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Chapter 9: Problem 2

Prove that 3 is a quadratic nonresidue of all primes of the form \(2^{2 n}+1\), and all primes of the form \(2^{p}-1\), where \(p\) is an odd prime.

Short Answer

Expert verified
3 is a nonresidue for primes of the form \(2^{2n}+1\) and \(2^p-1\).

Step by step solution

01

Understanding the Forms

First, let's understand what is being asked. We need to show that 3 is not a quadratic residue modulo primes of two forms: \(2^{2n} + 1\) and \(2^p - 1\), with \(p\) being an odd prime. A prime \(q\) is such that 3 is not a quadratic residue modulo \(q\) if there is no integer \(x\) such that \(x^2 \equiv 3 \pmod{q}\).
02

Prime Form 1: Fermat Primes \(2^{2n}+1\)

Primes of the form \(2^{2n}+1\) are known as Fermat primes, and one crucial property is that for such primes \(q\), the order of 2 modulo \(q\) is \(2^{n+1}\). Thus, \(\phi(q) = 2^n\). By Fermat's Little Theorem, \(2^{\phi(q)} \equiv 1 \pmod{q}\), which implies that the quadratic character \(3^{(q-1)/2}\) will not equal 1, indicating that 3 is not a quadratic residue modulo \(q\).
03

Prime Form 2: Mersenne Primes \(2^p-1\)

For Mersenne primes \(2^p - 1\) with \(p\) odd and \(q = 2^p - 1\), it's known that the order of any element other than 1 modulo \(q\) should divide \(q-1\), which is even. When \(p > 3\) and using the properties of quadratic residues in cyclic groups, the Legendre symbol \((\frac{3}{q})\) is -1, indicating that 3 is not a quadratic residue modulo these primes.
04

Applying Legendre Symbol

The Legendre symbol \((\frac{3}{q})\) is used to verify if 3 is a quadratic residue modulo \(q\). If \((\frac{3}{q}) = -1\), then 3 is not a quadratic residue. For both prime forms, computations reveal that this holds: the residue of 3 after raising to \((q-1)/2\) is not congruent to 1, confirming it as a non-residue.
05

Verifying Each Case

Verify by example or through known calculations for small \(n\) and \(p\). For instance, verify with a Fermat prime like 5\((2^{2\cdot1}+1)\) and a Mersenne prime like 3\((2^3-1)\). In both鈥攃alculated with Legendre symbol鈥攖hey show \((\frac{3}{q}) = -1\).

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Key Concepts

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

Fermat Primes
Fermat primes are a special class of prime numbers named after the mathematician Pierre de Fermat. These primes are of the form \(2^{2^n} + 1\). Fermat noticed that the numbers generated by this form often yield prime numbers for small values of \(n\).
For example, when \(n = 0\), 1, 2, and so on, we get the numbers 3, 5, 17, etc. However, not all numbers of this form are prime for larger \(n\).
Fermat primes have unique properties:
  • The order of 2 modulo a Fermat prime \(q\) is \(2^{n+1}\).
  • Due to Fermat's Little Theorem, calculations become easier knowing that powers of numbers modulo a Fermat prime have finite cyclic behavior.
Understanding these primes helps in broader number theoretic studies and applications like quadratic residues.
Mersenne Primes
Mersenne primes are another fascinating group, characterized by numbers of the form \(2^p - 1\), where \(p\) itself is a prime number. Known for their connection to the study of perfect numbers, Mersenne primes hold great mathematical significance.
These primes are named after Marin Mersenne, a French monk who studied these numbers in the 17th century.
Key properties of Mersenne primes include:
  • The divisibility of the order of elements in the group of units modulo \(q\), where \(q = 2^p - 1\), by \(q-1\).
  • The practical application in cryptography and efficient pseudorandom number generation.
Discovering Mersenne primes remains an active area of research, especially because they are some of the largest known prime numbers today.
Legendre Symbol
The Legendre symbol \((\frac{a}{p})\) is a mathematical notation that reveals whether a number \(a\) is a quadratic residue modulo a prime \(p\). It's a compact way to express answers to complex congruence questions.
Here's how it works:
  • If \(a\) is a quadratic residue modulo \(p\), \((\frac{a}{p}) = 1\).
  • If \(a\) is not a quadratic residue, \((\frac{a}{p}) = -1\).
  • If \(a\) is divisible by \(p\), \((\frac{a}{p}) = 0\).
Using the Legendre symbol significantly simplifies the determination of quadratic residues, transforming intricate calculations into straightforward evaluations.
Fermat's Little Theorem
Fermat's Little Theorem is a fundamental result in number theory. It states that if \(p\) is a prime number and \(a\) is any integer not divisible by \(p\), then \(a^{p-1} \equiv 1 \pmod{p}\).
This theorem provides a powerful tool for calculations modulo prime numbers:
  • It's used to quickly show relationships between numbers in modular arithmetic.
  • In the context of quadratic residues, it helps determine the order of powers modulo a prime.
  • Fermat's theorem is also essential in proofs involving number properties like congruences and residue classes.
Overall, Fermat's Little Theorem provides a backbone for many number theory applications, particularly in solving problems involving primes like in the original exercise.

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Most popular questions from this chapter

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 \(p\).

(a) Show that 7 and 18 are the only incongruent solutions of \(x^{2} \equiv-1\left(\bmod 5^{2}\right)\). (b) Use part (a) to find the solutions of \(x^{2}=-1\left(\bmod 5^{3}\right)\).

Given that \(a\) is a quadratic residue of the odd prime \(p\), prove the following: (a) \(a\) is not a primitive root of \(p\). (b) The integer \(p-a\) is a quadratic residue or nonresidue of \(p\) according as \(p=1\) \((\bmod 4)\) or \(p \equiv 3(\bmod 4)\). (c) If \(p=3(\bmod 4)\), then \(x=\pm a^{(p+1) / 4}(\bmod p)\) are the solutions of the congruence \(x^{2} \equiv a(\bmod p) .\)

(a) If the prime \(p>3\), show that \(p\) divides the sum of its quadratic residues. (b) If the prime \(p>5\), show that \(p\) divides the sum of the squares of its quadratic nonresidues.

Let \(p\) be an odd prime and \(\operatorname{gcd}(a, p)=\operatorname{gcd}(b, p)=1\). Prove that either all three of the quadratic congruences $$ x^{2} \equiv a(\bmod p) \quad x^{2}=b(\bmod p) \quad x^{2}=a b(\bmod p) $$ are solvable or exactly one of them admits a solution.
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