bookmark_borderEisenstein’s Proof of Quadratic Reciprocity

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 (ap)\left(\frac{a}{p}\right) which takes the values ±1\pm 1 depending on whether aa is a quadratic residue modulo pp. (ap)={0if pa1if a is a quadratic residue modulo p1if a is a quadratic non-residue modulo p\left(\frac{a}{p}\right) = \begin{cases}0 \quad \text{if }p|a \\ 1 \quad \text{if }a\text{ is a quadratic residue modulo }p \\ -1 \quad \text{if }a\text{ is a quadratic non-residue modulo }p\end{cases} Theorem (Quadratic Reciprocity Law): If pp and qq are distinct odd primes, then the quadratic reciprocity theorem states that the congruences x2q(mod p)x2p(mod q) \begin{aligned} x^2 \equiv q \quad (\text{mod }p) \\ x^2 \equiv p \quad (\text{mod }q) \end{aligned} are both solvable or both unsolvable unless both pp and qq leave the the remainder 33 when divided by 44. Written symbolically, (pq)(qp)=(1)(p1)(q1)/4\left(\frac{p}{q}\right)\left(\frac{q}{p}\right) = (-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)n\not\equiv 0 \;(\text{mod }p). Then nn is a quadratic residue modulo pp iff np121  (mod p)n^{\frac{p-1}{2}}\equiv 1 \; (\text{mod }p).

Proof: Fermat’s little theorem tells us that np11  (mod p)n^{p-1}\equiv 1 \;(\text{mod }p) whenever p∤np \not| n. Since, np11(np121)(np12+1)0  (mod p)n^{p-1}-1\equiv (n^{\frac{p-1}{2}}-1)(n^{\frac{p-1}{2}}+1)\equiv 0 \; (\text{mod }p) we have np12±1  (mod p)n^{\frac{p-1}{2}}\equiv \pm 1\; (\text{mod }p). Therefore, it suffices to prove that nn is a quadratic residue modulo pp if and only if np121  (mod p)n^{\frac{p-1}{2}}\equiv 1 \; (\text{mod }p).

Suppose that (np)=1\left(\frac{n}{p} \right)=1. Then, there is an integer xx such that nx2  (mod p)n\equiv x^2 \; (\text{mod p}). Now, we have np12xp11(np)  (mod p)n^{\frac{p-1}{2}}\equiv x^{p-1} \equiv 1 \equiv \left(\frac{n}{p} \right) \; (\text{mod }p)

Conversely, assume that np121  (mod p)n^{\frac{p-1}{2}}\equiv 1 \; (\text{mod }p). Let gg be a primitive root of pp. Then, we have ngk  (mod p)n\equiv g^k \; (\text{mod }p) for some integer kk. Since np12gk(p1)21  (mod p)n^{\frac{p-1}{2}}\equiv g^{\frac{k(p-1)}{2}} \equiv 1 \; (\text{mod }p), the order of gg must divide the exponent k(p1)2\frac{k(p-1)}{2}. Therefore, p1k(p1)2p-1 | \frac{k(p-1)}{2} and thus kk is an even integer. Let’s say k=2jk=2j for some integer jj. Then, we have ngkg2j(gj)2  (mod p)n \equiv g^k \equiv g^{2j} \equiv (g^j)^2 \; (\text{mod }p) This proves that nn is a quadratic residue modulo pp.

Lemma 2 (Gauss’s Lemma): For any odd prime pp, let aa be an integer that is co-prime to pp. Consider the integers S={a, 2a, 3a,,p12a}S = \left\{ a,\ 2a,\ 3a,\cdots, \frac{p-1}{2}a \right\} and their least positive residues modulo pp. Let nn be the number of these residues that are greater than p/2p/2. Then (ap)=(1)n\left(\frac{a}{p}\right)=(-1)^n

