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. Due to its subtlety, it has many formulations, but the most standard statement is:
Law of quadratic reciprocity — Let p and q be distinct odd prime numbers, and define the Legendre symbol as:
Then:
This law, together with its supplements, allows the easy calculation of any Legendre symbol, making it possible to determine whether there is an integer solution for any quadratic equation of the form x 2 ≡ ≡ --> a mod p {\displaystyle x^{2}\equiv a{\bmod {p}}} for an odd prime p {\displaystyle p} ; that is, to determine the "perfect squares" modulo p {\displaystyle p} . However, this is a non-constructive result: it gives no help at all for finding a specific solution; for this, other methods are required. For example, in the case p ≡ ≡ --> 3 mod 4 {\displaystyle p\equiv 3{\bmod {4}}} using Euler's criterion one can give an explicit formula for the "square roots" modulo p {\displaystyle p} of a quadratic residue a {\displaystyle a} , namely,
indeed,
This formula only works if it is known in advance that a {\displaystyle a} is a quadratic residue, which can be checked using the law of quadratic reciprocity.
The quadratic reciprocity theorem was conjectured by Euler and Legendre and first proved by Gauss,[1] who referred to it as the "fundamental theorem" in his Disquisitiones Arithmeticae and his papers, writing
Privately, Gauss referred to it as the "golden theorem".[2] He published six proofs for it, and two more were found in his posthumous papers. There are now over 240 published proofs.[3] The shortest known proof is included below, together with short proofs of the law's supplements (the Legendre symbols of −1 and 2).
Generalizing the reciprocity law to higher powers has been a leading problem in mathematics, and has been crucial to the development of much of the machinery of modern algebra, number theory, and algebraic geometry, culminating in Artin reciprocity, class field theory, and the Langlands program.
Quadratic reciprocity arises from certain subtle factorization patterns involving perfect square numbers. In this section, we give examples which lead to the general case.
Consider the polynomial f ( n ) = n 2 − − --> 5 {\displaystyle f(n)=n^{2}-5} and its values for n ∈ ∈ --> N . {\displaystyle n\in \mathbb {N} .} The prime factorizations of these values are given as follows:
The prime factors p {\displaystyle p} dividing f ( n ) {\displaystyle f(n)} are p = 2 , 5 {\displaystyle p=2,5} , and every prime whose final digit is 1 {\displaystyle 1} or 9 {\displaystyle 9} ; no primes ending in 3 {\displaystyle 3} or 7 {\displaystyle 7} ever appear. Now, p {\displaystyle p} is a prime factor of some n 2 − − --> 5 {\displaystyle n^{2}-5} whenever n 2 − − --> 5 ≡ ≡ --> 0 mod p {\displaystyle n^{2}-5\equiv 0{\bmod {p}}} , i.e. whenever n 2 ≡ ≡ --> 5 mod p , {\displaystyle n^{2}\equiv 5{\bmod {p}},} i.e. whenever 5 is a quadratic residue modulo p {\displaystyle p} . This happens for p = 2 , 5 {\displaystyle p=2,5} and those primes with p ≡ ≡ --> 1 , 4 mod 5 , {\displaystyle p\equiv 1,4{\bmod {5}},} and the latter numbers 1 = ( ± ± --> 1 ) 2 {\displaystyle 1=(\pm 1)^{2}} and 4 = ( ± ± --> 2 ) 2 {\displaystyle 4=(\pm 2)^{2}} are precisely the quadratic residues modulo 5 {\displaystyle 5} . Therefore, except for p = 2 , 5 {\displaystyle p=2,5} , we have that 5 {\displaystyle 5} is a quadratic residue modulo p {\displaystyle p} iff p {\displaystyle p} is a quadratic residue modulo 5 {\displaystyle 5} .
The law of quadratic reciprocity gives a similar characterization of prime divisors of f ( n ) = n 2 − − --> q {\displaystyle f(n)=n^{2}-q} for any prime q, which leads to a characterization for any integer q {\displaystyle q} .
Let p be an odd prime. A number modulo p is a quadratic residue whenever it is congruent to a square (mod p); otherwise it is a quadratic non-residue. ("Quadratic" can be dropped if it is clear from the context.) Here we exclude zero as a special case. Then as a consequence of the fact that the multiplicative group of a finite field of order p is cyclic of order p-1, the following statements hold:
For the avoidance of doubt, these statements do not hold if the modulus is not prime. For example, there are only 3 quadratic residues (1, 4 and 9) in the multiplicative group modulo 15. Moreover, although 7 and 8 are quadratic non-residues, their product 7x8 = 11 is also a quadratic non-residue, in contrast to the prime case.
