The Law of Quadratic Reciprocity
Proof: By Wilson’s Theorem
By Kylee Bracher
Portland State University
Department of Mathematics and Statistics
Summer, 2005
In partial fulfillment of the Masters of Science in Mathematics
Advisors:
John Caughman
Steve Bleiler
Table of Contents
1. Introduction and History . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2. Preliminary Definitions, Theorems and Examples . . . . . . . . . . . . . . . . . . . . . . . 3
3.
The Law of Quadratic Reciprocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .16
4.
Geometric Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .22
5.
Schmidt’s Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
6.
Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
2
1. Introduction and History
One of the high points in all of elementary number theory is the Law of Quadratic
Reciprocity which was first proven by Carl Friedrich Gauss on April 8, 1796 and
published in his work Disquisitiones Arithmeticae.
Throughout Gauss’s life he had a particular passion for “higher arithmetic” which
we know today as number theory. This is evident by his own authoritative statement that,
“Mathematics is the queen of the sciences, and higher arithmetic, the queen of
mathematics.” “Gauss’s Law of Quadratic Reciprocity which he called the “Golden
Theorem of Number Theory” is one of the gems of eighteenth and nineteenth century
mathematics.” (Beach and Blair, 265) The first statement of a theorem equivalent to this
law was found in the works of Euler without proof in the period 1744-1746. The result
was then rediscovered in 1785 by Adrien-Marie Legendre, who provided two proofs that
he believed to be complete but which later turned out to be invalid. In his first attempt at
a proof he assumed that for any prime p congruent to 1 modulo 8, there exists another
prime q congruent to 3 modulo 4 for which q is a quadratic residue, a result which is as
difficult to prove as the law itself. “He attempted another proof in his Essai sur la
Theorie des Nombres (1798); this one too contained a gap, since Legendre took for
granted that there are an infinite number of primes in certain arithmetical progressions (a
fact eventually proved by Dirichlet in 1837, using in the process very subtle arguments
from complex variable theory). (Burton, 235) In 1795, Gauss discovered the result
independently at the age of 18, and a year later provided the first complete proof. Gauss
kept returning to the Law of Quadratic Reciprocity, searching for proofs that would
generalize to higher reciprocity laws.
3
In all, Gauss published six proofs and no doubt discovered several others. He was
fully aware of its central importance and the role it would play in the further development
of number theory. Leopold Kronecker later referred to its proof as the “test of strength of
Gauss’ genius.”
To date there are over 200 known proofs of the Law of Quadratic Reciprocity.
This paper will explore two of these proofs. The first is a geometrical interpretation of
one of Gauss’s proofs. This proof is very visual and relies very heavily on what is known
today as Gauss’s Lemma. The second proof was found in an article by Hermann Schmidt
published in 1839 entitled Drei neue Beweise des Reciprocitatssatzes in der Theorie der
quadraischen (“Three New Proofs of the Theory of Quadratic Reciprocity) , which is in
turn a variant of the fifth proof by Gauss. As is known, Euler’s Criterion and the
theorems of Fermat and Wilson can be proved in a very simple manner by determining in
two ways a product. Schmidt shows that the same is true for the Law of Quadratic
Reciprocity. “The noteworthy feature of Schmidt’s proof is that is dispenses with
Gauss’s Lemma and depends on nothing more mysterious than the Chinese Remainder
Theorem and Wilson’s Theorem.” (Rousseau, 424)
4
2. Preliminary Definitions, Theorems, and Examples
Before we begin an examination of the Law of Quadratic Reciprocity we must
first define some terms and theorems that will be used throughout the discussion. We
will assume that the reader is familiar with some of the more elementary concepts in
Number Theory such as primes, relatively prime, Division Algorithm, Euclidean
Algorithm, and the Greatest Integer function. The reader may refer to Burton’s
Elementary Number Theory text as a reference for these and other elementary concepts.
Readers which are already familiar with Quadratic Reciprocity may want to skip to the
Geometric Proof section of this paper.
Definition 1: (Congruence) Let n be a positive integer. Integers a and b are said to be
congruent modulo n, denoted by a  b(mod n), if they have the same remainder when
divided by n.
Lemma 1: Let a, b, and n > 0 be integers. Then a  b(mod n) if and only if n | (a  b).
Proof: () If a  b(mod n) , then by definition a and b have the same remainder
when divided by n, so by the division algorithm a = nk + r and b = nm + r for some
integers k, m, and r with 0  r  n . Now if we solve both of these equations for r we
have
a  nk  b  nm

a  b  nk  nm
 n( k  m)
If a  b  n(k  m) then n | (a  b) which is what we wanted to show.
() Now assume that n | (a  b) . By the division algorithm there exists integers k
and m such that, a = nk + r1 and b = nm + r2 , where 0  r1  n and 0  r2  n .
Subtracting b from a we obtain the equation a  b  (k  m)n  (r 1r2 ) , and since it
was assumed n | (a  b) , it must be that n | (r1  r2 ) . But | r1  r2 | n and so
5
 n  r1  r2  n and the only multiple of n in this interval is 0. Thus we have that
r1  r2  0 or equivalently r1  r2 , and so a  b(mod n) which is what we wanted to
show.
▲
The reader may think that we do not need a new and slightly longer way to
say n | (a  b) . However the notation a  b(mod n) is useful because it suggests an
equation, and in fact congruences share many properties with equations. The next couple
of theorems list a number of ways in which congruences behave like equations.
Theorem 1: (Basic Properties for Congruences: Let p > 0 be a fixed integer and a, b,
c, and d be arbitrary integers. Then the following properties hold:
(1) a  a(mod n) .
(2) If a  b(mod n) , then b  a(mod n) .
(3) If a  b(mod n) and b  c(mod n) , then a  c(mod n) .
(4) If a  b(mod n) and c  d (mod n) , then a  c  b  d (mod n) and
ac  bd (mod n) .
(5) If a  b(mod n) , then a  c  b  c(mod n) and ac  bc(mod n) .
(6) If a  b(mod n) , then a k  b k (mod n) for any positive integer k.
Theorem 2: (Cancellation Law): If a is relatively prime to p and x and y are integers
such that xa  ya (mod p ) , then x  y (mod p ).
Proof: By definition xa  ya (mod p ) means p divides xa  ya  a( x  y ) . Given
that a and p are relatively prime p must divide (x – y) since the prime factorization is
unique by the Fundamental Theorem of Arithmetic. If p divides (x – y) then by
definition x  y (mod p ) which is what we wanted to show. ▲
6
Definition 2: (Least Residue): If n > 0 and r is the remainder when the Division
Algorithm is used to divide b by n, then r is called the least residue of b modulo n.
Examples: The least residue of 13 modulo 5 is 3. The least residue of 5 modulo 13 is 5.
The least residue of -13 modulo 5 is 2. The least residue of – 2 modulo 3 is 1. ▲
Lemma 2: If a is relatively prime to p, p > 0, then there exists a unique integer x modulo
n such that ax  1(mod n).
Proof: Let (a, n) = 1 then by the Euclidean Algorithm we can write 1 as a linear
combination of a and n, that is, ax  ny  1 for some integers x and y. Rearranging the
terms in this equation we have ax  1  n( y ) . Hence n | (ax  1) which implies that
ax  1(mod n) by the definition of congruence.
To prove uniqueness, suppose that ax  1(mod n) and ax'  1(mod n) . Then
ax  ax' (mod n)

n | (ax  ax' )

n | a( x  x' )

n | ( x  x' )
(since (a, n) = 1)
By definition of congruence if n | ( x  x' ) then x  x' (mod n) . Therefore the solution
x must be unique modulo n. ▲
Definition 3: Let a be any integer. Any integer x for which ax  1(mod n) is called the
inverse of a.
Definition 4: Let n be a positive integer, and let a be an integer such that n | a. Then a is
called a quadratic residue modulo n if the congruence
x 2  a(mod n)
is solvable, and a quadratic nonresidue otherwise.
7
Definition 5: (The Legendre Symbol): Suppose p is an odd prime not dividing a. We
a
define the symbol   to be 1 or – 1 according to whether a is a quadratic residue or
 p
quadratic nonresidue mod p. This is called the Legendre Symbol, after the French
a
mathematician Adrien Marie Legendre (1752 – 1833). We also call   the quadratic
 p
character of a with respect to p.
a
If    1 then a is a quadratic residue modulo p which means the
 p
congruence x 2  a(mod p) is solvable. In fact, if a is a quadratic residue modulo p then
the congruence has exactly two solutions modulo p. This fact is verified through the
following theorem and proposition.
Theorem 3: (Lagrange’s Theorem): If p is a prime and
f ( x)  an x n  an1 x n1       a1 x  a0 ,
an  0(mod p)
is a polynomial of degree n  1 with integral coefficients, then the congruence
f ( x)  0(mod p )
has at most n incongruent solutions modulo p.
Proof: See Burton page 191. ▲
Proposition 1: Let p be an odd prime and a be any integer such that p | a. If the
congruence x 2  a(mod p) is solvable, then there are exactly two incongruent solutions
modulo p.
Proof: Suppose that the congruence x 2  a(mod p) is solvable and let x1  b be one
of the solutions. Note b  0 since a  0. Then there must also be a second solution
8
x2  p  b since ( x2 ) 2  ( p  b) 2  (b) 2 (mod p) and (b) 2  b 2  a(mod p) . This
second solution is not congruent to the first. To see this, note that b  p  b(mod p)
implies that 2b  0(mod p) or equivalently b  0(mod p) which is impossible. Now
by Lagrange’s Theorem, these two solutions exhaust the incongruent solutions of
x 2  a(mod p) . Thus, if x 2  a(mod p) is solvable there are exactly two incongruent
solutions modulo p. ▲
For the remainder of this paper we will always assume that p is an odd prime such
that a is not divisible by p. Next we will look at an example that incorporates all of the
previous definitions.
Example 1: When a = 3 and p = 11, the quadratic character of 3 with respect to 11 is
3
equal to 1, or    1 , because there exists an x such that x 2  3(mod 11) is solvable. To
 11 
find x we simply square the numbers from 1 to 10, subtract 3, and see if the result is
divisible by 11. Thus we have that
12  3  2, not divisible by 11
2 2  3  1,
not divisible by 11
32  3  6,
not divisible by 11
4 2  3  13, not divisible by 11
52  3  22, divisible by 11. Thus, x  5(mod 11) is a solution.
6 2  3  33, divisible by 11. So, x  6(mod 11) is also a solution.
Recall by Proposition 1 that if a is a quadratic residue modulo p, then the
congruence has exactly two solutions. Thus, we can stop here. We should also note,
6  5(mod 11) . So we could write that the solutions to the congruence x 2  3(mod 11)
are exactly x  5(mod 11) . It is easy to see that this fact is true since,
52  3(mod 11)

