# Legendre-Teege Quadratic Reciprocity

Mark B. Villarino
Escuela de Matemática, Universidad de Costa Rica,
10101 San José, Costa Rica

##  1 Introduction

The law of quadratic reciprocity is one of the most celebrated theorems in all
of mathematics. Its fame rests not only on its technical use in the theory of
numbers but also on the influence it has had in the creation and development
of entire branches of mathematics such as commutative algebra and elliptic
curves. Franz Lemmermeyer’s important treatise _Reciprocity Laws_ [6] details
the history and mathematics of its proofs and is well worth perusing.

The quadratic reciprocity law also enjoys another extraordinary
characteristic. _More proofs of it have been published than of any other
mathematical theorem, except for Pythagoras’ theorem._ Indeed, no other
theorem (famous or not) even comes close. Lemmermeyer [1] cites 314 published
proofs up to 2015 and most likely a few more have appeared since.

So it is an historical irony and curiosity that the _first attempted_
published proof is _not_ on that list.

We briefly review its history.

The famous French mathematician, Adrien-Marie Legendre (1752-1833), published
a treatise on the theory of numbers [5] in which he:

  * •

introduced the “Legendre symbol”;

  * •

introduced the word “reciprocity;”

  * •

stated the quadratic reciprocity theorem in the form given in all texts today;

  * •

offered a proof that contained a “gap.”

His proof was a revision of an earlier (1785) proof but it had a new gap.

We formulate this new gap in the form of a lemma.

######  Lemma 1.1.

(Legendre’s Lemma) Given any prime of the form $a:=4n+1$ there exists a prime
of the form $4n+3$ of which $a$ is a quadratic non-residue.

For example $a=17$ is a quadratic nonresidue of $b=3$.

Legendre showed that he could complete his proof if he could prove his lemma.
Although he made several attempts he never succeeded. Indeed, it appears that
the lemma is much _harder_ to prove than the theorem, itself.

In 2011, Weintraub [12] published in this Monthly a very nice and detailed
account of Legendre’s attempted proofs of the quadratic reciprocity law , but
only cited, without proof, Legendre’s unproved lemma as “Hypothesis B.”

In the subsequent 220+ years after Legendre only two proofs of the lemma have
appeared. One is by a relatively unknown German mathematician, Hermann Teege
[11] in 1923, while the other was authored by Kenneth Rogers [8] and was
published in 1971. We will say more about these proofs later on.

The literature does not seem to contain an easily accessible, ab initio,
connected, detailed, rigorous and complete presentation of a proof of
Legendre’s Lemma. This paper is designed fill that gap.

In order to make our paper as self-contained as possible we will first outline
the major steps in Legendre’s original proof in its final form and then fill
the gap. He based his proof on

  * •

a theorem on quadratic forms which he rigorously proved; and

  * •

the theory of the Pell equation; and

  * •

Legendre’s (unproved) lemma; and

  * •

the multiplicativity of the legendre symbol and for any prime $p$, Euler’s
result $\left(\frac{-1}{p}\right)=(-1)^{\frac{p-1}{2}}.$

We will follow (and expand) Teege’s arrangement [10] of Legendre’s proof.

##  2 Legendre’s theorem on quadratic forms

The theorem proven by Legendre [5] §27, and which is today rather famous in
its own right, is the following:

######  Theorem 2.1.

If $a,b,c$ denote numbers which are relatively prime in pairs, not one of
which is zero nor divisible by a square, then the equation

$ax^{2}+by^{2}+cz^{2}=0$ (2.1)

has no solution in integers unless $-bc,-ac,-ab$ are, respectively, quadratic
residues of $a,b,c$, and at least one of them is positive and one of them is
negative. However, if these four conditions are fulfilled, then the equation
is solvable in integers.

$\hfill\blacksquare$

######  Corollary 2.2.

Using the Legendre symbol, if (2.1) has no solutions, and if $a$, $b$, $c$ be
primes or $1$, then at least one of the following equations must hold:

$\left(-\frac{bc}{a}\right)=-1\quad\text{or},\left(-\frac{ca}{b}\right)=-1\quad\text{or},\left(-\frac{ab}{c}\right)=-1.$

$\hfill\blacksquare$

Legendre’s proof of the quadratic reciprocity law also uses the following
simple lemma .

######  Lemma 2.3.

If the integers $a,b,c$ are all $\equiv 1\pmod{4}$, then the equation

$ax^{2}+by^{2}+cz^{2}=0$

is not solvable in integers.

###### Proof.

For, since the square of an odd number is of the form $8n+1$, it is easy to
see in this case that the value of $ax^{2}+by^{2}+cz^{2}$ is necessarily
either $\equiv 1$, or $\equiv 2$, or $\equiv 3\pmod{4}$ unless all the
quantities $x,y,z$ are simultaneously even. But this last possibility is
absurd since if we divide the equation by $2^{2\mu}$ where $2^{\mu}$ is the
highest power of $2$ dividing $x,y,z$ simultaneously, the new solution
$x^{\prime},y^{\prime},z^{\prime}$ will have at least one odd member. ∎

##  3 Pell’s equation

Legendre also uses the following consequence of the theory of Pell’s equation
[5] §49.

######  Theorem 3.1.

If $b$, $B$ and $\beta$ are prime numbers of the form $4n+3$ and $a$ is a
prime number of the form $4n+1$, then it is always possible to solve one of
the six equations

$aM^{2}-b\beta N^{2}=\pm 1,$ $bM^{2}-a\beta N^{2}=\pm 1,$ $\beta
M^{2}-abN^{2}=\pm 1.$

###### Proof.

It is well known that for every arbitrary positive integer $D$ the equation

$x^{2}-Dy^{2}=1$

is always solvable in distinct non zero integers $x,y$. Now, if $T$ and $U$
are the smallest positive solutions among them then

$T^{2}-DU^{2}=1$

or

$(T-1)(T+1)=DU^{2}.$

Moroever if $D\equiv 1\pmod{4}$, it is easy to see that $T$ must be odd and
$U$ even so that we can put

$\frac{T+1}{2}\cdot\frac{T-1}{2}=DU_{1}^{2}$

where $U=2U_{1}$ and the two factors on the left must be relatively prime
since their difference is $1$.

