# Property and Representation ofn-Order Pythagorean Matrix

Authors: Haizhou Song; Wang Qiufen
Journal: Mathematical Problems in Engineering
 (2020)

---

## Abstract

Here we study the character and expression ofn-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 (n≤10) simultaneously.


---

## Body

## 1. Introduction and Theme

If integersa,b,andc satisfy a2+b2=c2, then we call a,b,c 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 a,b,andcare coprime.It is well known that every Pythagorean vector is either of the formkm2−n22kmnkm2+n2 or of the form 2kmnkm2−n2km2+n2 with k,m,n∈Z. 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 vectora,b,c such that a2+b2=c2, Jesmanowicz’ [7] studied the Diophantine equation ax+by=cz 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.
IfF1=122212223, F2=−1−2−2212223, F3=122−2−1−2223, a,b,c is a 3-dimensional Pythagorean vector, and the vector satisfies a2+b2=c2, then a,b,cF1, a,b,cF2, and a,b,cF3 are still 3-dimensional Pythagorean vectors.Start with3,4,5 or 4,3,5 and multiply F1, F2, or F3 by it in any order any number of times. This yields another primitive Pythagorean vector x,y,z, that is, a triple of positive integers without a common factor satisfying x2+y2−z2=0. 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 Fj. Cha et al. [15] studied such trees that are applicable to any integral quadratic form.Generally, 3-order integral square matrixA satisfies the following condition:(i)
α=a,b,c is a 3-dimensional Pythagorean vector, then β=a,b,cA still is a Pythagorean vector(ii)
A2=1, then square matrix A is a 3-order Pythagorean matrix [16]LetT3 be a set which is constituted by all 3-order Pythagorean matrices, namely, T3=F∣F∈Z3×3,F is a three-order Pythagorean matrix}. Hence, we can calculate that the determinant values of F1, F2, and F3 are 1; in other words, F1, F2, and F3 are Pythagorean matrices, namely, F1∈T3, F2∈T3, and F3∈T3.Niu [17] researched algebraic properties of the set T3 and proposed the following theorem.Theorem 2.
T
3 constitutes a group about the matrix multiplication.In this paper, we further study algebraic properties and number-theoretic properties of the setT3. Is T3 a finitely generated group? If T3 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 T3 is a finitely generated group, and the generators of the finitely generated group T3 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 (n>3). Then, we get Theorem20.This paper is organized as follows. The goal of Section2 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 T3 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 (4<n). 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.
Ifa2+b2=c2 and a,b,andc are coprime numbers, then we call a,b,c a 3-dimensional primitive Pythagorean array and we call the correspondent vector a 3-order primitive Pythagorean vector [2].Lemma 1.
Ifα=a,b,c 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 matrixA∈T3 are as follows:(i)
Ifα=a,b,c is a 3-order Pythagorean vector, then β=a,b,cA is still a 3-order primitive Pythagorean vector(ii)
A2=1Lemma 3 (see [17]).
GivenB=10001000−1, then the necessary and sufficient condition of 3-order integral square matrix A∈T3 is ABA′=B.Lemma 4 (see [17]).
The necessary and sufficient condition of matrixA∈T3 is A′∈T3.Hence, we can get Lemma5 as follows from Lemma3.Lemma 5.
GivenA=aij∈Z3×3, then the necessary and sufficient condition of matrix A∈T3 is the following equations exist integer solutions aij:(1)a112+a212−a312=1,(2)a122+a222−a322=1,(3)−a132−a232+a332=1,(4)a11a12+a21a22=a31a32,(5)a11a13+a21a23=a31a33,(6)a12a13+a22a23=a32a33.


