-
The Law of Quadratic Reci-procity
An Honors Thesis ( ID 499)
By
Robert C. Matney
Thesis Director
Dr. Padmini Joshi
(j)
£ ·f
fJ·
__~J_c~L~~_{_I~
__
~_~_~~(~4:/~·~~_~~________ (advisor's
Ball State University
Muncie, Indiana
May, 1977
signature )
One of the most oriGinal, and nerhaps one of the
rr~ost
consenuential, discoveries in the eichteenth century in the
field of number theory is the Law of
(o~uadratic
Heci proci ty.
In the lanGUage of the period, if there exists an x such
r)
that
x~-p
is divisible by a, then n is said to be a Quadratic
residue of a; if no such x exists, p is said to be a ouao.ratic non-residue of q.
The Law of Quadratic decinrocity
states that, if n and a are odd nrimes, then:
In
(i)
each is a residue or non-residue of the other if
at least one of them is of t1:1e form 4N+l, and
(ii)
is a residue or non-residue of q according as
n is a non-residue or residue of p when both are
of the form 4N+3
180~\,
D
Lef!'endre develo;)ed a symbol to describe the rela-
tionship between nand c.
The symbol is defined as follows:
J1'or any number p and any odd nrirr.e q,
(E)
q
={
6
-1
if "D is a ("1uadrrltic residue of q
if a divides p
if 1) is a ouadratic non-residue of q
Stated in terms of the Legendre syrrbol, the Law of Quadratic
Reciprocity becomes:
For any distinct odd priITles n 8.nd q,
0-1 a-I
(2)(~)
= (-l)~ ~
CI
P
In 1783, .:":uler pave (C"ryroof of this theorem, but he
made an 8.ssUIDlJtion which, al tho'J.,f"h true, w[-:s not 1)roved to
be valid until rruch later.
LeFendre also fave a "Droof, but
he too made certain 8.ss1.lIT'Dtions which had not been nroved.
2
In 1949, however, the assumptions made by Euler were proved
by elementary means, and his proof can now be considered to
be complete.
The first complete proof of the law was given by Gauss.
He accomplished this in 1796, when he was only eighteen years
old.
Of this proof, Gauss wro:te: "For a whole year this
theorem tormented me and absorbed my great.est effoTts until,
at last, I obtained a proof • • •
He was so fascinated by
the result, which he called the "gem of higher arithmetic,"
that he went on to give six more proofs.
Since Gauss's time,.
the theorem has been proved in at least fifty different ways.
A large number o·f these are given by P. Bachmann in his work
Niedere Zahlentheorie.
In this paper, we shall consider seven of the numerous
proofs.
These are:
Gauss's First Proof
Gauss's Third Proof
Gauss's Fifth Proo·f
Lange's Second Proof
Frobenius' Proof
Eisenstein's Proaf
Lehmer's Proof
We shall also consider some interesting resul.ts which follow
from the discussion of the theorem.
Some of the proofs have
been altered slightly to provide for consistent notation
throughout the paper.
Before we can examine the proofs, we
will need certain preliminary results.
Many of the proo;fs
rely on Gauss's Lemma, so we shall first consider this very
important lemma.
G. Archibald. 2
The proof given here was presented by Ralph
3
Lemma 1 (Gauss t s Lemma):
Let p be an odd prime and let (p,q)
= 1.
If u denotes the
number of least positive residues ~reater than
¥
modulo p,
of the P;l integers in the set
p-l)
q, 2q, 3q, • • . , ( 2 q
(A)
then
PROOF:
Since (p,q)
= 1,
none of the intef'ers in (A) is congruent
to 0 and no two belong to the same residue class, modulo p.
Let
be the least positive residues greater than
¥'
and let
b l , b 2 , by • • • , b v
be the least nositive residues less than
¥.
Then
p-l
u + v =2
Each of
(B)
is positive and less than ~ , and each is in a different
residue class,
b.
J
modulo p.
=p-a.
1
for some j such that
Suppose that
(mod p)
1
~
j
~
v
4
Let a. and b. be the least residues, respectively, of Aq
J
~
and Bq, modulo p.
o :: a.
~
-Then
+ bj
~
Aq + Bq
Since A and B are positive and
A + B.
(e)
~
(A + B)q
'0-1
S
~
,
p does not divide
Consequently, the P'21 integers
p-al , p-a 2 , • • • , p-au ' b l , b 2 , • • • , b v
'0-1 and are, therefore, an arrangeare distinct numbers ~ ~
2
ment of the integers
1, 2, 3,
...,
p-l
2
The product of the integers in (e) is
• • •
Also,
..
(a) ( 2q) •
(D)
modulo p.
From (e) and (D), we f;et
'0-1
(P;l)! (q) 2
Ii (-1) u(P;l)!.
(mod p)
and so
p-l
(q) 2
_ (_l)u
(mod p)
Since
p-l
(9.) _ (q) 2
p
We get
(_l)u
(mod p)
5
and Gauss's Lemma is nroved.
Using this result, we can develop the followinp lemma, which
is necessary for Gauss's Third Proof and Eisenstein's Proof.
Lemma 2:
If P is an odd prime and q an odd integer such that (p,q)
then
(9.)
P
= (_l)M
where
PROOF:
If k is :in integer {Teater than 0 then
kq
= nQ
+ r
where
o
and Q is an
~ r
<
p
inte~er.
Thus
So
rJ
[-kqpJ = [ Q + P
since
=
.
Q
=1
6
Then
= p[~]
q
= p[~]
2q
(I)
+ r1
..•
•
•
•
•
•
•
(P;l)q
+ r2
•
.•
= p[(P;l)q]
+ r p-1
2
where
r.<
1.
O~
p
and
1
~i
:s;
n-1
2
Let
P
=1
A
=
+ 2 + 3 + • • • + P;l
2
=~
=
and let
• • • + au
B =
• • •
with the a. 's and b.'s the least
J
1.
in Gauss's
~ositive
residues as given
Le~ma.
If
M = [~] + [~] + • • • + [(P;l)~]
then
=A+ B
from Gauss's Lemma.
(1)
Pq
Since
=
By adding the eouations in (I) we e-et
pM + A + B
7
is an arrangement of the integers
1, 2, 3, •
p-l
. · , 2
we see that
(2)
P
= up
- A + B
Subtracting (2) from (1), we
= p(M
p(q -
1)
p(q -
1) s p(K - u)
~et
- u) + 2A
Thus
(mod 2)
Since p and q are both odd,
q
-
1
-
0
(mod 2)
p
-1
(mod 2)
- u
(mod 2)
and (3) becomes
0 ::
1,~
or
M == u
(mod 2)
and thus
By Gauss's Lemma,
Therefore
which corepletes the lemma.
