QUADRATIC FORMS, RECIPROCITY LAWS, AND PRIMES OF THE …math.uchicago.edu/~may/REU2018/REUPapers/Skenderi.pdf · and yare integers, began with Fermat, who conjectured necessary and

QUADRATIC FORMS, RECIPROCITY LAWS, AND PRIMES OF

THE FORM x2 + ny2

ALEKSANDER SKENDERI

Abstract. This paper demonstrates necessary and sufficient conditions, for

several values of n, for when a prime p can be written in the form x2 + ny2.In doing this, we provide an introduction to quadratic forms and elementary

genus theory. Moreover, we show how cubic and biquadratic reciprocity can

be used to tackle our problems in certain more complicated cases. The paperassumes a good knowledge of quadratic reciprocity, as well as basic abstract

algebra, such as familiarity with rings, fields, integral domains, and ideals.

Contents

1. Introduction 12. The Theory of Quadratic Forms 13. Elementary Genus Theory 94. Reciprocity Laws 124.1. Cubic Reciprocity 124.2. Biquadratic Reciprocity 19Acknowledgments 23References 23

1. Introduction

The question of when a prime can be written in the form x2 + ny2, where xand y are integers, began with Fermat, who conjectured necessary and sufficientconditions for primes of the form x2 + y2, x2 + 2y2, and x2 + 3y2. Several questionsimmediately come to mind: can we formulate similar results for other values of n?For how many values of n do we have such conditions? We will tackle the firstquestion, proving Fermat’s conjectures, in addition to addressing more advancedcases. Lagrange, Legendre, and Gauss made progress on this problem by first de-veloping a theory of quadratic forms. Gauss and Eisenstein worked on cubic andbiquadratic reciprocity, which we will use to prove results for the cases when n = 27or 64, respectively. As we shall soon see, the mathematics they developed is verybeautiful as well as a great source of mathematical inspiration.

2. The Theory of Quadratic Forms

The theory of quadratic forms, when used in conjunction with quadratic reci-procity, is exceedingly useful in giving necessary and sufficient conditions for whena prime may be represented in the form x2 +ny2. Only the notions of equivalence,

1

2 ALEKSANDER SKENDERI

the discriminant, and reduced form need to be developed in order to prove thefollowing beautiful results, which we will address at the end of this section:

(1) p = x2 + y2, if and only if p ≡ 1 (mod 4),

(2) p = x2 + 2y2, if and only if p ≡ 1, 3 (mod 8),

(3) p = x2 + 3y2, if and only if p = 3 or p ≡ 1 (mod 3),

(4) p = x2 + 7y2, if and only if p = 7 or p ≡ 1, 9, 11, 15, 23, 25 (mod 28).

This is already remarkable, as it not only fully resolves, but also extends upon theconjectures made by Fermat. Once we introduce the additional notion of genus, wewill prove similar results for many other cases, such as n = 6 or 10. Before tacklingour main question, we need to develop a theory of quadratic forms.

Definition 2.1. An integral quadratic form is a homogeneous polynomial of degree2 of the form f(x, y) = ax2 + bxy + cy2, where a, b, and c are integers.

Definition 2.2. A form ax2 + bxy+ cy2 is primitive if its coefficients a, b and c arerelatively prime. That is, gcd(a, b, c) = 1.

Definition 2.3. An integer m is represented by a form f(x, y) if the equationm = f(x, y) has an integer solution in x and y. Moreover, if x and y are relativelyprime, then we say that m is properly represented by f(x, y).

Thus, we are interested in the following question: when is a prime representedby the quadratic form x2 + ny2? Studying this question requires the notion ofequivalence of quadratic forms.

Definition 2.4. Two forms f(x, y) and g(x, y) are equivalent if there exist integersp, q, r and s such that ps− qr = ±1 and,

f(x, y) = g(px+ qy, rx+ sy).

If p, q, r and s can be chosen so that ps − qr = 1, then the equivalence is calledproper.

Notice that going from one form to an equivalent form amounts to nothing elsethan changing basis, and requiring that the determinant of the matrix

M =

[p qr s

]be ±1. An important property of equivalent forms, that follows directly from thedefinition (and is what motivates the definition), is that equivalent forms representthe same numbers. Moreover, properly equivalent forms properly represent thesame numbers. The following is a very useful lemma.

QUADRATIC FORMS, RECIPROCITY LAWS, AND PRIMES OF THE FORM x2 + ny2 3

Lemma 2.5. A form f(x,y) properly represents an integer m if and only if f(x,y)is properly equivalent to the form mx2 +Bxy + Cy2 for some B,C ∈ Z.

Proof. First, suppose that f(p, q) = m, where gcd(p, q) = 1. By the EuclideanAlgorithm, there exist integers r and s such that ps− qr = 1. By writingf(x, y) = ax2 + bxy + cy2, a short calculation shows that

f(px+ ry, qx+ sy) = f(p, q)x2 + (2apr + bps+ brq + 2cqs)xy + f(r, s)y2

= mx2 +Bxy + Cy2, where

B = 2apr + bps+ brq + 2cqs, and C = f(r, s).

Conversely, suppose that there exist integers p, s, q and r with ps − qr = 1 suchthat

f(px+ ry, qx+ sy) = mx2 +Bxy + Cy2,

for some integers B and C. Notice that (x, y) = (1, 0) gives f(p, q) = m, wheregcd(p, q) = 1, since otherwise we could not have ps− qr = 1.

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Definition 2.6. The discriminant D of a quadratic form f(x, y) = ax2 + bxy+ cy2

is D = b2 − 4ac.

Remark 2.7. Suppose that f(x, y) and g(x, y) are equivalent forms. That is,f(x, y) = g(px + qy, rx + sy), with discriminants D and D′, respectively. Then, asmall calculation shows that

(2.8) D = (ps− qr)2D′.Therefore, D = D′ since ps − qr = ±1. Thus, equivalent forms have the samediscriminant. However, forms with the same discriminant are not always equivalent,as we shall see in Remark 2.12.

Another important feature of the discriminant is that it has a strong effect onthe sign of the values represented by the corresponding quadratic form. Letf(x, y) = ax2 + bxy + cy2 be a form of discriminant D. Then, another littlecalculation verifies that

(2.9) 4af(x, y) = (2ax+ by)2 −Dy2.We call a form positive definite if it only represents non-negative integers, andnegative definite if it only represents non-positive integers. If D < 0, then theright-hand side is always non-negative. Thus, f(x, y) is positive definite if a > 0,and negative definite if a < 0.

Lemma 2.10. Let D ≡ 0, 1 (mod 4) be an integer and m an odd integer relativelyprime to D. Then m is properly represented by a primitive form of discriminant Dif and only if D is a quadratic residue modulo m.