## 3. Property ofT3

LetG be a set of 3-order integer square matrix A and satisfy the following two conditions:(i)
A∈T3(ii)
If anya, b is even in 3-dimensional primitive Pythagorean vector, then a′,b′,c′=a,b,cA is still a 3-dimensional primitive Pythagorean [2] vector and b′ is even.Thus,G⊂T3.For clearer expression, we use some signs. LetGt=A∣A=aij∈Z3×3, and A∈G, maxaij=t. Di (i=1,2,…,8) are the following diagonal matrices: D1=diag1,1,1, D2=diag−1,1,1, D3=diag1,−1,1, D4=diag1,1,−1, D5=diag−1,−1,1, D6=diag−1,1,−1, D7=diag1,−1,−1, D8=diag−1,−1,−1, and D9=010100001. F4=122212−2−2−3, LA1,A2,…,An express a finite group [18] about matrix multiplication, and Zn×n express a set with all integer elements of n-order square matrix.Theorem 3.
LetA=aij∈Z3×3 and A∈G, then a33=maxi,jaij.Proof.
BecauseG⊂T3 and A∈G, A∈T3. From Lemma 5, the above equalities (1)∼(6) are tenable.
From equality (3), we can get the following inequalities:(7)a33≥a13,a33≥a23,a33≥1,and if A∈T3, then A′∈T3; therefore,(8)a33≥a31,a33≥a32.
Whena11,a12,a21,anda22 are not equal to zero, we can obtain the following inequalities from equality (1):(9)a31≥a11,a31≥a21.
From equality (2), we can obtain the following inequalities:(10)a32≥a12,a32≥a22.
From (7)–(10), we know that equality a33=maxi,jaij is tenable.
When at least one of the numbers ina11 and a21 is equal to zero but all of the two numbers in a12 and a22 are not equal to zero, we can infer from equality (2) that inequality (10) is true.
When at least one of the numbers ina11 and a21 is equal to zero, we can get equality maxiai1=1, where a33≥1, so a33≥maxiai1; from inequalities (7), (8), (10), and a33≥maxiai1, we know that a33=maxi,jaij is tenable.
When at least one of the numbers ina12 and a22 is equal to zero but all of the two numbers in a11 and a21 are not equal to zero, we can infer from equality (1) that inequality (9) is true.
When at least one of the numbers ina12 and a22 is equal to zero, we can get equality maxiai2=1, where a33≥1, so a33≥maxiai2; from the inequalities (7)–(9), and a33≥maxiai2, we know that a33=maxi,jaij is tenable.
When at least one of the numbers ina12 and a22 is equal to zero and at least one of the numbers in a11 and a21 is equal to zero, we get maxiai1=1, where a33≥1, so a33≥maxiai1; moreover, we also get maxiai2=1, where a33≥1, so a33≥maxiai2. From inequalities (7), a33≥maxiai1 and a33≥maxiai2, we know that a33=maxi,jaij is tenable.Theorem 4.
IfA=aij∈Z3×3 and A∈G, then aii≡1 (mod 2) (i=1,2,3) and aii≡0 (mod 2) (i≠j;i=1,2,3;j=1,2,3).Proof.
BecauseA∈G, the above equalities (1)∼(3) are tenable.
From equality−a132−a232+a332=1, we get a33≡1 (mod 2), a13≡0 (mod 2), and a23≡0 (mod 2). From A∈T3, we know A′∈T3. So, a31≡0 (mod 2) and a32≡0 (mod 2) are tenable. Given α=a,b,c is an arbitrary primitive Pythagorean vector with even integer b, by Lemma 1, we know a and c are odd integers. Since A∈G, a′,b′,c′=a,b,cA is still a 3-order primitive Pythagorean vector with a even integer b′. By Lemma 1, we know a′ and c′ are odd integers. Since b is an even integer, a is an odd integer, a31≡0 (mod 2), and a′=aa11+ba21+ca31, we get a11≡1 (mod 2). From a11≡1 (mod 2), a31≡0 (mod 2), and equality (1), we know a21≡0 (mod 2). Since a is an odd integer, b is an even integer, a32≡0 (mod 2), and b′=aa12+ba22+ca32, we get a12≡0 (mod 2). From a12≡0 (mod 2), a32≡0 (mod 2). and equality (2), we obtain the equality a22≡1 (mod 2). From above, we know aii≡1 (mod 2) (i=1,2,3) and aij≡0 (mod 2) (i≠j;i=1,2,3;j=1,2,3) are tenable.Inference 1.
(1)G constitutes a group on matrix multiplication; (2) G is a subgroup of T3.Proof.
(1)
Since anyA or B∈G, A or B∈T3, so A∗B∈T3. Given α=a,b,c is an arbitrary 3-order primitive Pythagorean vector with even integer b because A∈G, so a,b,cA is also a 3-order primitive Pythagorean vector with second element which is an even integer, and B∈G, so a,b,cAB is a 3-order primitive Pythagorean vector with second element which is an even integer; thus, A∗B∈G, 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 A∈G, where A=aij∈Z3×3; we know aii≡1 (mod 2) (i=1,2,3) and aij≡0 (mod 2) (i≠j; i=1,2,3; j=1,2,3) are tenable by Theorem 4. Given α=a,b,c is an arbitrary 3-order primitive Pythagorean vector with even integer b, let a′,b′,c′=a,b,cA−1; since A∈T3, T3 constitutes a group on matrix multiplication, so A−1∈T3; accordingly, a′,b′,c′ is a primitive Pythagorean vector. One is an odd integer between a′ and b′ and another is an even integer by Lemma 1. Since a,b,c 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 a,b,c=a′,b′,c′A from the equality a′,b′,c′=a,b,cA−1. We can conclude that a=a′a11+b′a21+c′a31 because a is an odd integer and both a21 and a31 are even integer, so a′ is an odd integer while b′ must be an even integer. Accordingly, a′,b′,c′=a,b,cA−1 is a 3-order primitive Pythagorean vector with second element which is an even integer; hence, A−1∈G.
From above,G constitutes a group on matrix multiplication. And that is what we wanted to prove.
(2)
We can conclude thatG is a subgroup of T3 by G⊂T3 and G constitutes a group on matrix multiplication.