11 | 25  3

11 | 22
9
is clearly true. A similar argument would show that the congruence holds
for x  5(mod 11) .
Now it is a little deceptive to say that there are only two solutions to the
congruence because actually there are infinitely many solutions. It just happens that any
other solution will be congruent to  5(mod 11) . For example, 16 is also a solution to the
congruence since
162  3(mod 11)



11 | 16 2  3

11 | 253
This statement is true because 253 = (23)(11). Similarly 27, 38, 49, . . . . . , 5 + 11k
where k is any integer, are all solutions to the congruence x 2  3(mod 11) . The integers 16, -27, -38, . . . . . . . , - 5 – 11k where k is any integer are also solutions because each of
these integers are congruent to  5(mod 11) . ▲
The quadratic residues for a system can be found by squaring the numbers from 1
to p – 1 and reducing them modulo p as we will see in the following example.
Example 2: When p = 11, the quadratic residues can be found by squaring the numbers
from 1 to 10 and reducing them modulo 11.
Squaring the numbers from 1 to 5 yields:
12  1,2 2  4,32  9,4 2  5,52  3(mod 11).
Squaring the numbers from 6 to 10 yields the same residues in reverse order:
6 2   5  3,7 2   4  5,82   3  9,9 2   2  4,10 2   1  1(mod 11).
2
2
2
2
2
So the quadratic residues mod 11 are 1, 3, 4, 5, 9 and the quadratic nonresidues are 2, 6,
7, 8, 10. ▲
Since the numbers from 1 to
p 1
2
p 1
2
yield the same residues as the numbers from
to p, the quadratic residues can be found simply by squaring the numbers from 1 to
p 1 .
2
10
Theorem 4: Let p be an odd prime. Then half of the numbers between 1 and p – 1 are
quadratic residues of p, and half are nonresidues.
Proof: Any quadratic residue of p must be congruent modulo p to one of the
numbers 12 ,2 2 ,32 ,...., ( p  1) 2 , as was illustrated in the example above. Note that
p  1  1, p  2  2,....,
p 1
p 1
p 1
up

. So the squares of the numbers from
2
2
2
to p – 1 are congruent to the squares of the numbers
p 1
down to 1.
2
All we have left to show is that each of the squares of 1,2,3,........,
p 1
is not
2
congruent to any one of the other squares. We will do this by contradiction.
Suppose that there exist two squares such that i 2  j 2 (mod p) where i and j
are distinct and both in the range 1 to
p 1
. Then
2
i 2  j 2 (mod p )
 p (i 2  j 2 )
 p (i  j )(i  j )
 p (i  j )
or
p (i  j )
These last two statements are always false because both i and j lie within the
range 1  i, j 
p 1
. Thus, i – j and i + j are both too small to be divisible by p. So
2
we have shown that none of the squares are congruent mod p. Thus, there must be
exactly
p 1
quadratic residues of p. ▲
2
In general, the above theorem says that there are exactly
and exactly
p 1
2
p 1
2
quadratic residues
nonresidues of p.
11
The last theorem in this section, Euler’s Criterion, will be used throughout the
discussion of both proofs of the Law of Quadratic Reciprocity presented in the sections
Geometric Proof and Schmidt’s Proof below. In order to appropriately discuss the proof
of Euler’s Criterion we must first look at a preceding theorem known as Fermat’s Little
Theorem.
Theorem 5: (Fermat’s Little Theorem): Let p be any prime number. If a is any integer
such that p | a then a p 1  1(mod p).
Proof: Consider the set P = {1a, 2a, 3a, . . . . . . . , (p – 1)a}. By the Cancellation
Law each of these numbers must be distinct modulo p. So modulo p, the set P is the
same as the set of integers N = {1, 2, 3, . . . . . . . , (p – 1)}. So if we multiply the
elements of these two sets together, we would get the same result modulo p, that is,
1a  2a  3a  ......( p  1)a  1  2  3  ......( p  1)(mod p)
Regrouping the left hand side we obtain:
(1 2  3  ......( p  1))  a p1  1 2  3  ......( p  1)(mod p)
or equivalently
( p  1)!a p1  ( p  1)!(mod p)
To finish the proof we need only to cancel the common term of (p – 1)! from both
sides. Given that p is prime, p and (p – 1)! can have no factors in common. Thus by
the previous lemma we can divide both sides by (p – 1)! yielding a p 1  1(mod p)
which is what we wanted to show. ▲
Theorem 6: (Euler’s Criterion): Let p be an odd prime and a any integer such
that p | a . Then a is a quadratic residue modulo p if and only if a ( p1) / 2  1(mod p).
12
Proof: () Let a be a quadratic residue modulo p. Then there must exist an x such
that x 2  a(mod p) . Since it was assumed that p | a this implies p | x. Thus by
Fermat’s Little Theorem we have
 
a ( p 1) / 2  x 2
( p 1) / 2
 x p 1  1(mod p).
Hence if a is a quadratic residue modulo p then a ( p 1) / 2  1(mod p) which is what we
wanted to show.
Recall that if (r, p) = 1 and r has order  ( p ) modulo p (where  ( p ) is the largest
possible order), then r is called a primitive root of p.
() Now suppose that the congruence a ( p 1) / 2  1(mod p) holds and let r be a
primitive root of p, r exists because p is assumed to prime. If r is a primitive root of p
then a  r m (mod p) for some integer m, with 1  m  p  1. Thus it follows that
r m( p1) / 2  a ( p1) / 2  1(mod p).
Given that r is a primitive root of p the order of r is p – 1. By the previous
congruence r m( p 1) / 2  1(mod p) so p – 1 must divide m(p – 1)/2 which implies that m
 
must be an even integer. Let m = 2k, then r k
2
 r 2 k  r m  a(mod p),
making the integer r k a solution of the congruence x 2  a(mod p). Thus a is a
quadratic residue modulo p which is what we wanted to show. ▲
Example 4: Let a = 17. For which primes p is 17 a quadratic residue? We can test prime
p's manually given the formula above.
If we let p = 3, we have 17(3 − 1)/2 = 171 ≡ 2 (mod 3) ≡ -1 (mod 3), therefore 17 is not a
quadratic residue modulo 3.
Now if we let p = 13, we have 17(13 − 1)/2 = 176 ≡ 1 (mod 13), therefore 17 is a quadratic
residue modulo 13. As confirmation, note that 17 ≡ 4 (mod 13), and 22 = 4. ▲
13
The following results are direct consequences of Euler’s Criterion.
Lemma 3: The congruence x 2  1(mod p) is solvable if and only if p  1(mod 4).
Proof: By Euler’s Criterion, x 2  1(mod p) is solvable if and only if
(1) ( p1) / 2  1(mod p) . But (1) ( p1) / 2 is either 1 or -1 depending on whether (p – 1)/2
is even or odd, respectively. Since p > 2, (1) ( p1) / 2  1(mod p) if and only if (p – 1)/2
is even. But (p – 1)/2 is even if and only if p  1(mod 4) which is what we wanted to
show. ▲
Example 5: When p = 17, the congruence x 2  1(mod 17) is solvable since
17  4(4)  1 . The solutions to the congruence x 2  1(mod 17) are exactly
x  4(mod 17) . ▲
Corollary 1: Let p  3(mod 4) be an odd prime and a any integer such that p | a . Then
a is a quadratic residue modulo p if and only – a is a quadratic nonresidue modulo p.
Proof:  Let a be a quadratic residue modulo p. By Euler’s Criterion, if a is a
quadratic residue modulo p then a ( p1) / 2  1(mod p). So to show that –a is a quadratic
nonresidue we need only show that (a) ( p1) / 2  1(mod p). But
(a)( p1) / 2  (1)( p1) / 2 (a)( p1) / 2  1(mod p)
since (p – 1)/2 is odd. Hence – a is a quadratic nonresidue modulo p, which is what
we wanted to show.

Let –a be a quadratic residue modulo p. By Euler’s Criterion, if –a is a
quadratic residue modulo p then (a)( p1) / 2  1(mod p). So to show that a is a
quadratic nonresidue we need only show that (a)( p1) / 2  1(mod p). But
(a) p1 / 2  ((a)) ( p1) / 2  (1)( p1) / 2 (a)( p1) / 2  1(mod p)
since (p – 1)/2 is odd. Hence a is a quadratic nonresidue modulo p, which is what we
wanted to show. ▲
14
Corollary 2: Let p  1(mod 4) be an odd prime and a any integer such that p | a . Then
a and –a are either both quadratic residues modulo p or quadratic nonresidues modulo p.
Proof: First we will assume that a be a quadratic residues modulo p. By Euler’s
Criterion, if a is a quadratic residue modulo p then a ( p1) / 2  1(mod p). So to show
that –a is a quadratic residue modulo p we need only show that
(a)( p1) / 2  1(mod p). But
(a)( p1) / 2  (1)( p1) / 2 (a)( p1) / 2  1(mod p)
since (p – 1)/2 is even. Hence –a is a quadratic residue modulo p.
Now assume that a is a quadratic nonresidue modulo p. Then by Euler’s
Criterion, a ( p1) / 2  1(mod p). To Show that –a is also a quadratic nonresidue modulo
p we need only show that (a)( p1) / 2  1(mod p). But
(a)( p1) / 2  (1)( p1) / 2 (a)( p1) / 2  (a)( p1) / 2  1(mod p)
since (p – 1)/2 is even. Hence –a is a quadratic nonresidue modulo p, which is what
we wanted to show. ▲
15
3. The Law of Quadratic Reciprocity
We are now ready to present the statement of the Law of Quadratic Reciprocity.
Law of Quadratic Reciprocity: Let p and q be distinct odd primes. Then
 p 1   q 1 
 p  q 
     1 2  2  .
 q  p 
We should remark that by the statement of the Law of Quadratic Reciprocity
 q   p
    
