Abstract
In this paper, we study algebraic properties of lattice points of the arc on the conics especially for , which is the Fermat factorization equation that is the main idea of many important factorization methods like the quadratic field sieve, using arithmetical results of a particular hyperbola parametrization. As a result, we present a generalization of the forms, the cardinal, and the distribution of its lattice points over the integers. In particular, we prove that if , Fermat’s method fails. Otherwise, in terms of cardinality, it has, respectively, 4, 8, , , and lattice pointts if is an odd prime, with and being odd primes, with being prime, with being distinct primes, and with being odd primes. These results are important since they provide further arithmetical understanding and information on the integer solutions revealing factors of . These results could be particularly investigated for the purpose of improving the underlying integer factorization methods.
1. Introduction
Diophantine equations have been for many decades a very important subject of research in number theory, and lattice points on curves have been studied in the literature particularly by Gauss, and bounds on arcs of conics have also been studied since then (see [1–6]). However, the necessity of representing an integer as difference of two squares, i.e., for a given , finding nontrivial couples such that , appears in the literature as the main idea of many factorization methods (see [7, 8]) as suggested by Fermat (see [9, 10]). While being not hard to observe, its lattice points are easily computable if one knows the factorization of , and in contrast, this gets exponentially harder when it comes to special cases of , mainly when , where are large primes, in which case this problem becomes equivalent to factoring the parameter .
For this reason, one of fundamental research problems on conics is to find integral solutions of particular hyperbola parametrizations mainly over the integers, particularly when is a large semiprime, in which case if a computationally efficient algorithm is found, cryptosystems like RSA [11] would no longer be secured.
Reviewing the literature, some results on various hyperbola parametrizations and their applications have been studied. Particularly in [1], Javier and Jorge used ideals in quadratic field to find an upper bound for the number of lattice points on Pell’s equation , while in [12], Jin et al. used results from the forms of integral solutions of the hyperbola to solve the same equation where d is of the form . In [13], the author studied a special case of hyperbola and presented the forms of its integral points over , and in [14], Yeonok investigated some behaviors of integral points on the hyperbola to the generalizations of Binet formula and Catalan’s identity, while in [15], the authors gave an application of group law on affine conics to cryptography. Still, in the previous works, algebraic properties and distribution of lattice points and cardinalities on Fermat’s equation are not presented. More recently, in [16], Gilda et al. investigated algebraic and arithmetical properties on the group structure , mainly isomorphisms, integral solutions, and a description of a factorization method with no generalization to the Fermat factorization equation.
In this paper, we use the hyperbola parametrization introduced in [16] to study algebraic properties of lattice points and their distribution for Fermat’s factorization equation for which we find exact upper and lower bounds and we present the forms and cardinalities with a generalization of results for most of special cases of , using results from the particular hyperbola parametrization.
The article is organized as follows:(i)In Section 1, we give an introduction(ii) In Section 2, we present the particular hyperbola parametrization and related arithmetical results(iii)In Section 3, we present the application of the hyperbola parametrization to the study of lattice points on the Fermat equation(iv)In Section 4, In Section 4, we do a discussion on the likelihood of finding solutions to the Fermat factorization equation(v)In Section 5, we finally conclude
Here is a list of the commonly used nomenclature in this paper: : algebraic set of all rational points on . : algebraic set of all integral points on . : algebraic set of integral points on whose coordinates are greater or equal to . : an injective homomorphism from to . : the cardinal of . : set of divisors of . : the set of all prime divisors of . : the Fermat factorization equation. : the Kronecker symbol.
2. BN Hyperbola Parametrization
A conic is an algebraic set satisfying an equation of the form where . Setting the parametrization defined by , in the projective space , we have .
At infinity, setting and considering , we obtain , and the equivalence class , and hence one of the points at infinity is .
From now on, , denotes over the field Q denotes over and denotes over the integers, i.e., ; and and .
Proposition 1. Consider the application is defined , , byThen, is an abelian group with neutral element .
Proof. Let us consider the affine space .
is a hyperbola of equation , where , i.e., , with . SetGiven two points ,We easily verify that .We have bothWe denote as the above defined additive law. This addition law is strongly unified since point doubling does exist and is well defined.