The final lemrra which we will prove is necessary for the
discussion of Gauss's First Proof.
The proof given here
follows a presentation of Dirichlet and Dedekina. 3
8
Lemma 3:
If
0
=
4M + 1, there exists an odd nrime pt < q for which
c})
=
-1
PROOF:
He consider two cases; either q
= 8N
+ 5 or q
= 8N
+ 1.
Case 1:
If
q
= 8N
+
5 then
0+1 == 3
(mod 4)
~
can be
Consequently, since not all Tlrime factors of q+1
2
congruent to 1, mod 4, q+l must have a prime factor
2
pt
- 3
(Irod 4)
That is,
q + 1 _ 0
(mod p')
or
q _ -1
(mod u')
So
"0'-1
(~)
= (~)
= C-l)~ = -1
p
P
since p'
=3
(mod 4).
Therefore
(~)
= -1
p
Case 2:
Let a
=
8N + 1.
AssuJYe
q
is a quadratic residue of every
prime which does not exceed 2m + 1
< q.
Then q is also a
quadratic residue of every positive interer which is a
product of numbers, each
~
2m + 1.
9
If
• • 2m(2m + 1)
then
x
2
- q
solution.
(mod M)
'Then (k,M)
is solvable.
=1
Let
=k
x
be one
a.nd
• •
(A)
--
k(k 2 _ 12 )(k 2
-
0
(k 2
.' .'
22) •
m2 )
(k + m) (k + m - l)(k + m - 2) • • • (k - m)
(mod M)
since the product in (A) is divisible by M.
=1
since (k,M)
M
=
Therefore,
and
(m+l) (m+l)2 - 12)(m+l)2 _ 22) • •
the product
(B)
a ==
g _ 12
-1 .
(m+l)2 _ 12
m+l
q - 22
• •
(m+l)2 _ 22
•
.
q - m2
(m+l)2 _ m2
is an inte{':'er.
On the other hand, a is not an integer if
if
m <J<f<m + 1
m2
then
<
q
m
=
[JCf]
since
< (m + 1)2
and all of the fre.ctions in (B) are proper fractions.
Since
c ~ 17, we see that
4q
<
q2 - 2q + 1
8 < (q - 3)2
=
and
(q _ 1)2
Thus
2
ro: <
2m + 1
q - 1,
<
2 vq + 1
< q , and (when
q
=
=
[Vcr]),
q
Thus, for m, our orip'inal assumption is invalid.
if
m
8N + 1, there exists a prime
p' < 2¥iQ + 1
Consequently
for which
10
,From these two cases, if
prime p ,<
q
(~)
p
q
= 4M
+ 1, there exists an odd
for which
=
-1
which completes the
~roof.
With these results, we are now ready to consider the
reciprocity law.
We shall first exarrine Gauss's First Proof,
which he presented in his Disquisitiones Arithmeticae. 4
This
proof uses mathematical induction to achieve the desired
results, and is quite comnlex in nature.
Gauss considered
eight cases, but these can be combined to f'i ve rise to two
mutually exclusive cases.
The proof given here follows the
presentcdion {'i ven by Dirichlet and Dedekind. 5
GAUSS'S FIRST PROOF
Sirce we will be using mathematical induction, we will
uroceed as follows.
If the theorerr. is assumed true for all
pairs of odd primes less than a prime
q
then it will be
shown that the theorerr Vlill be true for a cOJTlbined with any
prime less than q.
Since
(~)
=
(~)
= -1
we see that the
theorem is true for the tvvo smallest odd primes.
In the course of the proof we use the fact that, on the
assurr.ption that the theorem is true for every pair of odd
primes less than a, the statement
-
P-l Q-l
(1)
(~)(~) = (_1)-2-"""2
11
will be true for all odd inteFers P and Q such that (p,Q)
and all prime factors of P and Q are less than q.
=1
The Law
of quadratic Reciprocity will be called the Theorem, and
statement (1) will be referred to as the Generalized Theorem.
The p-enera1ized theorem is proved by using the truth of the
original theorem for all nairs of odd
pri~es
one of which is
a factor of P and the other of Q.
We begin by assuming the theorem is true for every pair
of distinct odd primes less than q.
we see that, for
q
= 4M
Since
+ 3,
9-1
(=2)
q
= (-1) 2
=
(E)
q
_(E)
q
and either
(E.) = 1
(A)
q
For
q
=
(.=E.)
or
q
=1
4M + 1,
q-1
(::E)
q = (-1)
2 (E)
q
=
(E. )
q
and either
(E)
(::E)
1
q = q =
(B)
or
(c)
These can be combined into two cases:
At least one of (E)
and (::E)
is 1;
q
q
(A and B)
12
(ii)
=
q
Consider case (i) •
(E)
q
and (E)
4M + 1
q
=
-1 •,
(0)
Either
=1
Or
(.=.E)
q
=1
Or
~rove
In order to
the theorew, we need to show that
p-1 q-l
2 (E)
(.9.) = (-1) 2
(2)
p
q
Let w be either 'P or -p, so tha.t
(~)
= 1.
There are two
distinct solutions x of the congruence
x 2 :: w
(mod q)
o< x <
q
such that
If
Xo
is one solution, the other is
q - xO.
that one of the two solutions that is even.
o< e <
Let e be
Then
q
and
e2 - w
(3 )
=
fq
then w > 0 ,i.e. w = p. Then p - e
2
q (p - e ) , which is impossible.
So f > O.
If
f < 0
and
I
Consider the two 3.lternatives:
First, suppose
p~f.
From (3) we see that
2
>0
13
and since fq is a Quadratic residue, mod p, we have
(~) - (-f-)(~)
Iwl
Iwl
Iwl
1 -
and so
(';1)
=
(Twi-)
Now
0
<f < q
This is obvious if
fq
o < Iwl < q
=
>
p
e2
~
(q_1)2 + p
=
q
2
-
2q + 1 +
<
=
q2
-
2q
-
w
When
-
-p
Sinc e (f, w) = 1 ,
0
<
f
, we have the f:enera1ized theorem
=P
then
a(q-1)
Iwl-1 f-l
, we have
1
w-l f-1
f ) = (-l)~ ~
( -;w;-
=
=
p
and
w
w
q
(l!l)
= (~)
=
f
IWI
If
If
w
(/~' )(lfL) = (-1)
(5)
0 •
=
f < q - 1.
