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

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.

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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

5

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.

6

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

7

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

8

. 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.

14

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 . ▲

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

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.

24

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.

25

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

26

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

27

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, .

28

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

29

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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