Quadratic residues appear as entries in the following table, indexed by the row number as modulus and column number as root:
This table is complete for odd primes less than 50. To check whether a number m is a quadratic residue mod one of these primes p, find a ≡ m (mod p) and 0 ≤ a < p. If a is in row p, then m is a residue (mod p); if a is not in row p of the table, then m is a nonresidue (mod p).
The quadratic reciprocity law is the statement that certain patterns found in the table are true in general.
Another way to organize the data is to see which primes are residues mod which other primes, as illustrated in the following table. The entry in row p column q is R if q is a quadratic residue (mod p); if it is a nonresidue the entry is N.
If the row, or the column, or both, are ≡ 1 (mod 4) the entry is blue or green; if both row and column are ≡ 3 (mod 4), it is yellow or orange.
The blue and green entries are symmetric around the diagonal: The entry for row p, column q is R (resp N) if and only if the entry at row q, column p, is R (resp N).
The yellow and orange ones, on the other hand, are antisymmetric: The entry for row p, column q is R (resp N) if and only if the entry at row q, column p, is N (resp R).
The reciprocity law states that these patterns hold for all p and q.
Ordering the rows and columns mod 4 makes the pattern clearer.
The supplements provide solutions to specific cases of quadratic reciprocity. They are often quoted as partial results, without having to resort to the complete theorem.
Trivially 1 is a quadratic residue for all primes. The question becomes more interesting for −1. Examining the table, we find −1 in rows 5, 13, 17, 29, 37, and 41 but not in rows 3, 7, 11, 19, 23, 31, 43 or 47. The former set of primes are all congruent to 1 modulo 4, and the latter are congruent to 3 modulo 4.
Examining the table, we find 2 in rows 7, 17, 23, 31, 41, and 47, but not in rows 3, 5, 11, 13, 19, 29, 37, or 43. The former primes are all ≡ ±1 (mod 8), and the latter are all ≡ ±3 (mod 8). This leads to
−2 is in rows 3, 11, 17, 19, 41, 43, but not in rows 5, 7, 13, 23, 29, 31, 37, or 47. The former are ≡ 1 or ≡ 3 (mod 8), and the latter are ≡ 5, 7 (mod 8).
3 is in rows 11, 13, 23, 37, and 47, but not in rows 5, 7, 17, 19, 29, 31, 41, or 43. The former are ≡ ±1 (mod 12) and the latter are all ≡ ±5 (mod 12).
−3 is in rows 7, 13, 19, 31, 37, and 43 but not in rows 5, 11, 17, 23, 29, 41, or 47. The former are ≡ 1 (mod 3) and the latter ≡ 2 (mod 3).
Since the only residue (mod 3) is 1, we see that −3 is a quadratic residue modulo every prime which is a residue modulo 3.
5 is in rows 11, 19, 29, 31, and 41 but not in rows 3, 7, 13, 17, 23, 37, 43, or 47. The former are ≡ ±1 (mod 5) and the latter are ≡ ±2 (mod 5).
Since the only residues (mod 5) are ±1, we see that 5 is a quadratic residue modulo every prime which is a residue modulo 5.
−5 is in rows 3, 7, 23, 29, 41, 43, and 47 but not in rows 11, 13, 17, 19, 31, or 37. The former are ≡ 1, 3, 7, 9 (mod 20) and the latter are ≡ 11, 13, 17, 19 (mod 20).
The observations about −3 and 5 continue to hold: −7 is a residue modulo p if and only if p is a residue modulo 7, −11 is a residue modulo p if and only if p is a residue modulo 11, 13 is a residue (mod p) if and only if p is a residue modulo 13, etc. The more complicated-looking rules for the quadratic characters of 3 and −5, which depend upon congruences modulo 12 and 20 respectively, are simply the ones for −3 and 5 working with the first supplement.
The generalization of the rules for −3 and 5 is Gauss's statement of quadratic reciprocity.