'rhus
w
-p , from (5) we have
2
~
<
q , a.nd
14
=
-w-l f-l
(-1)~ -2-(=!)
=
-\,\,+1 f-l
(-1)~ -2-
=
f
-(w+l) f-l
(-1)
2
-2-
w-l f-l
= (_1)2 2
In both instances
w-l f-l
f)
( 'TWT
=
(_1)--2- --2-
Since e is even, from (3) we have
-w - 1 ~ fa - 1
(mod 4)
Therefore
=f
_ W~1
q
;1
(wod 2)
Letting
f' = -y
f-l
we
and
at
a-I
= 2""
ha.VE!
fa - 1
=
(2f t + 1)(20' + 1)
= 4f'qt
+ 2(f' + q')
and so
fa-l
~
=
2f'q' + f' + q'
rrherefore
w+l - --2f-l + ~
q-l
- --2-
(mod 2)
15
f'
ul ti plying by
vv-l
-2-
,
w-l w+l
- ""'2
-y -
we get
w-l f-l
w-l <1-1
-2- -y + -y 2'"
(mod 2)
~;nce w-l
d w+l are consecu t~"
........
ve"~n t ep'ers,
2- an . - 2
w-l w+l
is
-2-~
even, and thus
"'1-1 f-l
VI-I a-I
-2-~--Y2
(mod 2)
Therefore
w-l 0.-1
(-1-) = (_1)-2-- ~
(6)
w
whether
w = p
or
w =
-n.
The theorem '.Nill be established
for p and q if we show that (2) is valid.
If
w
=
from (4) and (6) we get
p ,
"0-1 \1-1
(~)
= (L)
= (-l)~ -y
p
n
Since
statement (2) is established for nand q when
If
w = -p , then since
we see that
(2) = (-1)
q
q
Then, from (4) and (6),
n
f
(.:.)
= (-)
=
p
p
-p-l q-l
2
2
(-1)
w
=
p
•
16
-(11+1) q-l
=
(-1)
2
2
2±1 !l::!.
= (-1) 2
2
0-1 0-1
0-1
= (-1) 2
2 (-1) 2
""""--
lJ-l a-I
..lo...-
~
= (-1) 2
2 (-1)
q
lJ-l a-I
= (-1)~ ~(£)
q
= -lJ
This gives us (2) when
w
for case (i) when (f,p)
= 1.
and corT'l)letes the proof
Consider now the instance where
f
p/f.
Then
= wf 1
where f1 is odd and
f1 <
Frorr (3) , we see that
q
w/e
i.e.
e
=
weI
,
where e is
1
even, and we get
-I
e 1 2w
(7)
ConseouHnt1y
1
(8)
w{f1
=
= flq
."
(-f,q)
Iw
Also (f ,e ) = 1 and we have
1 1
(~)
and
/.1 r
Now
tie may, therefore, assume the truth of (1), the e:enera1ized
theorem for
Iwl
and If~;
that is,
17
If
p
w -
then
fl
>
and
0
f
= (-1)
l
-1 p-l
2
2
From (8) we pet
and
(¥:-)
1
=1
Thus
= (-1)
f l -1 p-l
2
2
Conseauently,
f1-l
= (-1)
2
:£::!
p-l
fl+l .E::1.
2 (-1) 2 = (-1) 2
2
From (7), we see that
Since fl and
are both odd, both cannot be of the form
q
4N + 1 nor can both be of the form 4N + 3.
=q;1
fl~l
Since
(E)
q
= 1,
Conseouently,
(mod 2)
(2) is valid; that is,
p-l q-l
(~)
p
If
w
=
=
(-1) 2
-p ,then
fl
2 (E)
q
< o.
Since
Ifll
and we can assume that (1) holds for Ifll
<
q
and
and p.
p
'rhus
<q
,
18
-f1 -l n-l
= (-1)
2
2
fl+1 p-1
=
(-1)
2
2
From (8),
and
1 =
(lf~l)
so that
= (-fp 1 ) = ( .9.P )
(lfi-)
p
fl+1 p-l
2
2 (-E--)
If ,
1
=
(-1)
=
fl+1 p-1
(-1) 2
2 ( -1 )
=
fl+1
(-1) 2
=
(-1)
TtlT
E=!
-f1 -1
2 (-1)
2
fl+1 p-1 + fl+1
2
2
2
Since
=
1
(!! )
q
(E)
= (-1)
q q
we see that
(E)
q
=
(-1)
q
=
q-1
(-1) 2
19
As we showed above,
fl+l _ 0-1
(mod 2)
2
2
Thus
(-4-)
=
p
p-l q-l
~
(.9.) == (-1) 2
2 (-1) 2
p
and so
p-1 9-1
2 (~)
q
(.9.) == (-1) 2
p
This cOIDDletes the proof of
(2),
and establishes the theorem
for case (i).
Consider now the more difficult case (ii).
Let
q
= 4M
+ 1
and
there exists a prime
( q)
P'"
If
(~)
q
=1
p
<q
, with (~)
p'
<q
such that
= -1.
From Lemma 3,
= -1
then from case (i) the theorem could be used,
and
This, however, contradicts our assumption that
Thus,
and the theorem is valid for pt and 0; that is,
P.' = -1.
(q)
20
-
.E.:=1
(~)(~)
= (-1)
P
q
2
q-1
2
In order to complete the proof, we need to show that, if
p
<
f.
q , p
p'
(~)
, and
=
<%) = -1
-1 , then
; that is,
Since
ppt is
quadratic residue of q, and there exists a positive
'3.
even integer e such that
e 2 _ pp'
(10)
Here,
< q.
f
First, suppose
=
fq
We now consider three possibilities.
(f,pp')
= 1.
Then, from
(10),
<ljff) = 1
Likewise,
Since (1) is applicable to If I and pp', when
f-l "01)'-1
(p§')
When
f
<
= (p~,) = (_1)--2--
I
2
0 ,
Q)
<PPT
=
< f)
pp'
= (-1)
=
=
-1)( If I)
(pp' pPT
pp'-l
2
-f-1 "Op'-l
(_1)---2-
-f+l pp'-l
2
--y
(-1)
2
f >0
we have
21
=
-(f-l) pp"-l
(-1)
2
2
f-l pp'-1
= (-1)~
2
But, from (10),
-pp _ fq
Since
q
=1
(mod 4)
4),
we see that
"'Op'+1
2
np'+l
2
(mod
f _ -pp'
(mod
4)
and so
f-l
-2Also,
(mod 2)
pp' 1
pp'+l
2- and
2
are consecutive integers, so their
product is even.
