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3
votes
1answer
36 views

What is $\left[\frac{1}{2}(p-1)\right]! \;(\text{mod } p)$ for $p = 4k+1$?

Theorem #114 in Hardy and Wright says if $p = 4k+3$ then $$ \left[\frac{1}{2}(p-1)\right]! \equiv (-1)^\nu \mod p$$ where $\nu = \# \{ \text{non residues mod } p\text{ less than }p/2\}$. Is ...
0
votes
2answers
20 views

Quadratic reciprocity: $\left( \dfrac{-1}{p}\right) = (-1)^{\frac{p-1}{2}}$

Prove $\left( \dfrac{-1}{p}\right) = (-1)^{\frac{p-1}{2}}$, where $p$ is an odd prime, and the LHS is the legendre symbol. I've got $-1 = x^2 \pmod p \implies (-1)^{\frac{p-1}{2}} = x^{p-1} = 1 = ...
1
vote
2answers
85 views

Different formulations of the Law of Quadratic Reciprocity

The law of quadratic reciprocity is given as: $(\frac{p}{q})(\frac{q}{p}) = (-1)^{((p-1)/2)((q-1)/2)}$ Apparently we can say: $(\frac{p}{q}) = (\frac{q}{p})(-1)^{((p-1)/2)((q-1)/2)}$ and ...
0
votes
0answers
20 views

If $S=\sum (\frac{n}{p})\zeta^n$ then how to prove that $S^2=(\frac{-1}{p})p$? [duplicate]

Here $\zeta$ is a primitive $n$-th root of unity and ($\frac{n}{p}$) denotes the Legendre symbol. Can someone please give a proof of this fact? I tried writing $S^2$ as the product of two sums $S=\sum ...
1
vote
1answer
31 views

Determining prime numbers p which satisfy quadratic residues modulo p

I'm learning about quadratic reciprocity and I'm stuck on an exercise. It states : Determine the congruence characterizing all prime numbers p for the following integers such that they are quadratic ...
1
vote
1answer
23 views

Proof on the Legendre Symbol

I'm working on an exercise involving the Legendre Symbol. It gives me a hint but I'm not sure how to prove it. Let p and q be odd prime numbers with $p = q + 4a$ for some $a \in \mathbb{Z}$. Prove ...
0
votes
2answers
23 views

Find all primes $p > 11$ such that $11$ is a quadratic residue $\pmod p$.

Find all primes $p > 11$ such that $11$ is a quadratic residue $\pmod p$. (The answer should be in terms of congruence conditions mod a certain number.) I believe that we need to find all primes ...
1
vote
1answer
29 views

Equivalent statement to Quadratic reciprocity

One corollary of quadratic reciprocity is that $(\frac{p}q)$ depends only on the equivalence class of $q$ modulo $4p$. Is this enough to derive quadratic reciprocity. While equivalent does not really ...
1
vote
1answer
33 views

Computing Jacobi symbol $\left(\frac{47}{109}\right)$

Compute the Jacobi symbol $\left(\frac{47}{109}\right)$ using multiplicativity and quadratic reciprocity ...
1
vote
1answer
30 views

A conjecture about quadratic residues given $p \equiv 5 \pmod 8$ (Resolved)

Original Problem $p$ is a prime that is congruent to $5$ modulo $8$ and $a$ is a quadratic residue modulo $p$. Prove that excactly one of $x_1=a^{\frac{p+3}{8}},x_2=(2a)(4a)^{\frac{p-5}{8}}$ is the ...
1
vote
1answer
28 views

For which primes $p$ does $x^2\equiv3\pmod{p}$ have a solution?

attempt at solution: using Legendre symbol, what I did was notice that the equation has a solution if ($3/p$)=1 IFF ($p/3$)x(-1)^(p-1)/2 = 1 So, can I say that p has to satisfy two properties (4 ...
1
vote
0answers
26 views