 p  q 
if p  1 or q  1(mod 4) , while
q
 p
    
 p
q
if p  q  3(mod 4) .
Before we look at how we can actually prove this statement we will first look at a
few examples which illustrate the different ways in which the Law of Quadratic
Reciprocity may be used.
 31   171 
Example 6: Let p = 3 and q = 17 then  3  17    1 2  2   (1) 8  1 which is what is
 17  3 
expected since q = 17 is of the form 4k + 1. Thus by the Law of Quadratic Reciprocity
the congruences
x 2  3(mod 17)
and
x 2  17(mod 3)
are either both solvable or both insolvable. The easiest congruence to check would be
x 2  17(mod 3) since it reduces to the congruence x 2  2(mod 3) which is clearly not
solvable since 12  1  2  1 and 22  4  2  2 are not divisible by 3. Therefore both
of the congruences are insolvable and so  3   1 and  17   1 . ▲
 17 
3
16
The following examples illustrate how you can use the Law of Quadratic Reciprocity to
determine the quadratic character of p with respect to q, when the value of either p or q is
known.
Example 7: In this example we will determine the value of the quadratic character of 7
with respect to p when p is an odd prime different from 7. We should note that 7 is an
odd prime of the form 4k + 3 (since 7 = 4(1) + 3). Now by the Law of Quadratic
Reciprocity we have
  p 17 1 

4


 7   p
    (1) 
 p  7 
 p
  (1)
7
3 p 1
2
To determine the value of  7  we must consider two things:
 p
1) The value of  p  ;
7
2) The value of the exponent on -1.
To find the value of  p  we can find the least residues modulo 7 squaring the numbers
7
from 1 to 3  3  7  1  and reducing them mod 7. Squaring the numbers from 1 to 3 yields

2 
12  1,2 2  4,32  9  2(mod 7) . Thus, the quadratic residues mod 7 are 1, 2, 4 and the
quadratic nonresidues are 3, 5, 6.
Now with the previous work done we can determine the value of  p  for each
7
value of a from 1 to 6.
 p
If p  1(mod 7) , then    1, since 1 is a quadratic residue mod 7.
7
 p
If p  2(mod 7) , then    1, since 2 is a quadratic residue mod 7.
7
 p
If p  3(mod 7) , then    1, since 3 is a quadratic nonresidue mod 7.
7
 p
If p  4(mod 7) , then    1, since 4 is a quadratic residue mod 7.
7
17
 p
If p  5(mod 7) , then    1, since 5 is a quadratic nonresidue mod 7.
7
 p
If p  6(mod 7) , then    1, since 6 is a quadratic nonresidue mod 7.
7
Next we must note that since p is an odd prime then p is either of the form
4k + 1 or 4k + 3.
Case 1) p = 4k + 1 where k is any integer
p  4k  1  4k | ( p  1)  p  1(mod 4)
If p  1(mod 4) , then the exponent on -1 is
3 p  1 3(4k  1)  1 12k


 6k , an even number.
2
2
2
Case 2) p = 4k + 3 where k is any integer
p  4k  3  4k | ( p  3)  p  3(mod 4)
If p  3(mod 4) , then the exponent on -1 is
3 p  1 3(4k  3)  1 12k  6


 6k  3, an odd number.
2
2
2
The assumptions on p show that it is not divisible by 7 because we assumed that p
was odd prime different from 7 and it is not divisible by 4 because it either has a
remainder of 1 or 3 when divided by 4. Thus, p must be relatively prime to their product
(7)(4) = 28. So p must be congruent to one of 1, 3, 5, 9, 11, 13, 15, 17, 19, 23, 25, or 27
(all numbers between 1 and 28 that are relatively prime to 28).
If p  1(mod 28) , then
28 | (p – 1)
 (4)(7) | ( p  1)
 4 | ( p  1)
and
7 | ( p  1)
 p  1(mod 4) and
p  1(mod 4)  the exponent on -1 is even so (1)
p  1(mod 7)
3( p 1)
2
1
 p
p  1(mod 7)     1 .
7
18
3( p 1)
 7   p
Hence the value of the Legendre symbol is     (1) 2  (1)(1)  1.
 p  7 
This example illustrates how when one of the odd primes is of the form
4k + 1, then either both of the congruences are solvable or both are unsolvable, yielding
 p
7
the same quadratic characters for   and  . In this case, they are both solvable since
p
7


 7   p
      1.
 p  7 
Next we will illustrate the case when both p and q are of the form 4k + 3,
if p  3(mod 28) then
28 | ( p  3)
 (7)( 4) | ( p  3)
 4 | ( p  3) and
7 | ( p  3)
 p  3(mod 4) and
p  3(mod 4)  the exponent -1 is odd so (1)
3( p 1)
2
p  3(mod 7)
 1
 p
p  3(mod 7)     1 (since 3 is a quadratic residue mod 7)
7
7
Hence the value of the Legendre symbol for    (1)( 1)  1.
 p
 p
7
In this example the value of the quadratic characters for   and   are opposite, so
p
7


one of the congruences is solvable and one is not. You could continue in this fashion
checking all of the values 1, 3, 5, 9, 11, 13, 15, 17, 19, 23, 25, and 27. The results are
summarized below.
7
  =
 p
1
if p  1,3,9(mod 28)
-1
if p  5,11  13(mod 28)
19
7
From this it is easy to see that if for example p = 27 then    1. Thus, there must exist
p


an x such that x 2  7(mod 27) is solvable. After checking the values from 1 to 27 it can
be easily be verified that x  13(mod 27) are the solutions to the congruence since
132  7(mod 27)  27 | (132  7)  27 | 162
which is clearly true since 162 = (27)(6). ▲
Example 8: What about for larger values of p? Suppose that p = 1,999 and q = 3. Both
are of the form 4k + 3 so by The Law of Quadratic Reciprocity of the congruences
x 2  1,999(mod 3) and x 2  3(mod 1,999) only one is solvable. The first congruence
x 2  1,999(mod 3) reduces to x 2  1(mod 3) since 1,999 is congruent to 1 mod 3.
x 2  1(mod 3) is clearly solvable when x  2(mod 3) since
2 2  1(mod 3)

3 | (4  1)
 3|3
Therefore, by the Law of Quadratic Reciprocity the second congruence
x 2  3(mod 1,999) must have no solution. You could verify that this congruence has no
solution by testing each of the numbers from 1 to 999, but the Law of Quadratic
Reciprocity solves the problem without any need for testing.
Now let’s look at a case where one of the numbers p or q is of the form 4k + 1.
Suppose that p = 1,999 and q = 5. Then the law states that the congruences
x 2  1,999(mod 5) and x 2  5(mod 1,999) are either both solvable or unsolvable. The
first congruence x 2  1,999(mod 5) reduces to x 2  4(mod 5) since 1,999 is congruent to
4 mod 5. The congruence x 2  4(mod 5) is clearly solvable when x  2(mod 5) since
2 2  4(mod 5)

5 | (4  4)
 5|0.
By the Law of Quadratic Reciprocity the other congruence x 2  5(mod 1,999) must also
be solvable. In order to determine the solution by brute force we would have to square
the numbers between 1 and 999. It turns out that there is another method for finding the
solution without squaring all of the numbers from 1 to
p 1
. If p is of the form 4k + 3
2
20
(which in this case it is, 1,999  4  499  3) then one solution is x  a
p 1
4
. So for this
particular example the congruence x 2  5(mod 1,999) has the solution
x  5500 (mod 1,999). With the help of a calculator it can be checked that this yields
x  100(mod 1,999) , so the other answer is x  100  1,899(mod 1,999) . ▲
There are several other uses of Quadratic Reciprocity that may found in many
textbooks. The reader may want to refer to Burton’s text, Elementary Number Theory,
for further illustrations.
21
4. Geometric Proof
The geometric interpretation of Gauss’s proof of the Law of Quadratic Reciprocity relies
very heavily on what is known today as Gauss’s Lemma. So before we begin our
exploration of this proof we will first introduce and discuss this concept.
Lemma 4: Gauss’s Lemma: Let p be an odd prime and a an integer not divisible by p.
Consider the least residues of the integers ka for 1  k  p  1 , reduced to lie between  p
2
2
to p . If the number of these which are negative is s, then  a    1s .
 p
 
2
Proof: Let the least residues be ui  0 for 1  i  t and  v j  0 for 1  j  s, where
1  ui 
p 1
p 1
p 1
, 1 vj 
. We first claim, with proof to
, and s  t 
2
2
2
p  1

follow, that the set u1 , u 2 ,..., ut , v1 , v2 ,..., vs   1,2,3,....,
. Given that this is true
2 

we have
p 1
2
 ka  a
k 1
 p 1 


 2 
t
s
 p 1
s
s  p 1

!   1  ui  v j   1 
!(mod p)
 2 
 2 
i 1
j 1
 p 1
if we divide both sides by 
! we are left with
 2 
a
 p 1 


