Gauss's Lemma (Number Theory)

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Lemma

Let $p$ be an odd prime.

Let $a \in \Z: a \not \equiv 0 \pmod p$.

Let $S = \set {a, 2 a, 3 a, \ldots, \dfrac {p - 1} 2 a}$.

Let $n$ denote the number of elements of $S$ whose least positive residue modulo $p$ is greater than $\dfrac p 2$.


Then:

$\paren {\dfrac a p} = \paren {-1}^n$

where $\paren {\dfrac a p}$ 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:

$S' = \set {b_1, b_2, \ldots, b_m, c_1, c_2, \ldots, c_n}$

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 $S = \set {b_1, b_2, \ldots, b_m, p - c_1, p - c_2, \ldots, p - c_n}$

We need to show that $S = \set {1, 2, 3, \ldots, \dfrac {p - 1} 2}$.

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:

$p \equiv \paren {r + s} a \pmod p$ 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 $S = \set {1, 2, 3, \ldots, \dfrac {p - 1} 2}$.


Now, we multiply all the elements of $S$ together:

\(\ds \paren {\frac {p - 1} 2} !\) \(=\) \(\ds b_1 b_2 \cdots b_m \paren {p - c_1} \paren {p - c_2} \cdots \paren {p - c_n}\)
\(\ds \) \(\equiv\) \(\ds b_1 b_2 \cdots b_m \paren {-c_1} \paren {-c_2} \cdots \paren {-c_n}\) \(\ds \pmod p\)
\(\ds \) \(\equiv\) \(\ds \paren {-1}^n b_1 b_2 \cdots b_m c_1 c_2 \cdots c_n\) \(\ds \pmod p\)
\(\ds \) \(\equiv\) \(\ds \paren {-1}^n a \times 2 a \times 3 a \times \cdots \frac {p - 1} 2 a\) \(\ds \pmod p\)
\(\ds \) \(\equiv\) \(\ds \paren {-1} ^n a^{\paren {\frac {p - 1} 2} } \times \paren {\frac {p - 1} 2} !\) \(\ds \pmod p\) as the elements in $S'$ are congruent, in some order, to those in $S$


Now, from GCD with Prime:

$\gcd \set {p, \paren {\dfrac {p - 1} 2} !} = 1$

So from Cancellability of Congruences the term $\paren {\dfrac {p - 1} 2} !$ can be cancelled from both sides of the above congruence:

$1 \equiv \paren {-1}^n a^{\paren {\frac {p - 1} 2} } \pmod p$

Finally, from the definition of the Legendre symbol:

$\paren {\dfrac a p} \equiv a^{\paren {\frac {p - 1} 2} } \pmod p$

Multiplying both sides of the congruence by $\paren {-1}^n$:

$\paren {\dfrac a p} = \paren {-1}^n$

$\blacksquare$


Source of Name

This entry was named for Carl Friedrich Gauss.


Sources

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