Quadratic Reciprocity Law¶

Question

Given an odd prime \(p \in \mathbb{Z}\) and an integer \(a\), is there an efficient way to determine if the equation \(x^2 \equiv a \pmod p\) has a solution?

Definition

(Legendre Symbol)

\[ \left(\dfrac{a}{p}\right) = \begin{cases} 0 \enspace \textrm{if} \enspace a \equiv 0 \pmod p, \\ 1 \enspace \textrm{if} \enspace p \not \mid a, x^2 \equiv a \pmod p \enspace \textrm{has a solution}, \\ -1 \enspace \textrm{if} \enspace p \not \mid a, x^2 \equiv a \pmod p \enspace \textrm{has no solution}. \end{cases} \]

Lemma

(i) If \(a \equiv b \pmod p\), then \(\left(\dfrac{a}{p}\right) = \left(\dfrac{b}{p}\right)\).

(ii) \(\left(\dfrac{1}{p}\right) = 1\).

(iii) (Euler)

\[ \left(\dfrac{a}{p}\right) \equiv a^{\frac{p-1}{2}} \pmod p. \]

(iv) \(\left(\dfrac{a}{p}\right)\left(\dfrac{b}{p}\right) = \left(\dfrac{ab}{p}\right)\)

(v) Set \(a = -1\), then \(\left(\dfrac{-1}{p}\right) = (-1)^{\frac{p-1}{2}}\)

Proof

(iii) (a) If \(x^2 \equiv a \pmod p\), then \(a^{\frac{p-1}{2}} \equiv x^{p-1} \equiv 1 \pmod p\). So \(a^{\frac{p-1}{2}} \equiv \left(\dfrac{a}{p}\right) \pmod p\).
(b) Conversely, if \(a^{\frac{p-1}{2}} \equiv 1 \pmod p\), we need to prove that \(x^2 \equiv a \pmod p\) has a solution. Consider \(a = g^k, g\) is the generator of \(\mathbb{Z}/p\mathbb{Z}\). Then \(a^{\frac{p-1}{2}} \equiv g^{k\frac{p-1}{2}} \equiv 1 \pmod p\). Because the order of \(g\) is \(p-1\), so \(p - 1 \mid k\frac{p-1}{2}\). So \(k\) is even. Then \(a = g^{2k'}\). So \(x^2 \equiv a \pmod p\) has a solution.

Propsotion

\[ \left(\dfrac{2}{p}\right) \equiv (-1)^{\frac{p^2-1}{8}} \pmod p. \]

Proof

The idea is that \(2\) is "almost" a square in \(\mathbb{Z}[\sqrt{-1}]\).
By previous lemma, \(\left(\dfrac{2}{p}\right) \equiv 2^{\frac{p-1}{2}} \pmod p\). Then we have

\[ \left(\dfrac{2}{p}\right) \equiv (-(\sqrt{-1}) \cdot (1+\sqrt{-1})^2)^{\frac{p-1}{2}} \equiv (1+\sqrt{-1})^{p-1} \cdot (\sqrt{-1})^{-\frac{p-1}{2}} \pmod p. \]

So we have

\[ (1+\sqrt{-1}) \left(\dfrac{2}{p}\right) \equiv (1+\sqrt{-1})^p \cdot (\sqrt{-1})^{-\frac{p-1}{2}} \equiv (1+ (\sqrt{-1})^p) \cdot (\sqrt{-1})^{-\frac{p-1}{2}} \pmod p. \]

Note that we use \(p \mid C_p^{k}, k \neq 0, k \neq p\) when we simplified \((1+\sqrt{-1})^p \equiv 1+(\sqrt{-1})^p \pmod p\).

Then cheak the identity for the following cases: \(p = 8k+1, 8k+3, 8k+5, 8k+7\).

When \(p \equiv 1 \mod 8\) or \(p \equiv 7 \mod 8\), \(\dfrac{p^2-1}{8}\) is even, and \(\left(\dfrac{2}{p}\right) \equiv 1 \pmod p\).
When \(p \equiv 3 \mod 8\) or \(p \equiv 5 \mod 8\), \(\dfrac{p^2-1}{8}\) is odd, and \(\left(\dfrac{2}{p}\right) \equiv -1 \pmod p\).
So \(\left(\dfrac{2}{p}\right) \equiv (-1)^{\frac{p^2-1}{8}} \pmod p\).

Theorem

(Gauss)

\[ \left(\dfrac{p}{q}\right) \left(\dfrac{q}{p}\right) = (-1)^{\frac{p-1}{2} \cdot \frac{q-1}{2}}. \]

Theorem

(Hensel's lemma) Assume \(p\) is an odd prime, \(f(x) \in \mathbb{Z}[x]\). If \(f(x) \equiv 0 \pmod p\) has a solution \(x_1\), and \(f'(x_1) \not \equiv 0 \pmod p\). Then \(f(x) \equiv 0 \pmod {p^n}\) has a solution. More concretely, , the solution is given inductively by

\[ x_n \equiv x_{n-1}-f(x_{n-1}) \cdot f'(x_1)^{-1} \pmod {p^n}. \]

