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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Quadratic reciprocity: Tell if $c$ got quadratic square root mod $p$

I am reading the wiki article about Quadratic reciprocity and I don't understand how can I tell if some integer $c$ got quadratic root mod $p$? So far I am using brute search to find $y$ such that ...
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Fermat's little theorem and Euler's criterion

Is it possible to find the solution of this congruence by Fermat's little theorem and how ? $$15125^{2401}\pmod {72}$$ Can somebody tell me how to do it by Euler's criterion?
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Quadratic Reciprocity and Congruences

I have to find the result of congruences : $$(a)\left(\frac{34}{73}\right)$$ $$(b)\left(\frac{36}{73}\right)$$ $$(c)\left(\frac{1356}{2467}\right)$$ By the way,I found that Theorem of Quadratic ...
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If $p>3$ and $x^2+3y^2=p$, prove $\left(\frac{-3}{p}\right)=1$ and $p\equiv 1\pmod{6}$.

If $p>3$ and $x^2+3y^2=p$, prove $\left(\frac{-3}{p}\right)=1$ and $p\equiv 1\pmod{6}$. $$\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)$$ We know that ...
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Quadratic congruence with linear term

I'm trying to determine whether the congruence $x^2+x-1\equiv0\pmod{61}$ has a solution. I know how to do this without the linear term, for instance, $x^2-5\equiv0\pmod{61}$. I can solve it just by ...
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Irreducibility in a polynomial related to quadratic residues

From Romania TST 2004 Day 5 P3, I was introduced to the polynomial $$f(x)=\sum_{i=1}^{p-1} \left( \frac{i}{p} \right)x^{i-1}$$ This polynomial is clearly not irreducible - $x=1$ is a root. Even more ...
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Let $p> 7$, prove that $\left(\frac{2}{q}\right) = (-1)^{\frac{q^2-1}{8}}$ with $q$ an odd prime

Let $p> 7$, prove that $\left(\frac{2}{q}\right) = (-1)^{\frac{q^2-1}{8}}$. with $q$ an odd prime. We can by using the following verifications: $$\left(\frac{2}{p}\right) = ...
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If $n\in\Bbb Z^+$ is not a square, prove exist infinitely many primes $p$ such that $\left(\frac{n}{p}\right)=-1$.

If $n\in\Bbb Z^+$ is not a square, prove exist infinitely many primes $p$ such that $\left(\frac{n}{p}\right)=-1$. Note that if $p\nmid n$ and $n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots ...
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Prove that $\sum^{P}_{k=1} \lfloor\frac {ka}{p}\rfloor = \frac{p^{2} - 1}{8} + \mu(a,p)$ mod$(2)$

I can prove that $\sum^{P}_{k=1} \lfloor\frac {ka}{p}\rfloor = \mu(a,p)$ mod$(2)$ where $p$ is an odd prime, $P = \frac {p-1}{2}$, $a$ is an integer not divisible by $p$, and $\mu(a,p)$ is the ...
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verify if $8x^2 - 2x - 3 \equiv 0 \pmod{75}$ has solutions or not.

verify if $8x^2 - 2x - 3 \equiv 0 \pmod{75}$ has solutions or not. I tried to write the left term in a form $y^2 \equiv a \pmod{75}$, where $a \in \mathbb{Z}$ so then I can use quadratic repriocity ...
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Quadratic Reciprocity Equivalent Statements

How could I show that the commonly accepted statement of the Law of Quadratic Reciprocity is equivalent to Legendre's statement of Quadratic Reciprocity? Hints and pointers appreciated.
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Missing link in proof of quadratic reciprocity law

I am trying to figure out the proof of the quadratic reciprocity law in Serre's book A Course in Arithmetic. I think my question is possible to answer independent of the book though: Let $l$ and $p$ ...
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55 views

Gauss' original proof of quadratic reciprocity

Is the original proof of quadratic reciprocity due to Gauss available anywhere online? I've been looking for quite a while now, but with no results. Most papers seem not to include it because of it ...
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Technological incompetence [closed]

I am a high school student going into my junior year. Earlier this summer, I went to a math camp at Ohio State University, and one of the assignments was to prove quadratic reciprocity. I was told ...
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31 views

Quadratic reciprocity in the case $a=-1$

I am reading the proof the for odd prime $p$, $$ \left ( \frac{-1}{p} \right)_2 = (-1)^{\frac{p-1}{2}} = \begin{cases} 1 \hspace{2mm} \text{for} \hspace{2mm} p \equiv 1 \operatorname{mod} 4 \\ -1 ...
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28 views

Verifying quadratic reciprocity for the Jacobi symbol

I am trying to prove: If $m,n$ are odd coprime positive integers, then $$\Big(\frac mn\Big)\Big(\frac nm\Big)=(-1)^{\large\frac{m-1}2\frac{n-1}2},$$ where $\big(\frac mn\big)$ is the Jacobi ...
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55 views

What is the probability of having a quadratic residue?

Let $a$ be a random integer and $p$ be a prime. What is the probability of $\left(\frac{a}{p}\right)=1$, as $p$ goes to infinity. And similarly, what about $\left(\frac{a}{p}\right)=-1$ or $0$?
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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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44 views

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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Find all primes $p$ such that $\left( \frac{-19}{p} \right) = 1$

Find all primes $p$ such that -19 is a quadratic residue $\bmod p$. solution: We have that $(\frac{-19}{p})=(\frac{p}{19})$, so that if $-19$ is a quadratic residue modulo $p$, then $p$ is a ...
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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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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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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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42 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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48 views

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

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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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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49 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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1answer
62 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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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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29 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^ ...
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262 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 ...
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90 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) ...
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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 ...