Eisenstein's Lemma

Theorem
An unwieldy name for an unwieldy result.

Let $$p$$ be an odd prime.

Let $$a \in \Z$$ be an odd integer such that $$p \nmid a$$.

Let $$\left({\frac a p}\right)$$ be defined as the Legendre symbol.

Then:
 * $$\left({\frac a p}\right) = \left({-1}\right)^{\alpha \left({a, p}\right)}$$

where:
 * $$\alpha \left({a, p}\right) = \sum_{k=1}^{\frac {p-1}2} \left \lfloor {\frac {k a} p} \right \rfloor$$

and $$\left \lfloor {\frac {k a} p} \right \rfloor$$ is the floor function of $$\frac {k a} p$$.

Proof
We will borrow some ideas and techniques from the proof of Gauss's Lemma.

Let $$S = \left\{{a, 2a, 3a, \ldots, \frac {p-1} 2 a}\right\}$$.

Let us 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 $$b_m < \frac p 2 < c_1$$ and $$m + n = \frac {p-1}2$$.

According to the Division Theorem, we can divide any $$k a \in S$$ by $$p$$ and get:
 * $$k a = p \times \left \lfloor {\frac {k a} p} \right \rfloor + r$$

where $$r \in S'$$.

Let $$k$$ run from $$1$$ to $$\frac {p-1}2$$ and add up the resulting equations.

We get:
 * $$(1): \qquad \sum_{k=1}^{\frac {p-1}2} k a = \sum_{k=1}^{\frac {p-1}2} p \times \left \lfloor {\frac {k a} p} \right \rfloor + \sum_{i=1}^{m} b_i + \sum_{j=1}^{n} c_j$$

As in the proof of Gauss's Lemma, all the remainders are in fact distinct elements of $$S'$$.

From the proof of Gauss's Lemma again, we have that:
 * $$S'' = \left\{{b_1, b_2, \ldots, b_m, p-c_1, p-c_2, \ldots, p-c_n}\right\} = \left\{{1, 2, 3, \ldots, \frac {p-1} 2}\right\}$$.

Adding up all the elements of $$S''$$ we have:
 * $$\sum_{k=1}^{\frac {p-1}2} k = \sum_{i=1}^{m} b_i + \sum_{j=1}^{n} \left({p - c_j}\right) = \sum_{i=1}^{m} b_i + p n - \sum_{j=1}^{n} c_j$$.

Subtracting this equation from $$(1)$$ above gives us:
 * $$\sum_{k=1}^{\frac {p-1}2} k a - \sum_{k=1}^{\frac {p-1}2} k = \sum_{k=1}^{\frac {p-1}2} p \times \left \lfloor {\frac {k a} p} \right \rfloor + 2 \sum_{j=1}^{n} c_j - p n$$.

That is:
 * $$\left({a - 1}\right) \sum_{k=1}^{\frac {p-1}2} k = p \times \alpha \left({a, p}\right) + 2 \sum_{j=1}^{n} c_j - p n$$.

Now we are going to write this equation modulo $$2$$.

Note that:
 * $$\left({a - 1}\right) \sum_{k=1}^{\frac {p-1}2} k$$ is even, as $$a$$ is odd;
 * $$2 \sum_{j=1}^{n} c_j$$ is also of course even.

So:
 * $$p \alpha \left({a, p}\right) - p n \equiv 0 \pmod 2$$.

Since $$p$$ is odd, this amounts to saying $$\alpha \left({a, p}\right) \equiv n \pmod 2$$.

From the conclusion of Gauss's Lemma that $$\left({\frac a p}\right) = \left({-1}\right)^n$$, we have our result.