Property and Representation of n-Order Pythagorean Matrix

Abstract

Here we study the character and expression of n-order Pythagorean matrix using number theory. Theories of Pythagorean matrix are obtained. Using related algebra skills, we prove that the set which constitutes all n-order Pythagorean matrices is a finitely generated group of matrix multiplication and gives a generated tuple of this finitely generated group () simultaneously.

1. Introduction and Theme

If integers satisfy , then we call a Pythagorean array; if Pythagorean array is written in vector form, then we call it a Pythagorean vector [1]. A Pythagorean vector is called primitive [2] if and only if are coprime.

It is well known that every Pythagorean vector is either of the form or of the form with . Frisch and Vaserstein [3] pointed that there exists a parametrization of Pythagorean vectors by a single triple of integer-valued polynomials.

Estimates for the number of Pythagorean vectors with a given constraint are studied in [4–6]. Benito and Varona [4] found asymptotic estimates for the number of Pythagorean vectors with legs less than n. Omland [5] obtained the number of Pythagorean vectors with a given inradius. Okagbue et al. [6] gave statistical and algebraic properties of primitive Pythagorean vectors from the first 331 primitive Pythagorean vectors.

For any fixed primitive Pythagorean vector such that , Jesmanowicz’ [7] studied the Diophantine equation and conjectured the equation has a unique solution. The authors of [8–11] obtained some conclusions on Jesmanowicz’s conjecture.

The authors of [12–14] constructed the following three interesting matrices and obtained the following theorem.

Theorem 1. If , , , is a 3-dimensional Pythagorean vector, and the vector satisfies , then , , and are still 3-dimensional Pythagorean vectors.

Start with or and multiply , , or by it in any order any number of times. This yields another primitive Pythagorean vector , that is, a triple of positive integers without a common factor satisfying . Furthermore, every primitive Pythagorean vector can be obtained uniquely this way. In other words, all primitive Pythagorean vectors can be given a tree-order structure with each edge representing a multiplication by . Cha et al. [15] studied such trees that are applicable to any integral quadratic form.

Generally, 3-order integral square matrix A satisfies the following condition:(i) is a 3-dimensional Pythagorean vector, then still is a Pythagorean vector(ii), then square matrix A is a 3-order Pythagorean matrix [16]

Let be a set which is constituted by all 3-order Pythagorean matrices, namely, F is a three-order Pythagorean matrix}. Hence, we can calculate that the determinant values of , , and are 1; in other words, , , and are Pythagorean matrices, namely, , , and .

Niu [17] researched algebraic properties of the set and proposed the following theorem.

Theorem 2. constitutes a group about the matrix multiplication.

In this paper, we further study algebraic properties and number-theoretic properties of the set . Is a finitely generated group? If is a finitely generated group, then what are the generators of the finitely generated group? We prove our main theorem (Theorem 12). The theorem shows that is a finitely generated group, and the generators of the finitely generated group are given.

Furthermore, we also attempt to extend the Pythagorean vector and 3-order Pythagorean matrices to higher-order case and research algebraic properties and number-theoretic properties of the set formed from all n-order Pythagorean matrices (). Then, we get Theorem20.

This paper is organized as follows. The goal of Section 2 is to give some lemmas needed to prove the main conclusion of this paper. After we give some algebraic properties and number-theoretic properties of the set in Section 3, we prove our main theorem (Theorem 12) in Sections 4 and 5. Section 6 is devoted to the study of properties on n-order Pythagorean matrices (). Building on this, we prove our another main theorem (Theorem 20) in Section 7. Finally, in Section 8, we briefly discuss future work and prospects.

2. Some Preparations

Definition 1. If and are coprime numbers, then we call a 3-dimensional primitive Pythagorean array and we call the correspondent vector a 3-order primitive Pythagorean vector [2].

Lemma 1. If is a 3-order primitive Pythagorean vector, then there exist an odd integer and even integer between a and b, where c must be odd.

Lemma 2. The necessary and sufficient conditions of 3-order integral square matrix are as follows:(i)If is a 3-order Pythagorean vector, then is still a 3-order primitive Pythagorean vector(ii)

Lemma 3 (see [17]). Given , then the necessary and sufficient condition of 3-order integral square matrix is .