Now, if $D=b\cdot B$ where $b$ and $B$ are two prime numbers of the form
$4n+3$, and if the factorization of $U_{1}$ into two relatively prime factors
is given generally by

$U_{1}=M\cdot N,$

then it follows from the equation

$\frac{T+1}{2}\cdot\frac{T-1}{2}=b\cdot B\cdot U_{1}^{2}$

that all possible factorizations are given by the following four cases:

$\displaystyle 1.\quad\frac{T+1}{2}$ $\displaystyle=M^{2}$
$\displaystyle\frac{T-1}{2}$ $\displaystyle=b\cdot B\cdot N^{2}$
$\displaystyle 2.\quad\frac{T+1}{2}$ $\displaystyle=b\cdot B\cdot M^{2}$
$\displaystyle\frac{T-1}{2}$ $\displaystyle=N^{2}$ $\displaystyle
3.\quad\frac{T+1}{2}$ $\displaystyle=b\cdot M^{2}$
$\displaystyle\frac{T-1}{2}$ $\displaystyle=B\cdot N^{2}$ $\displaystyle
4.\quad\frac{T+1}{2}$ $\displaystyle=B\cdot M^{2}$
$\displaystyle\frac{T-1}{2}$ $\displaystyle=b\cdot N^{2}$

Of these Case $1$ is impossible because otherwise

$M^{2}-b\cdot B\cdot N^{2}=1$

which contradicts the minimality of the solution $T$ and $U$.

Similarly, Case $2$ is impossible because otherwise

$1=b\cdot BM^{2}-N^{2}$

and $-1$ would be a quadratic residue of a prime number of the form $4n+3.$

Two cases still remain:

$1=bM^{2}-B\cdot N^{2}$

and

$1=BM^{2}-b\cdot N^{2}$

one of which must occur.

∎

##  4 Legendre’s proof of the law of quadratic reciprocity

After these preparations we can now briefly sketch Legendre’s proof of the law
of quadratic reciprocity.

If we retain the earlier notation, so that $a,A$ are prime numbers of the form
$4n+1$ and $b,B$ are prime numbers of the form $4n+3$, then the theorem splits
into the following eight cases, where $\left(\frac{a}{b}\right)$ is the well
known Legendre symbol [5] §166.

######  Theorem 4.1.

$\displaystyle I.$ $\displaystyle\left(\frac{a}{b}\right)=-1$
$\displaystyle\Rightarrow$ $\displaystyle\left(\frac{b}{a}\right)=-1$
$\displaystyle II.$ $\displaystyle\left(\frac{b}{a}\right)=+1$
$\displaystyle\Rightarrow$ $\displaystyle\left(\frac{a}{b}\right)=+1$
$\displaystyle III.$ $\displaystyle\left(\frac{B}{b}\right)=+1$
$\displaystyle\Rightarrow$ $\displaystyle\left(\frac{b}{B}\right)=-1$
$\displaystyle IV.$ $\displaystyle\left(\frac{B}{b}\right)=-1$
$\displaystyle\Rightarrow$ $\displaystyle\left(\frac{b}{B}\right)=+1$
$\displaystyle V.$ $\displaystyle\left(\frac{a}{A}\right)=+1$
$\displaystyle\Rightarrow$ $\displaystyle\left(\frac{A}{a}\right)=+1$
$\displaystyle VI.$ $\displaystyle\left(\frac{a}{A}\right)=-1$
$\displaystyle\Rightarrow$ $\displaystyle\left(\frac{A}{a}\right)=-1$
$\displaystyle VII.$ $\displaystyle\left(\frac{a}{b}\right)=+1$
$\displaystyle\Rightarrow$ $\displaystyle\left(\frac{b}{a}\right)=+1$
$\displaystyle VIII.$ $\displaystyle\left(\frac{b}{a}\right)=-1$
$\displaystyle\Rightarrow$ $\displaystyle\left(\frac{a}{b}\right)=-1$

###### Proof.

The idea of Legendre’s proof is the following. One forms a suitable equation
(2.1) in which the coefficients are all $\equiv 1\pmod{4}$, which, by lemma
2.3, _cannot_ be solved, and uses Corollary 2.2 and the properties of the
Legendre symbol to show that the left hand side of the implication (I)-(VIII)
must therefore hold.

_Cases $I$ and $II$:_

We proceed from the _impossibility_ of the equation

$x^{2}+ay^{2}-bz^{2}=0$

Lemma 2.3 implies that we may not have, simultaneously,

$\left(\frac{-a}{b}\right)=+1\quad\text{and}\quad\left(\frac{b}{a}\right)=+1.$

Therefore, if

$\left(\frac{-a}{b}\right)=+1\quad\text{that
is}\quad\left(\frac{a}{b}\right)=-1,$

since the Legendre symbol can only have the values$+1$ and $-1$, it must
follow that $\left(\frac{b}{a}\right)=-1$, and if

$\left(\frac{b}{a}\right)=+1,$

it must follow that $\left(\frac{a}{b}\right)=+1.$

_Cases $III$ and $IV$:_

By the Theorem 3.1, one of the two equations

$bM^{2}-Bn^{2}=1$ $BM^{2}-bN^{2}=1$

is satisfied.

However, if $\left(\frac{B}{b}\right)=+1$, the equation $bM^{2}-BN^{2}=1$
cannot hold since otherwise $\left(\frac{-B}{b}\right)=+1$ or
$\left(\frac{B}{b}\right)=-1$ contrary to the hypothesis. Therefore the second
equation holds, from which it follows that $\left(\frac{-b}{B}\right)=+1$ or
$\left(\frac{b}{B}\right)=-1.$

On the other hand, if $\left(\frac{B}{b}\right)=-1$, then we must exclude the
second equation and the now valid first equation implies
$\left(\frac{b}{B}\right)=+1.$

_Cases $V$ and $VI$:_

_Choose a prime number $\beta$ of the form $4n+3$ such that
$\left(\frac{a}{\beta}\right)=-1.$_ Then, by case $I$ we must have
$\left(\frac{\beta}{a}\right)=-1.$ If we now consider the unsolvable equation

$x^{2}+ay^{2}-A\beta z^{2}=0$

we immediately see that we cannot have simultaneously that $(-a)$ is a residue
of $(A\beta)$ and $(A\beta)$ a residue of $a$.