Consequently,
q-l P12'-l
2
2
=0
==
12E'+l 12E'-l _ f-l 1212'-1
=2"
2
2
2
(mod 2)
Therefore
(p§t)
=
q-l 1212'-1
2
(-1) 2
In the first instance, then,
Second, suppose that
where
(fl,p)
= 1.
p'l f
p'e 1 2 _ p
.
Consequently, since
-.
(11) gives us
(9) is true.
and
Also, we have
(10) gives us
(11)
=1
=
(f,p)
e
=
= 1.
Then
f
=
p'f
1
p'e 1 ' and equation
22
p'e 2
( If 11 ) =
I
that is,
~
(If
.. 1 )
1:
.,
.,
=1
(~)
p
Now,
(p§ i )
Since (1) holds for
(~)
If11
and pp', we have
f
= (-1)
-1 ppt_l
12
2
(~)
Ifll -1 pp'-l
= (-1)
Thus
If
f1
>
0
...,'/e have
fl-l ppt_1
(~)
pp
= (-1)
2
f
=
2
(~)(~)
p
P
-1 pp'-l
(-1) ~
2
(E':)(g.)
p
p
Since the theorem is valid for p and pI,
f -1 ppt_1
pt_1 p-1
l
(~) = (-1) 2
2 (-1)
2
2
pp
=
If
fl
<
0
fl-l pp'-l + pt-1 p-1
(-1) 2
2
2
2
we h8.ve
2
2
23
-f:±-l E"P'-l
2
2
= (-1)
(-~)
(P%r)
(~) ( T ) (f~1)
1')-1
-f -l pp'-l
1
+ -2 + ~
2
2 CE.:) (~)
2
= (-1)
p
P
p-l
-f1 -l pp'-l
+ -2+ lcl~
2
2
2
2
= (-1)
-f1 -l EE'-l
E-1 E'+l
+ 2
2
2
2
= (-1)
From (11) , we know that
fl
=
-~
(mod 4)
and
fl;l _
"0;1.
-
-lJ
f q (mod 4)
1
., that
is,
(mod 2)
Therefore,
f -1 pp'-l
p'-l
= p+l
E2:!.
(PE'-l
+
2
~2+22-2
= P;l.p ,.E;l
_ 1
= P;l
P;l
=0
E.2)
2
(mod 2)
Since
fl+1 -~
2 := 2
(mod 2),
fl+l 1')1)'-1
;.
+ p-l p'+l
2
2
2
2
~
= p-1
2
-
PE'+P'
2
In both cases,
-
P
fl> 0
(p§') =
= p-l
2
EP'-l + p-l p'+l
2
2
2
-
,.p-l p+l = p-1 p+l
2
2 2
2 - 0
or
fl
<
0
(mod 2)
,
1
and (9) is true.
Since we did not use any slJecial "OroDerty of p', we can
interchange p and p' to nroduce the same result when
--------.---- -
..- . - -----
prf
24
and
( f , 1) ,) = 1.
Finally, let
np'lf; that is,
f = p~'f2.
Then
and
nn'e
~
22 - 1 =
(12)
c
From this,
and
-f 0)
1= ( ~
Thus
Since (1) holds forlf 2/and 1)'0', we have
If 2 1-1 npt-l
= (-1)
2
2
\vhen
f:2
>
0,
(p§')
= (-1)
pp'-1 + f -1 YJp'-l
2
~
2
f2+1 P12'-l
2
= (-1) 2
V'Ihen
f2
<
0
,
(p?)
= (-1)
'Op'-l + np'-l + -f -1 P12'-1
22
2
2
2
f2+1 P12'-1
2
2
= (-1)
From (12), since
f 2 a s f2 ~ -1
(mod 4),
._--_.__
...--- ....
_------....
e = PP'e 2
25
,-
f2~1 _ 0
(mod 2)
and therefore
f2~1 Pp;-l _ 0
(mod 2)
Consequently,
(p§d
= 1
and (9) is valid.
Irhis comuletes the proof of (9), and establishes the theorem
for case (ii).
Since in both cases (i) and (ii), the first
assu.IT:11tion is true, and since the theorem is true for the two
smallest odd nrimes, by com11lete mathematical induction, the
theOreI11 is tru,e for all odd 1)rimes.
'rhis comnletes (}auss' s
First Proof.
Gauss recopnized ths,t this first proof was extremely
corr.nlicated, and he continued to work on the theorem.
develoned six more proofs during his lifetime.
will be discussed here.
Gauss's 'rhird Proof.
by Harriet Griffin. 6
Two of these
The next proof we shall discuss is
The exnosi tion followed here is p,;i ven
(}AUSS'S 'rIURD PROOF
Let
1')
and
q
be distinct odd primes, with
q
less than 11.
Let
[%J +
N= [*J +
lVi
,-
=
He
[~J + • • •
[~J
+ [(
+ • .' • +
P;l )%J
[(q;l)~J
26
Then
2::l
q-l
2
(2) (-9.) = (-1)
q p
2
PROOF:
If
k
then
=
is
[SJ
p
1, 2, 3,
.-
• •
p-l
,
2
Also,
not an integer.
[%J = 0
since
,
q <p
and
[~J
~ 1.
p
If
[~J
80
= s
that
with
kq = sp + r
o<
r
<
p
then
(k+l)q
=
<
1
sp + r + q
But
r+q
p
Therefore
s
<
-
(k+l)q
p
<
8
+ 1
Further,
(n-l)q
2p
-~----------,---
=
=
pq -
p + p -
2p
q
27
=
=
since
TJ
a
-
< 2p
[~J =
•
n(q
2p
1)
+
1)
-
Ct
21)
q
=
-2
1
+
p
-
q
2p
q-1
2
Assume that, for
s
[(k+~)gJ
and
k <p-1
2
=
,
s + 1
'rhen
!f9.
p
< s + 1 < (k+1 )..9p
'rhus
k
< (s+l)p
<k
q
+ 1
'rherefore,
is the last term of IV; such that
[k~J =
s
'1:'hen the 12st terrn of
Rnd
I~" that has the value
s-l
30, for
the number of terms having the value
0-1
"""'2
'rheref ore,
11'1
=
[% ] + [~J +
• • • + [ ( P;l ) ~ ]
is
is
[S~J
·
28
=
O-([*J - [~J) +l-([~J - [*J)+
+
+
==
s.( [(S+~)PJ - [~J )+
q-l (n-l
[(¥)*J)
2
•••
2 •
-[~J
- [~J
• • • - [C"!21 )~J
+
CJ 21) (q;l)
== _ N + (0-1 ) ( q -1 )
2
2
Therefore,
M + N
By Lemma 2,
and
'rhus
p-l q-l
= (-1) 2
2
and so
which nroves the theorem.
