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### Article

Kyungpook Mathematical Journal 2023; 63(3): 345-353

Published online September 30, 2023

### On the Tarry-Escott and Related Problems for 2×2 matrices over ℚ

Supawadee Prugsapitak∗ and Walisa Intarapak, Vichian Laohakosol

Algebra and Applications Research Unit, Division of Computational Science, Faculty of Science, Prince of Songkla University, Hat Yai, Songkhla 90110, Thailand

Department of Mathematics, Faculty of Science, Kasetsart University, Bangkok 10900, Thailand
e-mail : fscivil@ku.ac.th

Received: October 31, 2022; Revised: July 2, 2022; Accepted: July 5, 2023

### Abstract

Reduced solutions of size 2 and degree n of the Tarry-Escott problem over M2() are determined. As an application, the diophantine equation αAn+βBn=αCn+βDn, where α,β are rational numbers satisfying α+β0 and n{1,2}, is completely solved for A,B,C,DM2().

Keywords: Diophantine Equation, Matrix Equation, Tarry-Escott Problem

### 1. Introduction

A Diophantine equation is an equation, usually with integral or rational coefficients, in which the sought-after unknowns are also integers. In 1989, Vaserstein [6] suggested solving classical problems of number theory substituting the ring by the ring M2() of 2×2 integral matrices. Some problems become easier and some give us interesting results. The Tarry-Escott problem is a classical problem in number theory which asks one to find two distinct multisets of integers {a1,,an} and {b1,,bn} such that

i=1naij= i=1nbij

for j=1,2,,k. We call n the size of the solution and k the degree. We abbreviate the above system by writing

{a1,,an}=k{b1,,bn}.

Solutions with k=n-1 are called ideal solutions. The Tarry-Escott problem has been extensively investigated in the literature; see for instance [1], [2] and also [5].

In 2006, Choudhry [3] introduced a matrix analog of the Tarry-Escott problem by considering the problem over M2(). The Tarry–Escott problem over Mm(R) for a given ring R can be stated as follows: given k,m,n and a ring R, two different multisets

A={A1,A2,,An}andB={B1,B2,,Bn},

where Ai,BiMm(R){0_}, constitute a non-trivial solution of the Tarry–Escott problem of size n and degree k over Mm(R) if

i=1nAij= i=1nBij  (j=1,2,,k),

abbreviated as {A1,,An}=k{B1,,Bn}. Choudhry [3] obtained, in parametric terms, two distinct pairs of matrices A1,A2 and B1,B2 in M2() such that A1n+A2n=B1n+B2n holds simultaneously for all integral values of n, whether positive or negative. This gives a non-trivial solution of the matrix analog of the Tarry-Escott problem of infinite degree and size 2. Using this solution, he obtained an arbitrarily long multigrade chain of matrices in M2() such that

A11n+A12n=A21n+A22n==Am1n+Am2n,

which also holds simultaneously for all integral values of n, whether positive or negative. Further, he obtained a parametric solution over M2() of the equation

A1n+A2n+A3n=B1n+B2n+B3n,

for all integral values of n. This solution leads to another arbitrarily long multigrade chain of matrices in M2().

In the present work, we present a different approach to obtain solutions of the Tarry-Escott problem over M2(); our approach also provides additional solutions different from those of Choudhry. As an application of our main result, general solutions, over M2(), are determined for the diophantine equation

αAn+βBn=αCn+βDn,

where α,β with α+β0 and n{1,2}.

### 2. Main Results

First, we prove an auxiliary result which will be used later.

Lemma 2.1. Let m and n be positive integers,

let Ai,BiMm()(i=1,,n), and let αi. If

i=1nαiAi= i=1nαiBiand i=1nαiAi2= i=1nαiBi2,

then

i=1nαi(Ai+C)= i=1nαi(Bi+C)and i=1nαi(Ai+C)2= i=1nαi(Bi+C)2

for any CMm().

Proof. Since i=1nαiAi= i=1nαiBi and i=1nαiAi2= i=1nαiBi2, it is easy to see that

i=1nαi(Ai+C)= i=1nαi(Bi+C)

and

i=1nαi(Ai+C)2=i=1nαiAi2+(i=1nαiAi)C+C(i=1nαiAi)+i=1nαiC2=i=1nαiBi2+(i=1nαiBi)C+C(i=1nαiBi)+i=1nαiC2=i=1nαi(Bi+C)2.

