Euler's Conjecture

Lemma 2.1   Let $a, b\in\mathbb{Q}$. Then for any $n\in\mathbb{Z}$,

$$\char93 \left((a,b)\cap \mathbb{Z}\right) \equiv \char93 \left((a... ...\mathbb{Z}\right) \equiv \char93 \left((a+2n,b)\cap \mathbb{Z}\right) \pmod{2}.$$

Proof. If $n>0$, then

$$(a,b+2n) = (a,b) \cup [b,b+2n),$$

where the union is disjoint. Let $[x]$ denote the least integer $\geq x$. There are $2n$ integers,

$$[b], [b]+1, \ldots, [b]+2n-1,$$

in the interval $[b,b+2n)$, so the assertion of the lemma is true in this case. We also have

$$(a,b-2n) = (a,b)\backslash [b-2n,b)$$

and $[b-2n,b)$ also contains exactly $2n$ integers, so the lemma is also true when $n$ is negative. The statement about $\char93 \left((a+2n,b)\cap \mathbb{Z}\right)$ is proved in a similar manner. $\qedsymbol$

The following proposition was first conjectured by Euler, based on extensive numerical evidence. Once we've proved this proposition, it will be easy to deduce the quadratic reciprocity law.

Proposition 2.2 (Euler's Conjecture)   Let $p$ be an odd prime and  $a\in\mathbb{N}$ a natural number with $p\nmid a$.

1. The symbol $\left(\frac{a}{p}\right)$ depends only on $p$ modulo $4a$.
2. If $q$ is a prime with $q\equiv -p\pmod{4a}$, then $\left(\frac{a}{p}\right) = \left(\frac{a}{q}\right)$.

Proof. To apply Gauss's lemma, we have to compute the parity of the intersection of

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

and

$$I = \left(\frac{1}{2}p,p\right) \cup \left(\frac{3}{2}p, 2p\right) \cup \cdots \cup \left(\left(b-\frac{1}{2}\right)p,bp\right),$$

where $b=\frac{1}{2}a$ or $\frac{1}{2}(a-1)$, whichever is an integer. (Why? We have to check that every element of $S$ that reduces to something in the interval $(-\frac{p}{2},0)$ lies in $I$. This is clear if $b=\frac{1}{2}a < \frac{p-1}{2}a$. If $b=\frac{1}{2}(a-1)$, then $bp+\frac{p}{2} > \frac{p-1}{2}a$, so $((b-\frac{1}{2})p,bp)$ is the last interval that could contain an element of of $S$ that reduces to $(-\frac{p}{2},0)$.) Also note that the integer endpoints of $I$ are not in $S$, since those endpoints are divisible by $p$, but no element of $S$ is divisible by $p$.

Dividing $I$ through by $a$, we see that

$$\char93 (S\cap I) = \char93 \left(\mathbb{Z}\cap \frac{1}{a}I\right),$$

where

$$\frac{1}{a} I = \left(\left(\frac{p}{2a},\frac{p}{a}\right) \cup... ...right) \cup \cdots \cup \left(\frac{(2b-1)p}{2a},\frac{bp}{a}\right) \right).$$

Write $p=4ac+r$, and let

$$J = \left(\left(\frac{r}{2a},\frac{r}{a}\right) \cup \left(\frac... ...ight) \cup \cdots \cup \left(\frac{(2b-1)r}{2a},\frac{br}{a}\right) \right).$$

The only difference between $I$ and $J$ is that the endpoints of intervals are changed by addition of an even integer. By Lemma 2.1,

$$\nu=\char93 \left(\mathbb{Z}\cap \frac{1}{a}I\right) \equiv \char93 (\mathbb{Z}\cap J)\pmod{2}.$$

Thus $\left(\frac{a}{p}\right)=(-1)^\nu$ depends only on $r$, i.e., only on $p$ modulo $4a$. WOW!

If $q\equiv -p\pmod{4a}$, then the only change in the above computation is that $r$ is replaced by $4a-r$. This changes $\frac{1}{a}I$ into

$$K = \left(\left(2-\frac{r}{2a},4-\frac{r}{a}\right) \cup \left(... ...cup \cdots \cup \left(4b-2-\frac{(2b-1)r}{2a},4b-\frac{br}{a}\right) \right).$$

Thus $K$ is the same as $-\frac{1}{a}I$, except even integers have been added to the endpoints. By Lemma 2.1,

$$\char93 (K\cap \mathbb{Z})\equiv \char93 \left(\left(\frac{1}{a}I\right)\cap \mathbb{Z}\right)\pmod{2},$$

so $\left(\frac{a}{p}\right) = \left(\frac{a}{q}\right)$, which completes the proof. $\qedsymbol$

The following more careful analysis in the special case when $a=2$ helps illustrate the proof of the above lemma, and is frequently useful in computations.

Proposition 2.3   Let $p$ be an odd prime. Then

$$\left(\frac{2}{p}\right) = \begin{cases}\hfill1 & \text{ if } p\equiv \pm 1\pmod{8}\\ -1 & \text{ if } p\equiv \pm 3\pmod{8} \end{cases}.$$

Proof. When $a=2$, the set $S = \{a,2a,\ldots,2\cdot\frac{p-1}{2}\}$ is

$$\{ 2, 4, 6, \ldots, p-1 \}.$$

We must count the parity of the number of elements of $S$ that lie in the interval $I=(\frac{p}{2}, p)$. Writing $p=8c+r$, we have

$$\char93 \left(I\cap S\right) =\char93 \left(\frac{1}{2}I \cap \mathbb{Z}\right) =\char93 \left(\left(\frac{p}{4},\frac{p}{2}\right)\cap \mathbb{Z}\right)$$

$$=\char93 \left(\left(2c+\frac{r}{4}, 4c+\frac{r}{2}\right)\cap \m... ...r93 \left(\left(\frac{r}{4}, \frac{r}{2}\right)\cap \mathbb{Z}\right) \pmod{2},$$

where the last equality comes from Lemma 2.1. The possibilities for $r$ are $1,3,5,7$. When $r=1$, the cardinality is 0, when $r=3, 5$ it is $1$, and when $r=7$ it is $2$. $\qedsymbol$