## 4. Expression ofG1 and G3

Theorem 5.
G
1
=
D
1
,
D
2
,
D
3
,
D
4
,
D
5
,
D
6
,
D
7
,
D
8.Proof.
From Theorems3 and 4, if A∈G and maxi,jaij=1, then A must be one of Di (i=1,2,…,8). Di∈G1, i=1,2,…,8, is easy to verify, so Theorem 5 is tenable.Theorem 6.
IfA=aij∈Z3×3, A∈G, and maxi,jaij=3, then a33=3, a31=a32=a13=a23=a12=a21=2, and a11=a22=1.Proof.
From Theorem4 we know, if A∈G and maxi,jaij=3, then a33=maxi,jaij=3. By A∈G, the former equations (1)∼(3) are tenable. By equation (3), equalities a13=a23=2 are true. From A∈T3, we get A′∈T3; hence, a31=a32=2 are true. From Theorem 3, we know that a11 and a22 only possibly are 1 or 3, while from a13=a23=a31=a32=2 and from equalities (1) and (2), we can conclude that a11=a22=1 are true. Then, equalities a12=a21=2 are also true. From above, Theorem 6 is tenable, and that is what we wanted to prove.Theorem 7.
G
3
=
A
A
=
D
i
F
1
D
j
,
D
i
∈
G
1
,
D
j
∈
G
1.Proof.
Both arbitraryDi (i=1,2,…,8) and Dj (j=1,2,…,8) also ∈G by Theorem 5. It is easy to prove that F1∈G. We obtain from Inference 1 (1) that DiF1Dj∈G. Also, it is easy to verify that the maximum absolute value of elements of matrix DiF1Dj is equal to 3. So, we get that DiF1Dj∈G3 for arbitrary Di (i=1,2,…,8) and Dj (j=1,2,…,8).
On the other hand, ifA∈G3, then a33=3, a11=a22=1, and a31=a32=a13=a23=a12=a21=2.
ForA, there must be Di and Dj, these three matrices produce a new matrix C=DiADj, let C=cij, 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 Di,Dj, and F1 belong to G, C=DiADj∈G; thus, C∈G3. Hence c33=3, c31=c32=c13=c23=c12=c21=2, and c11=c22=1. We obtain c11c12+c21c22=c31c32 by C∈G and Lemma 5, so cij>0 (i=1,2,3; j=1,2).
ForC, there must be Dk; they produce a new matrix H=CDk. Let H=hij. 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 h33=3, h31=h32=h13=h23=h12=h21=2, h11=h22=1, and h11h13+h21h23=h31h33. Note that hi1=ci1>0 (i=1,2,3), so hi3>0 (i=1,2,3); hence, H=F1.
So, there existDi, Dj, and Dk, they generate a new matrix F1=DiADjDk. Let P1=Di−1 and P2=DjDk−1. It is easy to verify that both P1 and P2 belong to G1, and A=P1F1P2.
From above,G3=AA=DiF1Dj,Di∈G1,Dj∈G1 is obtained.
The followingtheorem is obtained by direct verification.Theorem 8.
D
1
=
F
1
∗
F
1
−
1, D2=F1∗F2−1, D3=F1∗F3−1, D4=F1∗F4−1, D5=F2∗F3−1, D6=F2∗F4−1, D7=F3∗F4−1, and D8=F2∗F3−1∗F1∗F4−1.We can get Theorem9 by Theorems 7 and 8.Theorem 9.
(1)G3⊂LF1,F2,F3,F4; (2) G1⊂LF1,F2,F3,F4.


