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The Law of Quadratic ReciprocityProof: By Wilson’s Theorem
By Kylee Bracher
Portland State UniversityDepartment 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
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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.
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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)
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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 if they have the same remainder when
divided by n.
Lemma 1: Let a, b, and n > 0 be integers. Then if and only if
Proof: If , 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 . Now if we solve both of these equations for r we
have
If then which is what we wanted to show.
Now assume that . By the division algorithm there exists integers k and
m such that, a = nk + and b = nm + , where and . Subtracting b
from a we obtain the equation , and since it was assumed
, it must be that . But and so and the
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only multiple of n in this interval is 0. Thus we have that or equivalently
, and so which is what we wanted to show. ▲
The reader may think that we do not need a new and slightly longer way to say
. However the notation 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) .
(2) If , then .
(3) If and , then .
(4) If and , then and
.
(5) If , then and .
(6) If , then for any positive integer k.
Theorem 2: (Cancellation Law): If a is relatively prime to p and x and y are integers
such that , then
Proof: By definition means p divides . 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 which is what we wanted to show. ▲
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.
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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
Proof: Let (a, n) = 1 then by the Euclidean Algorithm we can write 1 as a linear
combination of a and n, that is, for some integers x and y. Rearranging the
terms in this equation we have . Hence which implies that
by the definition of congruence.
To prove uniqueness, suppose that and . Then
(since (a, n) = 1)
By definition of congruence if then . Therefore the solution
x must be unique modulo n. ▲
Definition 3: Let a be any integer. Any integer x for which is called the
inverse of a.
Definition 4: Let n be a positive integer, and let a be an integer such that Then a is
called a quadratic residue modulo n if the congruence
is solvable, and a quadratic nonresidue otherwise.
Definition 5: (The Legendre Symbol): Suppose p is an odd prime not dividing a. We
define the symbol to be 1 or – 1 according to whether a is a quadratic residue or
quadratic nonresidue mod p. This is called the Legendre Symbol, after the French
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mathematician Adrien Marie Legendre (1752 – 1833). We also call the quadratic
character of a with respect to p.
If then a is a quadratic residue modulo p which means the congruence
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
is a polynomial of degree with integral coefficients, then the congruence
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 If the
congruence is solvable, then there are exactly two incongruent solutions
modulo p.
Proof: Suppose that the congruence is solvable and let be one of
the solutions. Note since Then there must also be a second solution
since and . This
second solution is not congruent to the first. To see this, note that
implies that or equivalently which is impossible. Now
by Lagrange’s Theorem, these two solutions exhaust the incongruent solutions of
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. Thus, if 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
equal to 1, or , because there exists an x such that is solvable. To
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
not divisible by 11
not divisible by 11
not divisible by 11
not divisible by 11
divisible by 11. Thus, is a solution.
divisible by 11. So, 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,
. So we could write that the solutions to the congruence
are exactly . It is easy to see that this fact is true since,
is clearly true. A similar argument would show that the congruence holds for
.
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 . For example, 16 is also a solution to the
congruence since
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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 . The integers -
16, -27, -38, . . . . . . . , - 5 – 11k where k is any integer are also solutions because each of
these integers are congruent to . ▲
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:
Squaring the numbers from 6 to 10 yields the same residues in reverse order:
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 yield the same residues as the numbers from
to p, the quadratic residues can be found simply by squaring the numbers from 1 to
.
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 as was illustrated in the example above. Note that
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So the squares of the numbers from
up to p – 1 are congruent to the squares of the numbers down to 1.
All we have left to show is that each of the squares of is
not congruent to any one of the other squares. We will do this by contradiction.
Suppose that there exist two squares such that where i and j are
distinct and both in the range 1 to Then
or
These last two statements are always false because both i and j lie within the
range Thus, i – j and i + j are both too small to be divisible by p. So
we have shown that none of the squares are congruent mod p. Thus, there must be
exactly quadratic residues of p. ▲
In general, the above theorem says that there are exactly quadratic residues
and exactly nonresidues of p.
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.
