Editing Quadratic reciprocity (section)

==History and alternative statements==
The theorem was formulated in many ways before its modern form: Euler and Legendre did not have Gauss's congruence notation, nor did Gauss have the Legendre symbol.

In this article ''p'' and ''q'' always refer to distinct positive odd primes, and ''x'' and ''y'' to unspecified integers.

===Fermat===
Fermat proved<ref>Lemmermeyer, pp. 2&ndash;3</ref> (or claimed to have proved)<ref>Gauss, DA, art. 182</ref> a number of theorems about expressing a prime by a quadratic form:

:<math>\begin{align}
p=x^2+ y^2 \qquad &\Longleftrightarrow \qquad p=2 \quad \text{ or } \quad p\equiv 1 \bmod{4} \\
p=x^2+2y^2 \qquad &\Longleftrightarrow \qquad p=2 \quad \text{ or } \quad p\equiv 1, 3 \bmod{8} \\
p=x^2+3y^2 \qquad &\Longleftrightarrow \qquad p=3 \quad \text{ or } \quad p\equiv 1 \bmod{3} \\
\end{align}</math>

He did not state the law of quadratic reciprocity, although the cases −1, ±2, and ±3 are easy deductions from these and others of his theorems.

He also claimed to have a proof that if the prime number ''p'' ends with 7, (in base 10) and the prime number ''q'' ends in 3, and ''p'' ≡ ''q'' ≡ 3 (mod 4), then

:<math>pq=x^2+5y^2.</math>

Euler conjectured, and Lagrange proved, that<ref>Lemmermeyer, p. 3</ref>

:<math>\begin{align}
p &\equiv 1, 9 \bmod{20}\quad \Longrightarrow \quad p = x^2+5y^2 \\
p, q &\equiv 3, 7 \bmod{20} \quad \Longrightarrow \quad pq=x^2+5y^2
\end{align}</math>

Proving these and other statements of Fermat was one of the things that led mathematicians to the reciprocity theorem.

===Euler===
Translated into modern notation, Euler stated <ref>Lemmermeyer, p. 5, Ireland & Rosen, pp. 54, 61</ref> that for distinct odd primes ''p'' and ''q'':

# If ''q'' ≡ 1 (mod 4) then ''q'' is a quadratic residue (mod ''p'') if and only if there exists some integer ''b'' such that ''p'' ≡ ''b''<sup>2</sup> (mod ''q'').
# If ''q'' ≡ 3 (mod 4) then ''q'' is a quadratic residue (mod ''p'') if and only if there exists some integer ''b'' which is odd and not divisible by ''q'' such that ''p'' ≡ ±''b''<sup>2</sup> (mod 4''q'').

This is equivalent to quadratic reciprocity.

He could not prove it, but he did prove the second supplement.<ref>Ireland & Rosen, pp. 69&ndash;70. His proof is based on what are now called Gauss sums.</ref>

===Legendre and his symbol===
Fermat proved that if ''p'' is a prime number and ''a'' is an integer, 
:<math>a^p\equiv a \bmod{p}.</math>

Thus if ''p'' does not divide ''a'', using the non-obvious fact (see for example Ireland and Rosen below) that the residues modulo ''p'' form a [[field (mathematics)|field]] and therefore in particular the multiplicative group is cyclic, hence there can be at most two solutions to a quadratic equation:

:<math>a^{\frac{p-1}{2}} \equiv \pm 1 \bmod{p}.</math>

Legendre<ref>This section is based on Lemmermeyer, pp. 6&ndash;8</ref> lets ''a'' and ''A'' represent positive primes ≡ 1 (mod 4) and ''b'' and ''B'' positive primes ≡ 3 (mod 4), and sets out a table of eight theorems that together are equivalent to quadratic reciprocity:

{| class="wikitable"
|+ 
! Theorem
! width="200"|When
! width="200"|it follows that
|-
! I 
| <math>b^{\frac{a-1}{2}}\equiv 1 \bmod{a} </math>
| <math>a^{\frac{b-1}{2}}\equiv 1 \bmod{b} </math>
|-
! II 
| <math>a^{\frac{b-1}{2}}\equiv -1 \bmod{b} </math>
| <math>b^{\frac{a-1}{2}}\equiv -1 \bmod{a} </math>
|-
! III 
| <math>a^{\frac{A-1}{2}}\equiv 1 \bmod{A} </math>
| <math>A^{\frac{a-1}{2}}\equiv 1 \bmod{a} </math>
|-
! IV 
| <math>a^{\frac{A-1}{2}}\equiv -1 \bmod{A} </math>
| <math>A^{\frac{a-1}{2}}\equiv -1 \bmod{a} </math>
|-
! V 
| <math>a^{\frac{b-1}{2}}\equiv 1 \bmod{b} </math>
| <math>b^{\frac{a-1}{2}}\equiv 1 \bmod{a} </math>
|-
! VI 
| <math>b^{\frac{a-1}{2}}\equiv -1 \bmod{a} </math>
| <math>a^{\frac{b-1}{2}}\equiv -1 \bmod{b} </math>
|-
! VII 
| <math>b^{\frac{B-1}{2}}\equiv 1 \bmod{B} </math>
| <math>B^{\frac{b-1}{2}}\equiv -1 \bmod{b} </math>
|-
! VIII 
| <math>b^{\frac{B-1}{2}}\equiv -1 \bmod{B} </math>
| <math>B^{\frac{b-1}{2}}\equiv 1 \bmod{b} </math>
|}

He says that since expressions of the form

:<math>N^{\frac{c-1}{2}}\bmod{c}, \qquad \gcd(N, c) = 1</math>

will come up so often he will abbreviate them as:

:<math>\left(\frac{N}{c}\right) \equiv N^{\frac{c-1}{2}} \bmod{c} = \pm 1.</math>

This is now known as the [[Legendre symbol]], and an equivalent<ref>The equivalence is [[Euler's criterion]]</ref><ref>The analogue of Legendre's original definition is used for higher-power residue symbols</ref> definition is used today: for all integers ''a'' and all odd primes ''p''

:<math>\left(\frac{a}{p}\right) = \begin{cases} 0 & a \equiv 0 \bmod{p} \\ 1 & a \not\equiv 0\bmod{p} \text{ and } \exists x : a\equiv x^2\bmod{p} \\-1 &a \not\equiv 0\bmod{p} \text{ and there is no such } x. \end{cases} </math>

====Legendre's version of quadratic reciprocity====
:<math>\left(\frac{p}{q}\right) = \begin{cases}
 \left(\tfrac{q}{p}\right) & p\equiv 1 \bmod{4} \quad \text{ or } \quad q \equiv 1 \bmod{4} \\
-\left(\tfrac{q}{p}\right) & p \equiv 3 \bmod{4} \quad \text{ and } \quad q \equiv 3 \bmod{4}
\end{cases}</math>

He notes that these can be combined:

:<math> \left(\frac{p}{q}\right) \left(\frac{q}{p}\right) = (-1)^{\frac{p-1}{2}\frac{q-1}{2}}.</math>

A number of proofs, especially those based on [[Gauss's lemma (number theory)|Gauss's Lemma]],<ref>E.g. Kronecker's proof (Lemmermeyer, ex. p. 31, 1.34) is to use Gauss's lemma to establish that

:<math>\left(\frac{p}{q}\right) =\sgn\prod_{i=1}^{\frac{q-1}{2}}\prod_{k=1}^{\frac{p-1}{2}}\left(\frac{k}{p}-\frac{i}{q}\right)</math>

and then switch ''p'' and ''q''.</ref> explicitly calculate this formula.

====The supplementary laws using Legendre symbols====

:<math>\begin{align}
\left(\frac{-1}{p}\right) &= (-1)^{\frac{p-1}{2}} = \begin{cases} 1 &p \equiv 1 \bmod{4}\\ -1 & p \equiv 3 \bmod{4}\end{cases} \\
\left(\frac{2}{p}\right) &= (-1)^{\frac{p^2-1}{8}} = \begin{cases} 1 & p \equiv 1, 7 \bmod{8}\\ -1 & p \equiv 3, 5\bmod{8}\end{cases}
\end{align}</math>

From these two supplements, we can obtain a third reciprocity law for the quadratic character -2 as follows:

For -2 to be a quadratic residue, either -1 or 2 are both quadratic residues, or both non-residues :<math>\bmod p</math>.