Now let us consider the applicationThis application is an internal composition law since , .(i)Associativity: given 3 points , here we show that .Note that this can be shown either geometrically or analytically, but here we give just the analytic proof.
Consider .We clearly see by identification that .(i)Neutral element: . It is obvious to see that . Given any point , from (3), we have Hence, .(ii)Symmetric element: is the symmetric element of P. It is obvious to see that . Then, we have Hence, .(iii)Commutativity:Hence, .
Proposition 2. Let and . Then, the following map:defines a group homomorphism.
Proof. , . . Thus, .
Consider and .Set the inverse of in ; by definition, .
Then, ., (it is the relation ). Then, relations , , and () imply that X is a group morphism
Furthermore, . is then injective and since such that . is also surjective. is bijective.
Since is a homomorphism of group and bijective, then defines an isomorphism of groups. Thus, and are isomorphic.
Definition 1. Given an integer , we define as the set of all divisors of . That is to say, .
Example 1. : ; .
Proposition 3. (i)If , such that , then , .(ii)If , with , primes, then and where , , .(iii)More generally, if .
Proof. (i)If , , . . Also, .(ii)If , , where , , . It comes to verify that verifies the equation . As , being prime and , then , , . We have . Hence, .(iii)More generally, if , .
Proposition 4. If , then the following holds:(i).(ii).(iii).
Proof. (i) is straightforward since it results from the symmetry of with respect to the axis.(ii)At the point , we have , .(iii)From the symmetry of with respect to the axis, we obtain .
Lemma 1. , set ; then, if and only if .
Proof. Set ; then, which is square if and only if , which is impossible except for , since . This yields that , satisfies if and only if .
Remark 1. Over , is square only if either and are squares or such that , .
Theorem 1. Consider over denoted as . Then, , .
Proof. It is not difficult to see that . From the Remark 1, a given satisfies the condition only if either both and are squares or with , . Considering the fact that any integer can be written as a function of , that is to say , where , if is square, then such that . From Lemma 1, this holds if and only if . Now assume there exists ; in this case, such that . Then,and (19) is a square if and only if , which is absurd since .
Also, (19) is a square if and only if , which is once more absurd since . satisfying , .
If , , . Also, if , , , and thus .
Hence, , .
Proposition 5. If , then there exists an injective homomorphism More generally, there is an injective homomorphism
Proof. Set and consider , and and consider the following morphism: We clearly see that it is a morphism since it splits into morphisms and that are already known to be group morphisms from Proposition 2. It stays now to prove the injection. Given , such that where and . Then, Hence, and , hence the injection. This comes straightforward from .
Example 2. We consider the two hyperbolas plotted in Figure 1.(i):(ii):
Definition 2. A prime divisor of an integer is any prime number . We denote as the set of all prime divisors of and as the number of prime divisors of .
Example 3. 5 is a prime divisor of 40 since . and .
Proposition 6. Set , primes and consider . Set and as the cardinal of . Then, with . In this case, we have the induction relation and the sum of cardinals of given by the general term is .
Proof. , where , i.e., .
By induction on , we have For , (true). For , (true).Assume the relation to be true for , i.e., , and let us show the relation to be true for .By the same,By substituting (27) and (28), we obtainFor :
Example 4. (i), , and , knowing since . Thus, . Verification:(ii) and , knowing , and thus . Verification: and thus .(iii), , and knowing , and thus . Verification:and thus .
Theorem 2. (1) is prime if and only if .(2)If where and are distinct primes, then .(3)If where , and are distinct primes, then .(4)If where , , , and are distinct primes, then .(5).(6)Set as a prime number, and such that . Then, , and .(7)If , .More generally, .