Lemma 4 (see [17]). The necessary and sufficient condition of matrix is .

Hence, we can get Lemma5 as follows from Lemma3.

Lemma 5. Given , then the necessary and sufficient condition of matrix is the following equations exist integer solutions :

3. Property of

Let G be a set of 3-order integer square matrix A and satisfy the following two conditions:(i)(ii)If any a, b is even in 3-dimensional primitive Pythagorean vector, then is still a 3-dimensional primitive Pythagorean [2] vector and is even.

Thus, .

For clearer expression, we use some signs. Let , and , . () are the following diagonal matrices: , , , , , , , , and . , express a finite group [18] about matrix multiplication, and express a set with all integer elements of n-order square matrix.

Theorem 3. Let and , then .

Proof. Because and , . From Lemma 5, the above equalities (1)∼(6) are tenable.
From equality (3), we can get the following inequalities:and if , then ; therefore,When are not equal to zero, we can obtain the following inequalities from equality (1):From equality (2), we can obtain the following inequalities:From (7)–(10), we know that equality is tenable.
When at least one of the numbers in and is equal to zero but all of the two numbers in and are not equal to zero, we can infer from equality (2) that inequality (10) is true.
When at least one of the numbers in and is equal to zero, we can get equality , where , so ; from inequalities (7), (8), (10), and , we know that is tenable.
When at least one of the numbers in and is equal to zero but all of the two numbers in and are not equal to zero, we can infer from equality (1) that inequality (9) is true.
When at least one of the numbers in and is equal to zero, we can get equality , where , so ; from the inequalities (7)–(9), and , we know that is tenable.
When at least one of the numbers in and is equal to zero and at least one of the numbers in and is equal to zero, we get , where , so ; moreover, we also get , where , so . From inequalities (7), and , we know that is tenable.

Theorem 4. If and , then (mod 2) () and (mod 2) ().

Proof. Because , the above equalities (1)∼(3) are tenable.
From equality , we get (mod 2), (mod 2), and (mod 2). From , we know . So, (mod 2) and (mod 2) are tenable. Given is an arbitrary primitive Pythagorean vector with even integer b, by Lemma 1, we know a and c are odd integers. Since , is still a 3-order primitive Pythagorean vector with a even integer . By Lemma 1, we know and are odd integers. Since b is an even integer, a is an odd integer, (mod 2), and , we get (mod 2). From (mod 2), (mod 2), and equality (1), we know (mod 2). Since a is an odd integer, b is an even integer, (mod 2), and , we get (mod 2). From (mod 2), (mod 2). and equality (2), we obtain the equality (mod 2). From above, we know (mod 2) () and (mod 2) () are tenable.

Inference 1. (1) G constitutes a group on matrix multiplication; (2) G is a subgroup of .

Proof. (1)Since any A or B, A or B, so . Given is an arbitrary 3-order primitive Pythagorean vector with even integer b because , so is also a 3-order primitive Pythagorean vector with second element which is an even integer, and , so is a 3-order primitive Pythagorean vector with second element which is an even integer; thus, , that is, G is a closed operator of matrix multiplication. Matrix multiplication is clear to meet the combination of law, and 3-order unitary matrix is a unit element in G. Now, we prove that G is closed for inverse matrix. Arbitrary , where ; we know (mod 2) () and (mod 2) (; ; ) are tenable by Theorem 4. Given is an arbitrary 3-order primitive Pythagorean vector with even integer b, let ; since , constitutes a group on matrix multiplication, so ; accordingly, is a primitive Pythagorean vector. One is an odd integer between and and another is an even integer by Lemma 1. Since is an arbitrary 3-order primitive Pythagorean vector with even integer b, we know that a is an odd integer by Lemma 1. We obtain from the equality . We can conclude that because a is an odd integer and both and are even integer, so is an odd integer while must be an even integer. Accordingly, is a 3-order primitive Pythagorean vector with second element which is an even integer; hence, . From above, G constitutes a group on matrix multiplication. And that is what we wanted to prove.(2)We can conclude that G is a subgroup of by and G constitutes a group on matrix multiplication.

4. Expression of and

Theorem 5. .