 2 
 (1) s (mod p) .
 p 1 


2 
Now by Euler’s Criterion  a   a 
 p
Given that
a
(mod p) . Thus we have that    (1) s (mod p) .
 p
 a  and
(1) s will both yield values of either 1 or -1, it follows that
 
p
 
a
   (1) s (since p is prime) which is what we wanted to show.
 p
Thus, all we need is to prove the claim that the set of least residues
disregarding signs is exactly equal to the set of integers from 1 to p  1 , or
2
22
p  1
equivalently u1 , u 2 ,..., ut , v1 , v2 ,..., vs   1,2,3,....,
. To do this we must show that
2 

no two ui ' s are congruent, no two v j ' s are congruent and that there are no u i  v j .
We will do this by contradiction.
Let’s assume that two ui ' s are congruent then k1a  k2 a(mod p) with k1  k2
and assuming without loss of generality that k1  k2  0 (if k1  k2 we would be
talking about the same element in the set).
k1a  k 2 a(mod p)
 p | k 1a  k 2 a 
 p | a(k1  k 2 )
 p | (k1  k2 ) (since p | a )
This last statement is always false because both k1 and k2 fall within the range
1 k 
p 1
p2
. So 0  k1  k 2 
. Thus, k1  k2 is too small to be divisible by p.
2
2
Next we assume that two v j ' s are congruent then  k1a  k2 a(mod p) with
k1  k2 . Now
 k1a  k 2 a(mod p)
 p | (k1a  k 2 a)
 p | a(k 2  k1 )
 p | (k2  k1 ) (since p | a )
This last statement is again false by the same reasoning as above.
Finally, we will assume that u i  v j for some ui , v j , then k1a  k2 a(mod p) for
some k1 , k2 . So
k1a  k 2 a(mod p)
 p | (k1a  k 2 a)
 p | a(k1  k 2 )
 p | (k1  k2 ) (since p | a )
23
This last statement is also false because again both k1 and k2 fall with the range
1 k 
p 1
. So 2  k1  k2  p. Clearly no integer within this range is divisible by
2
p.
Thus, we have shown that no two integers in the set u1 , u2 ,..., ut , v1 , v2 ,..., vs  are
congruent to one another and since there are exactly
within the range 1, . . . . ,
p 1
elements in this set, each
2
p 1
, it must be that u1 , u 2 ,..., ut , v1 , v2 ,..., vs   1,2,3,...., p  1.
2
2 

Which is what we wanted to show. ▲
Example 9: Take p = 11 and a = 19, we first reduce each 19k for 1  k  5 mod 11 to lie
within the range

11
2
to 11 . Thus, we get
2
for k = 1,
19(1)  19  3(mod 11)
for k = 2,
19(2)  38  5(mod 11)
for k = 3,
19(3)  57  2(mod 11)
for k = 4,
19(4)  76  1(mod 11)
for k = 5,
19(5)  95  4(mod 11)
So the least residues for the first 5 multiples of 19 modulo 11 are -1, 2, -3, -4, 5,
which we should note are exactly the integers from 1 to 11 1 . From the chart above we
2
see there are 3 minus signs, so by Gauss’s Lemma  19    13  1 . So 19 is a quadratic
 11 
nonresidue mod11. Thus there does not exist an x such that x 2  19(mod 11) . ▲
In this next example we will just reverse the roles of 19 and 11.
Example 10: Now let p = 19 and a = 11, we first reduce each 11k for 1  k  9 mod 11
to lie within the range  19 to 19 . Thus, we get
2
for k = 1,
2
11(1)  11  8(mod 19)
24
for k = 2,
11(2)  22  3(mod 19)
for k = 3,
11(3)  33  5(mod 19)
for k = 4,
11(4)  44  6(mod 19)
for k = 5,
11(5)  55  2(mod 19)
for k = 6,
11(6)  66  9(mod 19)
for k = 7,
11(7)  77  1(mod 19)
for k = 8,
11(8)  88  7(mod 19)
for k = 9,
11(9)  99  4(mod 19)
So the least residues for the first 9 multiples of 11 modulo 19 are 1, -2, 3, 4, -5, 6,
-7, -8, 9, which in this case are exactly the set of integers from 1 to 19  1 . From the chart
2
above we see that there are 4 minus signs, so by Gauss’s Lemma
 19 
4
    1  1 .
 11 
In this
case, 11 is a quadratic residue mod 19. Thus, the congruence x 2  11(mod 19) is solvable.
In fact, the solutions to this congruence are exactly x  7(mod 19) since
7 2  11(mod 19)



19 | 7 2  11

19 | 38
which is clearly true. A similar argument would hold for x  7(mod 19). ▲
We are now ready to look at the geometric interpretation of Gauss’s proof. The
following proof is based on a proof presented in the The Jewel of Arithmetic: Quadratic
Reciprocity found online at www. webpages.com.
Proof of the Law of Quadratic Reciprocity: A point in the x, y plane will be called a
lattice point if both its coordinates are integers. We will count lattice points in
various regions.
Consider the rectangle R in the x, y plane with sides y = 1 and y 
vertical sides x = 1 and x 
q 1
and
2
p 1
. The total number of lattice points in or on the
2
25
p  1  q  1 
rectangle is equal to 

 (exactly equal to the exponent on (-1) in the
 2  2 
statement of the theorem). This is illustrated in the graph below where p = 11 and q =
11  1  19  1 
19, in this case there are 

  (5)(9)  45 lattice points in the rectangle.
 2  2 
Figure 1
Next we will use three lines to divide the rectangle into four regions. Let
q
L be the line y    x (or x =
 p
 p
  y )
q
q 1
L1 be the line y    x 
 p 2
 p
1
L2 be the line x    y 
2
q
q  q 
(or y    x    )
 p   2p 
Figure 2
26
We should note that there are no lattice points on any of these lines within the
rectangle because p | x and q | y since all of the coordinates x within the rectangle
are in the range 1  x 
p 1
q 1
and all of the coordinates y are in the range 1  y 
.
2
2
We will let A be the region above L1 where y   q  x  1 ,
 p 2
 
B be the region below L2 where x   p  y  1 ,
q
2
 
C1 be the region between L and L1 where  q  x  y   q  x  1 ,
2
 p
 p
 p
 p
1
C2 be the region between L and L2 where   y  x    y  .
2
q
q
This again is illustrated in the graph below for the case when p = 11 and q = 19.
Figure 3
If we let P(X) denote the number of lattice points in region X it should be that
 p  1  q  1 
P(A) + P(B) + P( C1 ) + P( C2 ) = 

.
 2  2 
Thus, the statement of the theorem could be written in the form
27
 p  q 
    (1) P ( A) P ( B ) P (C1 ) P (C2 )
 q  p 
We will first show that A and B contain the same number of lattice points so the sum
of P(A) and P(B) will always be even. Thus, they will have no effect on the exponent
of (-1). So the statement of the theorem can be reduced to the following expression
 p  q 
    (1) P (C1 ) P (C2 )
 q  p 
We will then show that C1 contains s1 points where  q   (1) s and that C2
1
 p
contains s2 where  p   (1) s . So we could again restate the theorem as
q
 
2
 p  q 
    (1) s1 (1) s2  (1) s1  s2
 q  p 
 p  1  q  1 
where, s1  s2  

(mod 2) .
 2  2 
To see that P(A) = P(B) note that the entire picture is symmetric about its center
point  p  1 , q  1 , as illustrated in the graph below.
 4
4 
Figure 4
28




To verify this claim, let f : x, y   x* , y * be defined by f ( x, y )   p  1  x, q  1  y 
 2

2
with the property that x, y   A  x* , y *  B. We will show that this mapping f is
a one-one correspondence between the lattice points of A and those in B.
Proof: To prove that f is a one-one correspondence between the lattice points of A
and B we must show that f is a one-one and onto mapping between the lattice points
of A and B.
We must first verify that f maps the lattice points of A to the lattice points in B.
Let x, y  be any lattice point in A, then x and y are both integer values such that
q 1
y    x  , .
 p 2
We wish to show that f ( x, y )   p  1  x, q  1  y   ( x  , y  ) is a lattice
 2
2

point in B. Observe first that f (x, y) will be a lattice point since both p and q were
assumed to be odd. By the definitions of regions A and B, the reader can easily check
that if y   q  x  1 then y    q  x   q . Hence ( x , y  )  B , which is what we wanted
 p
 p
2
2p
 
 
to show.
To show that f is one-one we must show that f(a, b) = f (c, d) implies that (a, b) = (c,
d). Now,
f ( a , b )  f (c, d )
q 1   p 1
q 1
 p 1


 a,
 b  
 c,
d
2
2
 2
  2


p 1
p 1
a 
c
2
2
ac
and
and
q 1
q 1
b 
d
2
2
b=d
 ( a , b )  ( c, d )
which is what we wanted to show. Therefore, f is one-one.
Next we must show that f is onto. To show that f is onto we must show for each
x , y  B there exists an x, y  A such that
*
*


f ( x, y)  x* , y * . Suppose
29
x , y  B and let ( x, y)   p 2 1  x , q 2 1  y
*
*
*
*


 , then (x, y)  A since by definition

of the region B y    q  x   q and thus the reader can easily verify through algebra
 
 p
2p
that this implies that y   q  x  1 . Hence (x, y)  A and (x, y) is clearly a lattice point
 p
2
 
since p and q were assumed to be odd. Further,
f (x, y)
 p 1 * q 1

 f
x ,
 y* 
2
 2

 p 1  p 1 *  q 1  q 1

 

 x ,

 y *  
 2
 2
 2

 2
 ( x* , y * )
 p 1 * q 1

x ,
 y *  , such that
Thus, there exists a lattice point x, y   A , namely 
2
 2



f ( x, y)  x* , y * . Therefore, f is onto. We have shown that f is one-one and onto.
Therefore, we have shown that f is a one-one correspondence between the lattice
points of A and those of B. Hence, P(A) = P(B).
Next we will examine the number of lattice points in C1 . Let 1  x  p  1 , Then
2
q
q
1
( x, y)  C1 if and only if   x  y    x  , by definition of the region C1 . The
2
 p
 p
interval has length 1 so it contains at most one integer y. There will be an integer y
2
in this interval if and only if there is an integer y'  y   q x  (where  q x  denotes the
p 
p 
greatest integer) in the interval
q 
q  1
q
q
x   x  y'  x   x 
p
p
p 
p  2
which is if and only if
q 
1 q
 x   x  1
2 p
p 
30
Multiplying through by p we get the equivalent statement
p
 qx 
2
q 
p  x  p .
p 
The integer in the middle is the remainder on dividing qx by p. The number of x’s in
the range 1  x  p  1 for which the remainder lies between p and p is exactly the
2
2
same integer s which occurs in Gauss’s Lemma satisfying (1) s   q . A similar
 p
1
argument could be used to show that the number of lattice points in C2 is the integer
 p
s2 which appears in Gauss’s Lemma such that (1) s2   . Thus we have,
q
 p  q 
    (1) s1 (1) s2  (1) s1  s2
 q  p 
which is exactly what we wanted to show. ▲
Example 11 (for p = 11 and q = 19): We should note that p and q are both of the
form 4k + 3. Thus, we must show that
 11  19 
    1.
 19  11 
We will proceed by counting lattice points in various regions. Consider the
rectangular region R in the x, y plane with sides y = 1 and y  q  1  19  1  9 and
2
2
vertical sides x = 1 and x  p  1  11  1  5 . The total number of lattice points in or on
2
2
the rectangle is equal to  q  1  p  1   (9)(5)  45 , as illustrated in the graph below.
 2  2 
Figure 5
31
Next we will use three lines to divide the rectangle into four regions. We will
let
L be the line y  19 x
11
L1 be the line
y
19
1
x
11
2
L2 be the line x  11 y  1  y  19 x  19 
19
2 
11
22 
We should note that no lattice points lie on any of these lines within the
rectangle because 11 | x when 1  x  5 and similarly 19 | y when 1  y  8 .
We will let
A be the region above L1 where y  19 x  1
11
2
B be the region above L2 where x  11 y  1
19
2
C1 be the region between L and L1 where 19 x  y  19 x  1
11
11
2
C2 be the region between L and L2 where 11 y  x  11 y  1
19
19
2
Figure 6
If we let P(X) denote the number of lattice points in Region X, it should be that,
 p  1  q  1 
P( A)  P( B)  P(C1 )  P(C 2 )  