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Theorem 5: (Fermat’s Little Theorem): Let p be any prime number. If a is any integer
such that then
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,
Regrouping the left hand side we obtain:
or equivalently
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
which is what we wanted to show. ▲
Theorem 6: (Euler’s Criterion): Let p be an odd prime and a any integer such that .
Then a is a quadratic residue modulo p if and only if
Proof: Let a be a quadratic residue modulo p. Then there must exist an x such
that . Since it was assumed that this implies Thus by
Fermat’s Little Theorem we have
Hence if a is a quadratic residue modulo p then which is what we
wanted to show.
Recall that if (r, p) = 1 and r has order modulo p (where is the largest
possible order), then r is called a primitive root of p.
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Now suppose that the congruence 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 for some integer m, with Thus it follows that
Given that r is a primitive root of p the order of r is p – 1. By the previous
congruence so p – 1 must divide m(p – 1)/2 which implies that
m must be an even integer. Let m = 2k, then
making the integer a solution of the congruence 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. ▲
The following results are direct consequences of Euler’s Criterion.
Lemma 3: The congruence is solvable if and only if
Proof: By Euler’s Criterion, is solvable if and only if
. But is either 1 or -1 depending on whether (p –
1)/2 is even or odd, respectively. Since p > 2, if and only if (p
– 1)/2 is even. But (p – 1)/2 is even if and only if which is what we
wanted to show. ▲
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Example 5: When p = 17, the congruence is solvable since
. The solutions to the congruence are exactly
. ▲
Corollary 1: Let be an odd prime and a any integer such that . 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 So to show that –a is a
quadratic nonresidue we need only show that But
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 So to show that a is a
quadratic nonresidue we need only show that But
since (p – 1)/2 is odd. Hence a is a quadratic nonresidue modulo p, which is what we
wanted to show. ▲
Corollary 2: Let be an odd prime and a any integer such that . 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 So to show
that –a is a quadratic residue modulo p we need only show that
But
since (p – 1)/2 is even. Hence –a is a quadratic residue modulo p.
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Now assume that a is a quadratic nonresidue modulo p. Then by Euler’s
Criterion, To Show that –a is also a quadratic nonresidue
modulo p we need only show that But
since (p – 1)/2 is even. Hence –a is a quadratic nonresidue modulo p, which is what
we wanted to show. ▲
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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
.
We should remark that by the statement of the Law of Quadratic Reciprocity
if or , while if .
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.
Example 6: Let p = 3 and q = 17 then which is
what is expected since q = 17 is of the form 4k + 1. Thus by the Law of Quadratic
Reciprocity the congruences
and
are either both solvable or both insolvable. The easiest congruence to check would be
since it reduces to the congruence which is clearly not
solvable since and are not divisible by 3. Therefore both of
the congruences are insolvable and so and . ▲
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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
To determine the value of we must consider two things:
1) The value of ;
2) The value of the exponent on -1.
To find the value of we can find the least residues modulo 7 squaring the numbers
from 1 to 3 and reducing them mod 7. Squaring the numbers from 1 to 3
yields . 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 for each
value of a from 1 to 6.
If , then since 1 is a quadratic residue mod 7.
If , then since 2 is a quadratic residue mod 7.
If , then since 3 is a quadratic nonresidue mod 7.
If , then since 4 is a quadratic residue mod 7.
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If , then since 5 is a quadratic nonresidue mod 7.
If , then since 6 is a quadratic nonresidue mod 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
If , then the exponent on -1 is
an even number.
Case 2) p = 4k + 3 where k is any integer
If , then the exponent on -1 is
an odd number.
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 , then
28 | (p – 1)
and
and
the exponent on -1 is even so
.
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Hence the value of the Legendre symbol is
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
the same quadratic characters for and In this case, they are both solvable since
Next we will illustrate the case when both p and q are of the form 4k + 3, if
then
and
and
the exponent -1 is odd so
(since 3 is a quadratic residue mod 7)
Hence the value of the Legendre symbol for
In this example the value of the quadratic characters for and are opposite, so 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.
