Last edited by Tajora
Thursday, July 16, 2020 | History

2 edition of Criteria for biquadratic residuacity modulo a prime p involving quaternary representaions of p found in the catalog.

Criteria for biquadratic residuacity modulo a prime p involving quaternary representaions of p

Kenneth S. Williams

# Criteria for biquadratic residuacity modulo a prime p involving quaternary representaions of p

## by Kenneth S. Williams

Written in English

Subjects:
• Numbers, Prime.,
• Congruences and residues.

• Edition Notes

Classifications The Physical Object Statement by Kenneth S. Williams, Christian Friesen, and Lawrence J. Howe. Series Carleton mathematical series,, no. 204 Contributions Friesen, Christian., Howe, Lawrence J. LC Classifications QA246 .W45 1983 Pagination 46 leaves ; Number of Pages 46 Open Library OL2943426M LC Control Number 84182662

It clearly follows from Theorem 4 that V > if p is a prime of the form 4n + 3, then the quadratic residues outnumber the nonresidues on the interval (0, p/2) [by a number divisible by 3 if p ≡ 3 (mod 8) and p ≠ 3].. This assertion, despite its simplicity, lies among some very deep results of number theory. It was obtained by us as a simple corollary of the fact that the number h. De nition When p is an odd prime number, de ne the Legendre sym-bol a p by a p = 8 >: +1; when a is a quadratic residue modulo p, 1; when a is a quadratic non-residue modulo p, 0; when p ja. Theorem (Euler’s criterion). When p is an odd prime, one has a p a(p 1)=2 (mod p): Proof. If a(p 1)=2 1 (mod p), then the desired conclusion.

quadratic residue[kwä¦dradik ′rezədü] (mathematics) A residue of order 2. Quadratic Residue a concept in number theory. A number a for which the congruence x2 ≡ a (mod m) has a solution is called a quadratic residue modulo m; in other words, a is a quadratic residue modulo m if for a certain integer x the number x2 − a is divisible by m. Lemma 5 Show that if p is an odd prime and a is an integer not divisible by p, then the congruence x2 ⌘ a mod p has either no solutions or exactly two incongruent solutions modulo p. Proof Assume that there is at least one solution, call it y. Let z be another solution to x2 ⌘ a mod p. Then, y2 ⌘ z2 mod p and thus (y z)(y + z) ⌘ 0 mod p.

Hence, (p - 1) ∣ (p - 1) k∕ 2, from which we conclude that 2 | k and k∕ 2 is an integer. Let b = g k∕ 2. Then b 2 ≡ g k ≡ a (mod p), so b is a square root of a modulo p, as desired. __ 65 Finding Square Roots Modulo Special Primes. The Euler criterion lets us test membership in QR p for prime p. Abstract Given any" 2 (0;1=2) and any positive integer s ‚ 2, we prove that for every prime p ‚ maxfs2(4=")2s;ssloglog(10s)g satisfying ’(p¡1)=(p¡1) • 1=2¡", where ’(k) is the Euler function, there are s consecutive quadratic non-residues which are not primitive roots modulo p. Mathematics Subject Classiﬂcation: 11A07, 11N35, 11N69 Key Words: quadratic residue, primitive root.

You might also like
Moving bog

Moving bog

Energy utilization by young grass carp fed different diets

Energy utilization by young grass carp fed different diets

Glen Canyon archaeological survey.

Glen Canyon archaeological survey.

Dateline--Saigon

Dateline--Saigon

Policing on American Indian reservations

Policing on American Indian reservations

Report for 1989.

Report for 1989.

collection of ancient glass 500 BC-500 AD

collection of ancient glass 500 BC-500 AD

Needs of rural schools regarding HIV education

Needs of rural schools regarding HIV education

Pet crafts

Pet crafts

Chronicle of the 20th century

Chronicle of the 20th century

Erwin Piscators Political Theatre

Erwin Piscators Political Theatre

The progressive revelation of Jesus as Savior, Lord, and King

The progressive revelation of Jesus as Savior, Lord, and King

Pastoral care in parishes without a pastor

Pastoral care in parishes without a pastor

Rugby - a tactical appreciation

Rugby - a tactical appreciation

### Criteria for biquadratic residuacity modulo a prime p involving quaternary representaions of p by Kenneth S. Williams Download PDF EPUB FB2

IMOmath: Problem solving with quadratic residues. From now on, unless noted otherwise, $$p$$ is always an odd prime and $$a$$ an integer. The quadratic residuosity problem in computational number theory is to decide, given integers and, whether is a quadratic residue modulo or not. Here = for two unknown primes and, and is among the numbers which are not obviously quadratic non-residues (see below).

The problem was first described by Gauss in his Disquisitiones Arithmeticae in This problem is believed to be. Modulo a prime, the product of two nonresidues is a residue and the product of a nonresidue and a (nonzero) residue is a nonresidue. The first supplement to the law of quadratic reciprocity is that if p ≡ 1 (mod 4) then −1 is a quadratic residue modulo p, and if p ≡ 3 (mod 4) then −1 is a nonresidue modulo p.

This implies the following. more than one random number modulo P. Introduction There is an extensive literature on the distribution of quadratic residues and nonresidues modulo a prime number.

Much of it is dedicated to the question of how small is the smallest quadratic nonresidue of a prime P congruent to ±1 modulo 8. In mathematics, a number q is called a quadratic residue modulo p if there exists an integer x such that: ${x^2}\equiv{q}\ (mod\ p)$ Otherwise, q is called a quadratic non-residue.

In effect, a quadratic residue modulo p is a number that has a square root in modular arithmetic when the modulus is law of quadratic reciprocity says something about quadratic residues and primes. criterion in terms of the parameters in the partitions p = 8/ + 1 = X2 4- Y2 = C2 + 2D2 for an odd prime q / p to be an octic residue (modp).

