## Abstract

In number theory and group theory, class numbers are used to quantify the “closeness” of a quadratic ring to being a unique factorization domain, where a class number of 1 implies unique factorization. Gauss conjectured that the only negative and even discriminants that have a class number of 1 are . This is the simplest example of a class number theorem, which attempts to find all discriminants associated with a given class number. In 1934, Gauss’ conjecture was generalized, using complex analysis, to the statement that there exist a finite number of discriminants for a given class number. In this paper, we use the theory of binary quadratic forms to present an alternative and more elementary proof of class number theorems. The paper explores group structures on quadratic curves in over . In particular, we define a general elliptical rotational group of rational points on a quadratic curve , denoted . This is a natural extension of the well-studied circle group, which is a geometric interpretation of through the rational points on a unit circle. The circle group can also be thought of as a group structure on the set of all Pythagorean triples. The general elliptical group is then used to show that the irreducible elements in relate to irreducible numbers in . Two elementary proofs for the class number theorem for are presented, which can then be generalized to all discriminants of the form . These proofs are used to classify the irreducible elements of for curves that use quadratic forms with the given discriminant.

**Keywords:** Rational Points, Quadratic Curves, Group Structures, Class numbers, Quadratic Forms

## Introduction

In this paper, we will explore the condition for the existence of rational points curves of the form , where and the group structures that they create. We will also explore the group structures of quadratic forms , namely the class group , using the principal forms with discriminant . Finally, we use the group structures of the rational points on quadratic curves to solve class number theorems using an elementary approach, which are solutions to the equation for some .

According to Shanks, Gauss conjectured that the only discriminants of the form such that are ^{1},^{2}. This corresponds to the forms and respectively. A more general conjecture pertaining to all negative discriminants was proven over 170 years later by Kurt Heegner, which was related refined by Harold Stark^{3},^{4}.

In 1934, Heilbronn and Linfoot showed that the number of solutions to is always finite for any when is negative, although this is still an open problem for positive . However, the machinery used by Heilbronn and Linfoot involves complex analysis and elliptical forms^{5}.

In this paper, we will use the group structures defined in the below sections to give a more elementary version of the proof for , which also has implications for the group structures on the rational points on ellipses.

A natural generalization of this simpler approach examines the rational points on different forms of varying degrees and finds a group structure on a manifold that has a similar connection, similar to the connection seen for Lie Groups^{6}. An example of such a proposed connection would be between quadratic forms of three variables and quadric surfaces.

## Definitions and Representations

A quadratic form is any polynomial , where is an arbitrary ring. This can be written as

where and . For example, the polynomial is a quadratic form in . Another representation of quadratic forms is the matrix form, given as

where and is an symmetric matrix defined by

For the case of , the matrix form is . This paper will concern itself with quadratic forms in , which are all polynomials of the form , where . The form is also denoted by . In this case, the related matrix is simply

The discriminant of a quadratic form is . As the form is also a form in , embedding the form in leads to rational points in .

By clearing the denominator to ensure all coefficients are integers, the equation becomes . Since this paper studies rational solutions, letting and transforms the equation into , which is the equivalent integer equation to .

### Some Conditions on a’, b’, and c’

This section will show a bijection between the rational solutions to any quadratic curve to one with positive, square-free, and pairwise coprime coefficients exist. If for some curve , there exists another curve such that has a rational point if and only if has one, this bijection will be denoted as for two quadratic curves .

**Lemma 1.2.1.** For any quadratic curve , , where has square-free coefficients.

**Proof.** Assume that the coefficients are not square-free, meaning where is square-free for . Then

is equivalent to

These two curves have a bijective map, as any rational point has the corresponding rational point .

**Lemma 1.2.2.** For any quadratic curve , , where has pairwise coprime coefficients.

**Proof.** Without loss of generality, we can assume that as if , then dividing each coefficient by ensures without changing the curve. Now assume that . Then, using the equivalent integer equation, we have that , implying . Since , , meaning . Furthermore, as every divisor of a square-free integer is square-free, implies . Letting and dividing by , becomes

Clearly, these two curves are related through the simple bijective map . Applying this argument for and proves the lemma.