Proof. If a primitive form f(x, y) of discriminant D properly represents m, then byLemma 2.5, f(x, y) is properly equivalent to the form mx2 + bxy + cy2 for someintegers b and c. By the above remark, these two forms have the same discriminantand so we see that D = b2 − 4mc ≡ b2 (mod m), as desired.Conversely suppose that D ≡ b2 (mod m). Since m is odd, we can assume Dand b have the same parity. From the fact that D ≡ 0, 1 (mod 4), we see thatD ≡ b2 (mod 4), and thus D ≡ b2 (mod 4m), because gcd(4,m) = 1. Therefore,


D = b2 − 4mc, for some c. Then, the form mx2 + bxy + cy2 properly representsm, has discriminant D, and has relatively prime coefficients, since m and D arerelatively prime.

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The lemma gives the following very useful corollary.

Corollary 2.11. Let n be an integer and p an odd prime not dividing n. Then(−np

)= 1 if and only if p is properly represented by a primitive form of

discriminant −4n.

Proof. By the properties of the Legendre symbol, we have(−4n

p

)=

(2

p

)2(−np

)=

(−np

),

and so −4n is a quadratic residue modulo p if and only if −n is a quadratic residuemodulo p. Applying Lemma 2.10 completes the proof.

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Remark 2.12. Notice that this corollary is relevant to our question of which primesare of the form x2 + ny2, since x2 + ny2 is primitive and has discriminant −4n.The problem is that there are many forms of the same discriminant. For example,x2 + 5y2 and 2x2 + 2xy + 3y2 are both forms of discriminant −20, but they arenot equivalent. Both forms are positive definite by (2.9), but x2 + 5y2 represents 1(let (x, y) = (1, 0)), whereas 2x2 + 2xy+ 3y2 does not. In fact the smallest positiveinteger that 2x2 + 2xy+ 3y2 represents is 2 (let (x, y) = (1, 0)). Therefore, we needto show that every quadratic form is equivalent to a very simple one.

Definition 2.13. A primitive positive definite form ax2 + bxy + cy2 is said to bereduced if |b| ≤ a ≤ c and b ≥ 0 if either |b| = a or a = c.

Primitive positive definite forms have a very important property, illustrated inthe following theorem.

Theorem 2.14. Every primitive positive definite form is properly equivalent to aunique reduced form.

The proof is long, tedious, and not very illuminating. See Theorem 2.8 of Cox[1] for the details.

Theorem 2.14 allows us to sometimes determine whether forms are merely equiv-alent, or actually properly equivalent. Consider reduced forms

ax2 ± bxy + cy2.

They are equivalent since we may go from one form to the other by sending (x, y) to(x,−y). However, since they are both reduced, Theorem 2.14 tells us that they arenot properly equivalent. Moreover, if we consider primitive positive definite forms

ax2 ± axy + cy2 with a ≤ c,

then only ax2 + axy + cy2 is reduced. By Theorem 2.14, the forms are properlyequivalent.


Definition 2.15. Two forms are said to be in the same class if they are properlyequivalent.

Theorem 2.16. Let D < 0 be fixed. Then the number of classes of primitivepositive definite forms of discriminant D, denoted by h(D), is finite; h(D) is infact equal to the number of reduced forms of discriminant D.

Proof. We first show that the number of reduced forms is finite. Let f(x, y) =ax2 + bxy+ cy2 be a reduced form of discriminant D < 0. Then, b2 ≤ a2 and a ≤ cimply that

−D = 4ac− b2 ≥ 4a2 − a2 = 3a2.

Therefore,

(2.17) a ≤√−D

3.

This inequality, along with the fact that |b| ≤ a, implies that there are only finitelymany choices for a and b. Since D = b2 − 4ac, there are also only finitely manychoices for c. Hence, there are only a finite number of reduced forms of discriminantD. By Theorem 2.14, the number of classes h(D) is finite, and is precisely thenumber of reduced forms of discriminant D.

�

Another consequence of Theorem 2.14 is that we have the following improvementto Corollary 2.11.

Theorem 2.18. Let n be a positive integer and p an odd prime not dividing n.

Then,

(−np

)= 1 if and only if p is represented by one of the h(−4n) reduced

forms of discriminant −4n.

Proof. By Corollary 2.11,

(−np

)= 1 if and only if p is properly represented by

a primitive form of discriminant −4n. Moreover, the proof of Lemma 2.10 showsthat we can find such a quadratic form whose coefficient in front of x2 is p. Hence,by (2.9), the form properly representing p is primitive positive definite, and byTheorem 2.16, this occurs if and only if p is represented by one of the h(−4n)reduced forms of discriminant −4n.

�

This theorem tells us something very interesting: whether a prime p is repre-sented by a reduced form of discriminant −4n is simply a matter of whether thereis a solution to the congruence x2 ≡ −n (mod p). In fact, we can obtain a moregeneral result (although the discriminant −4n is what interests us). First, we needa lemma, making heavy use of quadratic reciprocity.

Lemma 2.19. If D ≡ 0, 1 (mod 4) is a non-zero integer, then there is a unique

homomorphism χ : (Z/DZ)∗ −→ {±1} such that χ([p]) =

(D

p

)for odd primes p

not dividing D. Furthermore, χ([−1]) = 1 if D > 0 and χ([−1]) = −1 if D < 0.


Proof. The proof will be done in several steps. We first show that ifm ≡ n (modD),

where m,n are odd and positive, and D ≡ 0, 1 (mod 4), then

(D

m

)=

(D

n

). By

quadratic reciprocity we have that

(2.20)

(D

m

)= (−1)

(D−1)(m−1)4

(m

D

)and

(D

n

)= (−1)

(D−1)(n−1)4

(n

D

).

Consider the case when D ≡ 1 (mod 4) and D > 0. Now, m ≡ n (mod D) implies

that

(m

D

)=

(n

D

). Finally, from D ≡ 1 (mod 4) and m ≡ n ≡ 1 (mod 2), we

obtain the identity

(D − 1)(m− 1)

4≡ (D − 1)(n− 1)

4≡ 0 (mod 2).

Hence,

(D

m

)=

(D

n

).

Now, suppose D ≡ 0 (mod 4) and D > 0. Then, D = 2ka, where k ≥ 2 anda is odd and positive. Moreover, (2.20) implies that m ≡ n ≡ 1 (mod 4). By theproperties of the Jacobi symbol and quadratic reciprocity, we have(

D

m

)=

(2

m

)k(a

m

)= (−1)k·

m2−18

(a

m

)= (−1)k·

m−14 ·

m+12 (−1)

(m−1)4 ·(a−1)

(m

a

)= (−1)k·

m+12 ·(a−1)

(m

a

)=

(m

a

),

since both m and a are odd. Similarly,

(D

n

)=

(n

a

). Since m ≡ n (mod a), we

see that

(D

m

)=

(D

n

).

Consider the cases when D is negative. First, suppose D ≡ 0 (mod 4). We compute(D

m

)=

(−1

m

)(−Dm

)= (−1)

m−12

(−Dm

), and similarly

(D

n

)= (−1)

n−12

(−Dn

).

By analogous reasoning as before, m ≡ n ≡ 1 (mod 4). Therefore,

(−1)m−1

2 = (−1)n−12 = 1.

The fact that −D > 0 implies that

(−Dm

)=

(−Dn

). This shows that(

D

m

)=

(D

n

).