## 5. Representation ofG and T3

Theorem 10.
ArbitraryA=aij∈G if the maximum absolute value of elements of matrix A is equal to y and y>3; let Hi=AFi (i=1,2,3,4), then there must exist a matrix Hi, its maximum absolute values of elements are less than y.Proof.
SinceFi∈G (i=1,2,3,4) and A∈G, Hi=AFi∈G (i=1,2,3,4). So, we get that the maximum absolute value of elements of matrix A is y=a33 by Theorem 3, while the maximum absolute value of elements of matrix Hi is 2a31+2a32+3a33, matrix H2 is −2a31+2a32+3a33, matrix H3 is 2a31−2a32+3a33, and H4 is 2a31+2a32−3a33.
Cases ofa31≥0, a32≥0, and a33>0 are considered firstly.
SinceA∈G, we get A′∈G; thus, equality −a312−a322+a332=1 is tenable consequently. Also, because a33>3, a31≠0 and a32≠0; hence, it follows that a31>0 and a32>0; thus, a33>a31 and a33>a32. Finally, the inequality 2a31+2a32−3a33<a33 is true as a consequence.
We obtaina33−a31=1+a322/a33+a31 from −a312−a322+a332=1, while 1+a322/a33+a31<a32a31+a32a33/a33+a31=a32, so a33−a31<a32 or a33<a32+a31; thus, −a33<2a31+2a32−3a33 is true.
By inequalities2a31+2a32−3a33<a33 and −a33<2a31+2a32−3a33, we get 2a31+2a32−3a33<y; in other words, the maximum absolute value of elements of matrix H4 is less than y, and that is Theorem 10 which we wanted to prove.
In the same way, we can prove that, under the casesa31≥0, a32≤0, and a33>0, Theorem 10 is still valid.Theorem 11.
G
=
L
F
1
,
F
2
,
F
3
,
F
4, namely, G is a finitely generated group.Proof.
It can be seen thatLF1,F2,F3,F4⊂G.
Given arbitraryA=aij∈G; if a33≤3, then a33=1 or a33=3 by Theorem 4. When a33=1, there must be A∈G1; hence, it follows that A∈LF1,F2,F3,F4 by Theorem 9. When a33=3, there must be A∈G3; hence, it follows that A∈LF1,F2,F3,F4 by Theorem 9.
Ifa33>3, from Theorem 10, we know that there exist P1,P2,…,Pk, these matrices satisfy that Pi∈F1,F2,F3,F4 (i=1,…,k) and they produce a new matrix A∗P1∗P2∗⋯∗Pk 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 A∗P1∗P2∗⋯∗Pk∈G, so A∗P1∗P2∗⋯∗Pk∈LF1,F2,F3,F4; hence, A∈LF1,F2,F3,F4 and then G⊂LF1,F2,F3,F4.
From above, we getG=LF1,F2,F3,F4.Inference 2.
(1)G=LF1,D2,D3,D4; (2) G=LF1,D2,D4,D5; (3) G=LFi,D2,D3,D4 (i=2,3,4).Proof.
We only prove that equality (1) is tenable, and others may prove similarly.
It is obvious thatLF1,D2,D3,D4⊂G. Arbitrary A=aij∈G; we get A∈LF1,F2,F3,F4 by Theorem 11 and D2=F1∗F2−1, D3=F1∗F3−1, and D4=F1∗F4−1 by Theorem 8; hence, F2=D2−1∗F1, F3=D3−1∗F1, and F4=D4−1∗F1, so A∈LF1,D2,D3,D4.
From above,G=LF1,D2,D3,D4 is tenable.Definition 2.
IfH is a finitely generated group and X1,X2,…,Xn satisfy H=LX1,X2,…,Xn, then we call X1,X2,…,Xn is a generated tuple of H. If X1,X2,…,Xn is a generated tuple of H with least element, then we call X1,X2,…,Xn is a minimum generated tuple of H and we call n is cardinality of H, expressed as n=dH.Theorem 12.
T
3
=
L
F
1
,
D
2
,
D
4
,
D
9. In other words, T3 is a finitely generated group; furthermore, dT3=4.Proof.
It is easy to verify thatD9∈T3; hence, LF1,D2,D4,D9⊂T3.
Owing toD3=D9D2D9, G=LF1,D2,D3,D4⊂LF1,D2,D4,D9. Now, let arbitrary A=aij∈T3 and x,y,z=3,4,5A such that one in x,y is an even and another is an odd.(1)
First, considerx is odd and y is even.Equalityy=3a12+4a22+5a32 is obtained from x,y,z=3,4,5A. Because y is an even integer, both a12 and a32 are either even integer or odd integer.Givena,b,c is an arbitrary three-order primitive Pythagorean vector with an even integer b; note a′,b′,c′=a,b,c, so that a′,b′,c′ is also a three-order primitive Pythagorean vector and a,c are odd integer. Thanks to b′=aa12+ba22+ca32, where b is an even integer and a12anda32 are either odd integers or even integers simultaneously. From the previous reason, we know that b′ is an even integer. So, A∈G; hence, A∈F1,D2,D3,D4. In addition, LF1,D2,D3,D4⊂LF1,D2,D4,D9; therefore, A∈F1,D2,D4,D9.(2)
Now, considerx is even and y is odd.Letx′,y′,z′=x,y,zD9=3,4,5AD9, then x′ is an odd integer and y′ is an even integer. From A∈T3 and D9∈T3, we have AD9∈T3. Thus, AD9∈LF1,D2,D4,D9; consequently, A∈LF1,D2,D4,D9.From above, we getT3⊂LF1,D2,D4,D9; on this account, T3⊂LF1,D2,D4,D9 and dT3=4 are tenable.Inference 3.
T
3
=
L
F
1
,
D
2
,
D
4
,
D
9
=
L
F
1
,
D
3
,
D
4
,
D
9
=
L
F
1
,
F
2
,
F
4
,
D
9
=
L
F
1
,
F
3
,
F
4
,
D
9
=
L
F
2
,
F
3
,
F
4
,
D
9.