**Lemma 1.2.3.** For any quadratic curve , , where such that and are positive.

**Proof.** If only and are negative or only is negative, the only solution to is , which produces indeterminate results. If and either or is negative, then making the change of variables will make all coefficients positive. Finally, if one or three coefficients are negative, multiplying by -1 reduces it to the cases discussed above.

Using these three lemmas, we have a bijection to the set of quadratic curves with all coefficients to be positive, pairwise co-prime, and square-free.

### The Class Group

For any given quadratic form, there is a multitude of forms that have similar properties to it. For example, the forms and are equivalent under the substitution . Note that this change of variables can be represented by the matrix since . In general, a change of variables can be expressed as for some matrix . Therefore, for a quadratic form that has an associated matrix , the change of variables is equivalent to the matrix transformation . For defining the class group, we will be interested in transformations that leave the discriminant unchanged, meaning that . This implies that . However, the rest of the paper will only be concerned with the case where , or , which will be made clear from lemmas 1.3.4 and 1.3.8. For the rest of the paper, we will also use the following definitions:

**Definition 1.3.1.** Two forms and are said to be properly equivalent if and only if there exists some unique that transforms to , denoted .

**Definition 1.3.2.** A number is said to represent a form if there exists some such that . is said to be a principal representative if .

**Lemma 1.3.3.** Two equivalent forms represent the same set of numbers.

**Proof.** By definition, for two forms and to be equivalent, there exists some that encodes the transformation from to . Since forms a group, there exists an inverse matrix that encodes the transformation from to . Therefore, if represents , then , meaning describes the point such that , meaning for any that represents , it also represents . Similarly, for any that represents , it also represents , which proves the lemma.

We will now define the notion of principal forms, using the following lemma:

**Lemma 1.3.4.** is a principal representative of a form if and only if .

**Proof.** By definition, there must exist some such that . Now consider the matrix such that . The existence of is guaranteed since . By making the substitution and expanding terms, we get

Similarly, if , then letting and using Lemma 1.3.3 proves the lemma.

**Theorem 1.3.5.** Every form with a negative discriminant is properly equivalent to a form such that .

**Proof.** Let be a form with a negative discriminant. It can be assumed that is positive since . Now let denote the Hessian matrix, defined by

For a quadratic form , meaning has a maximum or minimum. Since , a minimum of exists in . Now let denote the smallest integer value of . From Lemma 1.4.3, denoted . Since is the minimum, by definition. Since , we have .

By letting , we have , meaning we can test values of with the same parity as . Since , , implying . Therefore, only values of satisfying this inequality need to be tested. Any form that meets all these properties is called primitive forms.

**Corollary 1.3.6.** There are a finite number of primitive discriminants for each negative .

This directly follows from .

**Example 1.3.7.**To calculate all primitive forms for , we must have , and . Since , must be even as well. The following table summarizes the results.

h | ac | (a,c) |

0 | 30 | (1,30), (2,15), (3,10), (5,6) |

31 | (1,31) | |

34 | (1,34),(2,17) | |

39 | (1,39),(3,13) |

Since we only consider solutions where , the only primitive forms with are , , , and . For example, by letting , we see that .

The class group, denoted is the group of principal forms with a fixed discriminant . The group operation is defined by considering the product where and are independent sets of variables and writing the product as , where are linear combinations of . See section 6.1 of Long’s work for more details on this multiplication^{7}. To check whether this operation is well-defined, we use the following lemma:

**Lemma 1.3.8.** If and , then .

**Proof:** Let be a primitive representative of and be a primitive representative of . Then by definition, represents . By choosing and such that is a primitive representative, . From Lemma 1.3.3, represents and represents , meaning represents , implying , proving the lemma.

Due to the “if and only if” nature of Lemma 1.3.4, the above lemma only works for proper equivalence. If the forms and were related by a change of variables such that , the multiplication of quadratic forms would not be well-defined.

**Example 1.3.9.** To demonstrate the above lemma, consider the forms and . By making the change of variables and , we see that and . To compute the product , we see that

Meaning that . However, by letting , , and computing their product gives

Which means that , demonstrating the lemma for this specific case.

**Example 1.3.10.** For , for each form , , where , meaning is the identity. Furthermore, . By letting , , every can be described using only and , meaning through the isomorphism for .