  45
 2  2 
32
Thus, the statement of the theorem could be rewritten as
 11  19 
P ( A )  P ( B )  P ( C1 )  P ( C2 )
    (1)
19
11
  
One could use a rigorous algebraic proof to show that the number P(A) =
P(B). However, for the purpose of this discussion we will use the graph to illustrate
this idea.
Figure 7
It is obvious from the graph that if you were to reflect region A about the line x =
3 and then again about the line y = 5, region A would lie right on top of region B.
Thus, each lattice point in region A would correspond to exactly one lattice point in
region B. Therefore, since f is one-one and onto it must be that P(A) = P(B). So the
sum of their lattice points will always equal an even number. Thus, they will have no
effect on the exponent -1 in the statement of the theorem. So we can omit them from
the exponent leaving  11  19   (1) P (C ) P (C ) .
1
2
 19  11 
Next we will show that the region C1 contains s1 lattice points where  19   (1) s
1
 11 
by showing that s1 is the exact same number that occurred in Gauss’s Lemma when
p = 11 and q = 19. Similarly, we will show that C2 contains s2 lattice points where
33
 11 
s
   (1) 2 by showing that s2 is the exact same number that occurred in Gauss’s
19
 
Lemma when p = 19 and q = 11. So we would have
 11  19 
s
s
s s
    (1) 1 (1) 2  (1) 1 2
 19  11 
where s1  s2 must be congruent to  p  1  q  1  mod 2.
 2  2 
Let 1  x  5 , by definition ( x, y )  C1  19 x  y  19 x  1 . Since the length of the
11
11
2
interval is only ½ it can only contain at most one integer y. There will be an integer y
in this interval if and only if there is an integer y   y  19 x  in the interval
 11 
19
19
19 
19  1
x   x  y  x   x  .
11
11
 11 
 11  2
which is if and only if
1 19
19 
 x   x  1
2 11
 11 
or multiplying through by 11 we get the equivalent statement
11
19 
 19 x  11 x   1
2
 11 
The integer in the middle, 19 x  1119 x  , is the remainder on diving 19x by 11.
 11 
Computing these remainders for the integers x in the range 1  x  5 , yields
for x = 1,
19 
19(1)  11 (1)  8
 11 
for x = 2,
19 
19(2)  11 (2)  5
 11 
for x = 3,
19 
19(3)  11 (3)  2
 11 
for x = 4,
19 
19(4)  11 (4)  10
 11 
for x = 5,
19 
19(5)  11 (5)  7
 11 
34
Thus, there are only three integers that lie within the range 11 and 11, so s1 = 3.
2
If we recall from example 3, that the least residues for 19 with respect to 11 were -1,
2, -3, -4, and 5, so in this case s = 3 (remember that s is the number of negative least
residues). We have s1 = s = 3. Thus, the number of lattice points in C1 is the exact
same number which occurred in Gauss’s Lemma for  19 . By counting the number
 11 
of lattice points in region C1 on the graph we see there are indeed 3. So
 19

3
  (1)  1. 
11


By using a similar argument we will show that the number of lattice points in C2
is the same as the number of minus signs we found earlier using Gauss’s Lemma for
p = 19 and q = 11. Let 1  y  8, by definition ( x, y)  C2 if and only if
11
11
1
y  x  y  . The interval again is only of length ½, so it will contain at most
19
19
2
one integer x. There will be an integer x in this interval if and only if there is an
integer x  x  11 y  in the interval
19 
11
11
 11 
 11  1
y   y   x  y   y  
19
19
19 
19  2
which is true if and only if
1 11
 11 
 y   y  1
2 19
19 
Multiplying through by 19 we obtain the equivalent statement
19
 11 
 11y  19 y   19
2
19 
The integer in the middle here is the remainder on dividing 11y by 19.
Computing these remainders for the integers y in the range 1  y  8 we see
for y = 1,
 11 
11(1)  19 (1)  11
19 
for y = 2,
 11 
11(2)  19 (2)  3
19 
35
for y = 3,
11 
11(3)  19 (3)  14
19 
for y = 4,
11 
11(4)  19 (4)  6
19 
for y = 5,
 11 
11(5)  19 (5)  17
19 
for y = 6,
 11 
11(6)  19 (6)  9
19 
for y = 7,
 11 
11(7)  19 (7)  1
19 
for y = 8,
11 
11(8)  19 (8)  12
19 
for y = 9,
 11 
11(9)  19 (9)  4
19 
As we see there are only four integers that lie within the range 19 to 19, so s2  4.
2
Again if we recall from example 4 that the least residues for 19 with respect to 11
were 1, -2, 3, 4, -5, 6, -7, -8, and 9 so in this case s = 4. So we have s2  s  4. Thus,
the number of lattice points in region C2 corresponds exactly to the number which
occurred in Gauss’s Lemma for  11 . Counting the number of lattice points on the
 19 
graph we see that region C2 does contain four lattice points. So  11   (1) 4  1.
 19 
Therefore, we have
 11  19 
4
3
7
    (1) (1)  (1)  1
 19  11 
which is exactly what we wanted to show. ▲
36
5. Schmidt’s Proof
We are now ready to begin our examination of Hermann Schmidt’s proof. As stated in
the introduction of this paper Schmidt’s proof uses some of the more elementary results
in Number Theory such as the Chinese Remainder Theorem and Wilson’s Theorem. So
we begin our discussion with these and a few related results.
Lemma 5: If (p, q) = 1, then x  a(mod pq) if and only if x  a(mod p) and
x  a(mod q ) .
Proof: () Let p and q be two integers such that (p, q) = 1 and suppose that there
exists integers x and a such that x  a(mod pq). By definition of congruence if
x  a (mod pq ) then pq | ( x  a ) . But if pq | ( x  a ) then p | ( x  a ) and q | ( x  a )
and so x  a (mod p ) and x  a(mod q) which is what we wanted to show.
() Let p and q be two integers such that (p, q) = 1 and suppose that there exists
integers x and a such that x  a(mod p) and x  a(mod q) . If x  a(mod p) then by
definition of congruence p | ( x  a ) which implies that there exists and integer r such
that (x – a) = rp. Similarly, if x  a(mod q) then by definition of congruence
q | ( x  a ) which implies that there exists an integer s such that (x – a) = sq. Now if
(x – a) = rp and (x – a) = sq then rp = sq and so p must divide sq. But if p | sq then
p | s since it was assumed that (p, q) = 1. Hence there must exist an integer t such
that tp = s. Thus (x – a) = sq = t(pq) which implies that pq | ( x  a ) and so
x  a(mod pq) which is what we wanted to show. ▲
37
Recall the Chinese Remainder Theorem, which assures the existence of a solution to
any system of linear congruences over moduli that are pairwise relatively prime. In this
paper, we only require a special case involving a pair of congruences, as stated in the next
theorem. We include a proof for the sake of completeness.
Theorem 7: (Chinese Remainder Theorem): Let p and q be positive integers, with (p,
q) = 1. Then the system of congruences
x  a (mod p )
x  b(mod q )
has a solution. Furthermore, any two solutions will be congruent modulo pq.
Proof: Let p and q be two positive integers such that (p, q) = 1. Since (p, q) = 1, by
the Euclidean Algorithm there must exist integers r and s such that rp + sq = 1. It
follows that the integers sq and rp satisfy
and
sq  1(mod p)
sq  0(mod q)
rp  0(mod p )
rp  1(mod q )
and so x = asq + brp must be a solution to the system of
congruences x  a (mod p ) and x  b(mod q ) . This can be seen by reducing modulo p
and then modulo q as follows,
x  asq  brp  a(1)  b(0)  a(mod p)
and
x  asq  brp  a(0)  b(1)  b(mod p)
Thus, if (p, q) = 1 the system of congruences x  a (mod p ) and x  b(mod q ) will
have a solution.
If x is a solution, then adding any multiple of pq to x will still be a solution since
x  k ( pq)  x(mod pq) for any integer k. Conversely if x1 and x 2 are two solutions of
the given congruences, then they must be congruent modulo p and modulo q, that is if
and
x1  a(mod p)
x1  b(mod q)
x2  a(mod p)
x2  b(mod q)
38
then x1  x2 (mod p) and x1  x2 (mod q) . Thus x1  x2 is divisible by both p and q
and so it is divisible by pq since by assumption (p, q) = 1. Therefore
x1  x2 (mod pq) which is what we wanted to show. ▲
Example 12: To solve the system
x  1(mod 3)
x  4(mod 5)
we first use the Euclidean Algorithm to write (2)(3) + (-1)(5) = 1. Then
x = (2)(3)(4) + (-1)(5)(1) = 19 is a solution and the general solution is x = 19 + 15t. The
smallest nonnegative solution is therefore 4, so we have x  4(mod 15) . ▲
Lemma 6: Let p and q be positive integers, with (p, q) = 1. If a is a quadratic residue
modulo p and a is a quadratic residue modulo q then a is a quadratic residue modulo pq.
Proof: Suppose that a is a quadratic residue modulo p and modulo q. If a is a
quadratic residue modulo p then there exists an x such that x 2  a(mod p) .
Considering the list of integers x, x + p, x + 2p, . . . . . , x + (q – 1)p we have that
x, x  p, x  2 p,......., x  (q  1) p  1,2,3,......., (q  1)(mod q)
in some order. Since a is also a quadratic residue modulo q at least one of the
numbers from this second list, call it y, when squared, must yield a modulo q. That
is, y 2  a(mod q). But y must be congruent to one of the numbers in the list
x, x + p, x + 2p, . . . . . , x + (q – 1)p. In other words, y  x  ip (mod q) for some
integer i. Now we have that y 2  a(mod q) and also y  x  ip (mod q)  x(mod p).
If y  x(mod p ) then y 2  x 2  a(mod p). Thus we have that y 2  a(mod q) and
y 2  a(mod p) so y 2  a(mod pq) by Lemma 5. Hence a is quadratic residue modulo
pq. ▲
Next we would like to recall Wilson’s Theorem which allows us to determine for
what integers p, p is a prime.
39
Theorem 8: (Wilson’s Theorem): Let p be an integer. Then p is a prime if and only if
( p  1)! 1(mod p).
Proof: If p = 2, then (2  1)!  1  1(mod 2) as desired. So assume that p is an odd
prime. For the rest of the proof we will assume that p is an odd prime.
() If p is an odd prime and G = {1, 2, . . . . , p – 1} then by Lemma 2 each a in G will
have a unique inverse modulo p. If a  b(mod p) , then multiplying both sides of this
congruence by a yields a 2  ab  1(mod p) . If a 2  1(mod p) then a 2  1  0(mod p) or
equivalently (a  1)( a  1)  0(mod p) . Given that p is prime,
(a  1)( a  1)  0(mod p) implies a  1(mod p) and so a = 1 or a = p – 1. Thus, 1 and
p – 1 are each their own inverses, but every other element of G has a distinct inverse, and
so if we collect the elements of G pairwise and multiply them all together we have
( p  1)!  1[2  3  .........  ( p  2)]( p  1)  (1)[(1)(1)......(1)]( p  1)(mod p)
but
(1)(1)(1)......( p  1)  ( p  1)  1(mod p)
Hence
( p  1)! 1(mod p ) .
() Now suppose by contradiction that the congruence holds for a composite number p.
If p is composite then p has a proper divisor d with 1 < d < p. If d < p then d also divides
( p  1)! since d must be one of the integers between 1 and p – 1. However, we assumed
the congruence ( p  1)! 1(mod p) holds so d must also divide (p – 1)! + 1 and so d
must divide 1 which a contradiction. Thus if the congruence holds p must be prime
which is what we wanted to show. ▲
Example 13: Let p = 11, then we should be able to pair the integers from 1 to
10  1(mod 11) , not inclusive, with their inverses. Indeed,
2  6  12  1(mod 11) , 3  4  12  1(mod 11) , 5  9  45  1(mod 11) , 7  8  56  1(mod 11).
40
Thus,
(11  1)! 1  2  3  5  6  7  8  9 10  1(2  6)(3  4)(5  9)(7  8) 10  1(mod 11). ▲
With Wilson’s Theorem and the Chinese Remainder Theorem established we will begin
our exploration of Schmidt’s proof, which may be found in his paper entitled Drei neue
Beweise des Reciprocitätssates in der Theorie der quadratischen Reste, beginning with
the following lemma.
Lemma 7: Let p, q be distinct odd primes. If r denotes any quadratic residue (mod pq),
then the congruence x 2  r (mod pq) has exactly four distinct solutions, xI through x IV
that satisfy:
(1) xI  xII   xIII   xIV (mod p)
(2) xI   xII  xIII   xIV (mod q)
In particular, two solutions have positive least residues modulo pq.
Proof: Assume r is a quadratic residue mod pq. Then the congruence
x 2  r (mod pq) has solutions mod pq. So by Lemma 5, this congruence has solutions
mod p and mod q. Indeed, by Proposition 1, x 2  r has exactly two incongruent
solutions mod p, call them c and –c. Similarly, x 2  r has exactly two incongruent
solutions mod q, call them d and – d.
Therefore x 2  r (mod pq) must have exactly four solutions xI , xII , xIII , and
x IV , each satisfying a pair of congruences as follows:
(i)
xI  c(mod p)
and
  d (mod q )
(ii) xII  c(mod p) and
  d (mod q )
(iii) xIII  c(mod p) and
  d (mod q )
(iv) xIV  c(mod p) and
  d (mod q )
By the Chinese Remainder Theorem xI through x IV are each uniquely determined
(mod pq). Moreover, xI through x IV are all distinct (mod pq) by construction.
41
Observe that x I   x IV and xII   xIII modulo pq and so the conditions of the
Lemma are satisfied. ▲
Definition 6: If r is any quadratic residue modulo pq then we say the four solutions of
x 2  r (mod pq) are conjugates.
Corollary 3: When the p and q are distinct odd primes then the number of quadratic
 p  1  q  1 
residues modulo pq is exactly 