Averages of $L(1,\chi)$

Let $(\frac{m}{n})$ denote the usual quadratic Jacobi symbol and $\mu(n)$ be the Moebius function. The series $$ \sum_{\substack{m,n \in \mathbb{N} \\ m,n\equiv 1 \mod{4}}} ...
0
votes
2answers
42 views

Evaluating quadratic residue $(6/p) $

Evaluating quadratic residue $(6/p) $ I want to find for which numbers mod 6 this residue is equal to 1 and for which it's equal to -1. I've tried splitting it into $(3/p)(2/p)$ and expressing it as ...
0
votes
0answers
51 views

For which n does a quadratic congruence have solutions e.g $x^2=83 (modn) $

For which n does a quadratic congruence have solutions e.g $x^2=83 \pmod n$ I've tried using the Chinese remainder theorem to break it down, this is what I have so far $x^2=83 \pmod2 $ has ...
0
votes
2answers
39 views

Quadratic Reciprocity

For which prime numbers p does the congruence $x^2+x+1\equiv0$ mod p have solutions? I am new to the topic of quadratic reciprocity and I know how to answer this question had it been for which prime ...
2
votes
1answer
45 views

3 is Square in finite field

Let $K$ be a finite field with $p^2$ elements. Show that $3$ is square in $K$. I know that 3 is sum of two squares. Thanks.
3
votes
0answers
74 views

Number Theory: Legendre Symbols

I have the following question. Calculate the Legendre symbol $\bigl(\frac{77}{5^{200}+1}\bigr)$. I know the following: $5^6\equiv1\pmod7$, $5^{10}\equiv1\pmod{11}$. Thus I approached this ...
1
vote
1answer
22 views

Quadratic nonresidues mod p

The question asks to find congruence conditions on prime $p$ such that $7$ is the least quadratic nonresidue mod p. Also, find the least such prime. I solved it for $1,2,3,4,5,6$ mod $p$ and got ...
0
votes
2answers
72 views

Ramification and Quadratic Reciprocity Law

I have a question regarding the follow problem: Show that the prime number 27644437 splits completely in $L = \mathbb{Q}(\sqrt{55})$. From what I understand. This deals with ...
6
votes
0answers
69 views

If $a$ is a quadratic residue modulo every prime $p$, it is a square - without using quadratic reciprocity.

The question is basically the title itself. It is easy to prove using quadratic reciprocity that non squares are non residues for some prime $p$. I would like to make use of this fact in a proof of ...
0
votes
1answer
27 views

Suplementary law and reciprocity

Let $p$ odd prime. If $\omega\in\overline{\mathbb{F_p}}$ is a 8-th primitive root of unity. Then if $\gamma=\omega+\omega^{-1}$. Why is i) $\gamma^p=\omega^ ...
0
votes
2answers
159 views

Reciprocal of 81 being the sequence of all natural numbers?

According to this document: http://www.answering-christianity.com/fakir60/81.htm describing the theory of scientist Peter Plichta, the reciprocal of 81 is: the ...
-1
votes
1answer
82 views

A Conjecture on The Generalization of Quadratic Reciprocity Law

Is there any way to prove the following conjecture regarding the Generalization of Quadratic Reciprocity Law. The statement being, $$ \left(\dfrac{a_1}{a_2}\right)\left(\dfrac{a_2}{a_3}\right) ...
1
vote
1answer
28 views

Exercise of Quadratic Reciprocity

This is an exercise in Burton : Prove that $$(5/p) =1\ iff\ p\equiv 1,\ 9,\ 11,\ or\ 19\ (20) $$ Note that $5=4+1$ so that $(5/p)=(p/5)$. In further $$ (p/5)^2=(5/p)(p/5) = (-1)^{1\cdot ...
0
votes
0answers
32 views

Find a criterion for the primes p such that (5/p) = 1. [duplicate]

Find a criterion for the primes p such that (5/p) = 1. I don't understand this question it is like Determine all primes P such that (5/p)=1 I appreciate any help
4
votes
3answers
144 views