In general, since the multiplication of forms is a commutative operation, is Abelian, meaning it can be decomposed into the product of cyclic groups.

### Prime Representatives

This section will look at the prime representatives of the forms in for a fixed , as the prime representatives can be used to find every representative of a given form in using. To motivate the study of prime representatives, we first start with a few lemmas:

**Lemma 1.4.1.** If is a primitive representative of a form , every divisor of is represented by another form

**Proof.** From Lemma 1.3.4, if is represented by , then

Therefore, for any such that , is represented by . Note that , meaning .

**Lemma 1.4.2.** For any prime representing a form that doesn’t divide , , where is the Legendre symbol.

**Proof.** Using Lemma 1.4.1, where . Since , . Since, , .

**Lemma 1.4.3.** A reduced form has order 2 in if and only if or

**Proof.** Let , with discriminant . Then represents . Note that this is a primitive representative of the identity element, implying . Therefore, has order 2 if and only if . Since any transformation taking to must fix the and coefficients, the transformation must be of the form or , where . However, if , the coefficient of in is , meaning no such transformation can exist. Therefore, the only valid transformation is , giving , implying that if . This proves the lemma.

These lemmas are the beginning of classifying what primes represent different forms in . For example, when , , so any primes representing these two forms must satisfying

. To analyze these expressions further, we use the following theorems without proof:

**Theorem 1.4.4.** Law of Quadratic Reciprocity: For odd primes

**Theorem 1.4.5.**

Therefore, we see that if , then for any odd prime ,

If , this can be used to drastically simplify the expression for in general. For example,

Using Theorem 1.5.4, the value of remains constant for representatives of the same form, and can be easily calculated using Theorem 1.4.5. Hence, only using the coefficients of and , we can derive what primes represent a given form in . However, this method can fail to distinguish between forms, most notably the forms and . If , then , meaning this method works best when for all . This is equivalent to , for some . In this case, it becomes possible to distinguish the primitive representatives for all using quadratic residues. Gauss conjectured that there existed 65 such for which this holds. We will explore some such that in Section 2.5.

The final piece necessary to find prime representatives is the following theorem:

**Theorem 1.4.6.** If and are odd primes, then , where is arbitrary.

**Example 1.4.7.** To find all prime representatives for the form , we first write

Using the coefficients of the quadratic form ( and ), we see that , , , meaning that for any prime representative of , , , . This means , , , implying . Hence, the primes are prime representatives of .

### 1.5 Legendre’s Theorem on Quadratic Curves

This section examines Legendre’s theorem on quadratic curves, giving the exact conditions for the solvability of an equation of the form , which is given below:

**Theorem 1.5.1.** The Diophantine equation has a solution if and only if , , and , and all three coefficients have the same sign.

Proving one direction of this theorem becomes trivial with a few simple lemmas presented below when combined with the results of Section 1.4.

**Lemma 1.5.2.** and are pairwise coprime.

**Proof.** It can be assumed without loss of generality that . If , then the triple

satisfies . Now assume that , meaning for some prime , . This means , implying . Since is square-free, . This violates the fact that , meaning no such exists. Therefore, . Repeating this argument for and gives the desired result.

**Lemma 1.5.3.** .

**Proof.** Assume that . Then, there would exist some prime such that , implying and . However, since , either or . However, if , then , contradicting Lemma 1.2.2, and if , then , contradicting Lemma 1.5.2. This proves the lemma.

A very similar argument is used to show . The final lemma we need is from Proposition 6.19 of Allen Hatcher’s “Topology of Numbers”^{8}.

**Theorem 1.5.4.** For any that represents a form , takes the same value for any odd prime dividing , given that .

If has a non-trivial solution , then by making the substitution , we see that , implying that is a representative of . Hence, from Lemma 1.4.2, for any , must take the same value for any representative. Since represents , for all , meaning . By making the substitution and and repeating the above procedure, we recover the other two conditions from Legendre’s theorem.

From Legendre’s theorem, we are able to know when a quadratic curve has a rational point, which is key to studying the group of rational points on a quadratic curve, explored in the next few sections below.