## 6. Property ofn-Order Pythagorean Matrix (n≥4)

Definition 3.
If integersa1,a2,…,an satisfy ∑k=1n−1ak2=an2, then a1,a2,…,an is designated as an n-order Pythagorean array, while vector-style expression a1,a2,…,an is named as an n-order Pythagorean vector.Definition 4.
If ann-order square matrix A meets the following two conditions, then we name A as n-order Pythagorean matrix.(1)
Ifα=a1,a2,…,an is an arbitrary n-order Pythagorean vector and β=a1,a2,…,anA is still a Pythagorean vector(2)
IfA2=1Definition 5.
Among ann-order Pythagorean vector a1,a2,…,an, a1,a2,…,an are coprime numbers; then a1,a2,…,an is named as an n-order primitive Pythagorean vector.
LetTn=FF∈Zn×n,Fis an−order Pythagoreanmatrix and B=diag1,1,…,1,−1 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 A∈Tn is ABA′=B.Proof.
Sufficiency condition.
IfA satisfies ABA′=B, then we can easily check that A2=1. Let α=a1,a2,…,an be an arbitrary n-order Pythagorean vector. In the expression of β=a1,a2,…,anA, we can simplify it. Let β=a11,a12,…,a1n. Clearly, a11,a12,…,a1n are all integers. Since ∑i=1n−1a1i2−a1n2=βBβ′=αABA′α′=αBα′=∑i=1n−1ai2−an2=0, β is an n-order Pythagorean vector. Thereby, A∈Tn.
Necessary condition.
IfA=aij∈Tn and α=c1,c2,…,cn is an n-order Pythagorean vector, then β=c1,c2,…,cnA is still an n-order Pythagorean vector, that is to say βBβ′=αABA′α′=0.
Hence, we can get the following equality:(11)∑i=1n−1a1i2−a1n2c12+∑i=1n−1a2i2−a2n2c22+⋯+∑i=1n−1ani2−ann2cn2+2∑i=1n−1a1ia2i−a1na2nc1c2+2∑i=1n−1a1ia3i−a1na3nc1c3+⋯+2∑i=1n−1a1iani−a1nannc1cn+2∑i=1n−1a2ia3i−a2na3nc2c3+⋯+2∑i=1n−1a2iani−a2nannc2cn+⋯+2∑i=1n−1an−1,iani−an−1,nanncn−1cn=0.
Letα=1,0,…,0,1, then substitute it in equality (11), and we can get the following equality:(12)∑i=1n−1a1i2−a1n2+∑i=1n−1ani2−ann2+2∑i=1n−1a1iani−a1nann=0.
Letα=1,0,…,0,−1, then substitute it in equality (11), and we can get the following equality:(13)∑i=1n−1a1i2−a1n2+∑i=1n−1ani2−ann2−2∑i=1n−1a1iani−a1nann=0.
Letα=0,1,0,…,0,1 and α=0,1,0,…,0,−1, then substitute it in equality (11) Then, we can get(14)∑i=1n−1a2i2−a2n2+∑i=1n−1ani2−ann2+2∑i=1n−1a2iani−a2nann=0,(15)∑i=1n−1a2i2−a2n2+∑i=1n−1ani2−ann2−2∑i=1n−1a2iani−a2nann=0.⋮
Letα=0,…,0,1,1, then substitute it in equality (11), and we obtain the following equation:(16)∑i=1n−1an−1i2−an−1n2+∑i=1n−1ani2−ann2+2∑i=1n−1an−1,iani−an−1,nann=0.
Letα=0,…,0,1,−1, then substitute it in equality (11), andwe obtain the following equation:(17)∑i=1n−1an−1i2−an−1n2+∑i=1n−1ani2−ann2−2∑i=1n−1an−1,iani−an−1,nann=0.
From equations (12) and (13), we obtain(18)∑i=1n−1a1i2−a1n2=−∑i=1n−1ani2−ann2,∑i=1n−1a1iani−a1nann=0.
From equations (14) and (15), we get(19)∑i=1n−1a2i2−a2n2=−∑i=1n−1ani2−ann2,∑i=1n−1a2iani−a2nann=0,⋮