Lastly, consider the case when D is negative and D ≡ 1 (mod 4). By quadraticreciprocity, (

D

m

)= (−1)

m−12 (−1)

(m−1)(−D−1)4

(m

−D

), and

(D

n

)= (−1)

n−12 (−1)

(n−1)(−D−1)4

(n

−D

).

But, m ≡ n (mod−D) and so

(m

−D

)=

(n

−D

). Also

(−1)m−1

2 (−1)(m−1)(−D−1)

4 = ((−1)m−1

2 )2(−1)−D−1

2 = (−1)−D−1

2 , and similarly

(−1)n−12 (−1)

(n−1)(−D−1)4 = (−1)

−D−12 .

This completes the last check that

(D

m

)=

(D

n

). We are done with the first step

of the proof.

We now show that for m odd and positive, χ([m]) =

(D

m

)gives a well-defined

homomorphism from (Z/DZ)∗ to {±1}. Notice that all the equivalence classes [k]in (Z/DZ)∗ may be represented by odd elements. If D ≡ 0 (mod 4), then noneof the even numbers in the set {0, 1, ..., D − 1} are invertible, and therefore allthe equivalence classes in (Z/DZ)∗ are represented by odd elements. Then, for asufficiently large integer a, the number m = k + a|D| is positive, still odd, and[m] = [k]. If D ≡ 1 (mod 4), then the equivalence classes can be represented byeither odd or even numbers. Let [k] be one such class. Because k and k+ |D| havedifferent parity, if we consider m = k + a|D| for a sufficiently large, then m is odd,positive, and [m] = [k]. Therefore, any two elements in an equivalence class maybe represented by odd and positive numbers m and n, where m ≡ n (mod D). Butwe showed that in this case, (

D

m

)=

(D

n

).

It follows that,

χ : (Z/DZ)∗ −→ {±1}

[m] 7−→(D

m

),

is well defined. The fact that χ is a homomorphism follows from the fact thatthe Jacobi symbol is multiplicative. The result for χ([−1]) follows from quadraticreciprocity, in a similar manner to what we did before. Moreover,

χ([p]) =

(D

p

)for odd primes p not dividing D. Therefore, χ is determined uniquely because itis a homomorphism, and every class contains a positive odd number, which is aproduct of odd primes.

�


We are now in a position to formulate our first major theorem.

Theorem 2.21. Let D ≡ 0, 1 (mod 4) be negative, and let χ : (Z/DZ)∗ −→ {±1}be as defined in Lemma 2.19. Then, for an odd prime p not dividing D, [p] is anelement of ker(χ) if and only if p is represented by one of the reduced forms ofdiscriminant D.

Proof. From the definition of χ,

[p] is an element of ker(χ) if and only if

(D

p

)= 1.

By Lemma 2.10, this is equivalent to being represented by a primitive positivedefinite form of discriminant D. By Theorem 2.14, this is equivalent to beingrepresented by one of the reduced forms of discriminant D.

�

Quadratic reciprocity and the fact that

(D

p

)= 1 imply that(

p

D

)= (−1)

(p−1)(D−1)4 .

Therefore, the above theorem tells us that there are congruences p ≡ α1, α2, ...(modD)which give necessary and sufficient conditions for when a prime p may be repre-sented by a reduced form of discriminant D. In particular, when the discriminantcontains only one reduced form, the theorem is exceptionally useful and easy to use.However, there are only finitely many values of n for which h(−4n) = 1 (see Theo-rem 2.18 of Cox [1]). Nevertheless, the theorem helps us make significant progresson the question of when p = x2 + ny2, as we will now show.

Theorem 2.22. Let x and y be integers. Then, for odd primes p,

(1) p = x2 + y2, if and only if p ≡ 1 (mod 4),


(3) p = x2 + 3y2, if and only if p = 3 or p ≡ 1 (mod 3),

(4) p = x2 + 7y2, if and only if p = 7 or p ≡ 1, 9, 11, 15, 23, 25 (mod 28).

Proof. (1) First we show that x2+y2 is the only reduced quadratic form of discrim-inant −4. One similarly shows that the other forms are the only reduced ones oftheir respective discriminants, and so those proofs are omitted. Let ax2 + bxy+ cy2

be a reduced form of discriminant −4. From (2.17), we have a ≤√

43 , which implies

that a = 0 or a = 1. But, a 6= 0 as otherwise D = b2 − 4ac = b2 = −4, which isimpossible. So, a = 1. By the definition of a reduced form, either b = 0 or b = 1.But, b2− 4c = −4, showing that b ≡ 0 (mod 2). This is only possible if b = 0. Thisthen implies that c = 1. Hence, x2 + y2 is the only reduced form of discriminant−4. Finally, from Theorem 2.21, p = x2 + y2 if and only if

1 =

(−4

p

)=

(−1

p

)(2

p

)2

=

(−1

p

)= (−1)

p−12 .


This occurs if and only p ≡ 1 (mod 4), as desired.

(2) x2 + 2y2 is the only reduced form of discriminant −8, and from Theorem2.21 we see that p = x2 + 2y2 if and only if

1 =

(−8

p

)=

(−1

p

)(−2

p

)= (−1)

p−12 (−1)

p2−18 .

Since p is odd, p ≡ 1, 3, 5, 7 (mod 8). Checking each case shows that

(−1)p−12 (−1)

p2−18 = 1 if and only if p ≡ 1, 3 (mod 8).

(3) The form x2 + 3y2 is the only reduced form of discriminant −12. Therefore,theorem 2.21 implies that p = x2 + 2y2 and p 6= 3 if and only if

1 =

(−12

p

)=

(−1

p

)(3

p

)= (−1)

p−12 (−1)

2(p−1)4

(p

3

)=

(p

3

).

But 1 is the only quadratic residue modulo 3. Notice also that if (x, y) = (0, 1),then x2 + 3y2 = 3. So, p = x2 + 3y2 if and only if p = 3 or p ≡ 1 (mod 3).

(4) The form x2 + 7y2 is the only reduced form of discriminant −28 and 7 isclearly represented by the form. Finally, from Theorem 2.21, p = x2 + 7y2 andp 6= 7 if and only if

1 =

(−28

p

)=

(−7

p

)= (−1)

p−12 (−1)

6(p−1)4

(p

7

).

There are two cases to consider. If p ≡ 1 (mod 4), then

(−1)p−12 (−1)

6(p−1)4

(p

7

)= 1, if and only if(

p

7

)= 1, which occurs if and only if

p ≡ 1, 2, 4 (mod 7).

If p ≡ 3 (mod 4), then

(−1)p−12 (−1)

6(p−1)4

(p

7

)= 1, if and only if(

p

7

)= −1, which occurs if and only if

p ≡ 3, 5, 6 (mod 7).

Thus, we solve six systems of two congruences and by the Chinese Remainder The-orem we see that p = x2+7y2 if and only if p = 7 or p ≡ 1, 9, 11, 15, 23, 25 (mod 28).