Proof. (i)(2), (3), (6), and (9) come straightforward from the Proposition 6. Nevertheless, we give other proofs for (2) and (3) using injective homomorphisms.(1)Assume prime; then, . The only injective homomorphisms in are giving the point and the trivial automorphism giving the point from Proposition 3. As and , . Now assume , and we know that and ; that is to say, . From Proposition 6, with , and since the cardinal , , i.e., , is prime.(2)If where and are distinct primes, then . We then have the following injective morphisms , , and the trivial automorphism . As and , then and ; in this case, , . , , and from the Proposition 3, . Hence, .(5)From Proposition 4, , ; then, ; as , except for the point which gives the point itself and the point , then each point on leads to 4 points on , i.e., .(6)By induction: if , , then from (2), (true). If , , from Proposition 3, since , we have the following injective morphisms , and the trivial automorphism yielding to respectively 3 points , , and . Thus, (true). Now assume the assumption to be true for , i.e., for with , . Let us prove that to be also true for . If . In this case, . To the injective morphisms obtained with from the induction hypothesis, there is now the new injective homomorphism got with the multiplication of N by Na. In this case, . One deduces from (11) that .(7)From Proposition 5, , it is clear that . By the same, we have injective homomorphisms obtained by composition and we have the injective homomorphisms for each case. Therefore, .Then, for each verifying this condition, we can express this condition according to the Kronecker symbol. Indeed for the injective homomorphism , if , there is no point since . If , we have a point since . Hence, . From (11), we deduce that .
Proposition 7. The cardinal of over is given by the sequence , and the sum of this sequence is .
Proof. These results are straightforward from Theorem 2. from the Proposition 6. Then, . Also, .
Plots of for different values of nare based on the dataset given by Table 1 for different distinct primes.
Comment on the plots on Figure 2 plotted with data from Table 1, from the first plot on the left corresponding to cardinals and sums of cardinals of , we observe that the number of solutions grow quasi-exponentially with the number of distinct prime factors o . In other words, the more distinct the prime factors of , the bigger the algebraic set of . Also, from the second plot (on the right), we observe that asymptotically, and have the same behavior. In other words, knowing over positive integers gives as much information as knowing over the whole integers.
(a)
(b)
Theorem 3. If , and are primes, then with , where , .
Proof. , with and primes, then and . By Proposition 6 and Theorem 2, . Furthermore, from Proposition 3, . Considering the addition law on given by Proposition 1 and setting , one verifies as well in the polynomial ring that , .
Thus, . It is not hard to see that is then a dimensional vector space with basis .
3. Application of Parametrization to the Lattice Points on Fermat Equation
In this section, we present results related to the lattice points on the arc of the hyperbola using results from parametrization.
Theorem 4. , if such that , then verifies . In this case, for positive lattice points, is the lower bound for .
Proof. Consider , with and which yields .Set ; then, .
and are squares, yielding to be also a square.
Then, there exists . Hence, , if , . Also, as , then , considering positive , .
From now on, we denote and as the algebraic set of over the integers.
Theorem 5. , ; then, .
Proof. Assume and .
From Theorem 1, , . From Theorem 4, , where verifies . If , then , taking into account the assumption. Since we work over the integers, we take the ceiling for . By the same, , . If , then , . By the same, , taking into account the assumption.
Hence, the bounds for , .
Lemma 2. Given the sequence
Any term of this sequence and, respectively, any number of this form cannot be represented as difference of two squares.
Proof. If , as , we have from Proposition 5, the following homomorphisms . From Theorem 2, (17), , and considering Proposition 5, we have ; where and , . Hence, 4 cannot be written as difference of two squares.
, and . By Proposition 3, since , then . Also, as , then .
By induction, for . From Theorem 2, . We have from Proposition 5 the following homomorphisms: , and from Proposition 3, the following points: , , , , and since , then . Thus, . . Hence, 10 cannot be represented as difference of two squares.
For . From Theorem 2, . We have from Proposition 5 the following homomorphisms: , and from Proposition 3, the following points: , , , , and since , then . Thus, . . Hence, 14 cannot be represented as difference of two squares.
Assume the assumption to be true for , i.e., for the term , and let us show that it is true for .Since . Set ; then, (true by assumption).
Hence, .
Theorem 6. Consider the Fermat–Diophantine equation . If , i.e., mod 4, then .
Proof. This result is straightforward from Lemma 2.
Proposition 8. If is an odd prime, then . In this case,
Proof. If is prime, from Theorem 2,(1).(2). From Proposition 5, through injective homomorphisms, we have .Now since is an odd prime, is even, and . Setting , (we first consider the positive values of and ), and as satisfies , . Then, over , . Taking into account symmetric properties of , . Hence, and .