From equations (16) and (17), we get(20)∑i=1n−1an−1i2−an−1n2=−∑i=1n−1ani2−ann2,∑i=1n−1an−1,iani−an−1,nann=0.Also, from equalities (18), (19), … , (20), we obtain the following equations:(21)∑i=1n−1a1i2−a1n2=∑i=1n−1a2i2−a2n2=⋯=∑i=1n−1an−1,i2−an−1,n2=−∑i=1n−1ani2−ann2,(22)∑i=1n−1a1iani−a1nann=∑i=1n−1a2iani−a2nann=⋯=∑i=1n−1an−1,iani−an−1,nann=0.
Substituting equalities (21) and (22) into equality (11), we can get the following equality:(23)2∑i=1n−1a1ia2i−a1na2nc1c2+⋯+2∑i=1n−1a1ian−1,i−a1nan−1,nc1cn−1+2∑i=1n−1a2ia3i−a2na3nc2c3+⋯+2∑i=1n−1a2ian−1,i−a2nan−1,nc2cn+⋯+2∑i=1n−1an−2,ian−1,i−an−2,nan−1,ncn−2cn−1=0.
Hence, we can get equation (24) by randomicity of α=c1,c2,…,cn and equality (23):(24)∑i=1n−1a1ia2i−a1na2n=⋯=∑i=1n−1a1ian−1,i−a1nan−1,n=∑i=1n−1a2ia3i−a2na3n=⋯=∑i=1n−1a2ian−1,i−a2nan−1,n=⋯=∑i=1n−1an−2,ian−1,i−an−2,nan−1,n=0.
For given∑i=1n−1a1i2−a1n2=k, we know equality (25) is valid by equalities (21), (22), and (24):(25)ABA′=kB.
Thanks toABA′=A2B=B=kB=knB, where k is an integer, k=1 or k=−1 when n is an even integer, but k=1 when n is an odd integer. Also, because matrices B and kB are congruence relationship from (6.15), k=−1 is inappropriate while k=1 is tenable; as a result, ABA′=B.
It i s easy to get Theorem14 by Theorem 13.Theorem 14.
GivenA=aij∈Zn×n, the necessary and sufficient condition of A∈Tn is that aij is integer solution of the following equations:(26)∑i=1n−1aik2−ank2=1,k=1,2,…,n−1,(27)−∑i=1n−1ain2+ann2=1,(28)∑i=1n−1aipaiq=anpanq,1≤p<q≤n;p,q∈Z.Lemma 6.
GivenA is an n-order integer square matrix, the necessary and sufficient condition of A∈Tn is A meets the following two cases:(1)
Ifα=a1,a2,…,an is an arbitrary n-order primitive Pythagorean vector, then β=αA is still an n-order primitive Pythagorean vector(2)
A2=1Lemma 7.
IfA∈Tn, then A−1∈Tn.Proof.
BecauseA∈Tn, ABA′=B; therefore, A−1BA−1′=B. Consequently, A−1∈Tn.Lemma 8.
GivenA∈Tn and C∈Tn, then AC∈Tn.Proof.
BecauseA∈Tn and C∈Tn, ACBAC′=ACBC′A′=ABA′=B; as a result, AC∈Tn.Lemma 9.
The necessary and sufficient condition ofA∈Tn is A′∈Tn.Proof.
If we want to prove Lemma9 is proper, we need to prove only that if A∈Tn, then A′∈Tn .
GivenA∈Tn, ABA′=B; hence, A′=B−1A−1B. Because both B and B−1 belong to Tn, we can get A−1∈Tn by Lemma 7 and we can conclude that A′∈Tn from Lemma 8.
We can easily get the following theorem by using the above lemmas.Theorem 15.
T
n compose a group about matrix multiplication.Theorem 16.
IfA=aij∈Zn×n and A∈Tn, then ann=maxi,jaij.Proof.
SinceA∈Tn, equations (6.16), (6.17), and (6.18) are tenable by Theorem 14.
From equality (27), we know ain≤anni=1,2,…,n−1 and 1≤ann are true.
FromA∈Tn, we get A′∈Tn, so(29)ani≤ann,i=1,2,…,n−1.
For arbitraryj (j=1,2,…,n−1), if maxa1j,a2j,…,an−1,j≤anj, then we get aij≤ann (i=1,2,…,n−1) by (29); if maxa1j,a2j,…,an−1,j>anj, then we can conclude that maxa1j,a2j,…,an−1,j≤1 by (26). Because 1≤ann, aij≤ann (i=1,2,…,n−1). Hence, we get(30)aij≤ann,i=1,2,…,n;j=1,2,…,n.That is to say ann=maxi,jaij is tenable.