1 if
= -1 if
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From this it is easy to see that if for example p = 27 then Thus, there must exist
an x such that is solvable. After checking the values from 1 to 27 it can
be easily be verified that are the solutions to the congruence since
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
and only one is solvable. The first congruence
reduces to since 1,999 is congruent to 1 mod 3.
is clearly solvable when since
Therefore, by the Law of Quadratic Reciprocity the second congruence
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
and are either both solvable or unsolvable. The
first congruence reduces to since 1,999 is congruent to
4 mod 5. The congruence is clearly solvable when since
.
By the Law of Quadratic Reciprocity the other congruence 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 If p is of the form 4k + 3
(which in this case it is, then one solution is So for this
particular example the congruence has the solution
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With the help of a calculator it can be checked that this yields
, so the other answer is . ▲
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.
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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 , reduced to lie between
to . If the number of these which are negative is s, then .
Proof: Let the least residues be for and for where
, and We first claim, with proof to follow,
that the set Given that this is true we
have
if we divide both sides by we are left with
.
Now by Euler’s Criterion . Thus we have that
.
Given that and will both yield values of either 1 or -1, it follows that
(since p is prime) which is what we wanted to show.
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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 or
equivalently To do this we must
show that no two are congruent, no two are congruent and that there are no
We will do this by contradiction.
Let’s assume that two are congruent then with
and assuming without loss of generality that (if we would be talking
about the same element in the set).
(since
This last statement is always false because both and fall within the range
So Thus, is too small to be divisible by p.
Next we assume that two are congruent then with
Now
(since
This last statement is again false by the same reasoning as above.
Finally, we will assume that for some , then for
some So
(since
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This last statement is also false because again both and fall with the range
So Clearly no integer within this range is divisible by p.
Thus, we have shown that no two integers in the set
are congruent to one another and since there are exactly elements in this set,
each within the range 1, . . . . , , it must be that
Which is what we wanted to show.
▲
Example 9: Take p = 11 and a = 19, we first reduce each 19k for mod 11 to lie
within the range to . Thus, we get
for k = 1,
for k = 2,
for k = 3,
for k = 4,
for k = 5,
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 . From the chart above we
see there are 3 minus signs, so by Gauss’s Lemma . So 19 is a quadratic
nonresidue mod11. Thus there does not exist an x such that . ▲
In this next example we will just reverse the roles of 19 and 11.
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Example 10: Now let p = 19 and a = 11, we first reduce each 11k for mod 11 to
lie within the range to . Thus, we get
for k = 1,
for k = 2,
for k = 3,
for k = 4,
for k = 5,
for k = 6,
for k = 7,
for k = 8,
for k = 9,
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 . From the
chart above we see that there are 4 minus signs, so by Gauss’s Lemma .
In this case, 11 is a quadratic residue mod 19. Thus, the congruence is
solvable. In fact, the solutions to this congruence are exactly since
19 | 38
which is clearly true. A similar argument would hold for ▲
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.
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Consider the rectangle R in the x, y plane with sides y = 1 and and
vertical sides x = 1 and The total number of lattice points in or on the
rectangle is equal to (exactly equal to the exponent on (-1) in the
statement of the theorem). This is illustrated in the graph below where p = 11 and q =
19, in this case there are lattice points in the rectangle.
Figure 1
Next we will use three lines to divide the rectangle into four regions. Let
L be the line (or x = )
be the line
be the line (or )
Figure 2
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We should note that there are no lattice points on any of these lines within the
rectangle because and since all of the coordinates x within the rectangle are
in the range and all of the coordinates y are in the range
We will let A be the region above where
B be the region below where
be the region between L and where
be the region between L and where
This again is illustrated in the graph below for the case when p = 11 and q = 19.
Figure 3
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If we let P(X) denote the number of lattice points in region X it should be that
P(A) + P(B) + P( ) + P( ) =
Thus, the statement of the theorem could be written in the form
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
We will then show that contains points where and that
contains where . So we could again restate the theorem as
where, .
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To see that P(A) = P(B) note that the entire picture is symmetric about its center
point as illustrated in the graph below.