Remark 2. If is an even prime, i.e., , then . This result is straightforward from Theorem 5, since .
Proposition 9. If , is prime. Then, . In this case,
Proof. If , is prime, from Theorem 2,(1).(2). From Proposition 5, through injective homomorphisms, we have . It is obvious to see that is not square . Now since is prime, then is even , is a multiple of 4. From Theorem 4, for this case, such that has a total of such terms since . Now considering the redundant terms each time , since i ranges from 0 to a, then each term is the same as because of the commutativity of the additive law. Then, we have exactly such terms after removing the redundant terms for each of the terms.Then, over , . Taking into account the symmetry of , , then over , we have .
Since , , then considering the positive values of and , and , .
Hence, .
Proposition 10. Let , with , odd primes. Then, . In this case,
Proof. Let odd primes, from Theorem 2,(1).(2).From Proposition 5, through injective homomorphisms, we have .
Now since , , are an odd, and are even. Then, means 4 divides and . Set , , since satisfies , and , (we first consider the positive values of and ), since satisfies , . Then, over , . Taking into account symmetric properties of , . Hence, and .
Proposition 11. Let with, odd primes, then , and more generally, for all distinct primes , with being the Kronecker symbol which is given by
Proof. We give a proof by induction.
For , if , (true) from Remark 2 and if , (true) from Proposition 8.
Now assume the proposition is true for , and let us prove it to be also true for .
For , . From the recurrence hypothesis, ; then, , hence the result.
Remark 3. If , , primes Set , then and . Indeed, . Since is prime, is even. Then, such that , . Since , From Theorem 5, .
Example 5. Solve . Here, . From Proposition 11, . Here, since : . In this case, . This algebraic set is empty over the integers.
Proposition 12. Let with and odd primes Then, . In this case,
Proof. From Proposition 5, through injective homomorphisms, we have . For each value of , the value of covers .
It is obvious to see that is not square and . Now since and are primes, then is even and , is a multiple of 4. From Theorem 4, for this case, such that . has a total of such terms since and . Now, considering the redundant terms each time and then this leads each term to be the same as , which is the same due to the commutativity of the common addition law, and then we have exactly such terms without any redundancy, since all of the terms represent all terms together with their doublet.
Then, over , . Taking into account the symmetry of , and , then over , we have .
Since , ; then, considering the positive values of and , and , . Hence,
Proposition 13. If , with odd primes. Then, . In this case,
Proof. We prove this by induction.
For , which is true, since from Proposition 9, .
For , which is also true, since from Proposition 8, .
Now assume the assumption to be true for , and let us prove it to be true for .
, taking into account the assumption .
From Proposition 5, through injective homomorphisms, we haveIt is obvious to see that is not square and . Now since is prime , then is even and is a multiple of 4. From Theorem 4, for this case, such that . ; then, considering the positive values of and , and . Since , then. Taking into account the symmetry of , and ; then, over , we have
4. Discussion
We have exposed the forms of the Fermat equation , dependently on the different forms of , for which we have proved the cardinal over the integers to be 0, 4, and 8, of the form or .
Over , has only one nontrivial solution for a RSA modulus .
Proposition 14. , with the probability .
Proof. Set , with odd primes. From Proposition 13, over , and from Theorem 5, the length of the interval is . .
5. Conclusion
In this paper, we have presented algebraic results on lattice points of the arc on the conics for , which is the Fermat factorization equation for which cardinals, forms of the algebraic set and exact upper and lower bounds are given using a particular hyperbola parametrization. These results provide further information on the structure of the algebraic set of this equation by exposing particularly the following.(i)The general forms of lattice points.(ii)The cardinals and the exact number of solutions.(iii)The distribution of its lattice points over the integers.
As a future work, we shall apply these results in the square sieving methods of factorization (mainly the quadratic sieve) and evaluate any resulting impact and performance.
Data Availability
The algorithms were developed in Python, and the source codes are available from the corresponding author upon request.
Conflicts of Interest
The authors declare that there are no conflicts of interest regarding the publication of this paper.