�

3. Elementary Genus Theory

What enabled us to prove Theorem 2.22 at the end of the last section wasultimately the fact that for those specific vales of n, h(−4n) = 1. However, weknow that this is only true in a few exceptional cases. Therefore, more theoryneeds to be developed to answer the question of when p = x2 + ny2, for othervalues of n. This is where genus theory comes in.


Definition 3.1. Two primitive positive definite forms of discriminant D are saidto be in the same genus if they represent the same values in (Z/DZ)∗.

Notice that forms in the same genus need not represent the same integers; theyonly need to represent the same congruence classes. It may be, for example, thattwo forms f(x, y) and g(x, y) of discriminant D both represent the congruence class[D − 1], but only one of the forms actually represents the integer D − 1.

Definition 3.2. Given D ≡ 0, 1 (mod 4), the principal form is defined byx2 − D

4 y2 if D ≡ 0 (mod 4),

x2 + xy + 1−D4 y2 if D ≡ 1 (mod 4).

Notice that when D = −4n, the principal form is x2 + ny2. We now state animportant lemma. The proof is somewhat long and tedious, so we omit it and referthe reader to Lemma 2.24 of Cox [1].

Lemma 3.3. Given a negative integer D ≡ 0, 1 (mod 4), let ker(χ) ⊂ (Z/DZ)∗ beas in Theorem 2.21, and let f(x, y) be a form of discriminant D. Then,

(1) The values in (Z/DZ)∗ represented by the principal form of discriminantD form a subgroup H ⊂ ker(χ).

(2) The values in (Z/DZ)∗ represented by f(x, y) form a coset of H in ker(χ).

Since the cosets are disjoint, the lemma tells us that different genera representdifferent values in (Z/DZ)∗. This is exactly what we were looking for now thath(−4n) > 1. This lemma allows us to refine Theorem 2.21, so that we may getexplicit congruence conditions for p = x2+ny2 in the case when the genus containingthe principal form (called the principal genus) contains only this form.

Definition 3.4. Let H be as in Lemma 3.3. Then if H ′ is a coset of H, we definethe genus of H ′ to consist of all forms of discriminant D which represent the valuesof H ′ modulo D.

Theorem 3.5. Let D ≡ 0, 1 (mod 4) be negative, and let H ⊂ ker(χ) be as inLemma 3.3. If H ′ is a coset of H in ker(χ) and p is an odd prime not dividing D,then [p] is an element of H ′ if and only if p is represented by a reduced form ofdiscriminant D in the genus of H ′.

Proof. By Theorem 2.21, [p] is an element of ker(χ) if and only if p is representedby a reduced form of discriminant D. By Lemma 3.3, this occurs if and only if [p]is in the coset H ′ ⊂ H represented by this reduced form. This is exactly the sameas saying that the reduced form is in the genus of H ′.

�

This is the most important result we have developed so far. It, along with thefollowing corollary, will be very useful in helping us formulate explicit congruenceconditions, as we did in the previous section.


Corollary 3.6. Let n be a positive integer and p an odd prime not dividing n.Then, p is represented by a form of discriminant −4n in the principal genus if andonly if, for some integer α, p ≡ α2 or α2 + n (mod 4n).

Proof. Notice that x2 +ny2 is always in the principal genus and so p is representedby a form in the principal genus if and only if p is represented by x2 + ny2 modulo4n. But,

x2 + ny2 ≡ x2 or x2 + n (mod 4n),

depending on whether y is even or odd.�

As was mentioned previously, when the principal genus consists only of x2 +ny2,we get explicit congruence conditions. We consider the following cases.

Theorem 3.7. Let x and y be integers and p an odd prime. Then,


(2) p = x2 + 10y2, if and only if p ≡ 1, 9, 11, 19 (mod 40),

(3) p = x2 + 13y2, if and only if p = 13 or p ≡ 1, 7 (mod 52),

(4) p = x2 + 15y2, if and only if p ≡ 1, 19, 31, 29 (mod 60),

(5) p = x2 + 21y2, if and only if p ≡ 1, 25, 37 (mod 84),

(6) p = x2 + 22y2, if and only if p ≡ 1, 9, 15, 23, 25, 31, 47, 49, 71, 81 (mod 88),

(7) p = x2 + 30y2, if and only if p ≡ 1, 31, 49, 79 (mod 120).

Proof. We prove the first statement. The other cases are handled similarly. Firstwe find all the reduced forms of discriminant −24. By (2.17), a reduced formax2 + bxy + cy2 of discriminant −24 satisfies

0 ≤ a ≤√

24

3=√

8.

So, a = 0, 1, 2. Since b2 − 4ac = −24, we see that a 6= 0. Let a = 1. By thedefinition of a reduced form, either b = 0 or b = 1. If b = 0, then c = 6 and weobtain x2 + 6y2. If b = 1, then 4c = 25, which is impossible. Considering a = 2, asimilar analysis shows the only reduced form is 2x2 + 3y2. Now, by Corollary 3.6,p is represented by a form in the principal genus if and only if

p ≡ α2 or α2 + 6 (mod 24).

Firstly, p ≡ α2 (mod 24) if and only if

p ≡ 1 (mod 8) and p ≡ 1 (mod 3),

which occurs if and only if p ≡ 1 (mod 24). Secondly, p ≡ α2 + 6 (mod 24) if andonly if

p ≡ α2 + 6 ≡ 1, 7 (mod 8) and p ≡ α2 + 6 ≡ α2 ≡ 1 (mod 3).


This occurs if and only if p ≡ 1, 7 (mod 24). So it suffices to show that 2x2 + 3y2

is not in the principal genus and then we are done by Theorem 3.5. But 2x2 + 3y2

represents 11 if we set x = 2 and y = 1. Thus, it cannot be in the principal genus,since the principal genus represents only the elements [1] and [7] in (Z/24Z)∗.Hence,

p = x2 + 6y2 if and only if p ≡ 1, 7 (mod 24).

�

In the previous section, we noted that there are only finitely many cases whenthe number of reduced forms h(−4n) of discriminant −4n is equal to 1. Thisprompts the following similar question: are there finitely many discriminants of theform −4n for which the principal genus consists of only the principal form? Theanswer to this question is yes; there are indeed only finitely many such discrim-inants. It turns out that all genera of forms of discriminant −4n consist of thesame number of classes. So, if the principal genus is to contain only the principalform, then all genera must consist of only one class. This is related to the studyof convenient numbers, which we have not discussed. There are only finitely manyconvenient numbers, and a positive integer n is a convenient number if and only ifthe genera of forms of discriminant −4n consist only of a single class. For a morethorough treatment of this topic, reference sections 3.B and 3.C of Cox [1].

4. Reciprocity Laws

In the previous two sections, we have made significant progress on the questionof when a prime p can be represented by the quadratic form x2 + ny2. In doingthis, we formulated necessary and sufficient conditions for numerous values of n.Besides developing some theory on quadratic forms and genera, the most advancednumber theoretic tool we have used has been quadratic reciprocity. As the discus-sion in the previous section highlighted, only certain cases can be solved with thesemethods. For other values of n, cubic and biquadratic reciprocity, which are in asense generalizations of quadratic reciprocity, need to be utilized. Just as quadraticreciprocity tells us when the congruence x2 ≡ a (mod p) has a solution, cubic andbiquadratic reciprocity determine when the respective congruences x3 ≡ a (mod p)and x4 ≡ a (mod p) have solutions. In this section, we will answer these questionsto show when a prime can be represented by the forms x2 + 27y2 or x2 + 64y2.