## 7. Expression ofn-Order Pythagorean Matrix (4≤n≤10)

GivenTnP=A∣A=aij∈Tnandmaxij=aij=p, F42=0111101111011112 and Fn2=En−400F42 when 5≤n≤10. Clearly Fn2∈Tn2. Now, we designate Wi (i=1,2,…,2n−1) as the following n-order square matrix, W1=diag1,1,…,1, W2=diag−1,1,…,1, W3=diag1,1,…,1,−1, W4=En1,2, W5=En1,3,…, Wn+1=En1,n−1, Wn+2=diag1,−1,1,…,1, Wn+3=diag1,1,−1,1,…,1,…, and W2n−1=diag1,…,1,−1,1. Here, Eni,j is an n-order square matrix produced by unitary matrix exchange row i and j. LW1,W2,…,W2n−1 represents the finitely generated group about matrix multiplication produced by W1,W2,…,W2n−1; LW1,W2,W3,Wn+2,Wn+3,…,W2n−1 represents the finitely generated group about matrix multiplication produced by W1,W2,W3,Wn+2,Wn+3,…,W2n−1; LW4,W5,…,Wn+1 represents the finitely generated group about matrix multiplication produced by W4,W5,…,Wn+1; LW1,W2,…,Wn+1 represents the finitely generated group about matrix multiplication produced by W1,W2,…,Wn+1. It is easy to know that LW1,W2,…,W2n−1=LW2,W3,W4…,Wn+1.Theorem 17.
T
n
1
=
L
W
1
,
W
2
,
…
,
W
n
+
1.Proof.
It is easy to validate thatW1,W2,…,Wn+1 belong to Tn1. So, we can conclude that LW1,W2,…,Wn+1⊂Tn1 by Theorem 15.
ConsideringA=aij∈Zn×n and A∈Tn1, equation ann=1 becomes true by Theorem 16. We can infer that ain=0 (i=1,2,…,n−1) from equation (27) and ann=1, and we can conclude that A′∈Tn1 by A∈Tn1, thereby ani=0 (i=1,2,…,n−1) are tenable. Thus, we can obtain the following conclusion by equation (26) and ani=0 (i=1,2,…,n−1).There exists only one 1 amonga11,a21,…,an−1,1.There exists only one 1 amonga12,a22,…,an−1,2.⋮There exists only one 1 amonga1,n−1,a2,n−1,…,an−1,n−1.We can get the other conclusion byA′∈Tn1.There exists only one 1 amonga11,a12,…,a1,n−1.There exists only one 1 amonga21,a22,…,a2,n−1.⋮There exists only one 1 amongan−1,1,an−1,2,…,an−1,n−1.
Accordingly, there existX1,X2,…,Xn−2 (X1,X2,…,Xn−2∈W1,W4,W5,…,Wn+1), and they cause C≜AX1X2⋯Xn−2≜cij and satisfy that cij=0 (i≠j) and cii=1 (i=1,2,…,n). For C, there exist Xn−1,Xn,…,X2n−1 (Xn−1,Xn,…,X2n−1∈W1,W2,W3,Wn+2,Wn+3,…,W2n−1), and they make CXn−1Xn⋯X2n−1=En. So, AX1X2⋯Xn−2Xn−1Xn⋯X2n−1=En. Consequently, A=X2n−1−1⋯X2−1X1−1. Hence, A∈LW1,W2,…,Wn+1 is true. From above, Tn1=LW1,W2,…,Wn+1 is tenable.
Alln which are mentioned hereinafter satisfy the condition of 4≤n≤10, and no longer explained.Theorem 18.
T
n
2
=
{
A
A
=
X
F
n
2
Y, both X and Y∈LW1,W2,…,W2n−1.Proof.
SupposeA=XFn2Y and X,Y∈LW1,W2,…,W2n−1, then we get A∈Tn because Fn2∈Tn2 and Wi∈Tn. If we express A=aij, then it is easy to check maxi,jaij=2. Therefore, A∈Tn2, namely, A∣A=XFn2Y, both X and Y∈LW1,W2,…,W2n−1}⊂Tn2.
IfA∈Tn2, then A can left or right multiply by matrix in LW1,W2,…,W2n−1; hence, matrix P1≜pijn×n is obtained, and it makes equations p22=2, pn,n−1=pn,n−2=pn,n−3=1=pn−1,n=pn−2,n=pn−3,n, and pnj=pjn=0 (j<n−3) tenable.
IfP1is partitioned into P1=P11P12P21P22, in which P11 is n−4-th-order square matrix and P22 is 4-th-order square matrix, then there exists only one number which absolute value is equal to 1 in the former n−4 column (include no. n−4 column) of P1 from equation (26). In a similar way, there exists only one number which absolute value is equal to 1 in the former n−4 row (include no. n−4 row) of P1 from equation (26). Use equation (28) and reductio ad absurdum, we can get P12=0 and P21=0, that is to say P1=P1100P22.
For matrixP2, it can be obtained by matrix P1 which left or right multiply by matrix in LW4,W5,…,Wn+1. It causes P2=R200Q, in which R2 is a diagonal matrix and the absolute value of diagonal elements is equal to 1.
For matrixP3, it can be obtained by matrix P2 which left or right multiply by matrix in LW1,W2,…,W2n−1. It causes P3=En−400Q.
For matrixP4, it can be obtained by matrix P3 which left or right multiply by matrix in LW1,W2,…,W2n−1. It causes P4=En−4000Uβ0α2, in which α=1,1,1 and β=1,1,1T. 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 P4, it can left or right multiply by matrix of LW4,W5,…,Wn+1. The result is matrix Fn2.
From above, matrixFn2 can be obtained by A which left or right multiply by matrix in LW1,W2,…,W2n−1. In other words, there exist X1,Y1∈LW1,W2,…,W2n−1; they cause X1AY1=Fn2. The other form is A=X1−1Fn2Y1−1. Let X=X1−1 and Y=Y1−1, then X,Y∈LW1,W2,…,W2n−1; so, A=XFn2Y. Therefore, Tn2⊂A∣A=XFn2Y, X and Y∈LW1,W2,…,W2n−1; it follows that Tn2=AA=XFn2Y, X and Y∈LW1,W2,…,W2n−1.Inference 4.