.
 2  2 
Proof: By the above lemma, each of the (p – 1)(q – 1) residues prime to pq belongs
to a set of four conjugates whose squares are congruent modulo pq to a common
quadratic residue. It follows that there are exactly
( p  1)( q  1)  p  1  q  1 



4
 2  2 
quadratic residues modulo pq. ▲
Remark
By the Lemma 7, it follows that for odd primes p  q , there will be four solutions to
x 2  1(mod pq) . If we consider the least residues of these solutions, two will be positive
and two will be negative. Clearly x  1 is one of the positive solutions. We henceforth
will use  to denote the remaining positive solution.
Lemma 8: Let p, q denote distinct odd primes. Let xI through x IV represent the four
conjugates stated in Lemma 7 satisfying:
(1) xI  xII   xIII   xIV (mod p)
(2) xI   xII  xIII   xIV (mod q)
Let a and b be two remainders such that a  xI (mod p) and b  xI (mod q) . Let r be any
quadratic residue modulo pq.
42
(i)
If p  q  3(mod 4) , then exactly one of the four solutions of
x 2  r (mod pq) is itself a quadratic residue, the other three are not.
(ii)
If p  1(mod 4) and q  3(mod 4) (or visa versa) then either exactly two of
the four solutions of x 2  r (mod pq) are quadratic residues or else none
of them are.
(iii)
If p  q  1(mod 4) then either all four solutions of x 2  r (mod pq) are
quadratic residues, or else none of them are.
Proof (i): Let p and q be odd primes such that p  q  3(mod 4) . Then by Corollary
1, either a or –a is a quadratic residue modulo p. Similarly either b or –b is a
quadratic residue modulo q. Thus we have the following four cases:
Case 1: (a and b are both quadratic residues)
Assume that a and b are quadratic residues modulo p and q respectively.
If a is a quadratic residue modulo p then by definition of a and (1) above, xI and
xII are both quadratic residues modulo p. Similarly if b is a quadratic residue modulo
q then by definition of b and (2) above, xI and xIII are quadratic residues modulo q.
Hence xI is a quadratic residue modulo pq by Lemma 6 and the others are not.
Case 2: (- a and b are quadratic residues)
Assume that –a and b are quadratic residues modulo p and q respectively.
If –a is a
quadratic residue modulo p then xIII and xIV are both quadratic residues modulo p by
definition (1) above. If b is a quadratic residue modulo q then xI and xIII are both
quadratic residues modulo q by definition (2) above. Hence xIII is a quadratic residue
modulo pq by Lemma 6 and the others are not.
43
Case 3: (a and –b are quadratic residues)
Assume that a and –b are quadratic residues modulo p and q respectively. If a is a
quadratic residue modulo p then xI and xII are both quadratic residues modulo p by
definition (1) above. If –b is a quadratic residue modulo q then xII and xIV are both
quadratic residues modulo q by definition (2) above. Hence xII is a quadratic residue
modulo pq by Lemma 6 and the others are not.
Case 4: (– a and –b are quadratic residues)
Assume that –a and –b are quadratic residues modulo p and q respectively. If –a is a
quadratic residue modulo p then xIII and xIV are both quadratic residues modulo p by
definition (1) above. If –b is a quadratic residue modulo q then xII and xIV are both
quadratic residues modulo q by definition (2) above. Hence xIV is a quadratic residue
modulo pq by Lemma 6 and the others are not.
Thus we have shown that when p  q  3(mod 4) in each of these cases only one of
the solutions xI through xIV is a quadratic residue modulo pq, which is what we
wanted to show.
Proof (ii): Without loss of generality, assume p  1(mod 4) and q  3(mod 4) . If
p  1(mod 4) then a and –a are either both quadratic residues or quadratic nonresidues
modulo p. Thus we have the following two cases:
Case 1: (a, –a and b or –b are quadratic residues)
Assume that a and –a are both quadratic residues modulo p then by definition (1) we
have that xI , xII , xIII , and xIV are all quadratic residues modulo p. Now
since q  3(mod 4) , if b is a quadratic residue modulo then by Corollary 1, – b is a
quadratic nonresidue and visa versa. If b is a quadratic residue modulo q then by
definition (2) above xI and xIII are quadratic residues modulo q. Thus by Lemma 6,
xI and xIII would be quadratic residues modulo pq and xII and xIV would not.
44
Similarly if – b is a quadratic residue modulo q then by definition (2) above xII and
xIV are quadratic residues modulo pq. Thus by Lemma 6, xII and xIV would be
quadratic residues modulo pq and xI and xIII would not. In either case exactly two of
the solutions xI through xIV are quadratic residues modulo pq.
Case 2: (a and –a are quadratic nonresidues)
Assume that a and –a are quadratic nonresidues modulo p then xI through xIV are
quadratic nonresidues modulo p. Therefore xI through xIV are quadratic nonresidues
modulo pq.
Thus we have shown that if p  1(mod 4) and q  3(mod 4) (or visa versa) then
exactly two of the conjugates are quadratic residues modulo pq or none of them are.
Which is what we wanted to show.
Proof (iii): Let p  q  1(mod 4) . If p  1(mod 4) then both a and – a are either both
quadratic residues or both quadratic nonresidues modulo p. Similarly, if
q  1(mod 4) both b and – b are either both quadratic residues or both quadratic
nonresidues modulo q. Thus we have the following two cases:
Case 1: (a, – a , b, – b are all quadratic residues) Assume that a, –a are quadratic
residues modulo p and b, – b are quadratic residues modulo q. Then all four
solutions are quadratic residues modulo p and modulo q. Hence by Lemma 6, all four
solutions would be quadratic residues modulo pq.
Case 2: (a, – a quadratic nonresidues or b, – b quadratic nonresidues)
Without loss of generality assume that a and –a are quadratic nonresidues modulo p.
Then all of the four solutions xI through xIV are quadratic nonresidues modulo p.
Hence all four solutions are quadratic residues modulo pq.
45
We have shown that if p  q  1(mod 4) then all four of the solutions are quadratic
residues modulo pq or none of them are, which is what we wanted to show. ▲
Example 14: This example illustrates the case when both p and q are congruent to 3
modulo 4. Let p = 3 and q = 7 then the (p – 1)(q – 1) =(2)(6) = 12 remainders prime to 21
are 1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20. Squaring each of these numbers and reducing
them modulo 21 we have
{– 1, 1, – 8, 8}
solve x 2  1(mod 21)
{– 2, 2, – 5, 5}
solve x 2  4(mod 21)
{– 4, 4, – 10, 10}
solve x 2  16(mod 21)
So we can easily see that there are exactly four remainders, two negative and two
