Gauss's Lemma (Number Theory)
== Lemma == Let $p$ be an odd prime. Let $a \in \Z: a \not \equiv 0 \pmod p$. Let $S = \left\{{a, 2 a, 3 a, \ldots, \dfrac {p - 1} 2 a}\right\}$. Let $n$ denote the number of elements of $S$ whose least positive residue modulo $p$ is greater than $\dfrac p 2$. Then: :$\left({\dfrac a p}\right) = \left({-1}\right)^n$ where $\left({\dfrac a p}\right)$ is the Legendre symbol. == Proof == First note that no two elements of $s$ are congruent modulo $p$: Let $r a \equiv s a \pmod p$ for some $r, s$ such that $1 \le s \le r \le \dfrac {p-1} 2$. As $a \perp p$, we have that $r a \equiv s a \implies r \equiv s \pmod p$ from Cancellability of Congruences. Now $r \equiv s \pmod p$ can happen only when $r = s$ as $0 \le r - s \le \dfrac {p-1} 2$. Next, no element of $S$ is congruent modulo $p$ to $0$. This is because $r a \equiv 0 \pmod p$ when $a \not \equiv 0$ requires $r \equiv 0$ which doesn't happen. Now, we create $S'$ from $S$ by replacing each element of $S$ by its least positive residue modulo $p$. Arranging $S'$ into increasing order, we get: :$S' = \left\{{b_1, b_2, \ldots, b_m, c_1, c_2, \ldots, c_n}\right\}$ where $b_m < \dfrac p 2 < c_1$ and $m + n = \dfrac {p-1}2$. As $p$ is an odd prime, $\dfrac p 2$ is not an integer so neither $b_m$ nor $c_1$ can be equal to $\dfrac p 2$. Now we let $S'' = \left\{{b_1, b_2, \ldots, b_m, p-c_1, p-c_2, \ldots, p-c_n}\right\}$ We need to show that $S'' = \left\{{1, 2, 3, \ldots, \dfrac {p-1} 2}\right\}$. Now: * all the elements of $S''$ are positive, as all the $c_j < p$ * the largest one is either $b_m$ or $p - c_1$ * there are $\dfrac {p-1} 2$ of them * they are all less than $\dfrac p 2$. So $S''$ contains $\dfrac {p-1} 2$ positive integers all less than $\dfrac p 2$. So all we need to do is show that they are distinct. As all the elements of $S'$ are distinct, the $m$ elements $b_1, b_2, \ldots, b_m$ are distinct. Similarly, so are the $n$ elements $p-c_1, p-c_2, \ldots, p-c_n$. So, we suppose $b_i = p - c_j$ for some $i, j$ and try to derive a contradiction. So: :$p = b_i + c_j \equiv r a + s a \pmod p$ for some $r, s$ such that $1 \le r, s \le \dfrac {p-1} 2$. But from: :$p \equiv \left({r+s}\right) a \pmod p$ and $a \not \equiv 0 \pmod p$ we apply Euclid's Lemma to get $r + s \equiv 0 \pmod p$. But this is impossible since $2 \le r + s \le p-1$. This establishes that $S'' = \left\{{1, 2, 3, \ldots, \dfrac {p-1} 2}\right\}$. Now, we multiply all the elements of $S''$ together: {{begin-eqn}} {{eqn | l = \left({\frac {p - 1} 2}\right)!       | r = b_1 b_2 \cdots b_m \left({p - c_1}\right) \left({p - c_2}\right) \cdots \left({p - c_n}\right)       | c =  }} {{eqn | o = \equiv       | r = b_1 b_2 \cdots b_m \left({-c_1}\right) \left({-c_2}\right) \cdots \left({-c_n}\right)       | rr= \pmod p       | c =  }} {{eqn | o = \equiv       | r = \left({-1}\right)^n b_1 b_2 \cdots b_m c_1 c_2 \cdots c_n       | rr= \pmod p       | c =  }} {{eqn | o = \equiv       | r = \left({-1}\right)^n a \times 2 a \times 3 a \times \cdots \frac{p - 1} 2 a       | rr= \pmod p       | c =  }} {{eqn | o = \equiv       | r = \left({-1}\right)^n a^{\left({\frac {p - 1} 2}\right)} \times \left({\frac {p - 1} 2}\right)!       | rr=\pmod p       | c = as the elements in $S'$ are congruent, in some order, to those in $S$ }} {{end-eqn}} Now, from GCD with Prime: :$\gcd \left\{ {p, \left({\dfrac {p - 1} 2}\right)!}\right\} = 1$ So from Cancellability of Congruences the term $\left({\dfrac {p - 1} 2}\right)!$ can be cancelled from both sides of the above congruence: :$1 \equiv \left({-1}\right)^n a^{\left({\frac {p - 1} 2}\right)} \pmod p$ Finally, from the definition of the Legendre symbol: :$\left({\dfrac a p}\right) \equiv a^{\left({\frac {p - 1} 2}\right)} \pmod p$ Multiplying both sides of the congruence by $\left({-1}\right)^n$: :$\left({\dfrac a p}\right) = \left({-1}\right)^n$  {{Namedfor|Carl Friedrich Gauss|cat = Gauss}} == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Definition:Legendre Symbol/Definition 1|next = Second Supplement to Law of Quadratic Reciprocity}}: $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $47 \ \text {a)}$