Immediate from Lemma 2.1 is

Corollary 2.2. Let m and n be positive integers and let

A={A1,A2,,An}andB={B1,B2,,Bn}

be subsets of Mm(). If A=2B, then for any matrix CMm() we have

A+C=2B+C

where

A+C={A1+C,A2+C,,An+C},B+C={B1+C,B2+C,,Bn+C}.

We next define equivalent solutions.

Definition 2.3. Let k, m and n be positive integers. Let

A={A1,,An},B={B1,,Bn},X={X1,,Xn},Y={Y1,,Yn}

be subsets of Mm(). We say that A=kB and X=kY are equivalent if there exist M and N in Mm() such that for all i,

Xi=MAi+NandYi=MBi+N.

Definition 2.4. Let k, m and n be positive integers. Let A={A1,,An},B={B1,,Bn} be subsets of Mm(). Then a solution A=kB is called a reduced solution if

i=1nAi= i=1nBi=0_.

The concept of being reduced is useful because of the next result.

Theorem 2.5. Let m and n be positive integers. Every solution of size n and degree 2 of the Tarry-Escott Problem over Mm() is equivalent to a reduced solution.

Proof. Let A={A1,,An} and B={B1,,Bn} be two subsets of Mm() such that A=2B. Now let X={X1,X2,,Xn} and Y={Y1,Y2,,Yn}, where Xi=AiS, Yi=BiS for i=1,,n and S=(A1++An)/n. It is easy to see that

i=1nXi= i=1nYi=0_.

Thus X=2Y is a reduced solution. Since A=2B and X=2Y, by Lemma 2.1, A=2B is equivalent to a reduced solution X=2Y.

We now consider the so-called symmetric solutions of the Tarry-Escott Problem over M2(); these are integral matrices X and Y satisfying

Xn+(X)n=Yn+(Y)n,

for all positive integers n. It suffices to show that X2=Y2. We first recall a result from [4].

Theorem 2.6. Let c. Suppose that X and Y are two elements in M2() such that XYYX. If X2+Y2=cI where I is the identity matrix, then tr(X)=tr(Y)=0 and detX+detY=c.

We prove now another auxiliary result.

Lemma 2.7. Suppose X and Y are nonzero elements in M2() and XY. Then X2=Y2andXY=YX if and only if there exist nonzero matrices A,BM2() such that

AB=BA=0_,X=A+B2andY=AB2.

Proof. Suppose X2=Y2 and XY=YX. Next, we let A = X+Y and B=X-Y. Then the results follows easily. For the converse, we suppose that X=(A+B)/2 and Y=(AB)/2 where AB=BA=0_. Then it is easy to see that XY=YX and X2=Y2. Hence the converse holds as desired.

From Lemma 2.7, in order to find commutative solutions of Xn+(X)n=Yn+(Y)n for all positive integer n, it suffices to solve for matrices A and B such that AB=BA=0_, and this leads us to our first main result.

Theorem 2.8. Suppose X and Y are nonzero elements in M2(). Then {X,X}=2{Y,Y} if and only if X,Y belong to one of the following two classes.

• 1. XYYX,X2=Y2, X=abcaandY=wxyw where

a,b,c,w,x,y are rationals such that a2+bc=w2+xy and (bxcx or axbw or aycw).

• 2. XY=YX,X2=Y2, and there exist nonzero matrices A,BM2() such that

AB=BA=0_,X=A+B2andY=AB2.

where A and B are of the following forms:

• (a) A=amacmc and B=wxwmxm where acmw0 and aw+xc=0,

• (b) A=00cd and B=w0cwd0 where cdw0,

• (c) A=0b0d and B=wbwd00 where bdw≠ 0,

• (d) A=000d and B=w000 where dw≠ 0,

• (e) A=00c0 and B=00y0 where cy≠ 0,

• (f) A=0b00 and B=0x00 where bx≠ 0.

Proof. Suppose X2=Y2.

Case 1: XYYX. Then by Theorem 2.6, tr(X)=tr(Y)=0 and detX+detiY=0. Let X=abca,Y=wxyw where a,b,c,w,x,y. Since detX+detiY=0, detX=detY. This implies that a2+bc=w2+xy as desired. Now note that XY=aw+byaxbwcwaycx+awandYX=aw+cxbwaxaycwby+aw. Thus we have bycx or axbw or aycw.