Proof: Since p∤ap\not | a, none of the integers in SS are congruent to 0 and no two of them are congruent to each other modulo pp. Let r1,,rmr_1, \cdots, r_m be the residues modulo pp smaller than p2\frac{p}{2}, and let s1,,sns_1, \cdots, s_n be the residues modulo pp greater than p2\frac{p}{2}. Then m+n=p12m+n=\frac{p-1}{2} and the integers r1,,rm,ps1,,psnr_1, \cdots, r_m, p-s_1, \cdots , p-s_n are all positive and less than p2\frac{p}{2}. Now, we will prove that no two of these integers are equal. Suppose that for some choice of ii and jj we have psi=rjp-s_i = r_j. We can choose integers uu and vv, with 1u,vp121 \leq u,v \leq \frac{p-1}{2} and satisfying siua  (mod p)rjva  (mod p) \begin{aligned} s_i &\equiv u a \; (\text{mod }p) \\ r_j &\equiv v a \; (\text{mod }p) \end{aligned} Now, we have si+rja(u+v)p0  (mod p)s_i+r_j \equiv a(u+v) \equiv p \equiv 0 \; (\text{mod }p) This implies that u+v0  (mod p)u+v \equiv 0 \; (\text{mod }p). However, this is not possible because 1u+vp11\leq u+v \leq p-1. Thus, we have proven that the numbers r1,,rm,ps1,psnr_1,\cdots, r_m, p-s_1, \cdots p-s_n are simply a rearrangement of the integers 1,2,,p121,2,\cdots, \frac{p-1}{2}. Their product is equal to (p12)!\left(\frac{p-1}{2} \right)!. Therefore, (p12)!=r1rm(ps1)(psn)(1)nr1rms1sn  (mod p)(1)na2a(p12)a  (mod p)(1)nap12(p12)!  (mod p) \begin{aligned} \left(\frac{p-1}{2}\right)! &= r_1 \cdots r_m (p-s_1)\cdots (p-s_n) \\ &\equiv (-1)^n r_1 \cdots r_m s_1\cdots s_n \; (\text{mod }p) \\ &\equiv (-1)^n a\cdot 2a\cdots \left(\frac{p-1}{2}\right)a \; (\text{mod }p) \\ &\equiv (-1)^n a^{\frac{p-1}{2}} \left( \frac{p-1}{2}\right)! \; (\text{mod }p) \end{aligned} The (p12)!\left( \frac{p-1}{2}\right)! term can be cancelled from both sides as p∤(p12)!p\not| \left( \frac{p-1}{2}\right)! . In other words, we have ap12(1)n  (mod p)a^{\frac{p-1}{2}}\equiv (-1)^n \; (\text{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\sin and Gauss’s lemma, it is easy to verify the following result:

Lemma: Let pp and qq be distinct odd primes and let A={αZ1αp12}A = \left\{ \alpha\in \mathbb{Z} | 1\leq \alpha \leq \frac{p-1}{2}\right\} be a half system modulo pp. Then, (qp)=αAsin(2πpqα)sin(2πpα)(1)\left(\frac{q}{p}\right) = \prod_{\alpha\in A}\frac{\sin\left(\frac{2\pi}{p}q\alpha\right)}{\sin\left(\frac{2\pi}{p}\alpha\right)} \quad\quad (1)

We start by examining the right hand side of equation (1)(1). The addition theorem for trigonometric functions yields sin2α=2sinαcosα\sin 2\alpha = 2\sin\alpha\cos\alpha and sin3α=sinα(34sin2α)\sin 3\alpha = \sin\alpha(3-4\sin^2\alpha). Induction shows that sinqα=sinαP(sinα)\sin q\alpha = \sin\alpha P(\sin\alpha) for all odd q1q\geq 1, where PZ[X]P\in \mathbb{Z}[X] is a polynomial of degree q1q-1 and highest coefficient (4)q12(-4)^{\frac{q-1}{2}}. Thus there exist aiZa_i \in \mathbb{Z} such that sinqzsinz=(4)q12((sinz)q1+aq2(sinz)q2++a0)=(4)q12ψ(X),where X=sinz\begin{aligned}\frac{\sin qz}{\sin z} &= (-4)^{\frac{q-1}{2}} \left( (\sin z)^{q-1}+a_{q-2} (\sin z)^{q-2}+\cdots + a_0 \right) \\ &= (-4)^{\frac{q-1}{2}} \psi(X), \quad \text{where }X=\sin z\end{aligned} Since ϕ(z)=sinqzsinz\phi(z)=\frac{\sin qz}{\sin z} is an even function, so is ψ(X)\psi(X), hence aq2==a1=0a_{q-2}=\cdots = a_1 = 0. Now ϕ(z)\phi(z) has zeros {±2πqβ, 1βq12}\left\{ \pm \frac{2\pi}{q}\beta, \ 1\leq \beta \leq \frac{q-1}{2}\right\}. Since ψ\psi is monic of degree q1q-1, we may write ψ(X)=βB(X2sin22πβq)\psi(X) = \prod_{\beta\in B}\left(X^2-\sin^2\frac{2\pi\beta}{q} \right) where B={1,,q12}B=\left\{1,\cdots,\frac{q-1}{2} \right\} is a half system modulo qq. Replacing XX by sinz\sin z, we get sinqzsinz=(4)q12βB(sin2zsin22πβq)(2)\frac{\sin qz}{\sin z} = (-4)^{\frac{q-1}{2}} \prod_{\beta\in B }\left( \sin^2z -\sin^2\frac{2\pi\beta}{q}\right) \quad \quad (2) Put z=2παpz=\frac{2\pi\alpha}{p} in equation (2)(2) and plug the result into equation (1)(1). (qp)=αA(4)q12βB(sin22παpsin22πβq)=(4)q12p12αAβB(sin22παpsin22πβq)(3) \begin{aligned} \left(\frac{q}{p}\right) &= \prod_{\alpha\in A} (-4)^{\frac{q-1}{2}} \prod_{\beta\in B} \left( \sin^2\frac{2\pi\alpha}{p} -\sin^2\frac{2\pi\beta}{q}\right)\\ &= (-4)^{\frac{q-1}{2} \frac{p-1}{2}} \prod_{\alpha\in A}\prod_{\beta\in B} \left( \sin^2\frac{2\pi\alpha}{p} -\sin^2\frac{2\pi\beta}{q}\right)\quad\quad (3) \end{aligned} Exchanging pp and qq on the right side of (3)(3) give rise to factor of (1)(p1)(q1)/4(-1)^{(p-1)(q-1)/4}. Therefore, (qp)=(1)(p1)(q1)/4(pq)(4)\left(\frac{q}{p}\right) = (-1)^{(p-1)(q-1)/4}\left(\frac{p}{q}\right) \tag{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

