divisors of the integer \(n^{2}+n+1\) that are different from 3 are of the form \(6 k+1\).

Verify that, for \(5040=2^{4} \cdot 3^{2} \cdot 5 \cdot 7, \lambda(5040)=12\) and \(\phi(5040)=1152\).

(a) Let \(p>5\) be prime. If \(R_{n}\) is the smallest repunit for which \(p \mid R_{n}\), establish that \(n \mid p-1\). For example, \(R_{8}\) is the smallest repunit divisible by 73 , and 8172 . [Hint: The order of 10 modulo \(p\) is \(n .]\) (b) Find the smallest \(R_{n}\) divisible by \(13 .\)

Assume that the order of \(a\) modulo \(n\) is \(h\) and the order of \(b\) modulo \(n\) is \(k\). Show that the order of \(a b\) modulo \(n\) divides \(h k ;\) in particular, if \(\operatorname{gcd}(h, k)=1\), then \(a b\) has order \(h k\).

Establish each of the statements below: (a) If \(a\) has order \(h k\) modulo \(n\), then \(a^{h}\) has order \(k\) modulo \(n\). (b) If \(a\) has order \(2 k\) modulo the odd prime \(p\), then \(a^{k}=-1(\bmod p)\). (c) If \(a\) has order \(n-1\) modulo \(n\), then \(n\) is a prime.
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