However, the first condition is fulfilled in case $V$ because then
$\left(\frac{-a}{A}\right)=+1$ and by the assumption about $\beta$ also
$\left(\frac{-a}{\beta}\right)=+1$ so that $(-a)$ must be a residue of
$(A\beta)$. Accordingly the second condition cannot also be fulfilled, i.e.,
we must have $\left(\frac{\beta A}{a}\right)=-1$ or, since
$\left(\frac{\beta}{a}\right)=-1$, also $\left(\frac{A}{a}\right)=+1.$

Case $VI$ can be reduced to Case $V$. For, if we could deduce the condition
$\left(\frac{A}{a}\right)=+1$ from the equation $\left(\frac{a}{A}\right)=-1$,
then the case just proved would imply $\left(\frac{a}{A}\right)=+1$, contrary
to the hypothesis.

_Cases $VII$ and $VIII$:,_

Here therefore $\left(\frac{a}{b}\right)=+1,$ we again as above _choose a
prime number $\beta$ of the form $4n+3$ so that
$\left(\frac{a}{\beta}\right)=-1$ _and therefore also
$\left(\frac{\beta}{a}\right)=-1.$ Then, by the theory of the Pell equation,
one can assume that one of the following six equations is fulfilled

$\pm 1=aM^{2}-b\beta N^{2},$ $\pm 1=bM^{2}-a\beta N^{2},$ $\pm 1=\beta
M^{2}-abN^{2}.$

for a suitable choice of sign on the left hand side.

Recalling that

$\left(\frac{a}{b}\right)=+1,\left(\frac{a}{\beta}\right)=-1,\left(\frac{\beta}{a}\right)=-1,$

we conclude that the following equations are impossible

$1.\quad+1=\beta M^{2}-abN^{2}\quad\text{which
assumes}\quad\left(\frac{\beta}{a}\right)=+1,$ $2.\quad-1=\beta
M^{2}-abN^{2}\quad\text{which assumes}\quad\left(\frac{\beta}{a}\right)=+1,$
$3.\quad-1=aM^{2}-b\beta N^{2}\quad\text{which
assumes}\quad\left(\frac{a}{b}\right)=-1,$ $4.\quad+1=aM^{2}-b\beta
N^{2}\quad\text{which assumes}\quad\left(\frac{a}{\beta}\right)=+1.$

Therefore, only two equations remain

$+1=bM^{2}-a\beta N^{2}$ $-1=bM^{2}-a\beta N^{2}$

each of which, whichever one holds, requires the relation
$\left(\frac{b}{a}\right)=+1.$ This proves Case (VII).

If $\left(\frac{b}{a}\right)=-1,$ then we must have
$\left(\frac{a}{b}\right)=-1,$ since if $\left(\frac{a}{b}\right)=+1,$ we
would have $\left(\frac{b}{a}\right)=+1,$ which is a contradiction. This
proves Case (VIII). ∎

##  5 The gap in the proof

Again we state the fundamental gap in Legendre’s proof, already mentioned in
the introduction, and which yet has to be filled. Namely, that _given a prime
number $a$ of the form $4n+1$ it is always possible to find a prime number
$\beta$ of the form $4n+3$, of which $a$ is a quadratic nonresidue. _ It is
easy to prove the existence of this auxiliary prime for prime numbers of the
form $8n+5$ since then $1+a$ is of the form $8n+6$, therefore divisible by a
prime of the form $4n+3$, of which $a$ is then $a$ is a nonresidue. But the
proof for the case $a=8n+1$ is profoundly more difficult.

Some forty years after Legendre, Dirichlet introduced brilliant new methods
into number theory in his famous determination of the _class number_ of binary
quadratic forms. On the basis of these investigations one _can_ provide the
necessary proof to complete Legendre’s demonstration.

Nevertheless, almost 125 years passed until (1923) when finally the German
mathematician H. Teege posthumously published the first rigorous proof of
Legendre’s lemma [11]. He based his proof on the (easy part (!)) of the
Dirichlet class number formulas mentioned above, and was able to obtain the
following much stronger result.

######  Theorem 5.1.

(Teege’s Theorem) It is always possible to find not only one, but infinitely
many prime numbers $\beta$ of the form $4n+3$ so that a given prime number $a$
of the form $8n+1$ is a quadratic nonresidue of them, and that there is always
at least one of them smaller than $a$.

We note that Gauss’ first proof of the quadratic reciprocity law proved the
existence of such a prime number $\beta$ but _without_ the condition that it
be of the form $4n+3.$

Then, some 48 years later (1971) K. Rogers published a proof [8] based on a
lemma of Selberg that the latter used in his elementary proof of the prime
number theorem for arithmetic progressions [9]. Just like Teege, Rogers also
proved that there are _infinitely many_ primes $\beta$ satisfying Legendre’s
lemma, but not the fact that one can find such a prime smaller than $a$.

These, apparently, are the only two proofs of Legendre’s lemma in the
literature.

We will elaborate a version of Teege’s proof and sketch the principle part of
Rogers’ proof.

##  6 Binary Quadratic Forms

We briefly review some facts about binary quadratic forms. See Landau [4] and
Mathews [7] for more details.

######  Definition 6.1.

A binary quadratic form (BQF) $Q(x,y)$ is a function

$Q(x,y):=ax^{2}+bxy+cy^{2}$

where $a,b,c$ are integers and will be denoted $(a,b,c)$. The quantity

$d:=b^{2}-4ac$

is the discriminant and $D:=\dfrac{d}{4}$ is the determinant. The discriminant
will always denote a nonsquare integer congruent to $0$ or $1$ modulo 4. It is
called a fundamental discriminant. If $d$ is divisible by $4$ then $b$ is
even.

######  Definition 6.2.

We say that a BQF $Q$ represents a number $n$ when there exist integers $x,y$
such that $Q(x,y)=n$. If $x$ and $y$ are relatively prime, we say that $Q$
properly represents $n$.

A famous theorem of Lagrange states the following.

######  Theorem 6.1.

The forms of given (fundamental) discriminant $d$ fall into classes of
mutually equivalent forms under linear substitutions of the form

$x=\alpha x^{\prime}+\beta y^{\prime},y=\gamma x^{\prime}+\delta y^{\prime}$
(6.1)

with integral coefficients $\alpha,\beta,\gamma,\delta$ satisfying
$\alpha\delta-\beta\gamma=1$. The number of classes, $h(d)$ (called the class
number), for a given discriminant, $d$, is finite.

$\hfill\blacksquare$

We give examples below.

##  7 Teege’s Identity

We can now state _Teege’s identity_ which is the basis of his proof and a
byproduct of Dirichlet’s researches.