bookmark_borderSimplification of Incomplete Beta Functions

In this post, we will derive closed form solutions for two incomplete beta integrals. 01dx1x9x6x3=π3(1)01dx1xx42x34=22π338(2) \begin{aligned} \int_0^1\frac{dx}{\sqrt{1-x}\sqrt[6]{9-x}\sqrt[3]x}&=\frac\pi{\sqrt3} \quad\quad (1)\\ \int_0^1\frac{dx}{\sqrt{1-x}\sqrt[4]x\sqrt[4]{2-x\sqrt3}} &= \frac{2\sqrt{2}\pi}{3\sqrt[8]{3}} \quad\quad (2) \end{aligned} These problems are from a math.stackexchange.com question posted by Vladimir Reshetnikov.

Alternate forms

The equations (1) and (2) can be expressed in terms of the incomplete beta function using some elementary hypergeometric identities. This makes our problems equivalent to proving that B(19;16,13)=3243πΓ(13)3(3)B(32;14,14)=23πΓ(14)2(4) \begin{aligned} B\left(\frac{1}{9};\frac{1}{6},\frac{1}{3}\right) &= \frac{\sqrt{3}}{2^{\frac{4}{3}}\pi}\Gamma\left(\frac{1}{3}\right)^3 \quad\quad (3)\\ B\left(\frac{\sqrt{3}}{2};\frac{1}{4},\frac{1}{4}\right) &= \frac{2}{3\sqrt{\pi}}\Gamma\left(\frac{1}{4}\right)^2 \quad\quad (4) \end{aligned} Proof of (3) B(19;16,13)=019x56(1x)23dx B\left(\frac{1}{9};\frac{1}{6},\frac{1}{3}\right) = \int_0^{\frac{1}{9}}x^{-\frac{5}{6}}\left( 1-x\right)^{-\frac{2}{3}} dx Substitute x=y31+y3 x=\frac{y^3}{1+y^3} in the above integral to get B(19;16,13)=30121y+y4dy B\left(\frac{1}{9};\frac{1}{6},\frac{1}{3}\right) =3\int_0^{\frac{1}{2}}\frac{1}{\sqrt{y+y^4}}dy We can now transform this integral into a beta integral using the substitution y=1z2+zy=\frac{1-z}{2+z}. B(19;16,13)=30111z3dz=B(13,12)=3243πΓ(13)3B\left(\frac{1}{9};\frac{1}{6},\frac{1}{3}\right) =3\int_0^1\frac{1}{\sqrt{1-z^3}}dz = B\left( \frac{1}{3},\frac{1}{2}\right)=\frac{\sqrt{3}}{2^{\frac{4}{3}}\pi}\Gamma\left(\frac{1}{3}\right)^3 This proves equations (1) and (3).