Quadratic Reciprocity (Gauss's statement). If q ≡ ≡ --> 1 mod 4 {\displaystyle q\equiv 1{\bmod {4}}} , then the congruence x 2 ≡ ≡ --> p mod q {\displaystyle x^{2}\equiv p{\bmod {q}}} is solvable if and only if x 2 ≡ ≡ --> q mod p {\displaystyle x^{2}\equiv q{\bmod {p}}} is solvable. If q ≡ ≡ --> 3 mod 4 {\displaystyle q\equiv 3{\bmod {4}}} and p ≡ ≡ --> 3 mod 4 {\displaystyle p\equiv 3{\bmod {4}}} , then the congruence x 2 ≡ ≡ --> p mod q {\displaystyle x^{2}\equiv p{\bmod {q}}} is solvable if and only if x 2 ≡ ≡ --> − − --> q mod p {\displaystyle x^{2}\equiv -q{\bmod {p}}} is solvable.
Quadratic Reciprocity (combined statement). Define q ∗ ∗ --> = ( − − --> 1 ) q − − --> 1 2 q {\displaystyle q^{*}=(-1)^{\frac {q-1}{2}}q} . Then the congruence x 2 ≡ ≡ --> p mod q {\displaystyle x^{2}\equiv p{\bmod {q}}} is solvable if and only if x 2 ≡ ≡ --> q ∗ ∗ --> mod p {\displaystyle x^{2}\equiv q^{*}{\bmod {p}}} is solvable.
Quadratic Reciprocity (Legendre's statement). If p or q are congruent to 1 modulo 4, then: x 2 ≡ ≡ --> q mod p {\displaystyle x^{2}\equiv q{\bmod {p}}} is solvable if and only if x 2 ≡ ≡ --> p mod q {\displaystyle x^{2}\equiv p{\bmod {q}}} is solvable. If p and q are congruent to 3 modulo 4, then: x 2 ≡ ≡ --> q mod p {\displaystyle x^{2}\equiv q{\bmod {p}}} is solvable if and only if x 2 ≡ ≡ --> p mod q {\displaystyle x^{2}\equiv p{\bmod {q}}} is not solvable.
The last is immediately equivalent to the modern form stated in the introduction above. It is a simple exercise to prove that Legendre's and Gauss's statements are equivalent – it requires no more than the first supplement and the facts about multiplying residues and nonresidues.
Apparently, the shortest known proof yet was published by B. Veklych in the American Mathematical Monthly.[4]
The value of the Legendre symbol of − − --> 1 {\displaystyle -1} (used in the proof above) follows directly from Euler's criterion:
by Euler's criterion, but both sides of this congruence are numbers of the form ± ± --> 1 {\displaystyle \pm 1} , so they must be equal.
Whether 2 {\displaystyle 2} is a quadratic residue can be concluded if we know the number of solutions of the equation x 2 + y 2 = 2 {\displaystyle x^{2}+y^{2}=2} with x , y ∈ ∈ --> Z p , {\displaystyle x,y\in \mathbb {Z} _{p},} which can be solved by standard methods. Namely, all its solutions where x y ≠ ≠ --> 0 , x ≠ ≠ --> ± ± --> y {\displaystyle xy\neq 0,x\neq \pm y} can be grouped into octuplets of the form ( ± ± --> x , ± ± --> y ) , ( ± ± --> y , ± ± --> x ) {\displaystyle (\pm x,\pm y),(\pm y,\pm x)} , and what is left are four solutions of the form ( ± ± --> 1 , ± ± --> 1 ) {\displaystyle (\pm 1,\pm 1)} and possibly four additional solutions where x 2 = 2 , y = 0 {\displaystyle x^{2}=2,y=0} and x = 0 , y 2 = 2 {\displaystyle x=0,y^{2}=2} , which exist precisely if 2 {\displaystyle 2} is a quadratic residue. That is, 2 {\displaystyle 2} is a quadratic residue precisely if the number of solutions of this equation is divisible by 8 {\displaystyle 8} . And this equation can be solved in just the same way here as over the rational numbers: substitute x = a + 1 , y = a t + 1 {\displaystyle x=a+1,y=at+1} , where we demand that a ≠ ≠ --> 0 {\displaystyle a\neq 0} (leaving out the two solutions ( 1 , ± ± --> 1 ) {\displaystyle (1,\pm 1)} ), then the original equation transforms into
Here t {\displaystyle t} can have any value that does not make the denominator zero – for which there are 1 + ( − − --> 1 p ) {\displaystyle 1+\left({\frac {-1}{p}}\right)} possibilities (i.e. 2 {\displaystyle 2} if − − --> 1 {\displaystyle -1} is a residue, 0 {\displaystyle 0} if not) – and also does not make a {\displaystyle a} zero, which excludes one more option, t = − − --> 1 {\displaystyle t=-1} . Thus there are
possibilities for t {\displaystyle t} , and so together with the two excluded solutions there are overall p − − --> ( − − --> 1 p ) {\displaystyle p-\left({\frac {-1}{p}}\right)} solutions of the original equation. Therefore, 2 {\displaystyle 2} is a residue modulo p {\displaystyle p} if and only if 8 {\displaystyle 8} divides p − − --> ( − − --> 1 ) p − − --> 1 2 {\displaystyle p-(-1)^{\frac {p-1}{2}}} . This is a reformulation of the condition stated above.