Eisenstein's Proof, which will be nresented later,
uses an annroach which is similar to this proof.
His proof,
however, makes use of a geometrical interpretation of the
numbers Mana. N.
-
The next nroof 'Nhich we shall consider is Gauss's
Fifth Proof.
The nresentation piven here is an adaptation
29
-
of the proof y:hich
W2.8
r'i ven by USDensky and Heaslett. 7 The
notatijn has been altered for consistency.
G.~USS'
S :PIFTH YiWOF
Let p and q be distinct odd primes.
'rhen
-0-1 0-1
(~)(~)
= (-l)~ ~
q
p
PROOF:
By Gauss's Lemma,
and
where
.~
is the nlJ.Ir."ber of elements in the set
(A)
q, 2q, 30,
• •
whose least Dositive residues modulo
~
2
,
are greater than
and v is the number of elements in the set
p, 2p, 3p, • • •
,
(a;l)p
whose least positive residues moclulo
%.
p
C1
are greater than
Then
The least uositive residue of any nurrber modulo p is either
o
(f)
(f' )
or one of the numbers
1, 2, 3,
n-l
...,~
. . . , p-l
30
-.
Similarly, the least T)osi ti ve residue of any number modulo a
is either 0 or one of the nrunbers
(F)
1, 2, 3,
(Ft)
q+l
0+3
~
2 '
2
. . . , q-l
. . . , 0-1
'
2
Correspondingly, the numbers
(c)
1, 2, 3, •
..,
none of which is (livisible by p and
be distributed in eight classes
Class 1:
9.S
Class 2:
follows:
Let their nUJl;ber be a.
Those numbers whose least nositive residues mod p
arEl in (f) and mod q in (F').
Class 3::
Let their nl.unber be b.
Those numbers whose le8st ';osi ti ve residues mod 1J
on.
are in (f') En<1 mod q in
4:
Those nlmbers whose least
Let their number be g.
~ositive
c:cre in (ft) ,nd mod a in (F').
Class 5:
mod
simul tsneously, C2n
'1'hose numbers whose least Dosi ti ve residu.es wod p
are in (f) <ind mod a in (F).
Class
(l
Those multi 1)le8 of
p
are in (ft).
(1
residues ITod D
Let their number be d.
wllose lecwt -positive residues
'1'hese 8,re the elements of (A).
Thus their nurr:ber is u.
Class 6:
Those multinles of
mod. p are in (f).
Class 7:
whose le8.st posi ti ve residues
.
n-l
Their number lS ~ - u.
Those multi1Jles of
mod a are in (F').
r
1J
whose least positive residues
rI'hese Cl.re the elements of (B).
Thus their number is v.
Class 3:
Those multi ')les of p whose least nosi ti ve residues
mod q are in
(F).
Their nurrber is 9-1 - v.
--------------_ -._---------..
2
31
4, and 7 comprise all of the numbers in (0) whose
Classes 2,
(F').
least Dositive residues mod q are in
(F'),
residue in
If r is a given
these numbers are
r, q + r, 2q + r, • • • (P;3)q + r
and the ineQuality
or
tq
is true for
< (p-3)q
-
t
2·
n-3
=~
+ (q + q-l _ r)
2
but not for
p-l
t
= -r
·
Since r can
n-l values, and each r gives p-l
take on ~
2 numbers, the number
of members in Classes 2, 4, and 7 is
number is also
( JJ)
b + d + v.
So
~
and q in the discussion above, we get
the numoer of members in Glasses
g + d + u
3, 4, and 5, which is
= p-l .9..::!
2
2
To each number x in (C) which is also in Class
(D)
But this
p-l ..-.a-I
b + d + v = .-....2
2
By interchanging
(2)
p-l.q-l
2
2 •
l?9.:!:1 E9.:!:l
2 '
2 '
...,
3 in the set
pq - 1
there corresponds the number pq - x whose least positive
residues mod p are in (f) and mod q in (F'), and vice versa.
Therefore in Class
3 there are exactly as many numbers as
there are elements of (D) whose least
~ositive
residues
32
(F').
mod p are in (f) and mod a in
-.
Sets (C) and (D) together form the set
1, 2, 3, • • • , pq - 1
(E)
Consequently, the number of members in Classes 2 and 3 is
the same as the nurr:ber of elements of (E) whose least positive
residues mod p are in (f) and mod q in O~,).
But to any pair
of such residues there corresponds a unic:ue number in
(E),
and since the nUIr:ber of pairs is P;l q;l , there are the
same number of elements in Classes 2 and
p-1 q-1
b+g=T 2
Adding (1) and (2) we get
b + g + u + v + 2d
= 2(P;1
So
p-l 0-1 + u + v + 2d
2~
or
u + v + 2d =
p-1 q-1
2
2
Thus
_ p-l q-1
u + v =
~ ~
(
mod 2
and so
(_l)u+v
By Gauss's
Le~~a,
=
lJ-l a-I
(-1) 2
2
~~
)
~)
3.
That is,
33
Therefore
and the theorem is proved.
These three proofs by Gauss show the fascination which
the theorem held for that ewinent ma.thematician.
Gauss was
not the only scholar to devote considerable effort to proving
the theorem.
of three
The next proof we shall consider is the second
~iven
by Lange.
due to Barnard and Child.
The diagrammatic illustration is
8
LANGE'S SECOND PROOF
If p and,
unless
0
are unequal oda, primes then
p ~ q ~ 3
(mod 4) , in which case
(~) = _(So)
q
p
PROOF:
Without loss of eenerality, we can aSSUJIle that
p
<
q.
Consider the sets
. .'
A
= {q,
2q, 3q,
•
,
(P;l)q}
B
= {p,
2p, 3p, • • •
,
(q;l)p}
represented by the uoints of division of two scales set
edge to edge.