So either :<math>\frac{p-1}{2} \text{ or } \frac{p^2-1}{8}</math> are both even, or they are both odd. The sum of these two expressions is

:<math>\frac{p^2+4p-5}{8}</math> which is an integer. Therefore,

:<math>\begin{align}
\left(\frac{-2}{p}\right) &= (-1)^{\frac{p^2+4p-5}{8}} = \begin{cases} 1 & p \equiv 1, 3 \bmod{8}\\ -1 & p \equiv 5, 7\bmod{8}\end{cases}
\end{align}</math>

Legendre's attempt to prove reciprocity is based on a theorem of his:

:'''Legendre's Theorem.''' Let ''a'', ''b'' and ''c'' be integers where any pair of the three are relatively prime. Moreover assume that at least one of ''ab'', ''bc'' or ''ca'' is negative (i.e. they don't all have the same sign). If 
::<math>\begin{align}
u^2 &\equiv -bc \bmod{a} \\
v^2 &\equiv -ca \bmod{b} \\
w^2 &\equiv -ab \bmod{c} 
\end{align}</math>
:are solvable then the following equation has a nontrivial solution in integers:
::<math>ax^2 + by^2 + cz^2=0.</math>

'''Example.''' Theorem I is handled by letting ''a'' ≡ 1 and ''b'' ≡ 3 (mod 4) be primes and assuming that <math>\left (\tfrac{b}{a} \right) = 1</math> and, contrary the theorem, that <math>\left (\tfrac{a}{b} \right ) = -1.</math> Then <math>x^2+ay^2-bz^2=0</math> has a solution, and taking congruences (mod 4) leads to a contradiction.

This technique doesn't work for Theorem VIII. Let ''b'' ≡ ''B'' ≡ 3 (mod 4), and assume

:<math>\left (\frac{B}{b} \right ) = \left (\frac{b}{B} \right ) = -1.</math>

Then if there is another prime ''p'' ≡ 1 (mod 4) such that

:<math>\left (\frac{p}{b} \right ) = \left (\frac{p}{B} \right ) = -1,</math>

the solvability of <math>Bx^2+by^2-pz^2=0</math> leads to a contradiction (mod 4). But Legendre was unable to prove there has to be such a prime ''p''; he was later able to show that all that is required is:

:'''Legendre's Lemma.''' If ''p'' is a prime that is congruent to 1 modulo 4 then there exists an odd prime ''q'' such that <math>\left (\tfrac{p}{q} \right ) = -1.</math>

but he couldn't prove that either. [[#Hilbert symbol|Hilbert symbol (below)]] discusses how techniques based on the existence of solutions to <math>ax^2+by^2+cz^2=0</math> can be made to work.

===Gauss===
[[Image:Disquisitiones-Arithmeticae-p133.jpg|200px|thumb|Part of Article 131 in the first edition (1801) of the ''[[Disquisitiones Arithmeticae|Disquisitiones]]'', listing the 8 cases of quadratic reciprocity]]

Gauss first proves<ref>Gauss, DA, arts 108&ndash;116</ref> the supplementary laws. He sets<ref>Gauss, DA, arts 117&ndash;123</ref> the basis for induction by proving the theorem for ±3 and ±5. Noting<ref>Gauss, DA, arts 130</ref> that it is easier to state for −3 and +5 than it is for +3 or −5, he states<ref>Gauss, DA, Art 131</ref> the general theorem in the form:

:If ''p'' is a prime of the form 4''n''&nbsp;+&nbsp;1 then ''p'', but if ''p'' is of the form 4''n'' + 3 then −''p'', is a quadratic residue (resp. nonresidue) of every prime, which, with a positive sign, is a residue (resp. nonresidue) of ''p''. In the next sentence, he christens it the "fundamental theorem" (Gauss never used the word "reciprocity").

Introducing the notation ''a'' R ''b'' (resp. ''a'' N ''b'') to mean ''a'' is a quadratic residue (resp. nonresidue) (mod ''b''), and letting ''a'', ''a''&prime;, etc. represent positive primes ≡ 1 (mod 4) and ''b'', ''b''&prime;, etc. positive primes ≡ 3 (mod 4), he breaks it out into the same 8 cases as Legendre:

{| class="wikitable"
|+ 
! width="50"|Case 
! width="80"|If 
! width="80"|Then 
|-
! 1) 
| ±''a'' R ''a''&prime; 
| ±''a''&prime; R ''a''
|-
! 2) 
| ±''a'' N ''a''&prime; 
| ±''a''&prime; N ''a''
|-
! 3) 
| +''a'' R ''b''<br>−''a'' N ''b'' 
| ±''b'' R ''a''
|-
! 4) 
| +''a'' N ''b''<br>−''a'' R ''b'' 
| ±''b'' N ''a''
|-
! 5) 
| ±''b'' R ''a'' 
| +''a'' R ''b''<br>−''a'' N ''b''
|-
! 6) 
| ±''b'' N ''a'' 
| +''a'' N ''b''<br>−''a'' R ''b''
|-
! 7) 
| +''b'' R ''b''&prime;<br>−''b'' N ''b''&prime; 
| −''b''&prime; N ''b''<br>+''b''&prime; R ''b''
|-
! 8) 
| −''b'' N ''b''&prime;<br>+''b'' R ''b''&prime; 
| +''b''&prime; R ''b''<br>−''b''&prime; N ''b''
|}