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 proved[5] (or claimed to have proved)[6] a number of theorems about expressing a prime by a quadratic form:
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
Euler conjectured, and Lagrange proved, that[7]
Proving these and other statements of Fermat was one of the things that led mathematicians to the reciprocity theorem.
Translated into modern notation, Euler stated [8] that for distinct odd primes p and q:
This is equivalent to quadratic reciprocity.
He could not prove it, but he did prove the second supplement.[9]
Fermat proved that if p is a prime number and a is an integer,
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 and therefore in particular the multiplicative group is cyclic, hence there can be at most two solutions to a quadratic equation:
Legendre[10] 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:
He says that since expressions of the form
will come up so often he will abbreviate them as:
This is now known as the Legendre symbol, and an equivalent[11][12] definition is used today: for all integers a and all odd primes p
He notes that these can be combined:
A number of proofs, especially those based on Gauss's Lemma,[13] explicitly calculate this formula.
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 : mod p {\displaystyle {\bmod {p}}} .
So either : p − − --> 1 2 or p 2 − − --> 1 8 {\displaystyle {\frac {p-1}{2}}{\text{ or }}{\frac {p^{2}-1}{8}}} are both even, or they are both odd. The sum of these two expressions is
Legendre's attempt to prove reciprocity is based on a theorem of his:
Example. Theorem I is handled by letting a ≡ 1 and b ≡ 3 (mod 4) be primes and assuming that ( b a ) = 1 {\displaystyle \left({\tfrac {b}{a}}\right)=1} and, contrary the theorem, that ( a b ) = − − --> 1. {\displaystyle \left({\tfrac {a}{b}}\right)=-1.} Then x 2 + a y 2 − − --> b z 2 = 0 {\displaystyle x^{2}+ay^{2}-bz^{2}=0} 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
Then if there is another prime p ≡ 1 (mod 4) such that
the solvability of B x 2 + b y 2 − − --> p z 2 = 0 {\displaystyle Bx^{2}+by^{2}-pz^{2}=0} 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:
but he couldn't prove that either. Hilbert symbol (below) discusses how techniques based on the existence of solutions to a x 2 + b y 2 + c z 2 = 0 {\displaystyle ax^{2}+by^{2}+cz^{2}=0} can be made to work.
Gauss first proves[14] the supplementary laws. He sets[15] the basis for induction by proving the theorem for ±3 and ±5. Noting[16] that it is easier to state for −3 and +5 than it is for +3 or −5, he states[17] the general theorem in the form:
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′, etc. represent positive primes ≡ 1 (mod 4) and b, b′, etc. positive primes ≡ 3 (mod 4), he breaks it out into the same 8 cases as Legendre:
In the next Article he generalizes this to what are basically the rules for the Jacobi symbol (below). Letting A, A′, etc. represent any (prime or composite) positive numbers ≡ 1 (mod 4) and B, B′, etc. positive numbers ≡ 3 (mod 4):
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,[18] a lemma similar to the one Legendre needed:
The proof of quadratic reciprocity uses complete induction.
These can be combined:
A number of proofs of the theorem, especially those based on Gauss sums[19] or the splitting of primes in algebraic number fields,[20][21] derive this formula.
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.
This can be proven using Gauss's lemma.
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 R 3; 3×7×11 R 5; 3×5×11 R 7; and 3×5×7 N 11, so there are an odd number of nonresidues.
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
and if both numbers are positive and odd (this is sometimes called "Jacobi's reciprocity law"):
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:
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
Example.