## Group of Rational Points

### Parameterization of Rational Points

Assuming that the curve has a rational point (See Section 1.5 for more details), using the natural embedding in the plane to generate every rational point along . First, construct a line through with slope , where . The equation for this line is or . The intersection between this line and can be found through substitution, as

or

Expanding this out and simplifying it gives

Since the line will always intersect the curve at , this quadratic must have a factor of . Furthermore, as satisfies , . Therefore,

Meaning either or

Using this, the expression for becomes

Clearly, if , then . However, for any rational point on , there exists a unique line connecting to the point which must have a rational slope, meaning there exists some that generates . Therefore, the above parameterization generates every rational point on , denoted .

### Group Action Motivation: The Circle Group

Before moving on to a general quadratic curve, we will first investigate the rational points on the unit circle. First of all, it is clear that all points on a circle form a group with the operation being rotation. The group structure is made more clear through the isomorphism , where mod , denoting the group formed by all real numbers .

**Theorem 2.2.1.** The rational points of a circle form a group under rotation.

**Proof.** Consider the rotation matrix,

Using the parameterization , to describe each point on the unit circle, given two rational points , where , the rotation operation is equivalent to the following matrix multiplication

(r_{3} – r_{4} r_{4} r_{3}) (r_{1} -r_{2} r_{2} r_{1}) = (r_{1}r_{3}-r_{2}r_{4} – r_{2}r_{3} – r_{1}r_{4} r_{2}r_{3} + r_{1}r_{4} r_{1}r_{2} – r_{2}r_{4})

Which suggests the following operation

Since the rational points form a subgroup of the entire circle, we only need to show closure and inverses exist. The point is the identity as and the inverse element of is given by , as . The operation is also closed since forms a field.

This group action for the unit circle, defined by is the simplest and a very well-studied group. Each rational point on the unit circle corresponds to an integer solution to . In the next section, we will generalize this to all curves of the form with a rational point.

### General Rotational Group of Rational Points

Consider a curve with a given rational point . will be defined as the generator of the group of rational points on , denoted . From Section 1.2, it can be assumed that , and are positive, meaning describes an ellipse. The counterclockwise parameterization of an ellipse is given by , for some . This means for some . From section 2.1, we can pick two additional rational points on , meaning that for some ,

We now define the action of “rotationally” adding two points and around the axis created by the origin, defined as rotating around ellipse by the angle between and . This is a generalization of the rotational action of the circle group. By considering the point at the angle , we see that

Similarly,

Therefore, we can define the group action as

The identity is , as the point is along the axis of rotation, meaning it adds an angle of zero, leaving the second point unchanged. This can also be verified algebraically. To find an inverse of a rational point , consider giving two linear equations of two variables, which can be solved. Furthermore, by definition, meaning the rational points on form a group under the operation .

**Example 2.3.1.** Consider the curve with rational point . Then , meaning that the group operation for rotationally adding two points around becomes

From section 2.1, points on (C) can be parameterized as follows:

By letting in these parameterizations, we get the points on . Hence,

Which is the point , and is on as

**Theorem 2.3.2.** For any , such that and have rational solutions for some fixed , there exists an isomorphism from to irrespective of the generators of the two groups.

To see why this theorem holds true, let the generator of be and the generator of be . For some rational points , , from the results of section 2.1, there exist some such that

Plugging these expressions into the group operation gives an extremely cumbersome algebraic expression, though the most important observation is that neither the definition of or group operation depend on , meaning that an isomorphism can be defined by “swapping” with , with , and with in both the parameterizations of and the group operation. Since the parameterizations of are unique, the resulting mapping from points on to .

\subsection*{Group Structures of }

For any rational points on a quadratic curve , there exists some that parameterize the points . Now let , , with such that . Then the triple defined as

which satisfies the equation , where . Using the definition of the group operation from section 2.3, we have that corresponds to

where are degree 4 polynomials. Since there must exist some that can parameterize this point, letting we see that

Therefore, the following claim is made:

\textbf{Claim 2.4.1.} A point is defined to be reducible if it can be expressed as , for some . If , , parameterize . Then is reducible if for some .

From the discussion above, it seems that the above claim should hold true, but in order to prove rigorously such a claim, we need to examine the structure of further. For that, the following theorem is very useful:

**Theorem 2.4.2.** Let , where and are pairwise coprime non-zero integers. Then the product can be written as in two distinct ways (up to multiplication by ), where and are integer polynomials of .