Proof of (4)

The proof of integrals (2) and (4) is more involved. Using the substitution x=4t21+4t2x=\frac{4t^2}{1+4t^2}, the integral transforms into the Weierstrass form: B(32;14,14)=032x34(1x)34dx=220αdt4t3+t B\left(\frac{\sqrt{3}}{2};\frac{1}{4},\frac{1}{4}\right) =\int_0^{\frac{\sqrt{3}}{2}}x^{-\frac{3}{4}}(1-x)^{-\frac{3}{4}}dx=2\sqrt{2} \int_0^\alpha \frac{dt}{\sqrt{4t^3+t}} where α=3+232\alpha=\frac{\sqrt{3+2\sqrt{3}}}{2}. Let (z)\wp(z) denote the Weierstrass Elliptic function with invariants g2=1g_2=-1 and g3=0g_3=0. (z)\wp(z) satisfies the differential equation: (z)2=4(z)3+(z)\wp'(z)^2=4\wp(z)^3+\wp(z) The half periods are ω1=1+i4L\omega_1=\frac{1+i}{4}L and ω2=1+i4L\omega_2=\frac{-1+i}{4}L where L=12πΓ(14)2L=\frac{1}{\sqrt{2\pi}}\Gamma\left(\frac{1}{4}\right)^2 is the Lemniscate constant. Now, we can express our integral as follows: B(32;14,14)=22(1(0)1(α))B\left(\frac{\sqrt{3}}{2};\frac{1}{4},\frac{1}{4}\right) =2\sqrt{2}\left(\wp^{-1}(0)-\wp^{-1}(\alpha) \right)Note that0dt4t3+t=122πΓ(14)2=L2\int_0^\infty \frac{dt}{\sqrt{4t^3+t}}=\frac{1}{2\sqrt{2\pi}}\Gamma\left(\frac{1}{4}\right)^2=\frac{L}{2} Therefore, 1(0)=L2\wp^{-1}(0)=\frac{L}{2}. To prove the claim we only need to prove that (L6)=α\wp\left(\frac{L}{6}\right)=\alpha. Using the addition theorem of (z)\wp(z) and the fact that (L2)=0\wp\left(\frac{L}{2}\right)=0 we get: (L6)=(L2L3)=14((L3)(L3))2(L3)=4(L3)3+(L3)4(L3)2(L3)=14(L3)(5) \begin{aligned} \wp\left(\frac{L}{6}\right) &= \wp\left(\frac{L}{2}-\frac{L}{3}\right)\\ &= \frac{1}{4}\left(\frac{\wp'\left(\frac{L}{3}\right)}{\wp\left(\frac{L}{3}\right)} \right)^2-\wp\left(\frac{L}{3}\right) \\ &= \frac{4\wp\left( \frac{L}{3}\right)^3+\wp\left(\frac{L}{3}\right)}{4\wp\left(\frac{L}{3}\right)^2}-\wp\left(\frac{L}{3}\right)\\ &= \frac{1}{4\wp\left(\frac{L}{3}\right)}\quad\quad\quad (5) \end{aligned} On the other hand, by the duplication theorem: (L3)=14(6(L6)2+12)24(L6)3+(L6)2(L6)(6) \begin{aligned} \wp\left(\frac{L}{3}\right)&= \frac{1}{4}\frac{\left(6\wp\left(\frac{L}{6}\right)^2+\frac{1}{2}\right)^2}{4\wp\left(\frac{L}{6}\right)^3+\wp\left(\frac{L}{6}\right)}-2\wp\left(\frac{L}{6}\right)\quad\quad\quad ({6}) \end{aligned} Combine equations (5) and (6) to get: 1+4(L6)2(6(L6)2+12)232(L6)48(L6)2=1    16(L6)424(L6)23=0\frac{1+4\wp\left(\frac{L}{6}\right)^2}{\left(6\wp\left(\frac{L}{6}\right)^2+\frac{1}{2} \right)^2-32\wp\left(\frac{L}{6}\right)^4-8\wp\left(\frac{L}{6}\right)^2}=1 \\ \implies 16\wp\left(\frac{L}{6}\right)^4-24\wp\left(\frac{L}{6}\right)^2-3=0 Since (L6)\wp\left( \frac{L}{6}\right) is positive, we conclude that (L6)=3+232\wp\left(\frac{L}{6}\right)=\frac{\sqrt{3+2\sqrt{3}}}{2} This completes our proof of equation (4) and (2).