2 is a residue modulo the primes 7, 23 and 31:
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 ( a m ) = − − --> 1 , {\displaystyle \left({\tfrac {a}{m}}\right)=-1,} then a is a non-residue modulo every prime in the arithmetic progression m + 4a, m + 8a, ..., if there are any primes in this series, but that wasn't proved until decades after Legendre.[26]
Eisenstein's formula requires relative primality conditions (which are true if the numbers are prime)
The quadratic reciprocity law can be formulated in terms of the Hilbert symbol ( a , b ) v {\displaystyle (a,b)_{v}} 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 ( a , b ) v {\displaystyle (a,b)_{v}} is 1 or −1. It is defined to be 1 if and only if the equation a x 2 + b y 2 = z 2 {\displaystyle ax^{2}+by^{2}=z^{2}} has a solution in the completion of the rationals at v other than x = y = z = 0 {\displaystyle x=y=z=0} . The Hilbert reciprocity law states that ( a , b ) v {\displaystyle (a,b)_{v}} , for fixed a and b and varying v, is 1 for all but finitely many v and the product of ( a , b ) v {\displaystyle (a,b)_{v}} 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 fields and this extension can rightly be considered a generalization of quadratic reciprocity to all global fields.
The early proofs of quadratic reciprocity are relatively unilluminating. The situation changed when Gauss used Gauss sums to show that quadratic fields are subfields of cyclotomic fields, and implicitly deduced quadratic reciprocity from a reciprocity theorem for cyclotomic fields. His proof was cast in modern form by later algebraic number theorists. This proof served as a template for class field theory, which can be viewed as a vast generalization of quadratic reciprocity.
Robert Langlands formulated the Langlands program, which gives a conjectural vast generalization of class field theory. He wrote:[27]
There are also quadratic reciprocity laws in rings other than the integers.
In his second monograph on quartic reciprocity[29] Gauss stated quadratic reciprocity for the ring Z [ i ] {\displaystyle \mathbb {Z} [i]} of Gaussian integers, saying that it is a corollary of the biquadratic law in Z [ i ] , {\displaystyle \mathbb {Z} [i],} but did not provide a proof of either theorem. Dirichlet[30] showed that the law in Z [ i ] {\displaystyle \mathbb {Z} [i]} can be deduced from the law for Z {\displaystyle \mathbb {Z} } without using quartic reciprocity.
For an odd Gaussian prime π π --> {\displaystyle \pi } and a Gaussian integer α α --> {\displaystyle \alpha } relatively prime to π π --> , {\displaystyle \pi ,} define the quadratic character for Z [ i ] {\displaystyle \mathbb {Z} [i]} by:
Let λ λ --> = a + b i , μ μ --> = c + d i {\displaystyle \lambda =a+bi,\mu =c+di} be distinct Gaussian primes where a and c are odd and b and d are even. Then[31]
Consider the following third root of unity:
The ring of Eisenstein integers is Z [ ω ω --> ] . {\displaystyle \mathbb {Z} [\omega ].} [32] For an Eisenstein prime π π --> , N π π --> ≠ ≠ --> 3 , {\displaystyle \pi ,\mathrm {N} \pi \neq 3,} and an Eisenstein integer α α --> {\displaystyle \alpha } with gcd ( α α --> , π π --> ) = 1 , {\displaystyle \gcd(\alpha ,\pi )=1,} define the quadratic character for Z [ ω ω --> ] {\displaystyle \mathbb {Z} [\omega ]} by the formula
Let λ = a + bω and μ = c + dω be distinct Eisenstein primes where a and c are not divisible by 3 and b and d are divisible by 3. Eisenstein proved[33]
The above laws are special cases of more general laws that hold for the ring of integers in any imaginary quadratic number field. Let k be an imaginary quadratic number field with ring of integers O k . {\displaystyle {\mathcal {O}}_{k}.} For a prime ideal p ⊂ ⊂ --> O k {\displaystyle {\mathfrak {p}}\subset {\mathcal {O}}_{k}} with odd norm N p {\displaystyle \mathrm {N} {\mathfrak {p}}} and α α --> ∈ ∈ --> O k , {\displaystyle \alpha \in {\mathcal {O}}_{k},} define the quadratic character for O k {\displaystyle {\mathcal {O}}_{k}} as
