The Quadratic Reciprocity Law

Let $p$ be an odd prime and let $a$ be an integer with $p\nmid a$. Set

$$\left(\frac{a}{p}\right) = \begin{cases} +1 & \text{if $a$ is a quadratic residue, and}\\ -1 & \text{otherwise}. \end{cases}$$

Proposition 1.1 implies that

$$\left(\frac{a}{p}\right) \equiv a^{(p-1)/2}\pmod{p}.$$

Also, notice that

$$\left(\frac{a}{p}\right)\cdot \left(\frac{b}{p}\right) = \left(\frac{ab}{p}\right),$$

because $\left(\frac{\cdot }{p}\right)$ is a homomorphism (see Remark 1.5).

The symbol $\left(\frac{a}{p}\right)$ only depends on the residue class of $a$ modulo $p$. Thus tabulating the value of $\left(\frac{a}{5}\right)$ for hundreds of $a$ would be silly. Would it be equally silly to make a table of $\left(\frac{5}{p}\right)$ for hundreds of primes $p$? Let's begin making such a table and see whether or not there is an obvious pattern. (To compute $\left(\frac{a}{p}\right)$ in PARI, use the command kronecker(a,b).)

$p$	$\left(\frac{5}{p}\right)$	$p$ mod 5
7	$-1$	2
11	$1$	1
13	$-1$	3
17	$-1$	2
19	$1$	4
23	$-1$	3
29	$1$	4
31	$1$	1
37	$-1$	2
41	$1$	1
43	$-1$	3
47	$-1$	2

The evidence suggests that $\left(\frac{5}{p}\right)$ depends only on the congruence class of $p$; more precisely, $\left(\frac{5}{p}\right)=1$ if and only if $p\equiv 1,4\pmod{5}$, i.e., $p$ is a square modulo $5$. However, when I think directly about the equation

$$5^{(p-1)/2} \pmod{p},$$

I see no way that knowing that $p\equiv 1,4\pmod{5}$ helps us to evaluate that strange expression! And yet, the numerical evidence is so compelling! Argh!

Based on such computations, various mathematicians found a conjectural explanation for this mystery in the 18th century. Finally, on April 8, 1796, at your age (age 19), Gauss proved their conjecture.

Theorem 2.1 (The Law of Quadratic Reciprocity)   Suppose that $p$ and $q$ are odd primes. Then

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

We will prove this theorem in the next lecture.

In the case considered above, this theorem implies that

$$\left(\frac{5}{p}\right) = (-1)^{2\cdot \frac{p-1}{2}} \left(\fra... ... if }p\equiv 1, 4\pmod{5}\\ -1 & \text{ if }p\equiv 2, 3\pmod{5}. \end{cases}$$

Thus the quadratic reciprocity law ``explains'' why knowing $p$ modulo $5$ helps in computing $5^{\frac{p-1}{2}}\pmod{p}$.

Here is a list of almost 200 proofs of Theorem 2.1:

    http://www.rzuser.uni-heidelberg.de/~hb3/rchrono.html