4.1. Cubic Reciprocity. Using Cubic Reciprocity, we will show that a prime pcan be represented by the form x2 + 27y2 if and only if p ≡ 1 (mod 3) and 2 is acubic residue modulo p. Before we can formulate the Law of Cubic Reciprocity, weneed to investigate some of the properties of the ring Z[ω] = {a + bω | a, b ∈ Z},where ω = −1+

√−3

2 is one of the cube roots of unity. We now review some basicdefinitions of abstract algebra.

Definition 4.1. Let R be an integral domain.

(1) An element u in R is called a unit if u divides 1.


(2) Two elements a and b in R are called associates if a = bu for some unit u.

(3) An element p in R is said to be irreducible if a | p implies that a is either aunit or an associate of p.

(4) A non-unit p in R is called prime if p 6= 0 and p | ab implies that p | a orp | b.

Definition 4.2. The ring R is said to be a unique factorization domain (UFD) ifevery non-zero element r in R, which is a non-unit, has the following properties:

(1) r can be written as a finite product of (not necessarily distinct) irreduciblespi of R. That is, r = p1p2...pn.

(2) if r = q1q2...qn is another factorization of r into irreducibles, then m = n,and there is some numbering such that pi and qi are associates fori = 1, 2, ..., n.

A very important property of the ring Z[ω] is that it is closed under complexconjugation. Indeed, let α = a+ bω be an element of Z[ω]. Then,

α = a+ bω = a+ bω2 = a+ b(−1− ω) = (a− b)− bω ∈ Z[ω].

This allows us to make the following definition.

Definition 4.3. Let α = a+ bω be an element of Z[ω]. We define the norm of α,N(α), by the formula

N(α) = αα = a2 − ab+ b2.

Remark 4.4. Notice that the norm is non-negative. If α = a+ bω, then

N(α) = a2 − ab+ b2 ≥ a2 + b2 − max(a2, b2) = min(a2, b2) ≥ 0.

Moreover, N(α) = 0 if and only if α = 0. One can also quickly verify that the normis multiplicative. That is, N(αβ) = N(α)N(β). An important fact about the ringZ[ω] is that it is a principal ideal domain (an integral domain where every ideal isprincipal, meaning that every ideal can be expressed as (r) = {ra | a ∈ R}). Sinceevery principal ideal domain (PID) is a UFD, the ring Z[ω] is a UFD. Moreover, thenotions of prime and irreducible coincide in Z[ω], since they coincide in any PID.For verifications of these facts, see Section 1.4 of Ireland and Rosen [2]. Knowingthis, we now find the units and primes of Z[ω].

Lemma 4.5. An element α contained in Z[ω] is a unit if and only if N(α) = 1.The units of Z[ω] are therefore ±1,±ω,±ω2.

Proof. If N(α) = 1, then αα = 1, implying that α is a unit. If α is a unit, thenthere exists an element β of Z[ω] such that αβ = 1, implying that N(α)N(β) = 1.Therefore, N(α) = 1 since both N(α) and N(β) are positive integers.

Now suppose that α = a+ bω is a unit. Then,

1 = a2 − ab+ b2, which is the same as

4 = 4a2 − 4ab+ 4b2 = (2a− b)2 + 3b2.


We therefore have two possibilities:

2a− b = ±1 and b = ±1, or

2a− b = ±2 and b = 0.

Solving the six equations gives that the units are ±1,±ω,±(1 + ω). These are thesame as ±1,±ω,±ω2, since 1 + ω + ω2 = 0.

�

We now turn our attention to primes in Z[ω]. Notice that a prime in Z may notnecessarily be a prime in Z[ω]. As an example, 7 = (3 + ω)(2 − ω), where neitherof the factors are units since,

N(3 + ω) = 7 and N(2− ω) = 7.

Therefore, primes in Z are customarily referred to as rational primes, whereasprimes in Z[ω] are often called Eisenstein primes. For the sake of brevity, we willrefer to primes in the ring Z[ω] simply as primes.

Lemma 4.6. If π is a prime in Z[ω], then there is a rational prime p such thatN(π) = p or p2. If N(π) = p, then π is not associated to a rational prime, whereasif N(π) = p2, then it is.

Proof. Since N(π) = ππ > 1 is some integer, it is a product of rational primes.Hence, π divides p for some rational prime p in the factorization of N(π). Thus,p = πγ for an element γ of Z[ω]. By the multiplicativity of the norm, we have

N(p) = N(π)N(γ) = p2.

Hence, either N(π) = p2 and N(γ) = 1 or N(π) = p, since π is by definition nota unit. In the former case, γ is a unit and so π is associated to p. In the latter,suppose π = uq where q is a rational prime and u is a unit. Then,

p = N(π) = N(u)N(q) = N(q) = q2.

This is impossible. Thus, π is not associated to a rational prime.�

Lemma 4.7. Suppose an element π in Z[ω] is such that N(π) = p is a rationalprime. Then, π is a prime in Z[ω].

Proof. Suppose π is not prime. Then π is also not irreducible and so π = ργ withN(ρ), N(γ) > 1. So,

p = N(π) = N(ρ)N(γ),

which is impossible since p is a rational prime. Thus, π is prime.�

We now have the necessary information to classify the primes in Z[ω].

Theorem 4.8. Let p be a rational prime. If p ≡ 2 (mod 3), then p is prime. Ifp ≡ 1 (mod 3), then p = ππ, where π is prime in Z[ω]. Finally, 3 = −ω2(1− ω)2,where 1− ω is prime in Z[ω].


Proof. Suppose that p ≡ 2 (mod 3) but that p is not prime. Then p = πγ whereN(π), N(γ) > 1. So, p2 = N(π)N(γ), implying that N(π) = p. Write π = a+ bω.Then,

p = N(π) = a2 − ab+ b2, and so

4p = (2a− b)2 + 3b2.

Therefore, p ≡ (2a − b)2 (mod 3). This is impossible since 1 is the only quadraticresidue modulo 3, but we have p ≡ 2 (mod 3). Hence, p is prime.

Now suppose that p ≡ 1 (mod 3). By quadratic reciprocity we have(−3

p

)=

(−1

p

)(3

p

)= (−1)

p−12

(p

3

)(−1)

(p−1)(3−1)4 =

(p

3

)=

(1

3

)= 1.

So, there exist integers a and b such that pb = a2 + 3. Hence, p divides

(a+√−3)(a−

√−3) = (a+ 1 + 2ω)(a− 1− 2ω).

If p were a prime in Z[ω], it would have to divide one of the factors. So, p wouldhave to divide either a + 1 or a − 1, and, in particular, 2. But this is impossiblesince p 6= 2 and so 2

p is not an integer. Thus, p = πγ, where π and γ are non-units.