for an arbitrary ideal a ⊂ ⊂ --> O k {\displaystyle {\mathfrak {a}}\subset {\mathcal {O}}_{k}} factored into prime ideals a = p 1 ⋯ ⋯ --> p n {\displaystyle {\mathfrak {a}}={\mathfrak {p}}_{1}\cdots {\mathfrak {p}}_{n}} define
and for β β --> ∈ ∈ --> O k {\displaystyle \beta \in {\mathcal {O}}_{k}} define
Let O k = Z ω ω --> 1 ⊕ ⊕ --> Z ω ω --> 2 , {\displaystyle {\mathcal {O}}_{k}=\mathbb {Z} \omega _{1}\oplus \mathbb {Z} \omega _{2},} i.e. { ω ω --> 1 , ω ω --> 2 } {\displaystyle \left\{\omega _{1},\omega _{2}\right\}} is an integral basis for O k . {\displaystyle {\mathcal {O}}_{k}.} For ν ν --> ∈ ∈ --> O k {\displaystyle \nu \in {\mathcal {O}}_{k}} with odd norm N ν ν --> , {\displaystyle \mathrm {N} \nu ,} define (ordinary) integers a, b, c, d by the equations,
and a function
If m = Nμ and n = Nν are both odd, Herglotz proved[34]
Also, if
Then[35]
Let F be a finite field with q = pn elements, where p is an odd prime number and n is positive, and let F[x] be the ring of polynomials in one variable with coefficients in F. If f , g ∈ ∈ --> F [ x ] {\displaystyle f,g\in F[x]} and f is irreducible, monic, and has positive degree, define the quadratic character for F[x] in the usual manner:
If f = f 1 ⋯ ⋯ --> f n {\displaystyle f=f_{1}\cdots f_{n}} is a product of monic irreducibles let
Dedekind proved that if f , g ∈ ∈ --> F [ x ] {\displaystyle f,g\in F[x]} are monic and have positive degrees,[36]
The attempt to generalize quadratic reciprocity for powers higher than the second was one of the main goals that led 19th century mathematicians, including Carl Friedrich Gauss, Peter Gustav Lejeune Dirichlet, Carl Gustav Jakob Jacobi, Gotthold Eisenstein, Richard Dedekind, Ernst Kummer, and David Hilbert to the study of general algebraic number fields and their rings of integers;[37] specifically Kummer invented ideals in order to state and prove higher reciprocity laws.
The ninth in the list of 23 unsolved problems which David Hilbert proposed to the Congress of Mathematicians in 1900 asked for the "Proof of the most general reciprocity law [f]or an arbitrary number field".[38] Building upon work by Philipp Furtwängler, Teiji Takagi, Helmut Hasse and others, Emil Artin discovered Artin reciprocity in 1923, a general theorem for which all known reciprocity laws are special cases, and proved it in 1927.[39]
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The Disquisitiones Arithmeticae has been translated (from Latin) into English and German. The German edition includes all of Gauss's papers on number theory: all the proofs of quadratic reciprocity, the determination of the sign of the Gauss sum, the investigations into biquadratic reciprocity, and unpublished notes. Footnotes referencing the Disquisitiones Arithmeticae are of the form "Gauss, DA, Art. n".
The two monographs Gauss published on biquadratic reciprocity have consecutively numbered sections: the first contains §§ 1–23 and the second §§ 24–76. Footnotes referencing these are of the form "Gauss, BQ, § n".
These are in Gauss's Werke, Vol II, pp. 65–92 and 93–148. German translations are in pp. 511–533 and 534–586 of Untersuchungen über höhere Arithmetik.
Every textbook on elementary number theory (and quite a few on algebraic number theory) has a proof of quadratic reciprocity. Two are especially noteworthy:
Franz Lemmermeyer's Reciprocity Laws: From Euler to Eisenstein has many proofs (some in exercises) of both quadratic and higher-power reciprocity laws and a discussion of their history. Its immense bibliography includes literature citations for 196 different published proofs for the quadratic reciprocity law.
Kenneth Ireland and Michael Rosen's A Classical Introduction to Modern Number Theory also has many proofs of quadratic reciprocity (and many exercises), and covers the cubic and biquadratic cases as well. Exercise 13.26 (p. 202) says it all
Count the number of proofs to the law of quadratic reciprocity given thus far in this book and devise another one.