Special cases have been given by von Lienen [11]. The proof of Theorem is based on the fact that a prime q ^ p is an e-th power (modp) if and only if the Gaussian period polynomial of degree e has. Prime moduli We are mainly interested in quadratic residues modulo a prime. Proposition Suppose pis an odd prime.

Then just (p 1)=2 of the numbers 1;2;;p 1 are quadratic residues mod p, and the same number are quadratic non-residues.

P- 1 modulo P. Denote by RN the number of occurrences of a quadratic residue followed by a nonresidue modulo P. We can similarly denote the number of occur- rences of any pattern of quadratic residues and nonresidues modulo P.

Exact formulas are known for RR, RN, NR, NN [1] (as expected, these are all approximately P/4). p, p prime, the square root of a is the solution of the quadratic congruence x2 ≡a (mod p).

This problem has an easy solution when p ≡3 (mod 4) namely x ≡±a p+1 4 (mod p). For the remaining primes p ≡ 1 (mod 4) there are explicit solutions when p ≡ 5 (mod 8) [2].

But the case p ≡ 1 (mod 8) is non-trivial. There are many. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up. If p $\equiv$ 3 (mod 4) with p prime, prove -1 is a non-quadratic residue modulo p.

I suppose this would not be true if p $\equiv$ 1 (modulo 4). To prove something is a non-square I find to be tricky. It's difficult to see any straightforward way to do this using only the definition of congruence for example.

root modulo p. Solution: Let a be a quadratic residue modulo the prime p, where p = 2k +1 for some prime p > 1. Then, by Euler’s criterion, a2k−1 = a(p−1)/2 ≡ −1 (mod p), and so upon squaring both sides we obtain a2k ≡ 1 (mod p). This means that the order of a modulo p is a divisor of 2k. If the order of a modulo p.

Efficiently distinguishing a quadratic residue from a nonresidue modulo $$N = p q$$ for primes $$p, q$$ is an open problem. This is exploited by several cryptosystems, such as Goldwassser-Micali encryption, or Cocks identity-based encryption.

More general variants of this problem underlie other cryptosystems such as Paillier encryption. the prime divisor property, which is equivalent to unique factorization).

Then phas a nontrivial factorizationp= inZ[i],sop2 = N(p) = N()N() =)N() = N() = p,i.e.,p= x2 + y2 wherex+yi. Again,onedirectionofTheoremsandareeasy. Ifp 5;7 mod8,thenp6= x2 +2y2 and if p 2 mod3, then p6= x2 + 3y2 (see Section ).

To prove the hard direction. If there exists a prime p [respectively infinitely many primes p] congruent to 3 modulo 8 and such that k is not a cube modulo p. Thanks in advance for proof or. Stack Exchange Network.

Stack Exchange network consists of Q&A communities including Stack Overflow. Very interesting book, although the way proofs are presented sometimes throws me off a little. Relationship between quadratic residues modulo a prime and quadratic residues modulo a prime power [closed] Ask Question Asked 4 years, 8 months ago.

Active 4 years, 8 months ago. Viewed 2k times 1. 1 $\begingroup$ Closed. This question is off. Quadratic Residues The Legendre symbol Let pbe an odd prime. For an integer a, we deﬁne the Legendre symbol a p:= +1, if ais a quadratic residue mod p, −1, otherwise.

Lemma ab p = a p b p. Proof. This is equivalent to saying that modulo p, the product of two quadratic residues (respectively nonresidues) is a quadratic residue, and the product of a. p =1 for each prime p and integer x, p ∤x.

Example 2. Since 2 is a quadratic residue modulo 7 (32 ≡ 2), and 3 is not, we have 2 7 =1 and 3 7 =−1. From now on, unless noted otherwise, p is always an odd prime and a an integer. We also denote p′ = p−1 2.

Clearly, a is a quadratic residue modulo p if and only if so is a+kp for some. Lemma 7 Let n = q1q2 qm be the unique prime factorization of n, where q i= p i, i > 0, each pi is a prime number, and p1 prime power qi of n.

Furthermore, a is a QR mod n if and only if the following conditions are met. (1) For each odd prime factor, pi, a is a QR. PropositionSuppose pis an odd prime; and suppose a2Z is coprime to p. Then ais a quadratic residue modulo pe (where e 1) if and only if it is quadratic residue modulo p.

Proof. The argument we gave above for quadratic residues modulo pstill applieshere. Lemma If: (Z=pe)!(Z=pe)timesis the homomorphism under which t7!t2 mod pe then ker. Theorem 2. If a6 0 (mod p), then there is some integer n 1 such that an 1 (mod p).

Proof. Consider the sequence a1;a2;a3; Since there are only nitely many numbers modulo p, by the Pigeonhole principle, there must be some numbers np). Since ahas an inverse modulo p, a nhas an inverse modulo p, so 1 am (mod p).

Thus, m.N must be composite. Since N is odd, it’s prime divisors are odd (and not of the form 4k+1). If p is a prime divisor of N, pjx2 + 1, so x2 + 1 p 0 and x2 p 1. Thus 1 is a quadratic modulo p, so p 1 mod 4 (by exercise ) and p = 4k + 1 for some k.

But then pjN and pjp p n implies pj4p 1 p n 2 (since p is one of those primes) and.numbers in {1,2,p−1} are quadratic nonresidues. Deﬁnition. Let pbe an odd prime, and let (a,p) = 1. The Legendre symbol a p is deﬁned by a p = ˆ 1 if is a quadratic residue mod p −1 if a is a quadratic nonresidue mod Note that a= 0 is disallowed (since (0,p) = p6= 1) even though x2 = 0 (mod p) has a solution.

As an easy example, 4