T
n
2
⊂
L
F
n
2
,
W
2
,
W
3
,
…
,
W
n
+
1.Theorem 19.
ArbitraryA=aij∈Tn, if maximum absolute value of A′s element is y; furthermore, y>2, and then there exist Qi∈LW2,W3,W4,…,Wn+1 (i=1,2,…,n) which make maximum absolute value of elements of matrix H=AQ1Q2⋯QnFn2 be less than y.Proof.
Clearly, there existQ1∈LW2,W3,W4,W5,…,Wn+1 which make the former n−1 elements of last row in matrix AQ1 be nonnegative, and the last element of last row in matrix AQ1 is equal to −y; there exist Q2,Q3,…,Qn∈LW2,W3,W4,W5,…,Wn+1 which make Q≜AQ1Q2⋯Qn≜qijn×n, of which qnn=−y, qni≥0 (i=1,…,n−1) and qn1≤qn2≤⋯≤qn,n−1. Let H=QFn2=hijn×n, and then we get hnn=qn,n−3⋅1+qn,n−2⋅1+qn,n−1⋅1−2y. Obviously, qn,n−3+qn,n−2+qn,n−1<3y.
Now we must prove thatqn,n−3+qn,n−2+qn,n−1>y is tenable when 4≤n≤10.
Whenqn,n−3=0, qn,n−3+qn,n−2+qn,n−1=∑i=1n−1qni>y is true. Now we must prove that qn,n−3+qn,n−2+qn,n−1>y is still true when 4≤n≤10 and qn,n−3>0. Otherwise, from qn,n−3+qn,n−2+qn,n−1≤y we can get qn,n−3+qn,n−2+qn,n−12≤y2. From 1+∑i=1n−1qni2=qnn2=y2, we can get 1+∑i=1n−4qni2+qn,n−32+qn,n−22+qn,n−12−∑i=n−3n−1qni2=y2−∑i=n−3n−1qni2≥0. Hence, 1+∑i=1n−4qni2≥2qn,n−3qn,n−2+qn,n−3qn,n−1+qn,n−2qn,n−1≥6qn,n−32. If qn,n−3>1, then n−4≥6; from previous agreement n≤10, we know that n−4=6; then from 1+∑i=1n−4qni2≥6qn,n−32 we get qn1=qn2=⋯=qn,n−1, so 9qn12=qnn2−1. But, it is out of question.
Ifqn,n−3=1 and qn,n−1≥2, then we get 1+n−4≥1+∑i=1n−4qni2≥22⋅1+1⋅1+1⋅1=8, so n≥11; it contradicts with previous agreement n≤10; if qn,n−3=1 and qn,n−1=1, then qni≤1 is true of arbitrary i≤n−1; from 1+∑i=1n−4qni2≥6qn,n−32 we know that n−3≥6, namely, n≥9 and qn,n−4=1; hence, matrix H can be expressed as the form of H=H11α1α2α3α4H16H211111−y. From equation (27), we know that only two elements’ absolute value of the former n−1 rows of n−ii=1,2,3,4 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,y<qn,n−3+qn,n−2+qn,n−1<3y is true; in other words, qn,n−3+qn,n−2+qn,n−1−2y<y is true, namely, hnn<y is tenable. Hence, Theorem 19 is established.Theorem 20.
T
n
=
L
F
n
2
,
W
2
,
W
3
,
W
4
,
…
,
W
n
+
1. In other words, Tn is a finitely generated group, and Fn2,W2,W3, W4,…,Wn+1 is a generated tuple of Tn.Proof.
ClearlyLFn2,W2,W3,W4,…,Wn+1⊂Tn. Arbitrary A∈Tn, and it can be written as A=aijn×n. If ann>2, then ∃Q1,Q2,…,Qn∈LW2,W3,W4,…,Wn+1, which make H1=AQ1Q2…QnFn2≜hij1n×n and hnn1<ann by Theorem 19. In other words, ∃X1∈LFn2,W2,W3,W4,…,Wn+1, which make H1=AX1≜hij1n×n, of which hnn1<ann; if hnn1>2, then we can apply the theorem time after time; hence, we get ∃X1,X2,…,Xk∈LFn2,W2,W3,W4,…,Wn+1, which make Hm+1=HmXm+1m=1,2,…,k−1≜hijmn×n and hnnk<hnnk−1<⋯<hnn1<ann. Because lower bound of hnnk is 1 or 2, which is to say there exist k, which make hnnk=2 or hnn1=1. If hnnk=2, then Hk∈Tn2; Tn2⊂LFn2,W2,W3,W4,…,Wn+1 is true by Theorem 18, so A∈LFn2,W2,W3,W4,…,Wn+1. If hnn1=1, then Hk∈LW1,W2,…,W2n−1 is true by Theorem 17, so Hk∈LFn2,W2,W3,W4,…,Wn+1.
From above,Tn⊂LFn2,W2,W3,W4,…,Wn+1 is true, accordingly Tn=LFn2,W2,W3,W4,…,Wn+1. This is what we want to prove.
We suppose thatTn (n≥11) still is a finitely generated group, but the presentation of Tn need to be further studied.


## 8. Future Work and Prospects

LetWn=a1,a2,…,ana1,a2,…,anisn‐orderprimitive Pythgorean vector.Start with3,4,5 or 4,3,5 and multiply F1, F2, or F3 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.SinceF1∈T3, F2∈T3, F3∈T3, Di∈T3 (1≤i≤9), and T3=LF1,D2,D4,D9, we get that every 3-order primitive Pythagorean vector can be obtained from multiplying F1,D2,D4, or D9 by 3,4,5 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 ofT3, we can obtain the another representation of W3; that is, we have that W3=a,b,c∣a,b,c=3,4,5∗F,∀F∈T3. Does Wn (n≥4) have a similar representation?In this paper, we have given the generators of the finitely generated groupTn (n≤10). Is Tn (n>10) a finitely generated group? If Tn (n>10) is a finitely generated group, what are the generators of Tn (n>10?These appear to be interesting questions, which we hope to take up in the near future.



---

*Source: 2857417-2020-03-24.xml*