######  Theorem 7.1.

(Teege’s Identity) Let $D>2$ be a nonsquare positive integer and $p$ a prime
which does not divide $D$. Then

$\boxed{\frac{h(D)\ln(T+U\sqrt{D})}{h(-D)\cdot 2\pi}=\lim_{s\rightarrow
1}\dfrac{\displaystyle\prod_{\left(\frac{D}{p}\right)=1}\frac{1+\frac{1}{p^{s}}}{1-\frac{1}{p^{s}}}}{2\displaystyle\prod_{\left(\frac{-D}{p}\right)=1}\frac{1+\frac{1}{p^{s}}}{1-\frac{1}{p^{s}}}}}$

where $h(D)$ and $h(-D)$ are the class numbers of binary quadratic forms of
determinant $\pm D$, $T,U$ is the minimum positive solution of the Pell
equation $T^{2}-DU^{2}=1$, and $\left(\dfrac{\pm D}{p}\right)$ is the Legendre
symbol.

######  Example 7.1.

Let $d=\pm 20$. Then $\pm D=\pm 5$. It is well-known that the primes $p$ with
$\left(\frac{5}{p}\right)=1$ are of the two forms $p=5m\pm 1$ and the primes
$p$ with $\left(\frac{-5}{p}\right)=1$ are of the four forms
$20n+1,20n+9,20n+3,20n+7$. Moreover (see the next two examples)
$h(5)=1,h(-5)=2$. Thus, Teege’s identity takes the form

$\boxed{\frac{\ln(9+4\sqrt{5})}{2\cdot 2\pi}=\lim_{s\rightarrow
1}\dfrac{\displaystyle\prod_{p=5m\pm
1}\frac{1+\frac{1}{p^{s}}}{1-\frac{1}{p^{s}}}}{2\displaystyle\prod_{p=20n+1,20n+9,20n+3,20n+7}\frac{1+\frac{1}{p^{s}}}{1-\frac{1}{p^{s}}}}}$

that is

$\boxed{\frac{\ln(9+4\sqrt{5})}{2\pi}=\frac{\frac{1+\frac{1}{11}}{1-\frac{1}{11}}}{\frac{1+\frac{1}{3}}{1-\frac{1}{3}}}\cdot\frac{\frac{1+\frac{1}{19}}{1-\frac{1}{19}}}{\frac{1+\frac{1}{7}}{1-\frac{1}{7}}}\cdot\frac{\frac{1+\frac{1}{29}}{1-\frac{1}{29}}}{\frac{1+\frac{1}{29}}{1-\frac{1}{29}}}\cdot\frac{\frac{1+\frac{1}{31}}{1-\frac{1}{31}}}{\frac{1+\frac{1}{37}}{1-\frac{1}{37}}}\cdots}$

The idea of Teege’s proof is this. We argue by contradiction. We let $D$_be a
prime for which Legendre’s lemma is false_. Then the left hand side of Teege’s
identity is a _finite number_ while the right hand side is shown to _diverge
to infinity_. Thus Legendre’s lemma is not false for any prime, that is,_it is
always true._ The hard part of the proof of the identity is the left-hand
side. Dirichlet showed that the numerator and denominator are measures of the
average number of distinct representations of a number $m$ by binary quadratic
forms of determinant $\pm D$, respectively, as $m\rightarrow\infty$.

It turns out that Teege’s identity is a consequence of two other results: the
first we call the _fundamental identity_ which we state later on after some
introductory examples; and the second is Dirichlet’s _class number formula._

Then we will devote the rest of our paper to proving the identity and
discussing the class number formulas, and apply them to prove Teege’s
identity. Finally we’ll use Teege’s identity to give a detailed proof of
Legendre’s Lemma. We will follow with a discussion of Rogers’ proof.

##  8 Dirichlet’s “fundamental identity”

We begin with two examples

######  Example 8.1.

Let $d:=-20$. Then it is well known that any BQF with $d:=-20$ is equivalent
to one of the two BFQ’s:

$Q_{1}(x,y):=x^{2}+5y^{2},\quad Q_{2}:=2x^{2}+2xy+3y^{2}.$

Thus, the class number

$h(-20)=2.$

The numbers properly and improperly represented by $Q_{1}$ include

$0,1,4,5,6,9,14,16,20,21,24,25,29,30,\cdots$

while the numbers properly and improperly represented by $Q_{2}$ include

$0,2,3,7,8,10,12,15,18,23,27,28,32,35,\cdots$

We will be interested in the subset of those integers represented by $Q_{1}$
and $Q_{2}$ which are _relatively prime_ to $d:=-20$, i.e. to

$1,3,7,9,21,23,27,29,41,43,47,49,61,63,67,69,81,\cdots.$

We will study the series (today called an _Epstein zeta function_)

$\sum_{x,y}\frac{1}{Q_{1}(x,y)^{s}}+\sum_{x,y}\frac{1}{Q_{2}(x,y)^{s}}=\frac{2}{1^{s}}+\frac{4}{3^{s}}+\frac{4}{7^{s}}+\frac{4}{9^{s}}+\frac{8}{21^{s}}+\cdots$

where $Q$ runs over the two BFQ’s $Q_{1}$ and $Q_{2}$ once and $x,y$ run over
all relatively prime pairs with $Q(x,y)$ relatively prime to $d$ , while the
numerator is t _he number of proper representations of the denominator by some
$Q(x,y)$ of determinant $-5$_. For example, $m=9$ is represented properly by
$Q_{1}$ by the four pairs $(x,y)=(2,1),(-2,1),(2,-1),(-2,-1)$ and $m=7$ is
represented properly by $Q_{2}$ by the four pairs
$(x,y)=(1,1),(-1,-1),(-1,2),(1,-2).$

######  Example 8.2.

Let $d:=+20$. Then it can be proved that any BQF with $d:=+20$ is equivalent
to the _single_ BQF

$Q(x,y):=x^{2}-5y^{2}.$

Therefore

$h(+20)=1.$

The numbers properly represented by $Q(x,y)$ which are prime to $d:=+20$ are

$1,9,11,19,29,31,41,49,59,61,71,79,81,89,99,\cdots,$

and we will study the Epstein zeta series

$\sum_{x,y}\frac{1}{Q(x,y)^{s}}=\frac{2}{1^{s}}+\frac{4}{9^{s}}+\frac{4}{11^{s}}+\frac{4}{19^{s}}+\cdots$

where again the numerator is the number of proper representations of the
denominator by the one $Q(x,y)$ of determinant $+5$.

These two examples illustrate the left side of a fundamental identity which we
now study.