In the next Article he generalizes this to what are basically the rules for the [[#Jacobi symbol|Jacobi symbol (below)]]. Letting ''A'', ''A''&prime;, etc. represent any (prime or composite) positive numbers ≡ 1 (mod 4) and ''B'', ''B''&prime;, etc. positive numbers ≡ 3 (mod 4):

{| class="wikitable"
|+ 
! width="50"|Case 
! width="80"|If 
! width="80"|Then 
|-
! 9) 
| ±''a'' R ''A'' 
| ±''A'' R ''a''
|-
! 10) 
| ±''b'' R ''A'' 
| +''A'' R ''b''<br>−''A'' N ''b''
|-
! 11) 
| +''a'' R ''B'' 
| ±''B'' R ''a''
|-
! 12) 
| −''a'' R ''B'' 
| ±''B'' N ''a''
|-
! 13) 
| +''b'' R ''B'' 
| −''B'' N ''b''<br>+''N'' R ''b''
|-
! 14) 
| −''b'' R ''B'' 
| +''B'' R ''b''<br>−''B'' N ''b''
|}

All of these cases take the form "if a prime is a residue (mod a composite), then the composite is a residue or nonresidue (mod the prime), depending on the congruences (mod 4)". He proves that these follow from cases 1) - 8).

Gauss needed, and was able to prove,<ref>Gauss, DA, arts. 125&ndash;129</ref> a lemma similar to the one Legendre needed:

:'''Gauss's Lemma.''' If ''p'' is a prime congruent to 1 modulo 8 then there exists an odd prime ''q'' such that:
::<math>q <2\sqrt p+1 \quad \text{and} \quad \left(\frac{p}{q}\right) = -1.</math>

The proof of quadratic reciprocity uses [[Mathematical induction#Complete induction|complete induction]].

:'''Gauss's Version in Legendre Symbols.''' 
::<math>\left(\frac{p}{q}\right) = \begin{cases} \left(\frac{q}{p}\right) & q \equiv 1 \bmod{4} \\ \left(\frac{-q}{p}\right) & q \equiv 3 \bmod{4} \end{cases}</math>

These can be combined:

:'''Gauss's Combined Version in Legendre Symbols.''' Let
::<math>q^* = (-1)^{\frac{q-1}{2}}q.</math>
:In other words:
::<math>|q^*|=|q| \quad \text{and} \quad q^*\equiv 1 \bmod{4}.</math>
:Then:
::<math> \left(\frac{p}{q}\right) = \left(\frac{q^*}{p}\right).</math>

A number of proofs of the theorem, especially those based on [[Gauss sum]]s<ref>Because the basic Gauss sum equals <math>\sqrt{q^*}.</math></ref> or the splitting of primes in [[algebraic number field]]s,<ref>Because the quadratic field <math>\Q(\sqrt{q^*})</math> is a subfield of the cyclotomic field <math>\Q(e^{\frac{2\pi i}{q}})</math></ref><ref>See [[#Connection with cyclotomic fields|Connection with cyclotomic fields]] below.</ref> derive this formula.

===Other statements===
The statements in this section are equivalent to quadratic reciprocity: if, for example, Euler's version is assumed, the Legendre-Gauss version can be deduced from it, and vice versa.

:'''Euler's Formulation of Quadratic Reciprocity.'''<ref>Ireland & Rosen, pp 60&ndash;61.</ref> If <math>p \equiv \pm q \bmod{4a}</math> then <math>\left(\tfrac{a}{p}\right)=\left(\tfrac{a}{q}\right).</math>

This can be proven using [[Gauss's lemma (number theory)|Gauss's lemma]].
 