34
-
q
I
f 1?
sp
- --
I
p
(¥)q
rq
2p
- - I
(Q;3)p
(s-l)p
t 1
(Q;l)p
Let u be the number of elements in A with negative absolute
<
q ,
between two consecutive members of B.
If
least residues, modulo D.
(s-l)p < rq
Since
p
each rq in A lies
< sp
with
and
If we let rq,
1
.s:
r
~ p-l
2
(s-l)p, and sp be the points A, B, and C,
respectively, then
IBAI
and
ICAI
are the least positive
and numerically least negative residues of rq, modulo p.
Thus the absolute least residue of ra, modulo p is neggtive
if
ICAI < IBAI ;
sp - rq
that is, if
1
< '2
P
So u is the number of pairs of positive integers (r,s) such
that
I
s < ~
o<
Let v
bE~
sp - rq
<
I
2P
the number of eleTI'ents in B 'ltvi th negB-tive absolute
least residues, modulo q.
Let (P;l)q and (q;l)p be the points X and Y respectively.
35
Then
=
( q-p)
2
< £..2
So,
<%
IXYI
IXY I
Therefore,
reuresents the absolute least resid_ue of
(q;l)p , modulo q, and is positive.
then
IXyl
Also, if
is the absolute least residue of k-p, modulo q.
Then for any sp in B with nerative absolute least residue,
modulo q, there exists an r such that
(r-l)q
<
sp
<
rq
with
1
< r <
p-l
- 2
and
-
Therefore v is the number of pairs of positive integers
(r,s) such that
r
1
<2'P
1
s<~
o
i.e.
< rq
- sp
1
<~
36
- 21 q < sn - rq <
0
Since
stl - rq
i
0
u + v is the number of pairs of interers (r,s) such that
r
< Ip
2
1
(c)
s <~
-
1
~
< stl
- rq
< 12P
Let
r + rt
and.
s + st
Since
o
1
< r < 2'P
1
O<s<~
we see that
o
1
< rt < 2P
o
<
Sf
< ~q
Also,
-----------,--
---------,-----
37
-.
r'
s
=
(q+l) - s'
2
and the third inequality in (0) becomes
- 12 < (Cl+l)
- s'
2
(p+l)
p -
~~
2
- r' q <'21 P
or
-
1
-q
2~
Adding (q;P)
_
t
< 1-p
2
1
- s'P - '2 Q + r'q
< ~p
we get
~P
< -s 'p
< ~q
+ r' q
Multiplying by -1, we have
1
- '2 q
< s' p
- r' q
< 1'2 P
Thus the pair of integers (r',s') satisfies the inequalities
in (C).
But
(r,s) = (r' ,st)
only when
r = r' =
p+l
4
q+l
s = st =
4
That is, only when
p - q
-
3 (mod 4)
Thus u + v is odd only when
otherwise.
p
From Gauss's Lerr.ma,
and
-
q
- 3
(rr.od 4), ancl is even
38
Thus
(D)
Tiul t i plying by (~), we Fe t
But
(.9.) (.9.) = 1 •
p p
Thus
And, since u + v is odd only when
(l?)
q
unless
=
p _ q _
3
(mod
4),
(.9.)
p
p:: q _ 3
(mod
4), in which case
This ,roves the theorem as it was originally stated.
To
show that this is equivalent to
E.=!
(E)
(E)(.9.) = (-1) 2
q P
q-l
2
we use the fact that, if p ~ q E 3 (mod 4)
1')-1 q-l
product ...........=- is odd, but not otherwise.
2
2
then the
Thus statements
(D) and (E) are eouivalent, and the two methods of stating
the theorem are the sarre.
The next proof which we shall consider was given by
Frobenius.
The rredification nresented here was given by
Daniel Shanks.
He feels that this proof is possibly the
simplest and most direct ureof of the Law of Quadratic
Reciprocity.9
39
FROB1~~HIUS t
If
p
= 2P
+ 1
and
q
= 2Q
PROOF
+ 1
are unequal odd primes then
PROOF:
Assume that
p
< q.
Let a be an odd integer,
0
<a < p
such that
qa
= pat
+ r
where
and r is even.
If u is the number of such a, then from
Gauss's Lemma,
We know that a, p, and q are odd and r is even.
must a1:30 be odd.
a, and
0
< a' < q
•
where v is the number of odd a' such that
+ r
with
<a <q
0< r <
o<a <p
o
l
Q
and a oC,d.
l
Further, at is unioue1y determined by
Then, b:r symrnetry,
pal = qa
Thus a
,
40
Consider the function
= qa
R(a,a')
pa'
where a is an element of the set
1, 3, 5, • • • , p -
2
and a' :is 8.n element of the set
1, 3, 5,
Ne l{now th8.t
0
..,
•
q
< r <"p.
o
< pa'
o
< q 8.
-
2
'rhus
<p
+ r - pa'
or
- 1)a'
<p
and so
O<R<lJ
Since there are u values for r, there are also u values for
-q
<R <
< -r < 0
-q
< C1 a
R such tha.t
0
Similarly,
1).
0
< r < q.
So
-q
< R < O.
3ince
or
-
Cia -
r
<0
which rreans that
-q
< qa
-
(q a
+ r)
-0
< qa
-
pa t
<0
-q
<R < 0
<0
and
so that
Thus there v values for R such thgt
(a,p)
=1
,
D
and q are nrimes, and a <p,
qa ~ pat
so th8t
41
oa - pa' I.
Thus
°
R I. 0, rmel there are u + v values of R such that
-(1
<R <p
If
then
are two such R.
Therefore,
Rl + R 2 =
= q (a l
+ (:12)
-
= q(p
-
= ap
-
+ qa 2
a~
- pq
So
Rl + R2 =
P
-
q
If, however,
a 1 = a 2 = p-l
= p
2
and
a-'1. ' -- a2' -.9.::12 -
Q-
then Pend Q are both odd, ana
R
=
qP -
-
p(~t +
- p(a - 1)
1)
C1
,)
(pal' + na
- 2
pQ
= q(P;I) _ p(O;I)
+ p
a ')
2
42
=E
-2 9-2
= (E-l)
=
P -
Q
R = P -
Q
~(1-1)
So
Thus u + v is even unless P c'nd Q are both odo..
is even unless P'J.nd Q a.re ooth odd.
But PQ
Therefore
But
'rhus
\F;hich -proves the theorem.
(2::1)
2
=
P
Also, since
and
(2.::1)
2 . -- Q
we see that
8 .. nd
the~le
t"vo foms of the theorem are erui valent.