The first step in the proof of our final identity is the following standard
counting result (see [7] §202).

Let

$w:=\begin{cases}6\quad\text{if}\quad d=-3,\\\ 4\quad\text{if}\quad d=-4,\\\
2\quad\text{if}\quad d<-4,\\\ 1\quad\text{if}\quad d>0.\end{cases}$

######  Lemma 8.1.

Suppose $n$ is an integer relatively prime to $2D$. If $n$ is divisible by
$\mu$ distinct primes each of which satisfies $\left(\frac{D}{p}\right)=1$,
but by no other primes, then $n$ can be represented in $w2^{\mu}$ distinct
ways by a primitive form of determinant $D$. Otherwise $n$ cannot be
represented by a primitive form of determinant $D$.

$\hfill\blacksquare$

For example if $Q(x,y)=3x^{2}+2xy+2y^{2}=7$ the theorem states that it has
$2^{1+1}$ solutions, and indeed the pairs $(x,y)=(\pm 1,\pm 1),(\pm 2,\mp 1)$
are the four solutions predicted.

By applying the above lemma we obtain the following result.

######  Theorem 8.2.

The following identity is valid

$\sum_{Q}\sum_{x,y}\frac{1}{Q(x,y)^{s}}=w\sum_{m=1}^{\infty}\frac{2^{\mu}}{m^{s}}=w\prod_{\left(\frac{D}{p}\right)=1}\frac{1+\frac{1}{p^{s}}}{1-\frac{1}{p^{s}}}$

where $Q$ runs over each class once and $x$ and $y$ run over all relatively
prime pairs with $Q(x,y)$ relatively prime to $D$.

###### Proof.

We have to prove the right-hand equality. We follow Dirichlet( [3] pp.
154-155) Let

$p_{1},p_{2},p_{2},\cdots$

be the primes in the above product. Then each $m$ in the above sum is of the
form

$p_{1}^{n_{1}}p_{2}^{n_{2}}p_{3}^{n_{3}}\cdots$

where the exponents $n_{1},n_{2},n_{3},\cdots$ are positive integers or zero,
and each $m$ is uniquely expressible in this form. If we now form infinite
series corresponding to these primes

$1+\frac{2}{p_{1}^{s}}+\frac{2}{p_{1}^{2s}}+\frac{2}{p_{1}^{3s}}+\cdots\frac{2}{p_{1}^{n_{1}s}}+\cdots$
$1+\frac{2}{p_{2}^{s}}+\frac{2}{p_{2}^{2s}}+\frac{2}{p_{2}^{3s}}+\cdots\frac{2}{p_{2}^{n_{2}s}}+\cdots$
$1+\frac{2}{p_{3}^{s}}+\frac{2}{p_{3}^{2s}}+\frac{2}{p_{3}^{3s}}+\cdots\frac{2}{p_{3}^{n_{3}s}}+\cdots$

and so on, the product of arbitrary terms of the first, second, third series,
etc., has the form

$\frac{2^{\mu}}{(p_{1}^{n_{1}}p_{2}^{n_{2}}p_{3}^{n_{3}}\cdots)^{s}}=\frac{2^{\mu}}{m^{s}}$

where $\mu$ is the number of primes $p$ actually dividing $m$, i.e., with non-
zero exponent $n$. Thus, by unique factorization, we get each term in our sum
exactly once.

On the other hand

$1+\frac{2}{p^{s}}+\frac{2}{p^{2s}}+\frac{2}{p^{3s}}+\cdots\frac{2}{p^{ns}}+\cdots=\frac{2}{p^{s}}\frac{1}{2-\frac{1}{p^{s}}}=\frac{1+\frac{1}{p^{s}}}{1-\frac{1}{p^{s}}}$

This completes the proof. ∎

Now we transform the identity into a final form. First we multiply both sides
by $\displaystyle\sum_{(n,2D)=1}\frac{1}{n^{2s}}$.

Then the left-hand side of the modified identity is a triply-infinite series
whose first term is of the form (say):

$\frac{1}{(an^{2}x^{2}+2bn^{2}xy+cn^{2}y^{2})^{s}}.$

But, if we put

$x^{\prime}:=nx,y^{\prime}:=ny$

then the modified left-hand side becomes a doubly infinite series

$\sum\frac{1}{(ax^{\prime 2}+2bx^{\prime}y^{\prime}+cy^{\prime 2})^{s}}$

where $x,y$ are no longer necessarily relatively prime.

We have now proven what Dirichlet called ([3],§88) the fundamental identity.

######  Theorem 8.3.

$\boxed{\sum_{Q}\sum_{x,y}\frac{1}{Q(x,y)^{s}}=k\displaystyle\sum_{(n,2D)=1}\frac{1}{n^{2s}}\cdot\prod_{\left(\frac{D}{p}\right)=1}\frac{1+\frac{1}{p^{s}}}{1-\frac{1}{p^{s}}}}$

where $Q(x,y)$ is relatively prime to $2D$, and, if $D>0$, then

$y>0,U\cdot(ax+by)>Ty$

where $T,U$ are the smallest positive integer solutions of $T^{2}-DU^{2}=1.$
Finally, $x,y$ run over all pairs of positive integers where $Q(x,y)$
satisfies the above conditions.

$\hfill\blacksquare$

##  9 Dirichlet’s class number formulas and Teege’s identity.

Let $d$ be the discriminant of the quadratic forms $Q(x,y)$ with class number
$h(d)$. The left hand side of the fundamental identity is an infinite series
whose general term is

$\frac{r_{Q}(m)}{m^{s}}$

where $r_{Q}(m)$ is the number of pairs $(x,y)$ of integers which satisfy

$Q(x,y)=m.$

and $s>1$. In 1839 ([2]) Dirichlet published a brilliant _tour de force_ of
elementary (though not simple) number theory and calculus in which he shows
that if the fundamental identity is multiplied by $(s-1)$, then _each side
tends to a finite limite as_ $s\rightarrow 1$. Moreover, surprisingly, each
summand on the left-hand side tends to the _same_ limit, which depends only on
$D$. Dirichlet evaluates this limit by _geometric_ arguments (area of an
_ellipse_ when $D<0$, area of a _hyperbolic sector_ when $D>0$). Since there
are $h(D)$ terms on the left-hand side, and when $s\rightarrow 1$ they all
tend to the same limit, the _left-hand side tends to a number of the form_

$h(D)\times(\text{this geometric limit})$

.