Proof

Prove by induction. Suppose

\[ \begin{cases} x_n \equiv x_{n-1}-f(x_{n-1}) \cdot f'(x_1)^{-1} \pmod {p^n} \ \textrm{is a solution,} \\ f'(x_n) \equiv f'(x_1) \pmod p .\end{cases} \]

We have \(f(x_{n+1}) \equiv 0 \pmod {p^{n+1}} \equiv 0 \pmod {p^n}\) and \(f(x_n) \equiv 0 \pmod {p^n}\). Expect \(x_{n+1} = x_n + p^ny_n\). Using Taylor, we have

\[ f(x_{n+1}) = f(x_n + p^ny_n) = f(x_n) + f'(x_n) \cdot p^ny_n + \alpha \cdot p^{2n}y_n^2 \equiv f(x_n)+f'(x_n) \cdot p^ny_n \pmod {p^{n+1}}. \]

Then \(\dfrac{f(x_n)}{p^n} + f'(x_n) \cdot y_n \pmod p\). Obviously \(y_n \equiv -\dfrac{f(x_n)}{p^n \cdot f'(x_1)} \pmod p\).
So \(x_{n+1} \equiv x_n-f(x_n) \cdot f'(x_1)^{-1} \pmod p\).

And \(f'(x_{n+1}) = f'(x_n+p^ny_n) = f'(x_n)+\beta \cdot p^ny_n \equiv f'(x_1) \pmod p\).

Collary

Assume \((a, p) = 1\), p is an odd prime. \(x^2-a \equiv 0 \pmod p\) has a solution iff \(x^2-a \equiv 0 \pmod {p^n}\) has a solution.

Proof

(i) \(x_0^2-a \equiv 0 \pmod {p^n}\). Then obviously \(x_0^2-a \equiv 0 \pmod p\).
(ii) If \(x_0^2-a \equiv 0 \pmod p\), then \((x_0, p) = 1\). Then for \(f(x) = x^2-a, f'(x_0) = 2x_0 \not \equiv 0 \pmod p\). By Hensel, \(x^2-a \equiv 0 \pmod {p^n}\).

Theorem

(Classical Chinese Remainder Theorem) Assume \((m, n) = 1\). Then for any \(a, b\),

\[ \begin{cases} x \equiv a \pmod m, \\ x \equiv b \pmod n \end{cases} \]

has s solution.

Collary

Assume \((m, n) = 1, (a, mn) = 1\). \(x^2-a \equiv 0 \pmod {mn}\) has a solution iff

\[ \begin{cases} x^2-a \equiv \pmod m \\ x^2-a \equiv \pmod n \end{cases} \]

has a solution.

Proof

(i) It's obvious that \(x^2-a \equiv 0 \pmod {mn} \Rightarrow \begin{cases} x^2-a \equiv \pmod m \\ x^2-a \equiv \pmod n \end{cases}\).
(ii)

Theorem

(Special case for \(p = 2\)) (i) \(x^2-a \equiv 0 \pmod 2\) always has a solution.
(ii) \(x^2-a \equiv 0 \pmod 4\) has a solution implies that \(a \equiv 0 \pmod 4\) or \(a \equiv 1 \pmod 4\).
(iii) Lemma: Assume \((a, 2) = 1\)(\(a\) is odd), \(n \geqslant 3\). \(x^2-a \equiv 0 \pmod {2^n}\) has a soltuion iff \(a \equiv 1 \pmod 8\).

Proof

(iii) (\(\Rightarrow\)) \(x_0^2-a \equiv 0 \pmod {2^n} \equiv 0 \pmod 8\), which means \(a \equiv x_0^2 \pmod 8\). And \(a\) is odd, so \(a \equiv 1 \pmod 8\).
(\(\Leftarrow\)) Assume \(a \equiv 1 \pmod 8\). If \(n = 3\), then \(x^2 \equiv a \pmod 8 \equiv 1 \pmod 8\), which implies that there is a solution \(x_0 = 1\).
Prove by induction: \(x^2 \equiv a \pmod {2^n}\) and \(x^2 \equiv a \pmod {2^{n+1}}\). So \(x_{n+1}^2-x_n^2 \equiv 0 \pmod {2^n}\), which means \((x_{n+1}-x_n)(x_{n+1}+x_n) \equiv 0 \pmod {2^n}\). Note that \((x_{n+1}-x_n)\) and \((x_{n+1}+x_n)\) has the same parity, so we only expect that \(x_{n+1} = x_n+2^{n-1}y_n\).
Then \(x_{n+1}^2 \equiv x_n^2 + 2^n \cdot x_ny_n + 2^{2n-2}y_n^2 \pmod {2^{n+1}}\). Note \(n \geqslant 3\), so we have \(2n-2 \geqslant n+1\). So \(x_{n+1}^2 \equiv x_n^2 + 2^n \cdot x_ny_n \pmod {2^{n+1}}\).
We still have \(x_n^2 \equiv a \pmod {2^n}\). Set \(a = x_n^2 + 2^ny_n\). Then \(x_{n+1}^2 \equiv a-2^ny_n+2^nx_ny_n \pmod {2^{n+1}} \equiv a-2^ny_n(x_n-1) \pmod {2^{n+1}}\). We know \(x_n\) is odd, so \(x_n-1\) is even. Then \(x_{n+1}^2 \equiv a \pmod {2^{n+1}}\).