In this post, we take a look at an interesting proof of the Quadratic Reciprocity theorem by Gotthold Eisenstein.
Definition: The Legendre symbol is a function (pa) which takes the values ±1 depending on whether a is a quadratic residue modulo p. (pa)=⎩⎪⎪⎨⎪⎪⎧0if p∣a1if a is a quadratic residue modulo p−1if a is a quadratic non-residue modulo p
Theorem (Quadratic Reciprocity Law): If p and q are distinct odd primes, then the quadratic reciprocity theorem states that the congruences x2≡q(mod p)x2≡p(mod q) are both solvable or both unsolvable unless both p and q leave the the remainder 3 when divided by 4. Written symbolically, (qp)(pq)=(−1)(p−1)(q−1)/4
We will start by proving two important results about quadratic residues that will be useful later on.
Lemma 1: Let n≡0(mod p). Then n is a quadratic residue modulo p iff n2p−1≡1(mod p).
Proof: Fermat’s little theorem tells us that np−1≡1(mod p) whenever p∣n. Since,
np−1−1≡(n2p−1−1)(n2p−1+1)≡0(mod p)
we have n2p−1≡±1(mod p). Therefore, it suffices to prove that n is a quadratic residue modulo p if and only if n2p−1≡1(mod p).
Suppose that (pn)=1. Then, there is an integer x such that n≡x2(mod p). Now, we have
n2p−1≡xp−1≡1≡(pn)(mod p)
Conversely, assume that n2p−1≡1(mod p). Let g be a primitive root of p. Then, we have n≡gk(mod p) for some integer k. Since n2p−1≡g2k(p−1)≡1(mod p), the order of g must divide the exponent 2k(p−1). Therefore, p−1∣2k(p−1) and thus k is an even integer. Let’s say k=2j for some integer j. Then, we have
n≡gk≡g2j≡(gj)2(mod p)
This proves that n is a quadratic residue modulo p.
Lemma 2 (Gauss’s Lemma): For any odd prime p, let a be an integer that is co-prime to p. Consider the integers
S={a, 2a, 3a,⋯,2p−1a}
and their least positive residues modulo p. Let n be the number of these residues that are greater than p/2. Then (pa)=(−1)n
Proof: Since p∣a, none of the integers in S are congruent to 0 and no two of them are congruent to each other modulo p. Let r1,⋯,rm be the residues modulo p smaller than 2p, and let s1,⋯,sn be the residues modulo p greater than 2p. Then m+n=2p−1 and the integers
r1,⋯,rm,p−s1,⋯,p−sn
are all positive and less than 2p. Now, we will prove that no two of these integers are equal. Suppose that for some choice of i and j we have p−si=rj. We can choose integers u and v, with 1≤u,v≤2p−1 and satisfying
sirj≡ua(mod p)≡va(mod p)
Now, we have
si+rj≡a(u+v)≡p≡0(mod p)
This implies that u+v≡0(mod p). However, this is not possible because 1≤u+v≤p−1. Thus, we have proven that the numbers r1,⋯,rm,p−s1,⋯p−sn are simply a rearrangement of the integers 1,2,⋯,2p−1. Their product is equal to (2p−1)!. Therefore,
(2p−1)!=r1⋯rm(p−s1)⋯(p−sn)≡(−1)nr1⋯rms1⋯sn(mod p)≡(−1)na⋅2a⋯(2p−1)a(mod p)≡(−1)na2p−1(2p−1)!(mod p)
The (2p−1)! term can be cancelled from both sides as p∣(2p−1)!. In other words, we have a2p−1≡(−1)n(mod p). This completes the proof of Gauss’s lemma.
We are now ready to prove the Quadratic reciprocity theorem.
Proof of the Quadratic Reciprocity Theorem:
Using the periodicity properties of sin and Gauss’s lemma, it is easy to verify the following result:
Lemma: Let
p and
q be distinct odd primes and let
A={α∈Z∣1≤α≤2p−1} be a half system modulo
p. Then,
(pq)=α∈A∏sin(p2πα)sin(p2πqα)(1)
We start by examining the right hand side of equation (1). The addition theorem for trigonometric functions yields sin2α=2sinαcosα and sin3α=sinα(3−4sin2α). Induction shows that sinqα=sinαP(sinα) for all odd q≥1, where P∈Z[X] is a polynomial of degree q−1 and highest coefficient (−4)2q−1. Thus there exist ai∈Z such that sinzsinqz=(−4)2q−1((sinz)q−1+aq−2(sinz)q−2+⋯+a0)=(−4)2q−1ψ(X),where X=sinz Since ϕ(z)=sinzsinqz is an even function, so is ψ(X), hence aq−2=⋯=a1=0. Now ϕ(z) has zeros {±q2πβ, 1≤β≤2q−1}. Since ψ is monic of degree q−1, we may write
ψ(X)=β∈B∏(X2−sin2q2πβ)
where B={1,⋯,2q−1} is a half system modulo q. Replacing X by sinz, we get sinzsinqz=(−4)2q−1β∈B∏(sin2z−sin2q2πβ)(2) Put z=p2πα in equation (2) and plug the result into equation (1). (pq)=α∈A∏(−4)2q−1β∈B∏(sin2p2πα−sin2q2πβ)=(−4)2q−12p−1α∈A∏β∈B∏(sin2p2πα−sin2q2πβ)(3) Exchanging p and q on the right side of (3) give rise to factor of (−1)(p−1)(q−1)/4. Therefore, (pq)=(−1)(p−1)(q−1)/4(qp)(4) which is the quadratic reciprocity law.
References
- Lemmermeyer, Franz. “Reciprocity Laws: From Euler to Eisenstein”. New York, Springer, 2000
- Burton, David M. “Elementary Number Theory”. New Delhi, Tata McGraw-Hill Publishing Company Limited, May 1 2006