Unfortunately, space prevents us from giving the proofs but they can be found
in Dirichlet’s text (§88-§104 [3]).

We cite his final result.

######  Theorem 9.1.

(Dirichlet’s Class Number Formulas) Let $s>1$. If $D<0$ then

$\boxed{\lim_{s\rightarrow
1}(s-1)\cdot\sum_{Q}\sum_{x,y}\frac{1}{Q(x,y)^{s}}=\frac{h\cdot\pi}{\sqrt{|D|}}}$

If $D>0$ then

$\boxed{\lim_{s\rightarrow
1}(s-1)\cdot\sum_{Q}\sum_{x,y}\frac{1}{Q(x,y)^{s}}=\frac{h}{2\sqrt{D}}\ln(T+U\sqrt{D}).}$

$\hfill\blacksquare$

The standard statement of the class number formulas, namely

$\frac{h\pi}{2\sqrt{|D|}}=\sum_{(n,2D)=1)}\frac{1}{n}\left(\frac{D}{n}\right),\quad\frac{h}{2\sqrt{D}}\ln(T+U\sqrt{D})=\sum_{(n,2D)=1)}\frac{1}{n}\left(\frac{D}{n}\right)$

where $\left(\frac{D}{n}\right)$ is the Jacobi symbol, has the infinite
products in the fundamental identity replaced by “$L$-series” which Dirichlet
obtains by applying the law of quadratic reciprocity. Of course, we had to
avoid that. Our paper has described only the “easy” half of his full
investigation.

Historically it is worth noting that Dirichlet investigated the class number
because his proof of the infinitude of primes in an arithmetic progression
required that he prove that a certain $L$-series always have a non-zero value
and it turns out that that series has a sum that is a non-zero multiple of a
class number.

We also point out that Rogers’ proof [8] of Legendre’s Lemma cites a
proposition in Selberg’s proof [9] of the prime number theorem for arithmetic
progressions and the proof of that proposition also counts lattice points in
domains bounded by conic sections, as did Dirichlet above, a technique
originated by Gauss.

###### Proof.

(Proof of Teege’s Identity)

If we divide the second class number formula in Theorem 9.1 by the first and
apply the fundamental identity Theorem 8.3 we finally obtain _Teege’s
identity_ Theorem 3.2. ∎

##  10 Teege’s proof of Legendre’s lemma

Teege’s proof has two steps:

  * •

first he transforms the fundamental identity by assuming $\pm D$ to be a prime
of the form $4n+1$. This allows for simplifying cancellations in the numerator
and denominator of Teege’s identity.

  * •

then he further assumes that $p$ is a prime for which Legendre’s lemma is
false, and obtains a contradiction.

We write the _fundmental identity_ in the form

$\boxed{\sum\frac{1}{Q(x,y)^{s}}=k\displaystyle\sum_{(n,2D)=1}\frac{1}{n^{2s}}\cdot\prod_{\left(\frac{D}{p}\right)=1}\frac{1+\frac{1}{p^{s}}}{1-\frac{1}{p^{s}}}}$

where $Q(x,y)$ is relatively prime to $2D$ and $k=2$ if $D<0$, while, if
$D>0$, then $k=1$ and

$y>0,U\cdot(ax+by)>Ty$

where $T,U$ are the smallest positive integer solutions of $T^{2}-DU^{2}=1.$
Finally, $x,y$ run over all pairs of positive integers.

_Now to the proof that fills the gap._

First suppose that $D$ is a _positive prime number $p$ of the form $4n+1$_.
Then

$\sum\frac{1}{Q(x,y)^{s}}=\displaystyle\sum_{(n,2D)=1}\frac{1}{n^{2s}}\cdot\prod_{\left(\frac{p}{A}\right)=1}\frac{1+\frac{1}{A^{s}}}{1-\frac{1}{A^{s}}}\prod_{\left(\frac{p}{B}\right)=1}\frac{1+\frac{1}{B^{s}}}{1-\frac{1}{B^{s}}}$
(10.1)

where $A$ runs over all primes of the form $4n+1$ of which $p$ is a quadratic
residue, and $B$ runs over all primes of the form $4n+3$ of which $p$ is a
quadratic residue.

Secondly suppose that _the same prime number $p=4n+1$ is now assumed to be a
negative determinant $-p$ equal to $D$._ Then

$\sum\frac{1}{Q_{1}(x,y)^{s}}=2\displaystyle\sum_{(n,2D)=1}\frac{1}{n^{2s}}\cdot\prod_{\left(\frac{-p}{A_{1}}\right)=1}\frac{1+\frac{1}{A_{1}^{s}}}{1-\frac{1}{A_{1}^{s}}}\prod_{\left(\frac{-p}{B_{1}}\right)=1}\frac{1+\frac{1}{B_{1}^{s}}}{1-\frac{1}{B_{1}^{s}}}$
(10.2)

where the sum on the left-hand side runs over all non-equivalent forms of
determinant $-p$ , while $A_{1}$ and $B_{1}$ on the right hand side run over
all primes of the form $4n+1$ and $4n+3$ respectively of which $-p$ is a
quadratic residue.

However, it is well-known that for a prime number $A$ of the form $4n+1$ $-p$
is a quadratic residue or non residue of $A$ according as $+p$ is a quadratic
residue or non residue of $A$ which implies that the set of primes $A$
_coincides_ with the set of primes $A_{1}$ whence

$\prod_{\left(\frac{p}{A}\right)=1}\frac{1+\frac{1}{A^{s}}}{1-\frac{1}{A^{s}}}=\prod_{\left(\frac{-p}{A_{1}}\right)=1}\frac{1+\frac{1}{A_{1}^{s}}}{1-\frac{1}{A_{1}^{s}}}.$

However this is _not_ the case for the set of primes $B$ because here the
Legendre symbol obeys the equation
$\left(\frac{-p}{B}\right)=-\left(\frac{p}{B}\right)$.

_We now prove Legendre’s lemma by contradiction_.

Suppose that there exists _no_ prime number $B$ of which $p$ is a quadratic
non residue. Then $+p$ is also be a residue of all the prime numbers $B$, and
therefore _all prime numbers $B$ of the form $4n+3$_ must appear in
$\prod\frac{1+\frac{1}{B^{s}}}{1-\frac{1}{B^{s}}}$. On the other hand, if
since $-p$ is a non residue of all prime numbers of the form $4n+3$, then _no
prime_ $B_{1}$ appears in quantity
$\prod\frac{1+\frac{1}{B_{1}^{s}}}{1-\frac{1}{B_{1}^{s}}}$ which means that
_it has to collapse down to_ $1$.