**Proof.** Expanding the product gives . Since and are coprime, only terms that contain can form . However, if we only choose the term for (in effect letting ), the remaining three terms can never be of the form . This is because must be of the form , meaning

Comparing terms shows that , , and . However, since and are non-zero, there will always be extra non-zero terms such as which are not present in the expansion of , leading to a contradiction. Therefore, must contain the terms and , implying . By comparing terms, and . Since we are interested in uniqueness up to multiplication by , we can assume that and , meaning

This implies that

meaning

Therefore, or .

**Corollary 2.4.3.**

This can be shown by applying the theorem twice.

As an example, consider and . Equating coefficients, we see that , , . Theorem 2.4.2 then states that

where , or , .

Theorem 2.4.2 is a special case of a more general form of the statement, which includes a term in both and , the proof of which is contained in Proposition 7.5 of Allen Hatcher’s “Topology of Numbers”. This will be referred to as the generalized version of Theorem 2.4.1. The following lemma using generalized Theorem 2.4.1 is constructed:

**Lemma 2.4.4.** If the discriminant of is even for where , .

**Proof.** Using the generalized Theorem 2.4.1, . Since is even, must be even implying is even. Next, we use the substitution , , which is equivalent to the matrix

. Note that , meaning that

.

To make use of the theory of quadratic forms introduced in section 1.4, we will need the following definition of a reducible representative of a quadratic form and a theorem relating and the quadratic forms.

**Definition 2.4.5.** A primitive representative of the quadratic form is said to be reducible if is the product of two or more primitive representatives of .

**Theorem 2.4.6.** Let be a curve of the form , where are positive. A point parameterized by where is irreducible if and only if is an irreducible representative of the quadratic form .

**Proof.** By using the principle of contraposition, the theorem statement is equivalent to proving a point is reducible if and only if is reducible. By definition, if a point is reducible, then is the product of (at least) principle representatives. Now if is reducible, and is not the identity in , then using Corollary 2.4.3, . From Theorem 2.3.2, there is an isomorphism mapping generated by to , where is the curve defined by and is generated by the point . By letting , , we see that generated by and generated by have the property that is generated by and is reducible by Claim 2.4.1. If is the identity and is a reducible representative, then . Hence, by letting , using Theorem 2.3.2 shows , where is generated by the point . Since is reducible , where . Since , we see that the point generated by

is reducible by Claim 2.4.1.

Using Theorem 2.4.6, the following claim is constructed:

**Claim 2.4.7.**

For any such that is prime, the point generated by is an irreducible element of . It might seem that these are all the irreducible elements, though by considering the form when , we see that despite 6 being composite, it is a primitive representative as and have no solutions. In fact, these two primes represent the other primitive form in , namely . However, Claim 2.4.7 can be made stronger when , using the following theorem:

**Theorem 2.4.8.** If , the collection of points parameterized by such that is prime describes all irreducible elements of .

**Proof.** Assume that for some , factors into , where . From Lemma 2.4.8, both and must be represented by a form in , but since there is only one form in , must be represented by . Therefore, is a reducible representative.

**Theorem 2.4.9.** Let be the set of primes represented by a form . If and if is the identity, the collection of points generated by such that or , where and , describe all irreducible elements of . If is not the identity, then the collection of points generated by such that where

describes all irreducible points.

**Proof.** Using Claim 2.4.7, if is prime, then it is a primitive representative. If is not prime, then from Lemma 2.4.8, every divisor of is represented by or . Now, if is the identity in and there exists some , that represents , then by definition is a primitive representative of , meaning that is reducible. Therefore, either is prime or all divisors of are represented by (except 1). However, if has more than three factors that represent , then by considering the divisor pair , both numbers are representatives of , meaning is reducible.

If is not the identity, using Lemma 2.4.8 again shows that every divisor of is represented by or . By applying Theorem 2.4.1, if contains three prime divisors that all represent , then is reducible. Furthermore, if contains two prime divisors, that represent , then represents . Hence, represents , meaning is reducible. Therefore, can only contain one prime divisor representing , which is equivalent to the theorem statement.