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Badminton – Mixed doubles at the 2018 Commonwealth GamesVenueCarrara Sports and Leisure Centre, Gold CoastDates10–15 AprilMedalists Chris AdcockGabby Adcock England Marcus EllisLauren Smith England Chan Peng SoonGoh Liu Ying Malaysia← 20142022 → Badminton at the2018 Commonwealth GamesSinglesmenwomenDoublesmenwomenMixeddoublesteamvte The mixed doubles badminton event at the 2018 Commonwealth Games was held from 10 …
Basketball team in Akita, AkitaJR East Akita PeckersLeaguesJapan Industrial and Commercial Basketball FederationFounded1947ArenaAkita Northern Gate Square (practice)LocationAkita, AkitaHead coachNarihiro KuromasaOwnershipEast Japan RailwayWebsitewww.jreast.co.jp/akita/branch/jrclub/basketball/index.html Home Away The JR East Akita Peckers are a Japanese semi-professional basketball team based in Akita, Akita that competes in the Japan Industrial and Commercial Basketball Federation .[1]&…
Artikel ini sebatang kara, artinya tidak ada artikel lain yang memiliki pranala balik ke halaman ini.Bantulah menambah pranala ke artikel ini dari artikel yang berhubungan atau coba peralatan pencari pranala.Tag ini diberikan pada Oktober 2022. Gereja Kelahiran Perawan MariaGereja kelahiran Santa MariaKoordinat: 50°03′10″N 14°29′15″E / 50.05278°N 14.48750°E / 50.05278; 14.48750LokasiPrahaNegaraRepublik CekoDenominasiKatolik RomaSitus webwww.centrumhostivar.cz/…
Mazmur 103Naskah Gulungan Mazmur 11Q5 di antara Naskah Laut Mati memuat salinan sejumlah besar mazmur Alkitab yang diperkirakan dibuat pada abad ke-2 SM.KitabKitab MazmurKategoriKetuvimBagian Alkitab KristenPerjanjian LamaUrutan dalamKitab Kristen19← Mazmur 102 Mazmur 104 → Mazmur 103 (disingkat Maz 103 atau Mz 103; penomoran Septuaginta: Mazmur 102) adalah sebuah mazmur dalam bagian ke-4 Kitab Mazmur di Alkitab Ibrani dan Perjanjian Lama dalam Alkitab Kristen. Menurut ayat pertamany…
Debbie Bowman-Sullivan Plaats uw zelfgemaakte foto hier Persoonlijke informatie Geboortedatum 4 juli 1963 Geboorteplaats Southport Nationaliteit Australië Lengte 1,65 m Sportieve informatie Discipline Hockey Olympische Spelen 1988, 1992 Portaal Sport Debbie Lee Bowman-Sullivan (Southport, 4 juli 1963) is een Australisch hockeyster. Bowman-Sullivan was in 1988 aanvoerder van de Australische ploeg die olympisch kampioen werd. Erelijst 1986 - 6e Wereldkampioenschap hockey[1&…
For other people with the same name, see Clyde Williams. Clyde WilliamsWilliams from 1903 HawkeyeBiographical detailsBorn(1879-03-24)March 24, 1879Shelby, Iowa, U.S.DiedMarch 20, 1938(1938-03-20) (aged 58)Sheldon, Iowa, U.S.Playing careerFootball1898–1901IowaBaseball1899–1902Iowa1902–1903Sioux Falls Canaries1904–1905Marshalltown Grays1906Marshalltown Brownies1907St. Paul Saints1907–1908Toledo Mud Hens1908–1910Des Moines Boosters Position(s)Quarterback (football)Second baseman, s…
Air Terjun TegenunganAir Terjun TegenunganLokasiDesa Kemenuh, Kecamatan Sukawati, Kabupaten Gianyar, Provinsi Bali, IndonesiaTipeHorsetailTinggi total15 meter (49 ft)Jumlah titik1 (Satu)Anak sungaiTukad Petanu Air terjun Tegenungan adalah air terjun yang terletak di Desa Kemenuh, Kecamatan Sukawati, Kabupaten Gianyar, berjarak 30 km dari Kota Denpasar. Air terjun ini memiliki ketinggian 15 meter. Meski tidak begitu tinggi, debit airnya sangat deras. Airnya juga bersih dan bening. Air t…
شانتاراكشيتا معلومات شخصية الميلاد القرن 8 تاريخ الوفاة سنة 788 (62–63 سنة)[1] مواطنة الهند الحياة العملية المهنة فيلسوف، ومترجم، وكاتب اللغات السنسكريتية تعديل مصدري - تعديل شانتاراكشيتا شانتاراكشيتا (Śāntarakṣita) هو عالم ومفكر بوذي من البرهمن عا…
العلاقات المكسيكية الزامبية المكسيك زامبيا المكسيك زامبيا تعديل مصدري - تعديل العلاقات المكسيكية الزامبية هي العلاقات الثنائية التي تجمع بين المكسيك وزامبيا.[1][2][3][4][5] مقارنة بين البلدين هذه مقارنة عامة ومرجعية للدولتين: وجه المقارنة