:'''Quadratic Reciprocity (Gauss; Fourth Proof).'''<ref>Gauss, "Summierung gewisser Reihen von besonderer Art", reprinted in ''Untersuchumgen uber hohere Arithmetik'', pp.463&ndash;495</ref> Let ''a'', ''b'', ''c'', ... be unequal positive odd primes, whose product is ''n'', and let ''m'' be the number of them that are ≡ 3 (mod 4); check whether ''n''/''a'' is a residue of ''a'', whether ''n''/''b'' is a residue of ''b'', .... The number of nonresidues found will be even when ''m'' ≡ 0, 1 (mod 4), and it will be odd if ''m'' ≡ 2, 3 (mod 4).

Gauss's fourth proof consists of proving this theorem (by comparing two formulas for the value of Gauss sums) and then restricting it to two primes. He then gives an example: Let ''a'' = 3, ''b'' = 5, ''c'' = 7, and ''d'' = 11. Three of these, 3, 7, and 11 ≡ 3 (mod 4), so ''m'' ≡ 3 (mod 4). 5×7×11&nbsp;R&nbsp;3;&nbsp;3×7×11&nbsp;R&nbsp;5;&nbsp;3×5×11&nbsp;R&nbsp;7;&nbsp; and&nbsp; 3×5×7&nbsp;N&nbsp;11, so there are an odd number of nonresidues.

:'''Eisenstein's Formulation of Quadratic Reciprocity.'''<ref>Lemmermeyer, Th. 2.28, pp 63&ndash;65</ref> Assume
::<math>p\ne q, \quad p'\ne q', \quad p \equiv p' \bmod{4}, \quad q \equiv q' \bmod{4}.</math>
:Then
::<math> \left(\frac{p}{q}\right) \left(\frac{q}{p}\right) =\left(\frac{p'}{q'}\right) \left(\frac{q'}{p'}\right).</math>

:'''Mordell's Formulation of Quadratic Reciprocity.'''<ref>Lemmermeyer, ex. 1.9, p. 28</ref> Let ''a'', ''b'' and ''c'' be integers. For every prime, ''p'', dividing ''abc'' if the congruence 
::<math>ax^2 + by^2 + cz^2 \equiv 0 \bmod{\tfrac{4abc}{p}}</math>
:has a nontrivial solution, then so does:
::<math>ax^2 + by^2 + cz^2 \equiv 0 \bmod{4abc}.</math>

:'''Zeta function formulation'''
:As mentioned in the article on [[Dedekind zeta function]]s, quadratic reciprocity is equivalent to the zeta function of a quadratic field being the product of the Riemann zeta function and a certain Dirichlet L-function

===Jacobi symbol===
{{main|Jacobi symbol}}
The [[Jacobi symbol]] is a generalization of the Legendre symbol; the main difference is that the bottom number has to be positive and odd, but does not have to be prime. If it is prime, the two symbols agree. It obeys the same rules of manipulation as the Legendre symbol. In particular

:<math>\begin{align}
\left(\frac{-1}{n}\right) = (-1)^{\frac{n-1}{2}} &= \begin{cases} 1 & n \equiv 1 \bmod{4}\\ -1 & n \equiv 3 \bmod{4}\end{cases} \\
\left( \frac{2}{n}\right) = (-1)^{\frac{n^2-1}{8}} &= \begin{cases} 1 & n \equiv 1, 7 \bmod{8}\\ -1 & n \equiv 3, 5\bmod{8}\end{cases} \\
\left( \frac{-2}{n}\right) = (-1)^{\frac{n^2+4n-5}{8}} &= \begin{cases} 1 & n \equiv 1, 3 \bmod{8}\\ -1 & n \equiv 5, 7\bmod{8}\end{cases}
\end{align}</math>

and if both numbers are positive and odd (this is sometimes called "Jacobi's reciprocity law"):

:<math> \left(\frac{m}{n}\right) = (-1)^{\frac{(m-1)(n-1)}{4}}\left(\frac{n}{m}\right).</math>

However, if the Jacobi symbol is 1 but the denominator is not a prime, it does not necessarily follow that the numerator is a quadratic residue of the denominator. Gauss's cases 9) - 14) above can be expressed in terms of Jacobi symbols:

:<math>\left (\frac{M}{p}\right) = (-1)^{\frac{(p-1)(M-1)}{4}} \left(\frac{p}{M}\right ),</math>

and since ''p'' is prime the left hand side is a Legendre symbol, and we know whether ''M'' is a residue modulo ''p'' or not.