### Class Number Problem

This section will give an elementary proof for a special case of the class number problem, which looks for the solutions to for a given and , for which Linfoot and Heilbronn showed there existed finitely many of . Only and even will be considered in our case, as outlined in the introduction. Before moving on to the proof, the concept of “multiple identities” will be introduced, which is central to defining the class group and hence solving class number problems. This will expand upon Sections 1.3 and 1.4, and give some of the special cases for which for and . As discussed in Section 1.4, forms in these class groups have prime representatives that are rather easy to compute, giving us a way to compute the irreducible elements of for where is defined as the curve .

Consider the forms for which . Using the method shown in Example 1.4.6, the possible forms are , , . By using the generalized version of Theorem 2.4.2 and Lemma 1.4.8, , , , meaning they act like an identity element. This leads to a problem when defining a class group since groups cannot have multiple identities. Instead, the class group is defined only to contain the principal form with discriminant defined below:

**Definition 2.5.1.** The principal form for a discriminant is defined as if or if

**Lemma 2.5.2.** A reduced form is in the principal class group if and only if

**Proof.** If , then where the coefficients of share no common factors. Therefore, for every that represents , . Note that if , then for some , where is the principal form. However, , meaning that every representative of is also divisible by . In general, every representative of is divisible by . From Lemma 1.3.3, every representative of is a representative of . However, since represents the principal form, must represent , leading to a contradiction.

Now we are ready to tackle the class number problem for and even

**Lemma 2.5.3.** The Diophantine equation has no solution for and any odd prime .

**Proof.** We know that since . Therefore, . This means must be odd since the only quadratic residues mod 4 are 0 and 1. Letting gives

Therefore, , where . Note that and . Using the fact that and , where are some polynomials in , we have the following identity:

Where is a polynomial in . This leads to a contradiction since must be odd and a multiple of 4 simultaneously. Therefore, no solutions to this Diophantine equation can exist.

**Theorem 2.5.4**

The only negative even discriminants such that are .

**Proof.**

Since is even, we assume that since . Now let . If contains two or more distinct prime divisors, then by letting where and , the form since using Theorem 2.4.2, which is the principal form. This leads to a contradiction since . Now if for some odd prime and , by letting , then . From Lemma 2.5.2, , this is a reduced form. This implies , which again is a contradiction. Hence, the only possibilities are being an odd prime or . If then consider the same construction with . Then, we must have . This always factors into two distinct odd factors unless is a power of 2, in which case we consider . This is valid since . Then which again factors into relatively prime divisor pairs. Hence, in this case. Finally, if is an odd prime, then let again, meaning . If is not a power of 2, then can factor into relatively prime divisor pairs. Hence, must be of the form . If , then by considering (which is valid since ), we have . Letting and using Lemma 2.4.4 shows that , another contradiction.

Summarizing the above, if (powers of 2) or (Mersenne Primes). Below are the calculations for finding primitive forms for the corresponding .

h | ac | (a,c) |

0 | 1 | (1,1) |

h | ac | (a,c) |

0 | 2 | (1,2) |

h | ac | (a,c) |

0 | 3 | (1,3) |

±2 | 4 | (2,2) |

h | ac | (a,c) |

0 | 4 | (1,4),(2,2) |

±2 | 5 | – |

h | ac | (a,c) |

0 | 7 | (1,7) |

±2 | 8 | (2,4) |

However, from Lemma 2.5.2, the forms , and are not in the principal class group. Hence for even if and only if .

Combining Theorem 2.5.3 with Theorem 2.4.8 can give very powerful results, as it can be used to exactly describe every irreducible element of if , , .

**Lemma 2.5.5**

For any , there exists no such such that for all odd primes .

**Proof.**

We first use the fact that there is always a prime residue less than ^{9}. Furthermore, since , we must consider the primes , which give the following residues: . Notice that the smallest is , which means both the primes need to be considered. The smallest solution that satisfies all these constraints is larger than , and we can check that .

**Theorem 2.5.6**

The only negative even discriminants such that are .

**Proof.**

Let for some . If has three distinct prime divisors , then leading to a contradiction. Therefore, can have up to two distinct prime divisors.

**Case 1: with odd**

Since , no additional forms can be in . Therefore, if has an odd divisor , then , meaning . Therefore, , . If , then the forms . These are reduced forms as . Hence , leading to a contradiction. Therefore, the only possibilities are and since , the only possibility in this case is .