positive, whose squares are congruent modulo 21. Observe that each set of four contains
exactly one quadratic residue. If we choose the remainders – 5 and 5 then –a = - 2 , a =
2, –b = - 5 and b =5. Thus our four conjugates will be
xI  5  2(mod 3)
xII  5  2(mod 3)
and
and
 5(mod 7)
 5(mod 7)
xIII  5  2(mod 3) and
 5(mod 7)
xIV  5  2(mod 3) and
 5(mod 7)
Since p = 3 is congruent to 3 mod 4 and so only one of the numbers
– 2 or 2 is a quadratic residue modulo 3. If we square the numbers 1 and 2 we see
12  1  2(mod 3) and 2 2  4  1  2(mod 3)
and so – 2 is a quadratic residue modulo 3 and 2 is not. Similarly q = 7 is congruent to 3
mod 4 and so only one of the numbers – 5 or 5 is a quadratic residue modulo 7. If we
square the numbers from 1 to 6 and reduce each modulo 7 we have
12  1(mod 7)
2 2  4(mod 7)
32  9  2  5(mod 7)
4 2  16  2  5(mod 7)
52  25  4(mod 7)
6 2  36  1(mod 7)
and so – 5 is a quadratic residue modulo 7 and 5 is not. Thus the conjugate x IV is the
only quadratic residue modulo 21. If x IV  5(mod 3) and xIV  5(mod 7) then
46
x IV  5(mod 21) and – 5 is a quadratic residue modulo 21 since  5  16(mod 21) and
4 2  16(mod 21). ▲
Example 15: This example illustrates the case when either p or q is congruent to 1
modulo 4. Let p = 3 and q = 5 then the (p – 1)(q – 1) =(2)(4) = 8 remainders prime to 15
are 1, 2, 4, 7, 8, 11, 13, 14. Squaring each of these numbers and reducing them modulo
15 we have
{– 1, 1, – 4, 4} solve x 2  1(mod 15)
{– 2, 2, – 7, 7} solve x 2  4(mod 15)
So we can easily see that there are exactly four remainders, two negative and two
positive, whose squares are the same or congruent modulo 15. Observe that the first set
of four contains exactly two quadratic residues and the other does not contain any. If we
choose the remainders – 4 and 4 then –a = - 1, a = 1, –b = - 4 and b =4. Thus our four
conjugates will be
xI  4  1(mod 3)
xII  4  1(mod 3)
and
and
 4(mod 5)
 4(mod 5)
xIII  4  1(mod 3) and
 4(mod 5)
x IV  4  1(mod 3) and
 4(mod 5)
Since p = 3 is congruent to 3 mod 4 only one of the numbers – 1 or 1 is a quadratic
residue modulo 3. If we square the numbers 1 and 2 we see
12  1(mod 3) and 2 2  4  1(mod 3)
and so 1 is a quadratic residue modulo 3 and -1 is not. Now q = 5 is congruent to 1 mod
4 and so either the numbers – 4 and 4 are both quadratic residues or both quadratic
nonresidues modulo 5. If we square the numbers from 1 to 4 and reduce each modulo 5
we have
12  1  4(mod 5)
2 2  4(mod 5)
32  9  4(mod 5)
4 2  16  1  4(mod 5)
47
and so – 4 and 4 are both quadratic residues modulo 5. Thus the conjugates x I and
xII must be quadratic residues modulo 15. If xI  4(mod 3) and xI  4(mod 5) then
xI  4(mod 15) and 4 is a quadratic residue modulo 15 since 2 2  4(mod 15). If
xII  4  1(mod 3) and xII  4(mod 5) then we can determine the quadratic residue
modulo 15 by using the method implied by the proof of the Chinese Remainder Theorem.
By the Euclidean algorithm (3)(2) + (5)(-1) =1 and so a solution to the system is
x = (3)(2)(-4) + (5)(-1)(1) = -29 and so the smallest nonnegative solution x  1(mod 15) .
Thus 1 is a quadratic residue modulo 15 since 12  1(mod 15). ▲
Example 16: This example illustrates the case when both p and q are congruent to 1
modulo 4. Let p = 5 and q = 13 then squaring the (p – 1)(q – 1) =(4)(12) = 48 remainders
prime to 65 and reducing each of modulo 65 we have
solve x 2  64(mod 65)
{– 1, 1, – 14, 14}
solve x 2  1(mod 65)
{– 8, 8, – 18, 18}
{– 2, 2, – 28, 28}
solve x 2  4(mod 65)
{– 11, 11, – 24, 24} solve x 2  56(mod 65)
{– 3, 3, – 23, 23}
solve x 2  9(mod 65)
{– 12, 12, – 27, 27} solve x 2  14(mod 65)
{– 4, 4, – 9, 9}
solve x 2  16(mod 65)
{– 16, 16, – 29, 29} solve x 2  61(mod 65)
{– 6, 6, – 19, 19}
solve x 2  36(mod 65)
{– 21, 21, – 31, 31} solve x 2  51(mod 65)
{– 7, 7, – 32, 32}
solve x 2  49(mod 65)
{– 17, 17, – 22, 22} solve x 2  29(mod 65)
So we can easily see that there are exactly four remainders, two negative and two
positive, whose squares are the same or congruent modulo 65. Observe also that each set
of four contains either exactly four or zero quadratic residues. If we choose the
remainders – 14 and 14 then –a = - 4 , a = 4, –b = - 1 and b =1. Thus our four conjugates
will be
xI  4(mod 5)
xII  4(mod 5)
and
and
 1(mod 13)
 1(mod 13)
xIII  4(mod 5) and
 1(mod 13)
x IV  4(mod 5) and
 1(mod 13)
48
Since p = 5 is congruent to 1 mod 4 either both of the numbers –4 and 4 are quadratic
residues modulo 5 or both quadratic nonresidues modulo 5. If we square the numbers
from 1 to 4 we see
12  1  4(mod 5)
2 2  4(mod 5) 32  9  4(mod 5)
4 2  16  1  4(mod 5)
and so – 4 and 4 are both quadratic residues modulo 5. Similarly q = 13 is congruent to 1
mod 4 and so either both of the numbers – 1 and 1 are quadratic residues modulo 13 or
both are quadratic nonresidues modulo 13. If we square the numbers from 1 to 6 and
reduce each modulo 7 we have
12  1(mod 13)
2 2  4(mod 13)
52  25  12  1(mod 13)
32  9(mod 13)
4 2  16  3(mod 13)
6 2  36  10(mod 13)
and so –1 and 1 are both quadratic residues modulo 13. Thus all four conjugates must be
quadratic residues modulo 65. If xI  4(mod 5) and xI  1(mod 13) then by the
Euclidean Algorithm 1 = (2)(13) + (-5)(5) and so a solution to the system is
x = (2)(13)(4) + (-5)(5)(1) = 79 which implies that the smallest nonnegative solution is
x  14(mod 65) . Thus 14 is a quadratic residue modulo 65. If xII  4(mod 5) and
xII  1(mod 13) then by the C.R.T a solution to this system is x = (2)(13)(4) + (-5)(5)(-1)
= 129 which implies that the smallest nonnegative solution is x  64(mod 65) . Thus 64 is
a quadratic residue modulo 65. If xIII  4(mod 5) and xIV  1(mod 13) then by the
C.R.T. a solution to this system is x = (2)(13)(-4) + (-5)(5)(1) = -129 which implies that
the smallest nonnegative solution is x  1(mod 65) and 1 is a quadratic residue modulo
65. Finally, if x IV  4(mod 5) and xIV  1(mod 13) then by the C.R.T. a solution to
this system is x = (2)(13)(-4) + (-5)(5)(-1) = -79 which implies that the smallest negative
solution is x  51(mod 65) and 51 is also a quadratic residue modulo 65. ▲
Throughout the discussion of the next few lemmas and corollary we will be
examining and referring to a particular product which we will define here.
49
Definition 7: We will define Ppq to be the product of positive remainders that are prime
to pq. That is,
 p 1


 p 1  p 1
  p 1  q  1 
 2 q  1
  i   p  i       
 1 p  i  
p  i
2
2

 i 1  i 1
  i 1 
 i 1



Ppq 
p 1
2
 iq
i 1
q 1
q 1
q 1
q 1 p 1 


[1  2  3  ( p  1)]( p  1)( p  2)  (2 p  1)    (2 p  1)  ( p 
 1) ( p 
 1)( p 
 2)    p 