Hence,

p2 = N(π)N(γ), showing that

p = N(π) = ππ, as desired.

Finally, notice that

x2 + x+ 1 = (x− ω)(x− ω2).

Substituting x = 1 gives

3 = (1− ω)(1− ω2) = (1 + ω)(1− ω2) = −ω2(1− ω)2.

Taking norms we obtain 9 = N(1 − ω)2, which is the same as 3 = N(1 − ω). Bythe previous lemma, 1− ω is prime. This completes the proof.

�

We are now in a position to investigate the properties of the ring Z[ω] morethoroughly. Recall that we split the elements of Z into the set of equivalence classesZ/pZ via the congruence relation a ≡ b (mod p). We may similarly introduce acongruence relation on Z[ω] given by α ≡ β (mod γ), which once again meansγ | (α − β). This forms the ring Z[ω]/γZ[ω]. The first natural question to ask ishow many elements comprise this ring.

Theorem 4.9. Let π be a prime in Z[ω]. Then Z[ω]/πZ[ω] is a finite field withN(π) elements.

Proof. We first show that Z[ω]/πZ[ω] is a finite field. Notice that πZ[ω] is a primeideal. Indeed, if αβ is an element of πZ[ω], then αβ = π(α′β′) for some α′ and β′

contained in Z[ω]. Since π is prime, it either divides α or β; suppose it divides α.Then α = πγ, which means that α is an element of πZ[ω], as desired. But Z[ω] isa principal ideal domain. Hence, since πZ[ω] is a non-zero prime ideal, it followsthat πZ[ω] is maximal and thus Z[ω]/πZ[ω] is a field.


Consider first the case where N(π) = ππ = p 6= 3. By the Chinese RemainderTheorem we have

(4.10) Z[ω]/pZ[ω] ∼=Z[ω]

πZ[ω]× Z[ω]

πZ[ω].

The canonical map Z ↪→ Z[ω] induces the map ZZ[p] ↪→

Z[ω]pZ[ω] where the equivalence

class [k] = k + pZ of ZZ[p] is sent to the equivalence class k + pZ[ω] of Z[ω]

pZ[ω] . This

injective map is well-defined because pZ = pZ[ω] ∩ Z. The ring Z[ω] is a moduleover Z with basis {1, ω}. If we show that 1 and ω are not equal modulo pZ[ω], then

it will follow that {1, ω} serves as a basis for the ring Z[ω]pZ[ω] , when seen as a module

over ZpZ . This will show that Z[ω]

pZ[ω] has p2 elements. By the previous theorem, 1−ωis prime. So, p divides 1 − ω if and only if pα = 1 − ω for some unit α. Takingnorms we see that p2 = N(1− ω) = 3, which is impossible. Therefore,∣∣∣∣ Z[ω]

pZ[ω]

∣∣∣∣ = p2.

By (4.10), we have ∣∣∣∣ Z[ω]

πZ[ω]

∣∣∣∣ = p = N(π),

since neither π nor π are units.

Suppose now that N(π) = ππ = p2. By Lemma 4.6, π is associated to p and so

Z[ω]

πZ[ω]∼=

Z[ω]

pZ[ω].

But we previously showed that the right hand side is of size p2, completing this case.

The final case to consider is Z[ω]/(1−ω)Z[ω]. We show that the set {[−1], [0], [1]}completely represents our ring. The ring therefore has N(1−ω) = 3 elements. Letα = a + bω. We know that 3 = −ω2(1 − ω)2 and so (1 − ω) | 3. Notice that α iscongruent to one of nine elements of Z[ω] modulo 3, since both a and b are congruenteither to 0, 1, or −1 modulo 3. The simplest case is when a ≡ b ≡ 0 (mod 3). Thisimplies that 3 | α, and so (1− ω) | α. This is the same as α ≡ 0 (mod 1− ω). Weillustrate one more case. The others are treated similarly. Suppose a ≡ 0 (mod 3)and b ≡ 1 (mod 3). Then, for some integers m and n, we have a = 3m andb = 3n+ 1. Hence,

α = −mω2(1− ω)2 − nω2(1− ω)2 + 1 = (1− ω)[−mω2(1− ω)− nω2(1− ω)] + 1,

which is the same as α ≡ 1 (mod 3). �

This theorem implies the following analog of Fermat’s Little Theorem.

Corollary 4.11. If π - α, then αN(π)−1 ≡ 1 (mod π).

Proof. The previous theorem showed that Z[ω]/πZ[ω] is a field, and so every non-zero element has an inverse. Since Z[ω]/πZ[ω] has N(π) elements, it follows that(Z[ω]/πZ[ω])∗ has N(π)−1 elements and the result follows by Lagrange’s theorem.

�


If the norm of π is not equal to 3, then the residue classes of 1, ω, and ω2 aredistinct in Z[ω]/πZ[ω]. Suppose, for example that ω ≡ ω2 (mod π). Then π dividesω − ω2 = ω(1 − ω). Since π is not a unit, we see that π divides 1 − ω. Because1−ω is prime, it follows that π and 1−ω are associated. Thus, N(π) = 3, which isimpossible. The other cases are similarly handled. Now, since {1, ω, ω2} is a cyclicgroup of order three, its order divides that of (Z[ω]/πZ[ω])∗. That is, 3 dividesN(π)− 1.

Lemma 4.12. Suppose π is a prime such that N(π) 6= 3 and π - α. Then, there

exists a unique m = 0, 1, or 2 such that αN(π)−1

3 ≡ ωm (mod π).

Proof. By Corollary 4.11, π divides αN(π)−1 − 1. But,

αN(π)−1 − 1 = (αN(π)−1

3 − 1)(αN(π)−1

3 − ω)(αN(π)−1

3 − ω2).

Since π is prime, it must divide one of the factors. In fact, it can only divide oneof the factors, since if it divided two it would have to divide their difference, whichis impossible as a consequence of the above discussion.

�

The lemma allows us to make the following definition.

Definition 4.13. Let π be a prime in Z[ω]. IfN(π) 6= 3, the cubic residue characterof α modulo π is given by:

(1)

(α

π

)3

= 0, if π | α .

(2)

(α

π

)3

= ωm, for m = 0, 1, or 2, according to αN(π)−1

3 ≡ ωm (mod π), if

π - α.

The cubic residue character is entirely analogous in the theory of cubic residuesto the Legendre symbol in the theory of quadratic residues. Consequently, we havemany similar results.

Lemma 4.14. Let α and β be elements of Z[ω] and π as in Definition 4.13. Then,

(1) αN(π)−1

3 ≡(α

π

)3

(mod π).

(2)

(αβ

π

)3

=

(α

π

)3

(β

π

)3

.

(3) If α ≡ β (mod π), then

(α

π

)3

=

(β

π

)3

.

(4)

(α

π

)3

= 1 if and only if x3 ≡ α (mod π) has a solution in Z[ω].

