In number theory, the law of quadratic reciprocity is a theorem about modular arithmetic that gives conditions for the solvability of quadratic equations modulo prime numbers. (Ref: http://en.m.wikipedia.org/wiki/Quadratic_reciprocity)

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Primality radius and quadratic reciprocity law

Given an integer $n>1$, I say that $r$ is a primality radius of $n$ if both $n-r$ and $n+r$ are primes. Goldbach's conjecture asserts that every integer greater than $1$ admits a primality radius. ...
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Describe all odd primes p for which 7 is a quadratic residue

I need to describe all odd primes $p$ for which $7$ is a quadratic residue. Now let $\left(\frac{a}{b}\right)$ be the Legendre Symbol. Then if $7$ is a quadratic residue $p$ we must have: ...
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quadratic reciprocity

I know $x^2\equiv-7\pmod7$ has solutions. How can I check if $x^2\equiv-7\pmod{49}$ has solutions? I know $-7\equiv42\pmod{49}$ but $49$ isn't a prime so I can't use Euler's criterion. How shall I do ...
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When is $2$ a quadratic residue mod $p$?

For which prime numbers $p$, is $2$ a quadratic residue modulo $p$. I know that $2$ is a quadratic reside iff $$2^{\frac{p-1}{2}} =1 \; \bmod \;(p) $$ so $$2^{p-1} =1 \; \mod \; (p). $$ But I ...
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Describe the set of odd primes such that $\left(\frac{-5}{p}\right) = 1$ (Legendre Symbol)

Okay, so $\left(\frac{-5}{p}\right) = 1$. I am assuming that I can start this by saying $\left(\frac{-5}{p}\right) = \left(\frac{5}{p}\right) \times \left(\frac{-1}{p}\right)$. There are well ...
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Quadratic Reciprocity

I've been asked to see if $x^2\equiv83$ $(\mathrm {mod} \ 101^{2000})$ has solutions. Now I know $x^2\equiv(\mathrm{mod} \ 101)$ has no solutions since the quadratic reside symbol ...
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Calculating the Legendre symbols $\left(\frac{295}{401}\right)$ and $\left(\frac{713}{1009}\right)$ using quadratic reciprocity

Evaluate the following Legendre symbols using quadratic reciprocity: $\left(\frac{295}{401}\right)$ $\left(\frac{713}{1009}\right)$ I know that can flip the numbers and reduce because both $401$ ...
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Find conditions for $\left(\frac{-3}{p}\right)=1$

$\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{p}{3}\right)^{\frac{p-1}{2}}$ $=\begin{cases}1,\:p\equiv 1\pmod{4}\text{ or ...
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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 ...
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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 = ...
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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 ...
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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 ...
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37 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 ...
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24 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 ...
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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 ...
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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 ...
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34 views

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

Compute the Jacobi symbol $\left(\frac{47}{109}\right)$ using multiplicativity and quadratic reciprocity ...
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1answer
32 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 ...
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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 ...
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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}}} ...
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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 ...
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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 ...
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44 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 ...
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50 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.
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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 ...
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23 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 ...
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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 ...
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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 ...
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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^ ...
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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 ...
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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) ...
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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 ...
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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
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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 ...
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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 ...
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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 ...
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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 ...
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365 views

Is every integer a quadratic residue mod some p?

Is every integer (say $d$) a quadratic residue mod some prime number $p$?
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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) = ...
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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 ...
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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 ...
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51 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.
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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 ...
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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)?
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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 ...
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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 ...
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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 ...
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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 ...
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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} ...
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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 ...