Elementary, direct proof of when $5$ is a quadratic residue mod $p$

$\newcommand{\kron}[2]{\left( \frac{#1}{#2} \right)}$ It's easy to use Quadratic Reciprocity to show that $\kron{5}{p} = \kron{p}{5} = 1$ when $p \equiv \pm 1 \pmod 5$, and is $-1$ when $p \equiv \pm ...
1
vote
0answers
32 views

Quadratic reciprocity for odd integers

I have a book that proves the three laws of quadratic reciprocity for primes i.e $\left(\frac{-1}p\right)=1$ if $p=1\mod4$ and $-1$ if $p=3\mod4$, where $p$ is prime, etc. But in the end the book ...
0
votes
1answer
52 views

finding all primes $p$ for which a given number is a quadratic residue

I have seen an exercise on the Apostol, but I haven't understood some passages. I would be very grateful if you could solve my doubts. The problem is Find all primes $p$ for which 3 is a ...
0
votes
0answers
22 views

For which positive odd integers $n,$ $(n,33)=1$ does the Jacobi symbol $\left(\frac{33}{n}\right) = 1$?

Attention: What modulus should we use to insure that n is odd? So I know that $\left(\frac{33}{n}\right) = \left(\frac{11}{n}\right)\left(\frac{3}{n}\right)$ then using the legendre properties ...
1
vote
2answers
80 views

factor 9997 using quadratic reciprocity

I looked up the factorization and it is $13\cdot769$, but I have no idea how quadratic reciprocity allows you to deduce this without knowing it. I thought maybe ...
7
votes
1answer
360 views

Is every integer a quadratic residue mod some p?

Is every integer (say $d$) a quadratic residue mod some prime number $p$?
1
vote
0answers
38 views

When can a congruence relation be transformed into quadratic reciprocity expressions?

When can a congruence relation $$p \equiv c_1, c_2, \ldots, c_r \mod{N}$$ be transformed back into quadratic reciprocity expressions $$\left (\frac{d_1}{p} \right) = \left (\frac{d_1}{p} \right) = ...
0
votes
2answers
102 views

Let p be an odd prime, q the smallest quadratic non residue (mod p). Prove q is prime.

So I have this problem; Let p be an odd prime and let q be the smallest positive integer which is a quadratic non residue (mod p). Prove q is a prime. So what I know is that, since q is the ...
2
votes
1answer
168 views

Quadratic reciprocity problem

How can I use quadratic reciprocity to prove that $-3$ is a quadratic residue $\pmod p$ if and only if $p=2$ or $p \equiv 1 \pmod 6$ and deduce that $\mathbb{Z}[\sqrt{-3}]/(p)\cong \mathbb{F}_p ...
0
votes
1answer
50 views

Quadratic congruences problem $(x^2-a)(x^2-b)(x^2-ab)$ ≡ 0 (mod p)

Let p be an odd prime and let a,b ∈ Z such that p ∤ a. Prove that the congruence $(x^2-a)(x^2-b)(x^2-ab)$ ≡ 0 (mod p) is always solvable. Not sure where to begin here.
3
votes
2answers
251 views

elementary proof that infinite primes quadratic residue modulo $p$

$p \gt 2$ is a prime, then there are infinite primes $q$ such that $q$ is a quadratic residue modulo $p$. With Dirichlet's theorem on arithmetic progressions, the problem is easy! How about ...
1
vote
1answer
82 views

Analytic proof of quadratic reciprocity

Is there any proof of quadratic reciprocity that is more analytic than those described on Wikipedia (http://en.wikipedia.org/wiki/Proofs_of_quadratic_reciprocity)?
3
votes
1answer
132 views

Consecutive quadratic residues

I was studying quadratic reciprocity laws and came across the following question: Is it true that for every $k \in \mathbb{N} $ there exists a prime $p$ such that $1,2,...,k$ are all quadratic ...
1
vote
1answer
50 views

Quadratic Reciprocity as a consequence of Eisenstein Reciprocity

I was recently looking at the wikipedia page on Eisenstein Reciprocity, which says it "extends Quadratic Reciprocity." However, though the two do seem to be related, I don't completely understand how ...
2
votes
1answer
91 views