Мелані Джой Народилася 2 вересня 1966(1966-09-02) (57 років)Громадянство СШАДіяльність психологиня, зоозахисник, викладачка університетуГалузь соціальна психологія[1], веганізм[1] і Карнізм[1]Alma mater Гарвардський університетУніверситет СейбрукЗнання мов
U.S. Senate election in New York 1916 United States Senate election in New York ← 1911 November 7, 1916 1922 → Nominee William M. Calder William F. McCombs Party Republican Democratic Alliance American Popular vote 839,314 605,933 Percentage 54.32% 39.22% County ResultsCalder: 40–50% 50–60% 60–70%McCombs: 40–50% …
Baruch MarzelLahir23 April 1959Boston, Massachusetts, ASTempat tinggalHebron, Tepi BaratWarga negaraIsraelDikenal atasAktivitas KahanisPartai politikKach (dulu)Front Nasional Yahudi (2004–2012)Otzma Yehudit (current) Baruch Meir Marzel (Ibrani: ברוך מאיר מרזל, lahir 23 April 1959) adalah seorang politikus dan aktivis Israel.[1][2] Ia adalah seorang Yahudi Ortodoks yang berasal dari Boston[3] yang sekarang menetap di komunitas Yahudi Hebron di Tel Rumeida den…
Irish Liberal politician, lawyer and judge For other people with the same name, see Samuel Walker. Sir Samuel Walker, 1st BaronetIn The Sketch, 4 November 1896Born(1832-06-19)19 June 1832Gore Port, Finea, IrelandDied13 August 1911(1911-08-13) (aged 79)Dublin, IrelandBurial placeMount Jerome CemeteryEducationTrinity College DublinOccupation(s)Politician, lawyer, judgePolitical partyLiberalSpouses Cecilia Charlotte Greene (m. 1855; died 1880)…
Campaign for a new worker's international Part of a series onSocialism HistoryOutline Development Age of the Enlightenment French Revolution Revolutions of 1848 Socialist calculation debate Socialist economics Ideas Calculation in kind Collective ownership Cooperative Common ownership Critique of political economy Economic democracy Economic planning Equal liberty Equal opportunity Free association Freed market Industrial democracy Input–output model Internationalism Labor-time calculation Lab…
Chinese politician (born 1968) In this Chinese name, the family name is Zhong. Zhong Shaojun钟绍军Director of the General Office of the Central Military CommissionIncumbentAssumed office August 2017CMC ChairmanXi JinpingPreceded byQin ShengxiangDirector of the Chairman of the Central Military Commission OfficeIncumbentAssumed office June 2013CMC ChairmanXi JinpingPreceded byWu Zhiming Personal detailsBorn (1968-10-19) 19 October 1968 (age 55)[1]Kaihua County, Zhejiang, Ch…
City in Arkansas, United StatesBrookland, ArkansasCity SealLocation of Brookland in Craighead County, Arkansas.Coordinates: 35°54′9″N 90°34′54″W / 35.90250°N 90.58167°W / 35.90250; -90.58167CountryUnited StatesStateArkansasCountyCraigheadArea[1] • Total7.82 sq mi (20.25 km2) • Land7.80 sq mi (20.20 km2) • Water0.02 sq mi (0.04 km2)Elevation262 ft (80 m)Population…
TV series or program Saint Philip Neri: I Prefer HeavenWritten byGiorgio MariuzzoMonica Zapelli Giacomo CampiottiMario RuggeriDirected byGiacomo CampiottiStarringGigi ProiettiComposerMarco FrisinaOriginal languageItalianProductionProducerLuca BernabeiCinematographyMarco OnoratoEditorsRoberto Missiroli Alessandro LucidiRunning time200 min.Original releaseNetworkRai 1Release2010 (2010) Saint Philip Neri: I Prefer Heaven (Italian: Preferisco il Paradiso) is a 2010 Italian television movie writ…
For the film version, see The Tsarevich (1933 film). Richard Tauber Der Zarewitsch (The Tsarevich) is an operetta in three acts by Franz Lehár. The German libretto by Heinz Reichert [de] and Bela Jenbach is based on the play of the same name by Polish author Gabriela Zapolska. Lehár composed the work, one of his later operettas, as a vehicle for Richard Tauber, the acclaimed Austrian tenor. The work received its first performance at the Deutsches Künstlertheater [de]…
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