We now divide (10.1) by (10.2). Then we see the that the products involving
$A$ and $A_{1}$ _cancel_. Moreover, if we multiply the numerator and
denominator on the left hand side by $s-1$. we obtain

$\boxed{\frac{\sum\frac{s-1}{Q(x,y)^{s}}}{\sum\frac{s-1}{Q_{1}(x,y)^{s}}}=\frac{1}{2}\prod\frac{1+\frac{1}{B^{s}}}{1-\frac{1}{B^{s}}}}$
(10.3)

where the primes $B$ are all primes of the form $4n+3$.

_This equation is the goal of all of the previous developments._

We now take the limit as $s\rightarrow 1$. _Teege’s identity_ shows that the
limit of the left hand side is the _finite number_

$=\frac{h(p)\ln(T+U\sqrt{p})}{h(-p)\cdot 2\pi}$

while the limit of the right-hand side is the infinite product

$\displaystyle\frac{1}{2}\prod\frac{1+\frac{1}{B}}{1-\frac{1}{B}}$
$\displaystyle>$ $\displaystyle\frac{1}{2}\prod\frac{1}{1-\frac{1}{B}}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\prod\left(1+\frac{1}{B}+\frac{1}{B^{2}}+\frac{1}{B^{3}}+\cdots\right)$
$\displaystyle=$ $\displaystyle\frac{1}{2}\sum\frac{1}{\prod B}$
$\displaystyle>$ $\displaystyle\frac{1}{2}\sum\frac{1}{B}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left(\frac{1}{3}+\frac{1}{7}+\frac{1}{11}+\frac{1}{19}+\cdots\right)$

where the denominators are all the primes of the form $4n+3$ and it is well
known that the series _diverges_ This contradiction proves Legendre’s Lemma.
$\hfill\blacksquare$

##  11 Teege’s refinement

We will prove that the prime $B$ of the form $4n+3$ whose existence is
guaranteed by Legendre’s Lemma _can be taken to be smaller than_ $p$. This
latter condition was never mentioned by Legendre and is original with Teege.

###### Proof.

We give an expanded arrangement of Teege’s own proof which uses infinite
descent.

By Legendre’s lemma there exists a prime number $B$ such that the Legendre
symbol $\left(\dfrac{-p}{B}\right)=+1$, which means and there also exists an
equation

$x^{2}+p=B\cdot b^{\prime}.$ (11.1)

for some integer $b^{\prime}$.

First, if $B<p$, then we are done.

Second, if $B>p$, then _we claim:_

$b^{\prime}<B.$

In fact,

$x<B\Rightarrow x^{2}+p=Bb^{\prime}<B^{2}+p\Rightarrow
b^{\prime}<B+\frac{p}{B}<B+1\Rightarrow b^{\prime}\leqslant B.$

so either $b^{\prime}=B$ or we have the strict inequality $b^{\prime}<B$. But,
if

$b^{\prime}=B\Rightarrow x^{2}+p=B^{2}\Rightarrow p=(B-x)(B+x)\Rightarrow
B-x=1,B+x=p$

and this means

$B=\frac{p+1}{2}<p$

since $p>1$ and this is a contradiction. Thus we _cannot_ have $b^{\prime}=B$.
This proves our claim. $\hfill\blacksquare$

Now we will apply Fermat’s infinite descent.

Since $x$ is even and $p$ is of the form $8n+1$, $b^{\prime}$ is of the form
$4n+3$, and so $b^{\prime}$ must have a prime factor $B^{\prime}$ of the form
$4n+3$, whether or not $p$ and $b^{\prime}$ have a common factor $p$, such
that $\left(\dfrac{-p}{B^{\prime}}\right)=+1$. Thus there exists an equation

$x_{1}^{2}+p=B^{\prime}\cdot b^{\prime\prime}$

where $B^{\prime}<B$ and $b^{\prime\prime}$ is again smaller than $B^{\prime}$
and of the form $4n+3$, from which we can in the same way deduce that there
exists a third prime number $B^{\prime\prime}$ of the form $4n+3$ and smaller
than $B^{\prime}$ for which $\left(\dfrac{-p}{B^{\prime\prime}}\right)=+1$. By
repeating this process we obtain a sequence of decreasing prime numbers

$B>B^{\prime}>B^{\prime\prime}>\cdots$

of the form $4n+3$ of which $p$ is a nonresidue and finally one smaller than
$p$. Therefore we have proven Gauss’ lemma that _there always exists a prime
number $q<p$ for which a given prime number $p=8n+1$ a quadratic nonresidue
_but with the refinement that at _least one of these prime numbers $q$ must
have the form $4n+3$._ ∎

We note that a very brief numerical search showed _no_ example in which $B>p$.

##  12 Rogers’ proof of Legendre’s Lemma.

We now add some remarks on Rogers’ paper [8]. He rectifies Legendre’s original
proof (1785) by using the following lemma of Selberg [9].

######  Theorem 12.1.

If $D$ is a nonsquare integer, then

$\sum_{p\leqslant x,\left(\frac{D}{p}\right)=1}\frac{\ln p}{p}=\frac{1}{2}\ln
x+O(1),$

where $p$ runs over prime values only.

$\hfill\blacksquare$

Although Rogers never addresses Legendre’s lemma directly, he uses Selberg’s
result and some elegant mathematics to single out the contribution of distinct
arithmetic progressions of primes to Selberg’s total sum and in particular to
prove the following theorem.