We shall now consider two geometric[:l Droofs of the
theorem.
The fir8t of these, by Eisenstein, provides us
wi th a {,;eometrical inter1)retation of Gauss t s Third Proof.
The variation
~iven
Harriet Griffin.
need the
followin~
here is based on a presentation by
Before we begin the proof, ho.,.;ever, . .ve
definition:
43
Lattice Point:
A tloint (x,y) is called. a lattice point if and only if
x and yare both integers.
Ne can now give the "9roof as stateo by Griffin.
EISENSTEIN'S PROOF
Let p and
q
be distinct odd primes.
Then
From Lemma 2,
and
where
rf( =
[~J +
• • • +
N =
[~J
• • • +
+
Then
Consider the rectanr;le R with vertices
0(0,0)
B(E
2 '
9.)
2
C(o , ~)
10
44
0(0, ~)
9.)
2 ' 2
B(E
(0, q;l)
(P;l , 0)
0(0,0)
The diagonal OB has eauation
Y
= -pO,x
·
( q,p )
,andSlnce
= 1,
there are no lattice 1')oints on the open line segment OB.
Let Tl be the interior of R below OB, and let
~2
be the
interior of R above OB.
Consider a point (k, 0) on the x-axis, "'lith k an integer.
For
i.e.
< k -< p-l
2
1
-
the number of lattice points directly above the point in Tl
is [~J.
Thus the number of lattice uoints in Tl is
Like'llJise, for j an integer,
°<
,-
j
< 9.2
i.e.
1 < j
-
< 0,-1
-
2
the number of lattice noints in T2 directly to the right of
the point (O,j) is [~J.
Thus the number of lattice 'Points
45
[~J
+
[~J
+ • • •
[(q;l)~J = N
Hence the total number of lattice points in R is li: + N.
'0-1) (a-I)
But there are also exactly ( '2
2" lattice points in R.
Thus,
Nt + N
=
p-l q-l
2
2
Therefore,
which ryroves the theorem.
The final nroof of the theorem which we shall consider
was -presented by D. H. Lehmer in 1957.
It is, therefore,
one of the most recent proofs available.
The vRriation
given here was presented by George Andrews. 11
LEJmmR'S PROOF
If panel q are distinct odd primes, then
(E)
q
unless
p :: q
(12 )
q
)
= (£.
p
-
")
..)
(mod 4) , in vthich case
= -C Sp )
PROOF:
Let u
l
clenot the number of inteeers in the set
q, 2 q , 3 q,
• • • ,
( P;l ) q
46
with least 7)ositive residues greater than ~ , modulo p.
"-
Let u 2 denote the number of integers in the set
p, 2p, 3p, • •
.,
wi th lea.st nosi tive residues greater than ~ , modulo q.
Then, by Gauss's Lemma,
and
u
(9.) == (-1) 1
p
Thus,
== (-1)
~+u
2
But
if and only if
So
if and only if
To prove the theorem, we need to show that u l + u 2 is odd
if and only
p:: a ::
3
(mod 4) •
Consider the hexagon H with vertices OBCDEF as follows:
c(E q(p-l»)
0(0,0)
2'
E(P(o.-l)
.-
2q
:1)
'2
----------_._--
2p
47
E
D
c
o
B
y
=
~
P
_L
is the ecuation of
2p
ox
1
y = .;.:.:
p + -2
o
is the equation of EF.
lie~
If the point (x,y)
BC~
<x <~
in H then
o <y < ~
and
(A)
~_L<y<9.!.+l
p
2p
.
P
2
Let (a,b) be a lattice inside H.
(P;l _ C~,
0,;1 _b)
o
Then the uoint
is also a lattice noint, for if
<a < ~
then
0>
-a
>-
~
and
so tha.t
---------------
48
or
o
< 12±1:
2
-
a
But since a is an integer,
p+l -
n
a <_p-l< .:.
222
Thus
Similarly, if
o
<b
<~
then
A.lso, if
we have
-b
> _~
_ 1p
p
and
0+1
~
-
b
> 9+21
_ ~ _ 1
p
2
which Gives us
or
q+l _ b
2
> 9..E
~p
_
S@:. + .9......
D
~D
_ L
2n
49
in other V'lOrds
.9.±! _ b
(1)
2
> 9.P (p+l
2
- a) _ !L
2p
Likevvise, if
then
-b
aa +-"-(1
_ ........
<
P
2p
and
oa
- -p + .92p
so
0+1
~
-
o
Qa
b< 1)(q+l) +-----2p
21)
P
and we have
0+1 _ b
< pCHn+g
q+l
b
< Q(p+l)
2p
b
< P9. ( p+21
2
-naa
2p
or
~
-
+
! - C18.
2
p-
so that
(2)
S1.±! _
2
-
a) + ~
Co
Thus, from (1) and (2) we have
O<.2±.!-a<E
2
2
s. (p+21
_ a) _
1)
8,nd
%-< 9.±1:.
_p
2
_
b
< 9.P ( p~ 1
-
8, )
+ ~
Thus for each lattice T)oint (8" b) in H, there corresnonds
-
"
" t
a 1 8 tt lce
nOln
(p+1
2
-
a '0+1
2 - b) also in
~{.
50
Further,
= (P~l
(a,b)
_ a,q;l - b)
if and only if
a --
p+l
2
-
b
a
=(1+1
2 - b
so ths.t
=
2a
-p+l
2
and
2b
-
=
0.+ 1
2
and
4b
=
a + 1
and
4a = p + 1
or, in other words,
p ~ q ~
3
(mod 4)
Thus
_
( 8,b ) ~/ (p+l
2
lJ.nl e s s p
_
a ::
3
3.,
0+1 _ b)
2
( m 0 c1 4).
But if
p ~ q -
3
( mod 4),
in addition to the nairs
_ ~ q+l _ b)
) , (~
( a, b
2
'--, 2
"there
a
-I
P:l
8-nd
b.j
9.:1
point, namely (P4l,CJ.~1).
in H is octet if sr..cl or..ly if
, Y/e p'et one more lattice
Thus the number of lattice Doints
n -
Oc
=3
(mod 4).
Since (q,n) = 1, there are no lattice points on ODe
(2", b) is a lattice noint below OD, then
aa
p
and
If
51
~
C 8. -
< bp < Cia
Thus
- % < bn
AdelinE
J,
<0
we pet
%< bp
(B)
qa
- q (a - 1)
<q
Thus bl) has a le2st I)ositi.ve residue f;reater than
%,
mod Q.