Proof. (1) This follows immediately from the definition.(2) Notice that(

αβ

π

)3

≡ (αβ)N(π)−1

3 ≡ αN(π)−1

3 βN(π)−1

3 ≡(α

π

)3

(β

π

)3

(mod π).


Since the residue classes of 1, ω, and ω2 are distinct, we see that(αβ

π

)3

=

(α

π

)3

(β

π

)3

.

(3) If α ≡ β (mod π), then(α

π

)3

≡ αN(π)−1

3 = βN(π)−1

3 ≡(β

π

)3

(mod π).

Therefore, (α

π

)3

=

(β

π

)3

.

(4) See Section 7.1 of Ireland and Rosen [2]. The result holds because the multi-plicative group of any finite field is cyclic.

�

Remark 4.15. Let a be an integer. When p ≡ 1 (mod 3), part (4) of Lemma 4.14tells us something interesting about the congruence x3 ≡ a (mod p). By Theorem4.9, we know that Z/pZ is isomorphic to Z[ω]/πZ[ω]. Consequently, for p - a, wesee that

x3 ≡ a (mod p) is solvable in Z if and only if

(a

π

)3

= 1.

Before stating the Law of Cubic Reciprocity, we need the following definition:

Definition 4.16. A prime π in Z[ω] is said to be primary if π ≡ ±1 (mod 3).

One can quickly check that exactly two of the associates ±π,±ωπ,±ω2π areprimary by writing π = a+ bω and checking explicit cases.

Theorem 4.17 (Law of Cubic Reciprocity). Let π1 and π2 be primary primes withN(π1), N(π2) 6= 3 and N(π1) 6= N(π2). Then,(

π1π2

)3

=

(π2π1

)3

.

The proof is not beyond the scope of the paper, but requires the introduction ofGauss and Jacobi sums, in addition to being very long. See Section 9.4 of Irelandand Rosen [2]. We now turn our attention to the main problem of this section.

Theorem 4.18. Let p be a prime. Then, p = x2+27y2 if and only if p ≡ 1 (mod 3)and 2 is a cubic residue modulo p.

Proof. Suppose p = x2 + 27y2. Then, p ≡ x2 ≡ 1 (mod 3). It remains to show that2 is a cubic residue modulo p. Let π = x + 3

√−3y be an element of Z[ω] (notice

that√−3 = 1 + 2ω). Then, p = ππ, which implies that π is prime, by Lemma 4.7.

By Remark 4.15,

2 is a cubic residue modulo p if and only if

(2

π

)3

= 1.

We know that 2 is a primary prime. Also,

π = x+ 3√−3y = x+ 3(1 + 2ω)y = x+ 3y + 6ωy ≡ x ≡ ±1 (mod 3).


The last congruence holds because if x were divisible by 3, then p would also bedivisible by 3. We therefore have that π is primary. By cubic reciprocity, it sufficesto show that (

π

2

)3

= 1.

Now, (π

2

)3

≡ πN(2)−1

3 ≡ π (mod 2).

So, we just need π ≡ 1 (mod 2). But,

π = x+ 3y + 6ωy ≡ x+ 3y ≡ x+ y (mod 2).

Thus, π ≡ 1 (mod 2), since x and y have opposite parity, because p = x2 + 27y2 isodd.

Now suppose that p ≡ 1 (mod 3) and 2 is a cubic residue modulo p. By Theorem4.8, we can write p = ππ, where π is an element of Z[ω]. We can assume that πis primary, as a result of the little discussion following Definition 4.16. Henceπ = a+ 3bω for some integers a and b. So,

4p = 4ππ = 4(a2 − 3ab+ 9b2) = (2a− 3b)2 + 27b2.

Therefore, all we need to do is verify that b is even, since then both sides of theabove equation are divisible by four. Since 2 is a cubic residue we have,(

2

π

)3

= 1.

By cubic reciprocity, (π

2

)3

= 1.

But as we previously saw, (π

2

)3

≡ π ≡ 1 (mod 2).

Hence, a+ 3bω ≡ 1 (mod 2), implying that a is odd and b is even. This completesthe proof.

�

4.2. Biquadratic Reciprocity. Similarly to how we were able to use cubic reci-procity to answer the question of when a prime p = x2+27y2, we will use biquadraticreciprocity to show that p = x2 + 64y2 if and only if p ≡ 1 (mod 4) and 2 is a bi-quadratic residue modulo p. The development of biquadratic reciprocity is almostentirely analogous to that cubic reciprocity. The main difference is that instead ofconsidering the ring Z[ω], we consider the ring Z[i] = {a + bi | a, b ∈ Z}. Noticethat i is a fourth root of unity, playing the same role as ω, which is a third rootof unity. The ring Z[i] (commonly called the Gaussian Integers) is also a PID, andtherefore a UFD. Moreover, if α is an element of Z[i], we define the norm of α byN(α) = αα = a2 + b2. As before, N(α) = 1 if and only if α is a unit. There-fore, the units of Z[i] are ±1,±i. The following results are proved in an entirelysimilar fashion as the results of the previous section, and so their proofs are omitted.


Lemma 4.19. Let p be a prime in Z. Then,

(1) If p = 2, we can write 2 = i3(1 + i)2, where 1 + i is prime in Z[i].

(2) If p ≡ 1 (mod 4) then there is a prime π ∈ Z[i] such that p = ππ and theprimes π and π are not associated in Z[i]

(3) If p ≡ 3 (mod 4), then p remains prime in Z[i].

Theorem 4.20. Let π be an irreducible in Z[i]. Then the ring Z[i]/πZ[i] is a finitefield with N(π) elements.

Remark 4.21. Part (2) of Lemma 4.19 tells us that if p ≡ 1 (mod 4), then thereexists a prime π contained in Z[i] such that N(π) = p. The proof of Theorem 4.20illustrates, in the exact same way as the proof of Theorem 4.9, that in this caseZ/pZ is isomorphic to Z[i]/πZ[i]. This isomorphism will be very useful later on.

Corollary 4.22. If π - α, then αN(π)−1 ≡ 1 (mod π).

Lemma 4.23. If π - α and N(π) 6= 2, then there exists a unique integer m,0 ≤ m ≤ 3 such that

αN(π)−1

4 ≡ im (mod π).

Just as in the cubic case, the above lemma allows us to make the followingdefinition.

Definition 4.24. Let π be an irreducible with N(π) 6= 2. Then, the biquadratic(or quartic) residue character of α modulo π is given by

(1)

(α

π

)4

= 0, if π | α, and

(2)

(α

π

)4

= im, where m is determined by the above lemma, if π - α.

The biquadratic residue character has the following useful properties, entirelyanalogous to those of the cubic residue character.

Lemma 4.25. Let α and β be elements of Z[i], and let π be as in the previousdefinition.

(1) If π - α, then

(α

π

)4

= 1 if and only if x4 ≡ α (mod π) has a solution in Z[i].

(2)

(αβ

π

)4

=

(α

π

)4

(β

π

)4

.

(3) If α ≡ β (mod π), then

(α

π

)4

=

(β

π

)4

.

Before we state the Law of Biquadratic Reciprocity, we need the following defi-nition.