######  Theorem 12.2.

Let $D$ be a nonsquare. Then

$\sum_{p\leqslant x,p=4n+3,\left(\frac{D}{p}\right)=-1}\frac{\ln
p}{p}=\frac{1}{4}\ln x+O(1),$

###### Proof.

We sketch Rogers’ nice proof. Let

$s_{i}:=\sum_{p\leqslant x,p=4n+i}\frac{\ln p}{p},\quad\text{for}\quad i=1,3,$

and let $s_{i}^{+},s_{i}^{-}$ denote the corresponding sums over $p$ for which
$\left(\frac{D}{p}\right)=\pm 1$, respectively. First we note that Selberg’s
lemma and the Mertens’ classical result

$\sum_{p\leqslant x}\frac{\ln p}{p}=\ln x+O(1)$

together imply

$\sum_{p\leqslant x,\left(\frac{D}{p}\right)=-1}\frac{\ln p}{p}=\frac{1}{2}\ln
x+O(1).$

Now

$\displaystyle s_{1}^{+}+s_{3}^{+}$ $\displaystyle=$
$\displaystyle\sum_{p\leqslant x,p=4n+1,\left(\frac{D}{p}\right)=1}\frac{\ln
p}{p}+\sum_{p\leqslant x,p=4n+3,\left(\frac{D}{p}\right)=1}\frac{\ln p}{p}$
$\displaystyle=$ $\displaystyle\sum_{p\leqslant
x,\left(\frac{D}{p}\right)=1}\frac{\ln p}{p}=\frac{1}{2}\ln x+O(1)$

by Selberg’s lemma. A similar computation in which replacing $D$ by $-D$
leaves $s_{1}^{\pm}$ invariant while it interchanges $s_{3}^{+}$ and
$s_{3}^{-}$ and shows

$s_{1}^{-}+s_{3}^{-}=s_{1}^{+}+s_{3}^{-}=s_{1}^{-}+s_{3}^{+}=\frac{1}{2}\ln
x+O(1)$

Thus we have four linear equations in the four unknowns $s_{i}^{\pm}$, and
solving them we obtain

$s_{i}^{\pm}=\frac{1}{4}\ln x+O(1).$

The case $s_{3}^{-}$ is the equation stated in the theorem

∎

Now we let $x\rightarrow\infty$ in Theorem 12.2. This proves the following
result.

######  Corollary 12.3.

There exist infinitely many primes $p\equiv 3\pmod{4}$ for which
$\left(\frac{D}{p}\right)=-1$

In particular, if we put $D$ equal to a prime of the form $4n+1$ we reach our
goal.

$\hfill\blacksquare$

######  Corollary 12.4.

For every prime $a$ of the form $4n+1$ there exist infinitely may primes $b$
of the form $4n+3$ for which $\left(\frac{a}{b}\right)=-1$

$\hfill\blacksquare$

Of course, this is (a strong form of ) Legendre’s Lemma.

Rogers actually proves a much stronger result which we do not need; we have
extracted only a special case from the first part of his investigation.

It would seem that Rogers’ paper has suffered the same fate as Teege’s.
Although it appeared in a distinguished journal, the Mathematical Reviews and
Zentralblatt fur Mathematik und ihre Grenzgebiete only reproduce the
introduction in his paper, while there is no comment on the mathematical
content and, with the exception of Lemmermeyer’s book [6] we have not found
any references to his paper in the almost half-century since its publication
in any of the papers/textbooks that deal with the first proof of the
reciprocity theorem.

We remark that if one writes out in full detail the proof of Selberg’s lemma,
it is almost as long and complicated as the proof of Dirichlet’s class number
formulas. Nevertheless, Rogers refers to it as “…a simple lemma…”

Finally we note that Selberg used his identity to prove the non-vanishing of
an L-series just as Dirichlet used his class-number formulas. But, up to now,
the latter have had an enormous influence on the subsequent development of
algebraic number theory and commutative algebra while the former, as of yet,
has not.

##  13 Retrospect

Apparently Teege wrote his first attempted proof of Legendre’s Lemma [10]
before 1914, although it was not published until 1920, perhaps because of the
intervention of WW I. But it contained an important error which he repaired in
his second attempt [11] which was published four years later. This second
paper also contained the first appearance of Teege’s refinement (as he,
himself, writes with pride.) The magazine’s editor writes in a footnote that
the manuscript of Teege’s second paper was ready for printing although the
author was deceased. (The paper appeared with a dagger, $\dagger$, along side
the author’s name). Then, as noted earlier, unacountably his paper was later
forgotten. Given Teege’s identity (which Teege never explicitly sets out but
is an easy consequence of his reasoning: “…by a stroke of the pen” as G.
Watson wrote of an implicit identity of Ramanujan) his proof of Legendre’s
lemma is quite elegant and deserves to be known.

We see that the proof of Teege’s identity, itself, is long and complex because
it depends on (the easy part (!)) of Dirichlet’s class number formulas and the
proof of the latter is quite involved. Nevertheless, we are compensated by the
fact that its use in the proof of Legendre’s Lemma _is_ elementary and quite
beautiful and elevates Legendre’s attempted proof to the list of complete
proofs of the quadratic reciprocity law.

## References

  * [1] O. Baumgart, The Quadratic Reciprocity Law, A Collection of Classical proofs, second edition Birkhauser, New York, 2015.
  * [2] L. Dirichlet, Recheerches sur diverses applications de l’analyse infinitesimale a la theorie des nombres, Crelle  19,21 (1839,1840), 324-369; 1-12 and 134-155.
  * [3] L. Dirichlet, Lectures on Number Theory Translated by John Stilwell American Mathematical Society, New York, 1999.
  * [4] E. Landau, Elementary Number Theory, Chelsea, New York, 1958
  * [5] A-M. Legendre, Theorie des Nombres, Chez Firmin Didot Freres, Libraires, Paris, 1830.
  * [6] F. Lemmermeyer, Reciprocity Laws. From Euler to Eisenstein, Springer, New York, 2000.
  * [7] G.B. Mathews, Theory of Numbers, Chelsea, New York, 1961
  * [8] K. Rogers, Legendre’s Theorem and Quadratic Reciprocity, Journal of Number Theory  6 (1974), 339-344.
  * [9] A. Selberg, An elementary proof of the prime-number theorem for arithmetic progressions, Canad. J. Math. 2 (1950), 66-78.
  * [10] H. Teege, Uber den Legendreschen Beweis des sogenannten Reziprozitatsgesetzes in der Lehre von den quadratischen Resten und seine Vervollstandigung durch den Nachweis, dass jede Primzahl von der Form 8n+1 quadratischer Nichtrest unendlich vieler Primzahlen von der Form 4n+3 ist, Mitt. Math. Ges. Hamburg 5 (1920), 6-19.
  * [11] H. Teege, Richtigstellung eines fruheren Beweises fur den Satz, dass es fur jede Primzahl p von der Form 4n+1 unendlich viele Primzahlen von der Form 4n+3 gibt, von denen p quadratischer Nichtrest ist un Herleitung des Satzes, dass mindestens eine unter ihnen kleiner als p ist. Mitt. Math. Ges. Hamburg 6 (1924), 100-106.
  * [12] S.H. Weintraub, On Legendre’s Work on the Law of Quadratic Reciprocity, Amer. Math. Mo. 118 (2011), 210-216.