Proof. From Theorems 3 and 4, if and , then A must be one of (). , , is easy to verify, so Theorem 5 is tenable.

Theorem 6. If , , and , then , , and .

Proof. From Theorem 4 we know, if and , then . By , the former equations (1)∼(3) are tenable. By equation (3), equalities are true. From , we get ; hence, are true. From Theorem 3, we know that and only possibly are 1 or 3, while from and from equalities (1) and (2), we can conclude that are true. Then, equalities are also true. From above, Theorem 6 is tenable, and that is what we wanted to prove.

Theorem 7. .

Proof. Both arbitrary () and () also by Theorem 5. It is easy to prove that . We obtain from Inference 1 (1) that . Also, it is easy to verify that the maximum absolute value of elements of matrix is equal to 3. So, we get that for arbitrary () and ().
On the other hand, if , then , , and .
For A, there must be and , these three matrices produce a new matrix , let , among six elements of column 1 and column 2 in matrix C, there are two elements at most less than zero, and these less than zero elements are not in the same row. Since , and belong to G, ; thus, . Hence , , and . We obtain by and Lemma 5, so (; ).
For C, there must be ; they produce a new matrix . Let . Column vectors 1 and 2 of H are exactly the same as column vectors 1 and 2 of C, and one element at most in column vector 3 of H is less than zero. So, we get , , , and . Note that (), so (); hence, .
So, there exist , , and , they generate a new matrix . Let and . It is easy to verify that both and belong to , and .
From above, is obtained.
The following theorem is obtained by direct verification.

Theorem 8. , , , , , , , and .

We can get Theorem 9 by Theorems 7 and 8.

Theorem 9. (1) ; (2) .

5. Representation of G and

Theorem 10. Arbitrary if the maximum absolute value of elements of matrix A is equal to y and ; let (), then there must exist a matrix , its maximum absolute values of elements are less than y.

Proof. Since () and , (). So, we get that the maximum absolute value of elements of matrix A is by Theorem 3, while the maximum absolute value of elements of matrix is , matrix is , matrix is , and is .
Cases of , , and are considered firstly.
Since , we get ; thus, equality is tenable consequently. Also, because , and ; hence, it follows that and ; thus, and . Finally, the inequality is true as a consequence.
We obtain from , while , so or ; thus, is true.
By inequalities and , we get ; in other words, the maximum absolute value of elements of matrix is less than y, and that is Theorem 10 which we wanted to prove.
In the same way, we can prove that, under the cases , , and , Theorem 10 is still valid.

Theorem 11. , namely, G is a finitely generated group.

Proof. It can be seen that .
Given arbitrary ; if , then or by Theorem 4. When , there must be ; hence, it follows that by Theorem 9. When , there must be ; hence, it follows that by Theorem 9.
If , from Theorem 10, we know that there exist , these matrices satisfy that () and they produce a new matrix by multiplication of matrices, and the maximum absolute value of elements of the new matrix is less than or equal to 3. Indeed, the new matrix , so ; hence, and then .
From above, we get .

Inference 2. (1) ; (2) ; (3) ().

Proof. We only prove that equality (1) is tenable, and others may prove similarly.
It is obvious that . Arbitrary ; we get by Theorem 11 and , , and by Theorem 8; hence, , , and , so .
From above, is tenable.

Definition 2. If H is a finitely generated group and satisfy , then we call is a generated tuple of H. If is a generated tuple of H with least element, then we call is a minimum generated tuple of H and we call n is cardinality of H, expressed as .

Theorem 12. . In other words, is a finitely generated group; furthermore, .

Proof. It is easy to verify that ; hence, .
Owing to , . Now, let arbitrary and such that one in is an even and another is an odd.(1)First, consider x is odd and y is even. Equality is obtained from . Because y is an even integer, both and are either even integer or odd integer. Given is an arbitrary three-order primitive Pythagorean vector with an even integer b; note , so that is also a three-order primitive Pythagorean vector and are odd integer. Thanks to , where b is an even integer and are either odd integers or even integers simultaneously. From the previous reason, we know that is an even integer. So, ; hence, . In addition, ; therefore, .(2)Now, consider x is even and y is odd. Let , then is an odd integer and is an even integer. From and , we have . Thus, ; consequently, . From above, we get ; on this account, and are tenable.