Prove that S consists of the quadratic residues mod (p) and T consists of the quadratic non-residues mod (p)

All the followings are $\bmod$ $p$ Let p be an odd prime. Suppose that the set $X = \{ 1,2, . . . , p-1\}$ can be written as the union of two nonempty subsets $S$ and $T$, where $S \neq T$, such that ...
2
votes
1answer
207 views

Quadratic Reciprocity Joke

I found a joke on a site made by user Zev Chonoles on quadratic reciprocity, the joke is as follows: $$\text{Quadratic reciprocity: } \left(\frac{p}{q}\right)=\left(\frac{q}{p}\right), \text{ up to ...
2
votes
1answer
39 views

Jacobi symbol $\left(\frac{(n+1)/2}{n}\right)$

Let $(\frac{a}{n})_J$ be Jacobi symbol defined by \begin{equation} \left(\frac{a}{n}\right)_J=\left(\frac{a}{p_1}\right)^{e_1}\left(\frac{a}{p_2}\right)^{e_2}\cdots\left(\frac{a}{p_k}\right)^{e_k} ...
1
vote
2answers
43 views

Number of elements that exist so $b^3 \equiv a\pmod n$, when n is composed of p and q who are prime numbers

Given $2$ prime numbers,$ p$ and $q$, that are both not even, and $3$ doesn't divide $p-1$ or $q-1$, and $n=pq$, how many elements in $Z^*_n$ exists that has $b$ such that $b^3\equiv a\pmod n$ . I ...
5
votes
6answers
406 views

What is the point of quadratic residues?

What is the most motivating way to introduce quadratic residues? Are there any real life examples of quadratic residues? Why is the Law of Quadratic Reciprocity considered as one of the most ...
0
votes
1answer
156 views

Legendre symbols (p/q) = (a/q) [duplicate]

Suppose q and p are odd primes and p = q+4a for some integer a. From the properties of the Legendre symbol and the Law of Quadratic Reciprocity prove that the following two identities hold. (p/q) = ...
0
votes
2answers
55 views

$2^{4n+1} \equiv 1 \pmod{8n+7}$, this has been bugging me

Here is the question: Suppose that $p$ is an odd prime. The law of quadratic reciprocity says that $x^2\equiv 2\pmod p$ has a solution. if $p\equiv1 \text{ or } 7 \pmod 8$. Prove that ...
0
votes
2answers
79 views

If $x^2 \equiv a \bmod p$ has a solution, does it necessarily have infinitely many solutions? [closed]

If not, how would you prove it for a specific case? Such as, for example, $x^2 \equiv 2 \bmod 7$ ?
1
vote
1answer
85 views

Suppose $p$ and $q$ are odd primes and $p = q + 4a$ for some $a$. Prove that $(\frac ap) = (\frac aq)$ holds. [duplicate]

This problem also had me prove that $(\frac pq) = (\frac aq)$, but I've already managed to do that. I've tried messing around with the Law of Quadratic Reciprocity but can't get anything. I've also ...
0
votes
2answers
32 views

Quadratic character of 3

Using the QRL prove that, for any odd prime $p$, $(3/p) = 1$ if $p$ is congruent to $1$ or $11 \pmod{12}$. Using the Quadratic reciprocity law, $(3/p)(p/3)=(-1)^{(3-1)(p-1)/4}$, I get that the ...
1
vote
1answer
41 views

Prove that $\left(\frac2p\right) = 1$ if $p \equiv 1,7 \pmod 8$ and $\left(\frac2p\right) = -1$ if $p \equiv 3,5 \pmod 8$ using ring theory

Let $p$ be an odd prime number and let $\alpha = [X] \in R=\mathbb F_p[X]/\langle X^4+1\rangle$, and $y = \alpha + \alpha^{-1}$ I've proven: 1) $\alpha$ is a primitive eight root of unity in $R$. ...