Figure 4
To verify this claim, let f : be defined by
with the property that 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
be any lattice point in A, then x and y are both integer values such that
. We wish to show that is a
lattice 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
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check that if then . Hence , which is what we
wanted 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,
and
and b = 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
there exists an such that Suppose
and let , then (x, y) since by definition of the region
B and thus the reader can easily verify through algebra that this
implies that . Hence (x, y) and (x, y) is clearly a lattice point since p
and q were assumed to be odd. Further,
f (x, y)
Thus, there exists a lattice point , namely , such that
30
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 Let Then
if and only if by definition of the region The
interval has length so it contains at most one integer y. There will be an integer y
in this interval if and only if there is an integer (where denotes the
greatest integer) in the interval
which is if and only if
Multiplying through by p we get the equivalent statement
.
The integer in the middle is the remainder on dividing qx by p. The number of x’s in
the range for which the remainder lies between and p is exactly the
same integer s which occurs in Gauss’s Lemma satisfying A similar
argument could be used to show that the number of lattice points in is the integer
which appears in Gauss’s Lemma such that Thus we have,
which is exactly what we wanted to show. ▲
31
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
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 and
vertical sides x = 1 and . The total number of lattice points in or
on the rectangle is equal to , as illustrated in the graph
below.
Figure 5
Next we will use three lines to divide the rectangle into four regions. We will
let
be the line
be the line
be the line
We should note that no lattice points lie on any of these lines within the
rectangle because when and similarly when .
We will let
A be the region above where
32
B be the region above where
be the region between L and where
be the region between L and where
Figure 6
If we let P(X) denote the number of lattice points in Region X, it should be that,
Thus, the statement of the theorem could be rewritten as
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
33
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 .
Next we will show that the region contains lattice points where
by showing that is the exact same number that occurred in Gauss’s
Lemma when p = 11 and q = 19. Similarly, we will show that contains lattice
points where by showing that is the exact same number that occurred
in Gauss’s Lemma when p = 19 and q = 11. So we would have
where must be congruent to mod 2.
34
Let , by definition Since the length of
the 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 in the interval
.
which is if and only if
or multiplying through by 11 we get the equivalent statement
The integer in the middle, , is the remainder on diving 19x by 11.
Computing these remainders for the integers x in the range , yields
for x = 1,
for x = 2,
for x = 3,
for x = 4,
for x = 5,
Thus, there are only three integers that lie within the range and 11, so = 3.
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 = s = 3. Thus, the number of lattice points in is the exact
35
same number which occurred in Gauss’s Lemma for By counting the number
of lattice points in region on the graph we see there are indeed 3. So
By using a similar argument we will show that the number of lattice points in
is the same as the number of minus signs we found earlier using Gauss’s Lemma for
p = 19 and q = 11. Let by definition if and only if
The interval again is only of length ½, so it will contain at most
one integer x. There will be an integer x in this interval if and only if there is an
integer in the interval
which is true if and only if
Multiplying through by 19 we obtain the equivalent statement
The integer in the middle here is the remainder on dividing 11y by 19.
Computing these remainders for the integers y in the range we see
for y = 1,
for y = 2,
for y = 3,
for y = 4,
for y = 5,
36
for y = 6,
for y = 7,
for y = 8,
for y = 9,
As we see there are only four integers that lie within the range to 19, so
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 Thus,
the number of lattice points in region corresponds exactly to the number which
occurred in Gauss’s Lemma for Counting the number of lattice points on the
graph we see that region does contain four lattice points. So
Therefore, we have
which is exactly what we wanted to show. ▲
37
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 if and only if and
.
Proof: Let p and q be two integers such that (p, q) = 1 and suppose that there
exists integers x and a such that By definition of congruence if
then . But if then and and so
and 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 and . If then by
definition of congruence which implies that there exists and integer r such
that (x – a) = rp. Similarly, if then by definition of congruence
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 then
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 and so
which is what we wanted to show. ▲
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
38
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
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
and so x = asq + brp must be a solution to the system of congruences
and . This can be seen by reducing modulo p and then modulo q as
follows,
and
Thus, if (p, q) = 1 the system of congruences and will have
a solution.