Definition 4.26. A non-unit α in Z[i] is called primary if α ≡ 1 (mod (1 + i)3).


Theorem 4.27 (Law of Biquadratic Reciprocity). Let π and λ be relatively primeprimary elements of Z[i], then(

π

λ

)4

=

(λ

π

)4

(−1)N(π)−1

4 ·N(λ)−14 .

The proof, just as in the cubic case, requires extensive use of Gauss and JacobiSums. See Section 9.9 of Ireland and Rosen [2].

It is very interesting to notice how similar the statement of biquadratic reci-procity is to that of quadratic reciprocity. The relationship between quadratic andbiquadratic characters will be used in proving the following theorem, which detailsthe biquadratic character of 2. The reader may also see how the theorem is provenusing the supplementary laws of biquadratic reciprocity, by looking at Section 4.Bof Cox [1].

Theorem 4.28. If π = a+ bi is a primary prime in Z[i], then(2

π

)4

= iab2 .

Proof. Let p ≡ 1 (mod 4) be a rational prime. By Lemma 4.19, there exists a primeπ = a + bi such that Nπ = a2 + b2 = p. Without loss of generality, let a be oddand b be even. We first prove that

(4.29)

(a

p

)= 1.

By quadratic reciprocity, we have(a

p

)=

(p

a

)(−1)

(p−1)(a−1)4 =

(p

a

),

since a is odd and p ≡ 1 (mod 4). But,

b2 ≡ a2 + b2 ≡ p (mod a), and so

(a

p

)=

(p

a

)= 1, as desired.

We now show that (a+ b

p

)= (−1)

(a+b)2−18 .

By quadratic reciprocity,

(4.30)

(a+ b

p

)=

(p

a+ b

)(−1)

((a+b)2−1)(p−1)4 =

(p

a+ b

).

Notice that

2p = (a+ b)2 + (a− b)2. If a+ b = q1q2...qn is the prime factorization of a+ b, then

2p ≡ (a− b)2 (mod (a+ b)) implies that 2p ≡ (a− b)2 (mod qi) for 1 ≤ i ≤ n.

Therefore, (2p

a+ b

)=

(2

a+ b

)(p

a+ b

)= (−1)

(a+b)2−18

(p

a+ b

)= 1.


This implies that, (p

a+ b

)= (−1)

(a+b)2−18 ,

and so by (4.30), (a+ b

p

)= (−1)

(a+b)2−18 , as desired.

Now we show that

(4.31)

(a+ b

p

)=

(i

π

)4

iab2 .

Indeed, (a+ b

p

)= (−1)

(a+b)2−18 = (i2)

(a+b)2−18 ≡ i

(a+b)2−14 ≡ i

2p−(a−b)2−14 ≡

≡ ip−14 · i

p−(a−b)24 =

(i

π

)4

iab2 (mod π).

Since both are units we have, (a+ b

p

)=

(i

π

)4

iab2 .

We now need to relate

(a+ b

p

)to some other biquadratic residue character. Since

(a+ b)2 ≡ 2ab (mod p), we have

(a+ b)p−12 ≡ (2ab)

p−14 (mod p).

By Euler’s Criterion for the Legendre symbol and the definition of the biquadraticresidue character, this implies that

(4.32)

(a+ b

p

)=

(2ab

π

)4

.

A quick check shows that 2ab ≡ 2a2i (mod π). Thus by Lemma 4.25 we see that(2ab

π

)4

=

(2a2i

π

)4

=

(2i

π

)4

(a

π

)2

4

.

But, (a

π

)2

4

≡ ap−12 ≡

(a

p

)(mod π).

Since both are units,

(a

π

)2

4

=

(a

p

). Therefore,

(a

π

)4

= 1 if and only if

(a

p

)= 1,

and so we have

(2ab

π

)4

=

(2i

π

)4

, by (4.29).

Using this fact, (4.31), and (4.32), we obtain(i

π

)4

iab2 =

(2i

π

)4

=

(i

π

)4

(2

π

)4

.


This implies that (2

π

)4

= iab2 .

�

We are now ready to state and prove the main theorem of this section.

Theorem 4.33. If p is prime, then p = x2 + 64y2 if and only if p ≡ 1 (mod 4)and 2 is a biquadratic reside modulo p.

Proof. Let p ≡ 1 (mod 4) be prime. Then, by Lemma 4.19, there exists a primaryprime π = a + bi such that p = a2 + b2 = ππ. Notice that because π is primary,a is odd and b is even. From Remark 4.21, we know that Z/pZ is isomorphic toZ[i]/πZ[i]. Therefore, Theorem 4.28 and part (1) of Lemma 4.25 show that(

2

π

)4

= iab2 = 1 if and only if x4 ≡ 2 (mod p) has a solution in Z.

But this occurs if and only if b is divisible by 8, which immediately implies theresult.

�

The two main results proven in sections 4.1 and 4.2 share many similarities.Notice that for p = x2 + 27y2, we can write 27 as 33, p is congruent to 1 modulo 3,and cubic reciprocity (related to the number 3) is used; for p = x2 + 64y2, we canwrite 64 as 43, p is congruent to 1 modulo 4, and biquadratic reciprocity (relatedto the number 4) is used. Moreover, 2 is a cubic or biquadratic residue modulo p inboth cases, respectively. Perhaps a theory of quintic residues is needed to provideanalogous hypotheses for when p = x2 + 53y2 = x2 + 125y2, and so on for the moregeneral problem of p = x2 +n3y2. As a first step to investigating this problem, onemight try to see if something similar holds in the easier case of quadratic reciprocity.That is, is it true that p = x2 + 23y2 = x2 + 8y2 if and only if p ≡ 1 (mod 2) and2 is a quadratic residue modulo p? Unfortunately, the answer to this question isno, since 7 6= x2 + 8y2, but 2 is a quadratic residue modulo 7. Nevertheless, thisquestion might still be an interesting problem for future investigation.

Acknowledgments

It is my pleasure to thank my mentors Billy Lee and Karl Schaefer for theirinvaluable help and guidance throughout my process of studying and writing thispaper. I am very thankful for their having introduced this topic to me, whichprovided me with much enjoyment and a truly great learning experience. I wouldalso like to thank Professor Peter May for all the effort he puts into organizing thiswonderful REU program.

References

[1] David A. Cox. Primes of the Form x2 + ny2: Fermat, Class Field Theory, and Complex

Multiplication. John Wiley and Sons, Inc., Hoboken, NJ, Second edition, 2013.

[2] Kenneth Ireland and Michael Rosen. A Classical Introduction to Modern Number Theory.Springer-Verlag New York, Inc., New York, NY, Second edition, 1990.

[3] David S. Dummit and Richard M. Foote. Abstract Algebra. John Wiley and Sons, Inc., Hobo-

ken NJ, Second edition, 2004.

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QUADRATIC FORMS, RECIPROCITY LAWS, AND PRIMES OF THE …math.uchicago.edu/~may/REU2018/REUPapers/Skenderi.pdf · and yare integers, began with Fermat, who conjectured necessary and