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 \left(\frac{a}{p}\right) which takes the values \pm 1 depending on whether a is a quadratic 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 p and q are distinct odd primes, then the quadratic reciprocity theorem states that the congruences \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 p and q leave the the remainder 3 when divided by 4. Written symbolically, \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\not\equiv 0 \;(\text{mod }p). Then n is a quadratic residue modulo p iff n^{\frac{p-1}{2}}\equiv 1 \; (\text{mod }p).

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

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

Conversely, assume that n^{\frac{p-1}{2}}\equiv 1 \; (\text{mod }p). Let g be a primitive root of p. Then, we have n\equiv g^k \; (\text{mod }p) for some integer k. Since n^{\frac{p-1}{2}}\equiv g^{\frac{k(p-1)}{2}} \equiv 1 \; (\text{mod }p), the order of g must divide the exponent \frac{k(p-1)}{2}. Therefore, p-1 | \frac{k(p-1)}{2} and thus k is an even integer. Let’s say k=2j for some integer j. Then, we have n \equiv g^k \equiv g^{2j} \equiv (g^j)^2 \; (\text{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 = \left\{ a,\ 2a,\ 3a,\cdots, \frac{p-1}{2}a \right\}
and their least positive residues modulo p. Let n be the number of these residues that are greater than p/2. Then \left(\frac{a}{p}\right)=(-1)^n

**Proof:** Since p\not | a, none of the integers in S are congruent to 0 and no two of them are congruent to each other modulo p. Let r_1, \cdots, r_m be the residues modulo p smaller than \frac{p}{2}, and let s_1, \cdots, s_n be the residues modulo p greater than \frac{p}{2}. Then m+n=\frac{p-1}{2} and the integers
r_1, \cdots, r_m, p-s_1, \cdots , p-s_n
are all positive and less than \frac{p}{2}. Now, we will prove that no two of these integers are equal. Suppose that for some choice of i and j we have p-s_i = r_j. We can choose integers u and v, with 1 \leq u,v \leq \frac{p-1}{2} and satisfying
\begin{aligned}
s_i &\equiv u a \; (\text{mod }p) \\
r_j &\equiv v a \; (\text{mod }p)
\end{aligned}
Now, we have
s_i+r_j \equiv a(u+v) \equiv p \equiv 0 \; (\text{mod }p)
This implies that u+v \equiv 0 \; (\text{mod }p). However, this is not possible because 1\leq u+v \leq p-1. Thus, we have proven that the numbers r_1,\cdots, r_m, p-s_1, \cdots p-s_n are simply a rearrangement of the integers 1,2,\cdots, \frac{p-1}{2}. Their product is equal to \left(\frac{p-1}{2} \right)!. Therefore,
\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 \left( \frac{p-1}{2}\right)! term can be cancelled from both sides as p\not| \left( \frac{p-1}{2}\right)! . In other words, we have 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 and Gauss’s lemma, it is easy to verify the following result:

**Lemma:**Let p and q be distinct odd primes and let A = \left\{ \alpha\in \mathbb{Z} | 1\leq \alpha \leq \frac{p-1}{2}\right\} be a half system modulo p. Then, \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). The addition theorem for trigonometric functions yields \sin 2\alpha = 2\sin\alpha\cos\alpha and \sin 3\alpha = \sin\alpha(3-4\sin^2\alpha). Induction shows that \sin q\alpha = \sin\alpha P(\sin\alpha) for all odd q\geq 1, where P\in \mathbb{Z}[X] is a polynomial of degree q-1 and highest coefficient (-4)^{\frac{q-1}{2}}. Thus there exist a_i \in \mathbb{Z} such that \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 \phi(z)=\frac{\sin qz}{\sin z} is an even function, so is \psi(X), hence a_{q-2}=\cdots = a_1 = 0. Now \phi(z) has zeros \left\{ \pm \frac{2\pi}{q}\beta, \ 1\leq \beta \leq \frac{q-1}{2}\right\}. Since \psi is monic of degree q-1, we may write \psi(X) = \prod_{\beta\in B}\left(X^2-\sin^2\frac{2\pi\beta}{q} \right) where B=\left\{1,\cdots,\frac{q-1}{2} \right\} is a half system modulo q. Replacing X by \sin z, we get \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=\frac{2\pi\alpha}{p} in equation (2) and plug the result into equation (1). \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 p and q on the right side of (3) give rise to factor of (-1)^{(p-1)(q-1)/4}. Therefore, \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