If x is a solution, then adding any multiple of pq to x will still be a solution since
for any integer k. Conversely if and are two solutions of
the given congruences, then they must be congruent modulo p and modulo q, that is if
and
then and . Thus is divisible by both p and q and
so it is divisible by pq since by assumption (p, q) = 1. Therefore
which is what we wanted to show. ▲
39
Example 12: To solve the system
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 . ▲
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 .
Considering the list of integers x, x + p, x + 2p, . . . . . , x + (q – 1)p we have that
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, 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, for some
integer i. Now we have that and also If
then Thus we have that and
so 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.
Theorem 8: (Wilson’s Theorem): Let p be an integer. Then p is a prime if and only if
Proof: If p = 2, then 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.
40
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 , then multiplying both sides of this
congruence by a yields . If then or
equivalently . Given that p is prime,
implies 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
but
Hence
.
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
since d must be one of the integers between 1 and p – 1. However, we assumed
the congruence 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
, not inclusive, with their inverses. Indeed,
, , ,
Thus,
▲
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.
41
Lemma 7: Let p, q be distinct odd primes. If r denotes any quadratic residue (mod pq),
then the congruence has exactly four distinct solutions, through
that satisfy: (1)
(2)
In particular, two solutions have positive least residues modulo pq.
Proof: Assume r is a quadratic residue mod pq. Then the congruence
has solutions mod pq. So by Lemma 5, this congruence has solutions mod p and mod
q. Indeed, by Proposition 1, has exactly two incongruent solutions mod p, call
them c and –c. Similarly, has exactly two incongruent solutions mod q, call
them d and – d.
Therefore must have exactly four solutions , , , and
, each satisfying a pair of congruences as follows:
(i) and
(ii) and
(iii) and
(iv) and
By the Chinese Remainder Theorem through are each uniquely determined
(mod pq). Moreover, through are all distinct (mod pq) by construction.
Observe that and 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
are conjugates.
Corollary 3: When the p and q are distinct odd primes then the number of quadratic
residues modulo pq is exactly .
42
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
quadratic residues modulo pq. ▲
RemarkBy the Lemma 7, it follows that for odd primes , there will be four solutions to
. If we consider the least residues of these solutions, two will be positive
and two will be negative. Clearly 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 through represent the four
conjugates stated in Lemma 7 satisfying:
(1)
(2)
Let a and b be two remainders such that and . Let r be any
quadratic residue modulo pq.
(i) If , then exactly one of the four solutions of
is itself a quadratic residue, the other three are not.
(ii) If and (or visa versa) then either exactly two of
the four solutions of are quadratic residues or else none of
them are.
(iii) If then either all four solutions of are
quadratic residues, or else none of them are.
43
Proof (i): Let p and q be odd primes such that . 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, and
are both quadratic residues modulo p. Similarly if b is a quadratic residue modulo q
then by definition of b and (2) above, and are quadratic residues modulo q.
Hence 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 and are both quadratic residues modulo p by
definition (1) above. If b is a quadratic residue modulo q then and are both
quadratic residues modulo q by definition (2) above. Hence is a quadratic residue
modulo pq by Lemma 6 and the others are not.
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 and are both quadratic residues modulo p by
definition (1) above. If –b is a quadratic residue modulo q then and are both
quadratic residues modulo q by definition (2) above. Hence 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 and are both quadratic residues modulo p by
44
definition (1) above. If –b is a quadratic residue modulo q then and are both
quadratic residues modulo q by definition (2) above. Hence is a quadratic residue
modulo pq by Lemma 6 and the others are not.
Thus we have shown that when in each of these cases only one of
the solutions through is a quadratic residue modulo pq, which is what we
wanted to show.
Proof (ii): Without loss of generality, assume and . If
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 , , , and are all quadratic residues modulo p. Now since
, 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 and are quadratic residues modulo q. Thus by Lemma 6,
and would be quadratic residues modulo pq and and would not.
Similarly if – b is a quadratic residue modulo q then by definition (2) above and
are quadratic residues modulo pq. Thus by Lemma 6, and would be
quadratic residues modulo pq and and would not. In either case exactly two of
the solutions through are quadratic residues modulo pq.
Case 2: (a and –a are quadratic nonresidues)
Assume that a and –a are quadratic nonresidues modulo p then through are
quadratic nonresidues modulo p. Therefore through are quadratic nonresidues
modulo pq.