The formulas listed in the preceding section are true for Jacobi symbols as long as the symbols are defined. Euler's formula may be written

:<math>\left(\frac{a}{m}\right) =\left(\frac{a}{m \pm 4an}\right), \qquad n \in \Z, m\pm4an>0. </math>

'''Example.'''

:<math>\left (\frac{2}{7} \right ) = \left (\frac{2}{15} \right ) = \left (\frac{2}{23} \right ) = \left (\frac{2}{31} \right ) = \cdots = 1.</math>

2 is a residue modulo the primes 7, 23 and 31:

:<math>3^2 \equiv 2 \bmod{7}, \quad 5^2 \equiv 2 \bmod{23}, \quad 8^2 \equiv 2 \bmod{31}.</math>

But 2 is not a quadratic residue modulo 5, so it can't be one modulo 15. This is related to the problem Legendre had: if <math>\left (\tfrac{a}{m} \right) = -1,</math> then ''a'' is a non-residue modulo every prime in the arithmetic progression ''m'' + 4''a'', ''m'' + 8''a'', ..., if there ''are'' any primes in this series, but that wasn't proved until decades after Legendre.<ref>By [[Peter Gustav Lejeune Dirichlet]] in 1837</ref>

Eisenstein's formula requires relative primality conditions (which are true if the numbers are prime)

:Let <math>a, b, a', b'</math> be positive odd integers such that:
::<math>\begin{align}
\gcd &(a,b) =\gcd(a',b')= 1 \\
&a \equiv a' \bmod{4} \\
&b \equiv b' \bmod{4}
\end{align}</math>
:Then
::<math> \left(\frac{a}{b}\right) \left(\frac{b}{a}\right) =\left(\frac{a'}{b'}\right) \left(\frac{b'}{a'}\right).</math>

===Hilbert symbol===
The quadratic reciprocity law can be formulated in terms of the [[Hilbert symbol]] <math>(a,b)_v</math> where ''a'' and ''b'' are any two nonzero rational numbers and ''v'' runs over all the non-trivial absolute values of the rationals (the Archimedean one and the ''p''-adic absolute values for primes ''p''). The Hilbert symbol <math>(a,b)_v</math> is 1 or −1. It is defined to be 1 if and only if the equation <math>ax^2+by^2=z^2</math> has a solution in the [[Completion (ring theory)|completion]] of the rationals at ''v'' other than <math>x=y=z=0</math>. The Hilbert reciprocity law states that <math>(a,b)_v</math>, for fixed ''a'' and ''b'' and varying ''v'', is 1 for all but finitely many ''v'' and the product of <math>(a,b)_v</math> over all ''v'' is 1. (This formally resembles the residue theorem from complex analysis.)

The proof of Hilbert reciprocity reduces to checking a few special cases, and the non-trivial cases turn out to be equivalent to the main law and the two supplementary laws of quadratic reciprocity for the Legendre symbol. There is no kind of reciprocity in the Hilbert reciprocity law; its name simply indicates the historical source of the result in quadratic reciprocity. Unlike quadratic reciprocity, which requires sign conditions (namely positivity of the primes involved) and a special treatment of the prime 2, the Hilbert reciprocity law treats all absolute values of the rationals on an equal footing. Therefore, it is a more natural way of expressing quadratic reciprocity with a view towards generalization: the Hilbert reciprocity law extends with very few changes to all [[global field]]s and this extension can rightly be considered a generalization of quadratic reciprocity to all global fields.
<!--

The next example should be beefed up a bit and moved to [[Jacobi symbol]]. There it would be possible to see much more clearly how the algorithm looks just like the Euclidean algorithm. ~~~~

For example taking ''p'' to be 11 and ''q'' to be 19, we can relate <math>\left(\frac{11}{19}\right)</math> to <math>\left(\frac{19}{11}\right)</math>, which is <math>\left(\frac{8}{11}\right)</math> or <math>\left(\frac{-3}{11}\right)</math>. To proceed further we may need to know supplementary laws for computing <math>\left(\frac{3}{q}\right)</math> and <math>\left(\frac{-1}{q}\right)</math> explicitly. For example,

:<math>\left(\frac{-1}{q}\right) = (-1)^{(q-1)/2}.</math>

Using this we relate <math>\left(\frac{-3}{11}\right)</math> to <math>\left(\frac{3}{11}\right)</math> to <math>\left(\frac{11}{3}\right)</math> to <math>\left(\frac{2}{3}\right)</math> to <math>\left(\frac{-1}{3}\right)</math>, and can complete the initial calculation.

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