)
2
2
2
2
2 



p 1
1  2q  3q   
q
2
Lemma 9: Let p and q be distinct odd primes.
(i)
If p  q  1(mod 4) then Ppq (mod pq)  Ppq (mod p)  Ppq (mod q)  {1,1} .
(ii)
Otherwise,
Ppq (mod pq)  { , }
and
Ppq (mod p)   Ppq (mod q)  {1,1} .
Proof (i): If we assume that both p and q  1(mod 4) then both +1 and -1 are
quadratic residues modulo p and modulo q. So by Lemma 6, – 1 is a quadratic
residue modulo pq. Thus by Lemma 7 we have four remainders modulo pq whose
squares are congruent
– 1 (mod pq), of these, two are positive, we will denote
one by  , and the other by   , where  2  1(mod pq). Thus we have four
remainders 1,  ,  , and   whose squares and product are  1(mod pq). For
each of the remaining positive remainders a of pq there exists a unique remainder b
such that ab  1(mod pq) and so the entire product Ppq  1(mod pq ).
Proof (ii): In the case where (p – 1)/2 or (q – 1)/2 is odd, that is when either p or q is
congruent to 3 modulo 4, then -1 is a quadratic nonresidue modulo pq. Grouping all
of the remaining positive remainders we have ab  1 , a' b'  1, etc. when reduced
modulo pq. Thus, in all of the remaining cases we have that Ppq   (mod pq). ▲
50
Corollary 4: Let p and q be distinct odd primes. Then
( Ppq (mod p ))  ( Ppq (mod q ))  (1)
Proof: The exponent
 p 1 q 1 
 2  2 1


.
p 1 q 1

 1 is even if and only if p and q  1(mod 4) . So
2
2
this corollary holds by the work above. ▲
Lemma 10: Let p and q be distinct odd primes. Then
 p 1 q 1 

2 2 

 q  p 
( Ppq (mod p))  ( Ppq (mod q))    (1) 
 p  q 
.
Proof: In order to determine the signs on Ppq (mod p) and Ppq (mod q ) we will
proceed by looking at the following product:
1  2  3    ( p  1)  ( p  1)( p  2)    (2 p  1)( 2 p  1)    ( p 
q 1
 1)
2
 p 1  p 1
 p 1  q  1 

   i   p  i   
 1 p  i 

 i 1  i 1
 i 1  2

(1)
If we consider this product modulo p then we obtain
 p 1  p 1
  p 1  q  1 
  p 1  p 1   p 1 
  i   p  i       
 1 p  i     i   i       i   ( p  1)!( p  1)!  ( p  1)!(mod p)

 i 1  i 1
  i 1  2
  i 1  i 1   i 1 
Now by Wilson’s Theorem we have that ( p  1)! 1(mod p) and so
( p  1)!( p  1)!  ( p  1)! (1)( 1)    (1)  (1) ( q1) / 2 (mod p)
Hence
 p 1  p 1
  p 1  q  1 

  i   p  i       
 1 p  i   (1) ( q 1) / 2 (mod p).

 i 1  i 1
  i 1  2

Comparing this product to Ppq we note that both products contain the same number of
 q 1
factors, namely (p – 1) 
 . We can derive the product in (1) from the product
 2 
p 1
2
p 1
Ppq by multiplying by 1  q  2q    
q   iq and dividing by
2
i 1
51
p 1
2
 q  1  q  1   q  1 p  1 
 q 1
 1 p 
 2      p 

 p
  
p i .
2
2
2
2  i 1  2 


 
That is,
 p 1
 p 1

p 1
p 1

 
 p 1
  p 1  q  1 
 2 q  1



2
2
 1 p  i  
p  i  

    i   p  i       

iq
iq

  i 1  2
 i 1 2

   i 1  i 1



 


i 1
i 1






Ppq p 1

p 1
p 1
 2
 



2
2
 q 1
  q  1  p  i  



iq

 2
 
  2  p  i 



i 1
 i 1 
 

 i 1 


 p1  p1
 p1  q  1 

   i   p  i   
 1 p  i 

 i 1  i 1
 i 1  2

But
 p 1  p 1
  p 1  q  1 

  i   p  i       
 1 p  i   (1) ( q 1) / 2 (mod p).

 i 1  i 1
  i 1  2

Thus
(1) ( q 1) / 2
p 1


2


p 1
iq


q  2q    
q

2
  Ppq
 Ppq  p 1 i 1
(mod p)
p 1
 2

1 2     
 q 1


2
  2  p  i 

 i 1 

or equivalently
(1) ( q 1) / 2  Ppq q 
( p 1) / 2
(mod p)
or
q
(1) ( q 1) / 2  Ppq  (mod p)
 p
or multiplying both sides by  q  we have
 p
(I)
q
Ppq  (1) ( q 1) / 2  (mod p) .
 p
52
If we were to do the same analysis interchanging the roles of p and q we would obtain
the congruence
(II)
 p
Ppq  (1) ( p 1) / 2  (mod q) .
q
If we multiply (I) and (II) we obtain
p 1 q 1

 q   p
      (1) 2 2  Ppq (mod p)  Ppq (mod q)
 p  q 
Which is what we wanted to show. ▲
The result of the Law of Quadratic Reciprocity thus follows directly, as follows.
Proof of the Law of Quadratic Reciprocity: By Corollary 3 we have that
( Ppq (mod p ))  ( Ppq (mod q ))  (1)
 p 1 q 1 
 2  2 1


and by the previous Lemma 8 we have
 p 1 q 1 

2 2 

 q  p 
just shown that ( Ppq (mod p))  ( Ppq (mod q))    (1) 
 p  q 
. Thus we have
that
 p 1 q 1 

2 2 

 q  p 
  (1) 
 p  q 
 p 1 q 1 
 2  2 1

 (1) 
Dividing both sides of this equality by (1)
p 1 q 1

1
2 2
.
and subtracting exponents, the
reader could easily verify that this expression simplifies to
 q  p 
    (1)
 p  q 
p 1 q 1

2 2
.
Which is what we wanted to show. ▲
Example 17: This next example illustrates Lemma 7 for the case where p = 5 and q =13.
Let p = 5 and q = 13. Then
P65 
1  2  3  4  6  7  8  9 11 12 13 14 16 17 18 19  21  22  23  24  26  27  28  29  31  32
13  26
 1 2  3  4  6  7  8  9 1112 14 16 17 18 19  21 22  23  24  27  28  29  31 32
53
Given that 5 and 13 are both congruent to 1 modulo 4 we know that both -1 and 1 are
quadratic residues modulo 65. We can see this is true by the congruences below
12  1(mod 65)
and
14 2  1(mod 65)
82  1(mod 65)
and
18 2  1(mod 65)
Thus we have the four positive remainders 1,  ,  ,   = 1, 14, 8, (8)14  18(mod 65)
whose product 1  8 14 18  2016  1(mod 65). Now we will pair each of the following
positive remainders with their unique inverses as follows:
2  32  64  1(mod 65)
3  22  66  1(mod 65)
4 16  64  1(mod 65)
6 11  66  1(mod 65)
7  28  196  1(mod 65)
9  29  261  1(mod 65)
12  27  324  1(mod 65)
17  23  391  1(mod 65)
19  24  456  1(mod 65)
21  31  651  1(mod 65) .
Thus considering the product P65 modulo 65 yields
P65  1 2  3  4  6  7  8  9 1112 14 16 17 18 19  21 22  23  24  27  28  29  31 32
 1  (2  32)(3  22)( 4 16)(6 11)(7  28)(9  29)(12  27)(17  23)(19  24)( 21  31)  8 14 18
 (1)( 1)(1)( 1)(1)(1)(1)( 1)(1)(1)(1)(mod 65)
(since 1  8 14 18  2016  1(mod 65) )
 1(mod 65) . ▲
Example 18: This next example illustrates Lemma 7 for the case when either p or q is
congruent to 3 modulo 4. Let p = 5 and q = 11. Then
P55 

1  2  3  4  6  7  8  9 11 12 13 14 16 17 18 19  21  22  23  24  26  27
11  22
1  2  3  4  6  7  8  9 12 14 16 17 18 19  21  23  24  27
Given that 11 is congruent to 3 modulo 4 we know that 1 is a quadratic residue modulo
55 and -1 is not. We can easily see from the following congruences that 1 is a quadratic
residue modulo 55
12  1(mod 55)
and
212  441  1(mod 55) .
54
Thus we have the two positive remainders 1 and  = 1 and 21 whose squares are
 1(mod 55). Now we will pair each of the following positive remainders with their
unique inverses as follows:
2  27  54  1(mod 55)
3 18  54  1(mod 55)
4 14  56  1(mod 55)
6  9  54  1(mod 55)
7  8  56  1(mod 55)
12  23  276  1(mod 55)
13 17  221  1(mod 55)
16  24  384  1(mod 55)
19  26  494  1(mod 55)
Thus considering the product P55 modulo 55 yields
P55  1 2  3  4  6  7  8  9 12 13 14 16 17 18 19  21 23  24  26  27
 1  (2  27)(3 18)( 4 14)(6  9)(7  8)(12  23)(13 17)(16  24)(19  26)  21
 (1)( 1)( 1)(1)( 1)(1)(1)(1)( 1)( 1)  21(mod 65)
 21(mod 65) .
Now  21  1(mod 5) and  21  1(mod 11) . Thus P55 (mod 5)  P55 (mod 11)  (1)(1)  1
and so exactly one of the congruences x 2  5(mod 11) and x 2  11(mod 5) and the other is
not. It might be easiest to check the congruence x 2  11(mod 5) since it reduces to
x 2  1(mod 5) . By Lemma 3, the congruence x 2  1(mod 5) is solvable since 5  1(mod 4) .
Therefore, by the Law of Quadratic Reciprocity, the congruence x 2  5(mod 11) is not
solvable. ▲
55
6. Conclusion
The Law of Quadratic Reciprocity has been an important theme in the
development of number theory. Since the time of Gauss, many other mathematicians
such as Cauchy, Jacobi, and Kronecker have all devised their own proofs of the law to
determine for themselves why such a deep reciprocal relationship should exist between
simple congruences linking any pair of odd primes.
To date there are over 200 published proofs for the Law of Quadratic Reciprocity.
As was mentioned earlier Gauss provided several proofs himself, eight to be exact, which
were all motivated by his desire to generalize the law to higher powers. The search for
higher reciprocity laws eventually led to the development of a new field of mathematics
known as algebraic number theory.
56
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