45
Thus we have shown that if and (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 . If then both a and – a are either both
quadratic residues or both quadratic nonresidues modulo p. Similarly, if
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 through are quadratic nonresidues modulo p.
Hence all four solutions are quadratic residues modulo pq.
We have shown that if 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
{– 2, 2, – 5, 5} solve
{– 4, 4, – 10, 10} solve
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
46
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
and
and
and
and
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
and
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
and so – 5 is a quadratic residue modulo 7 and 5 is not. Thus the conjugate is the
only quadratic residue modulo 21. If and then
and – 5 is a quadratic residue modulo 21 since and
▲
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
{– 2, 2, – 7, 7} solve
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
47
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
and
and
and
and
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
and
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
and so – 4 and 4 are both quadratic residues modulo 5. Thus the conjugates and
must be quadratic residues modulo 15. If and then
and 4 is a quadratic residue modulo 15 since If
and 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 .
Thus 1 is a quadratic residue modulo 15 since ▲
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
{– 1, 1, – 14, 14} solve {– 8, 8, – 18, 18} solve
{– 2, 2, – 28, 28} solve {– 11, 11, – 24, 24} solve
48
{– 3, 3, – 23, 23} solve {– 12, 12, – 27, 27} solve
{– 4, 4, – 9, 9} solve {– 16, 16, – 29, 29} solve
{– 6, 6, – 19, 19} solve {– 21, 21, – 31, 31} solve
{– 7, 7, – 32, 32} solve {– 17, 17, – 22, 22} solve
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 and
and
and
and
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
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
and so –1 and 1 are both quadratic residues modulo 13. Thus all four conjugates must be
quadratic residues modulo 65. If and 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
. Thus 14 is a quadratic residue modulo 65. If and
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 . Thus
49
64 is a quadratic residue modulo 65. If and 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 and 1 is a quadratic residue modulo
65. Finally, if and 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 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.
Definition 7: We will define to be the product of positive remainders that are prime
to pq. That is,
Lemma 9: Let p and q be distinct odd primes.
(i) If then .
(ii) Otherwise,
and .
Proof (i): If we assume that both p and 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 Thus we have four
50
remainders 1, and whose squares and product are For
each of the remaining positive remainders a of pq there exists a unique remainder b
such that and so the entire product
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 , , etc. when reduced
modulo pq. Thus, in all of the remaining cases we have that ▲
Corollary 4: Let p and q be distinct odd primes. Then
Proof: The exponent is even if and only if p and . So
this corollary holds by the work above. ▲
Lemma 10: Let p and q be distinct odd primes. Then
Proof: In order to determine the signs on and we will
proceed by looking at the following product:
(1)
If we consider this product modulo p then we obtain
Now by Wilson’s Theorem we have that and so
51
Hence
Comparing this product to we note that both products contain the same number
of factors, namely (p – 1) . We can derive the product in (1) from the product
by multiplying by and dividing by
.
That is,
But
Thus
or equivalently
52
or
or multiplying both sides by we have
(I) .
If we were to do the same analysis interchanging the roles of p and q we would obtain
the congruence
(II) .
If we multiply (I) and (II) we obtain
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
and by the previous Lemma 8 we
have just shown that Thus we
have that
.
53
Dividing both sides of this equality by and subtracting exponents, the
reader could easily verify that this expression simplifies to
.
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
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
and
and
Thus we have the four positive remainders 1, = 1, 14, 8,
whose product Now we will pair each of the following
positive remainders with their unique inverses as follows:
.
Thus considering the product modulo 65 yields
(since
)
. ▲
54
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
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
and .
Thus we have the two positive remainders 1 and = 1 and 21 whose squares are
Now we will pair each of the following positive remainders with their
unique inverses as follows:
Thus considering the product modulo 55 yields
.
Now and . Thus
and so exactly one of the congruences
and and the other is not. It might be easiest to check the
congruence since it reduces to . By Lemma 3, the
congruence is solvable since . Therefore, by the Law of
Quadratic Reciprocity, the congruence 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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