Conversely, if on has a 1e8.st nositive residue greater than
%,
modulo
holds.
0.,
then we can find 2-n integer a such that (B)
But bp is<'n element of the set
...,
(1-1)
«-2n-
and has least Dositive residue greater than
There c;,re u
%,
modulo o.
of these integers, and thus there are u')c.. lattice
2
points in H below OD.
Similarly, we find u
there ;:::,re u
+ u
l
if and only if
2
l
l8.ttice points in H above OD.
lattice "':Joints in H, and u l + u 2 is odd
n::
q
== 3
(mocl 4), Yvhich Droves the theorem.
Thus
(E.) == ( .9. )
q
p
unless
p
-
('
-
3
Thus
(mod 4) , in which cO.se
(E.) == -(~)
1)
q
Wi th this nroof, we com''"llete our eJ.iscu8sion of the Law
of Quaclratic Reci T)roci tv.
There 2.re, however, three basic
results which nrise from a disc1)ssion of the Legendre Symbol.
The proofs of these corollaries are Quite
si~ple,
statements theJ'YIsel ves [-.re very important.
The first -oroof
we shall give deals Vii th the n8.ture of nuadr8.tic
The second 2.nd third are concerned "'d th the
Corollary 1:
=1
then
if and only if
p-l
r2
(mod p)
_ 1
P;iOOF:
If R is a primitive root of p, then
p-l
p-l
1)(R2
(R~
+
1) _ 0
(mod p)
and thus
p-l
R 2
_ xl
(mod p)
If
r
-
R2k
(mod p)
then
-0-1
r~
= R(P-l)k
_ 1
(mod p)
----------------
..
resid~les.
0u8.c1r~ltic
of -1 and 2 resnectively.
If p is an odd nrime and if (r,p)
yet the
_•..._._----
n8.tv.re
53
and we see that
--- n-1
r
2
1
(mod n)
1
(mod p)
SO
(E)
P
1
=
if and only if
p-1
r 2
-
Corollary 2:
If
p
is an odd prime then
(-1) = 1
P
if and only if
(mod 4)
p =1
PROOF:
;{e
lcnow thet
p-1
(-1)
- 1
2
p
if and only if
-
1
(mod n)
( ffiOO_
4) •
Thus, by Corollary 1,
(-1) = 1
p
if ana. only if
p-1
(-1) 2
- 1
(moel p)
that is, if and only if
-
n :: 1
(mod 4)
,-------_._----_._--------
54
Corollary 3:
..;....;;.~=--.~
for any odd nrime p.
PROOF:
Je
ltnOVl that
¥- 2J
[
p
=
1
[~J
= 0
Therefore, if
k <p;l
then
[2~J
=
.,.
0
In Lernm9. 3, therefore, if
then
= 2
G
M = 0, 20ncl in Dlace
of congruence (3) we h8ve
P(2 - 1) E 0 - u
(mod 2)
Thus
p _ -u _ u
(mod 2)
cmd so
Cl.nd
These corollaries, together
L8,W, fo:nn r:n imDortcmt neTt of
~ith
t~le
the )uadrRtic Recinrocity
study of (lu8.dr:::tic residues.
55
The pro lJer 2.nplication of these DroY)erties cc::') 'lrovic1e mp.ny
solutions to difficult nroblems.
ized to Give the
Fin;t Proof.
~}ener8.lized
The theorem can he general-
L8W, which \,,'e saw in G,auss's
The properties can be used to determine the
odd primes of which a given inteeer is a nuadratic residue
or non-residue.
limi tJ_ess.
'fhe ap"Cllic8.tions of the theorem are 21,lrrost
This is why the Law of Juadratic ileci'':lroci t~T has
occunied the attention of so many ereat mC'>the):laticians.
The proofs -presented here e,re only
2,
small
the innumerable Y)roofs, botl'l elementary ano,
theorem.
s8JYl~llin~
cOTr~plex,
of
of the
The best exY)osition on more of these proofs is the
boolc by Bachmann which was noted e8,rlier.
The other refer-
ences given at the end of thts pS:lJer 2,re also of special
interest to any study of this tyne.
56
.FoorNOTES
lCarl Friedrich Gauss, DisC1uisitiones Arithmeticae
(New Haven and London: Yale University Press, 1966) p. 87.
2Ralph G. Archibald, ~ Introduction to ~ Theoa of
Numbers (Columbus, Ohio: Charles E. N;erril-PUblishing Co:),
pp. 135-137.
3r-oid., -op. 277-278.
4 G~uss, pp. 89-98.
5Archibald, pp. 272-281.
6Harriet Griffin, Elementary Theory of Numbers (New
York: McGraw-Hill Book Comnany, 1954), np:-146-l47.
7 J .. V. Us-pensky and !'Il.A. Heaslett, Elementa§if Number
Theory (New York: KcGraw-Hill Book Company, 1939, pp. 284293 •.
8 S • Barnard and J .IVl e Child, Advanced Al~ebra (London:
and Company, Limited, 1939), pp. 14 -149.
Mc~l.lillan
9Daniel Shanks, Solved and Unsolved Problems in Number
Theory (Washington: Spartan Books, 1962), pp. 40-47:
lOGTr1Off"1n,
pp~
144- 145 •
11C~eorge E. Andrews, Number Theory (Philadelphia: VV.B.
Saunders Company, 1971), P~. 120-123.
57
-BIBLIOGRAPHY
Andrews, George E. Number Theory.
Saunders Company, 1971.
Philadelphia: W.B.
Archibald, Ralph G. An Introduction to the Theory of
Nwnbers. Columbus, Ohio, Charles E. ~erril Publishing
Company, 1970.
Bachmann, P. Niedere Zahlentheorie. Leinzig, B.G. Teubner,
kathematische ~issenschaften, 1902.
Barnard, S. and J.M. Child Advanced Al~ebra.
I\~cN:illan and Compe,ny, Limited, 193 •
London,
Gauss, Carl Friedrich Disquisitiones Arithrneticae. (Translated by Arthur A. Clarke, S.J.), New Haven and London,
Yale University Press, 1966.
Griffin, Harriet Elementary Theory of Numbers. New York
and London, W:cGraw-Hill Book Company, 1954.
Shanks, Daniel Solved and Unsolved Problems in Number
T1:.eory. Washington,-Spartan Books, 1962:Uspensky, J.V. and ~.A. Heaslett Elementary Number Theory.
New York and London, McGraw-Hill Book Company, 1939.
-