Inference 3. .

6. Property of n-Order Pythagorean Matrix ()

Definition 3. If integers satisfy , then is designated as an n-order Pythagorean array, while vector-style expression is named as an n-order Pythagorean vector.

Definition 4. If an n-order square matrix A meets the following two conditions, then we name A as n-order Pythagorean matrix.(1)If is an arbitrary n-order Pythagorean vector and is still a Pythagorean vector(2)If

Definition 5. Among an n-order Pythagorean vector , are coprime numbers; then is named as an n-order primitive Pythagorean vector.
Let and in this chapter. Then, we have the following theorem.

Theorem 13. A is an n-order integer matrix, and the necessary and sufficient condition of is .

Proof.  Sufficiency condition. If A satisfies , then we can easily check that . Let be an arbitrary n-order Pythagorean vector. In the expression of , we can simplify it. Let . Clearly, are all integers. Since , β is an n-order Pythagorean vector. Thereby, . Necessary condition. If and is an n-order Pythagorean vector, then is still an n-order Pythagorean vector, that is to say .Hence, we can get the following equality:Let , then substitute it in equality (11), and we can get the following equality:Let , then substitute it in equality (11), and we can get the following equality:Let and , then substitute it in equality (11) Then, we can getLet , then substitute it in equality (11), and we obtain the following equation:Let , then substitute it in equality (11), andwe obtain the following equation:From equations (12) and (13), we obtainFrom equations (14) and (15), we getFrom equations (16) and (17), we getAlso, from equalities (18), (19), … , (20), we obtain the following equations:Substituting equalities (21) and (22) into equality (11), we can get the following equality:Hence, we can get equation (24) by randomicity of and equality (23):For given , we know equality (25) is valid by equalities (21), (22), and (24):Thanks to , where k is an integer, or when n is an even integer, but when n is an odd integer. Also, because matrices B and are congruence relationship from (6.15), is inappropriate while is tenable; as a result, .
It i s easy to get Theorem 14 by Theorem 13.

Theorem 14. Given , the necessary and sufficient condition of is that is integer solution of the following equations:

Lemma 6. Given A is an n-order integer square matrix, the necessary and sufficient condition of is A meets the following two cases:(1)If is an arbitrary n-order primitive Pythagorean vector, then is still an n-order primitive Pythagorean vector(2)

Lemma 7. If , then .

Proof. Because , ; therefore, . Consequently, .

Lemma 8. Given and , then .

Proof. Because and , ; as a result, .

Lemma 9. The necessary and sufficient condition of is .

Proof. If we want to prove Lemma 9 is proper, we need to prove only that if , then .
Given , ; hence, . Because both B and belong to , we can get by Lemma 7 and we can conclude that from Lemma 8.
We can easily get the following theorem by using the above lemmas.

Theorem 15. compose a group about matrix multiplication.

Theorem 16. If and , then .

Proof. Since , equations (6.16), (6.17), and (6.18) are tenable by Theorem 14.
From equality (27), we know and are true.
From , we get , soFor arbitrary j (), if , then we get () by (29); if , then we can conclude that by (26). Because , (). Hence, we getThat is to say is tenable.

7. Expression of n-Order Pythagorean Matrix ()

Given , and when . Clearly . Now, we designate () as the following n-order square matrix, , , , , ,…, , , ,…, and . Here, is an n-order square matrix produced by unitary matrix exchange row i and j. represents the finitely generated group about matrix multiplication produced by ; represents the finitely generated group about matrix multiplication produced by ,; represents the finitely generated group about matrix multiplication produced by ; represents the finitely generated group about matrix multiplication produced by . It is easy to know that .

Theorem 17. .

Proof. It is easy to validate that belong to . So, we can conclude that by Theorem 15.
Considering and , equation becomes true by Theorem 16. We can infer that () from equation (27) and , and we can conclude that by , thereby () are tenable. Thus, we can obtain the following conclusion by equation (26) and (). There exists only one 1 among . There exists only one 1 among .  There exists only one 1 among . We can get the other conclusion by . There exists only one 1 among . There exists only one 1 among .  There exists only one 1 among .Accordingly, there exist (), and they cause and satisfy that () and (). For C, there exist (), and they make . So, . Consequently, . Hence, is true. From above, is tenable.
All n which are mentioned hereinafter satisfy the condition of , and no longer explained.