In this case, , meaning no additional forms can be in . Since , . Therefore, if has an odd divisor , then , leading to a contradiction. This means , , implying . From Lemma 2.5.2, , meaning . For to be a power of 2, . However, if , then and if , then , where . Hence, the only possibilities are .

**Case 3: **

In this case, . From Lemma 2.5.2, , where and odd. However, if , then , as these are distinct forms from Lemma 1.5.3. Hence, , implying either or is even. Now if , then if , , then , a clear contradiction. Finally, if and , , then , meaning , where . If , then , leading to a contradiction. If , then , but clearly the equation has no solutions for . Finally, if and is even, then implying for some , . This is only possible when , which leads to a contradiction. Finally, if is odd, then , which is another contradiction. Hence, the only possible values for are .

**Case 4: **

Since for , , we only consider . If , . However, if is not prime, for any odd such that , leading to a contradiction. Hence, for some . If , then the reduced forms , meaning that is the only possibility when . Hence the only possibilities are .

**Case 5: m is a non-Mersenne prime, **

Clearly, the form . Since is not a Mersenne prime, where and odd. If , then by lemma 1.5.3, are distinct forms, meaning . Hence . Furthermore, if , , then , implying . Therefore, , where is prime. The only satisfying this condition is . If and if , , then as that implies . However, , meaning must be prime. The only satisfying both conditions is . If and , ,then . Assuming , then . If , , contradicting the primality of . If , , another contradiction. Hence , implying either for some prime or , . However, for even , and if is odd, . Hence, must also be prime. The only that satisfies all three conditions is . Hence .

**Case 6: m is a non-Mersenne prime, **

For each primality condition discussed in Case 5, there are restrictions on the values for modulo a given prime . For example, being prime implies that . The general primality condition in Case 5 is that is prime for some odd and is prime for even . This means for any , , where . For example, when , the conditions are equivalent to . This statement is equivalent to for all , which contradicts Lemma 2.5.5. Hence there are no solutions in this case.

**Case 7: m is a Mersenne prime**

If is a Mersenne prime, then for , . For , and when , , meaning when is a Mersenne prime.

**Case 8: , is prime**

Since , , there cannot be any other forms in . Therefore, must also be prime. If and has an odd divisor , then , meaning is also a prime. The only that satisfies these two conditions is . If , and if has some divisor , then we have another contradiction. Furthermore, implies , leading to a contradiction, and also leads to a contradiction, meaning must also be prime. There exists no satisfying all three conditions. If , then cannot have any odd divisors. If there was some odd divisor , if , then and if , then . Hence, must also be prime. Note that the general primality condition is both being prime or being prime, where . This can be transformed into the more general statement for all that is prime. However, from lemma 2.5.4, this has no solutions for . It can be checked that satisfies all four primality conditions, as , , , . Hence, the only possibilities are .

By combining the results of the above theorem and theorem 2.4.9 with the calculations shown in example 1.4.7, we can calculate the irreducible elements of , where , , for any listed in the above theorem, given contains a rational point.

**Example 2.5.7.** For the irreducible elements of a curve where has a rational point, we first calculate the prime representatives of both forms in . From the results of section 1.5, any and any along with . Using theorem 2.4.9, irreducible representatives of include , and . Since , the points parameterized by are all irreducible elements of irrespective of the generator. Note that the listed values for are also the irreducible elements of any curve defined by , where is a binary quadratic form with discriminant .

## Conclusion

In conclusion, by defining the group of rational points on a given quadratic curve as

for and as a generator, we showed there is an explicit connection between the group structure of and the irreducible elements of a quadratic form . Using this connection, we can describe the irreducible elements for quadratic forms , such that where is the discriminant of the quadratic form . Future work could include generalizing the proofs given in Section 2.5 to class numbers that are not of the form .

## Acknowledgements

I would like to thank Ph.D. student Daniel Epelbaum for his mentorship and feedback during the learning, writing, and editing process for this paper. Without his support and inspiration for this paper, it would not have been possible. I would also like to thank Jose Morin of the Lumiere program for his helpful feedback during the writing process, and the Lumiere program in general for their full support. Finally, I thank an anonymous reviewer from the journal for their guidance and feedback, which has strengthened this paper further.

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