Theorem 18. , both X and .

Proof. Suppose and , then we get because and . If we express , then it is easy to check . Therefore, , namely, , both X and .
If , then A can left or right multiply by matrix in ; hence, matrix is obtained, and it makes equations , , and () tenable.
If is partitioned into , in which is -th-order square matrix and is 4-th-order square matrix, then there exists only one number which absolute value is equal to 1 in the former column (include no. column) of from equation (26). In a similar way, there exists only one number which absolute value is equal to 1 in the former row (include no. row) of from equation (26). Use equation (28) and reductio ad absurdum, we can get and , that is to say .
For matrix , it can be obtained by matrix which left or right multiply by matrix in . It causes , in which is a diagonal matrix and the absolute value of diagonal elements is equal to 1.
For matrix , it can be obtained by matrix which left or right multiply by matrix in . It causes .
For matrix , it can be obtained by matrix which left or right multiply by matrix in . It causes , in which and . We know that only two elements’ absolute values are equal to 1 in every row (column) of U by equation (26), and only two elements in every row (column) of U are equal to 1 by equation (28). For matrix , it can left or right multiply by matrix of . The result is matrix .
From above, matrix can be obtained by A which left or right multiply by matrix in . In other words, there exist ; they cause . The other form is . Let and , then ; so, . Therefore, , X and ; it follows that , X and .

Inference 4. .

Theorem 19. Arbitrary , if maximum absolute value of s element is y; furthermore, , and then there exist () which make maximum absolute value of elements of matrix be less than y.

Proof. Clearly, there exist which make the former elements of last row in matrix be nonnegative, and the last element of last row in matrix is equal to ; there exist which make , of which , () and . Let , and then we get . Obviously, .
Now we must prove that is tenable when .
When , is true. Now we must prove that is still true when and . Otherwise, from we can get . From , we can get . Hence, . If , then ; from previous agreement , we know that ; then from we get , so . But, it is out of question.
If and , then we get , so ; it contradicts with previous agreement ; if and , then is true of arbitrary ; from we know that , namely, and ; hence, matrix H can be expressed as the form of . From equation (27), we know that only two elements’ absolute value of the former rows of columns in matrix H is equal to 1, and the rest elements are equal to zero. And this conclusion is incompatible with equation (6.18).
From above, is true; in other words, is true, namely, is tenable. Hence, Theorem 19 is established.

Theorem 20. . In other words, is a finitely generated group, and , is a generated tuple of .

Proof. Clearly . Arbitrary , and it can be written as . If , then , which make and by Theorem 19. In other words, , which make , of which ; if , then we can apply the theorem time after time; hence, we get , which make and . Because lower bound of is 1 or 2, which is to say there exist k, which make or . If , then ; is true by Theorem 18, so . If , then is true by Theorem 17, so .
From above, is true, accordingly . This is what we want to prove.
We suppose that () still is a finitely generated group, but the presentation of need to be further studied.

8. Future Work and Prospects

Let .

Start with or and multiply , , or by it in any order any number of times, and all 3-dimensional primitive Pythagorean vectors can be formed trees which Cha et al. [15] call Berggren trees.

Since , , , (), and , we get that every 3-order primitive Pythagorean vector can be obtained from multiplying , or by in any order any number of times. Can all 3-dimensional primitive Pythagorean vectors be formed a Berggren tree starting with a primitive Pythagorean vector?

Using the definition and properties of , we can obtain the another representation of ; that is, we have that . Does () have a similar representation?

In this paper, we have given the generators of the finitely generated group (). Is () a finitely generated group? If () is a finitely generated group, what are the generators of (?

These appear to be interesting questions, which we hope to take up in the near future.

Data Availability

All data used to support the findings of this study are included within the article.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

This research was supported by the Scientific Research Foundation of Huaqiao University (10HZR26) and the Natural Science Foundation of Fujian Province (Z0511028).