Case 2: XY=YX. By Lemma 2.7, there exist A,BM2() such that AB=BA=0_. Suppose A=abcd and B=wxyz. Thus we have the following system of equations:

aw+by=aw+cx=0
ax+bz=bw+xd=0
cw+dy=ay+cz=0
cx+dz=by+dz=0

Since AB=0_, this implies that detA=0 or detB=0. We may assume that detA=0. Thus ad-bc=0.

Case 2.1: abcd0. Since ad-bc=0, a/b=c/d. Let m=b/a. By (2.1), y=w/m. By (2.2), z=x/m. Thus A=amacmc and B=wxwmxm as desired.

Case 2.2: ad = bc =0. Thus there are 4 cases to consider.

Case 2.2.1: a=b=0. Then (2.1)-(2.4) imply that

cx=xd=cw+dy=cz=cx+dz=dz=0.

Since at least one of c and d is nonzero, we have x=z=0. If cd0 then cw+dy=0. Then y=cw/d. Thus we obtain solutions of the form A=00c0 and B=w0cwdd where cdw0.

Case 2.2.1(i): c=0. Then y=0. The fact that A,B are nonzero matrices implies dw0. Thus we obtain solutions of the form A=000d and B=w000 where dw0.

Case 2.2.1(ii): d=0. Then w=0. Again since A,B are nonzero matrices, we have cy0. So we obtain solutions of the form A=00c0 and B=00y0 where cy0.

Case 2.2.2: a=c=0. Then (2.1)-(2.4) imply that

by=bz=bw+dx=dy=dz=by+dz=0.

Since at least one of b and d is nonzero, we have y=z=0.

Case 2.2.2(i): bd0. Then bw+dx=0 and x=bw/d. Thus we obtain solutions of the form A=0b0d and B=wbwd00 where bdw0.

Case 2.2.2(ii): b=0. Then x=0. The fact that A,B are nonzero matrices implies dw0. Thus we obtain solutions of the form A=000d and B=w000 where dw0.

Case 2.2.2(iii): d=0. Then w=0. Again since A and B are nonzero matrices, we have bx0. So we obtain solutions of the form A=0b00 and B=0x00 where bx0.

For the case b=d=0 and c=d=0, we proceed similarly and obtain solutions as shown in the previous cases. The converse is easily checked.

We next provide an example.

Example 2.9. Let A=1212,B=4422. It is easy to see that AB=BA=0_. Next, we let X=(A+B)/2 and Y=(AB)/2. Then

X=5/233/22,Y=3/211/20.

By Lemma 2.7, XY = YX and X2=Y2. Thus {X,X}=n{Y,Y} for any positive integer n. Moreover, since X and Y are nonsingular matrices, {X,X}=n{Y,Y} for all integer n.

We proceed now to our final task which is to solve

α1X1n+α2X2n=α1Y1n+α2Y2n

for n=1 and 2 over M2(). We need two more auxiliary results.

Lemma 2.10. Let m be a positve integer. Let A1,A2,B1,B2 be matrices in Mm(). Let α1 and α2 be nonzero rational numbers such that α1+α20. If

α1A1n+α2A2n=α1B1n+α2B2n

for n=1,2 then there exist Ai and Bi in Mm() such that

α1A 1+α2A 2=0_=α1B 1+α2B 2andα1A 12+α2A 22=α1B 12+α2B 22.

Proof. Suppose α1A1n+α2A2n=α1B1n+α2B2n for n=1, 2 and α1α2(α1+α2)0.

For i=1, 2, we define

S=α1A1+α2A2,Ai=AiSα1+α2,Bi=BiSα1+α2.

Then α1A 1+α2A 2=α1B 1+α2B 2=0. Since α1A1n+α2A2n=α1B1n+α2B2n for n=1, 2, by Lemma 2.1, we have α1A 12+α2A 22=α1B 12+α2B 22.

Lemma 2.11. Let m be a positive integer. Let α1 and α2 be nonzero rational numbers such that α1+α20. There exist A1,A2,B1,B2Mm() such that α1A1+α2A2=α1B1+α2B2=0_ and α1A12+α2A22=α1B12+α2B22 if and only if A12=B12.

Proof. Suppose α1A1+α2A2=α1B1+α2B2=0_, A2=(α1A1)/α2 and B2=(α1B1)/α2. Then

(α1+α12/α2)A12=α1A12+α2A22=α1B12+α2B22=(α1+α12/α2)B12.

Thus A12=B12.

For the converse, suppose A2=(α1A1)/α2 and B2=(α1B1)/α2. Then

α2A2+α1A1=α1B1+α2B2=0_.

Since A12=B12, we obtain that

α1A12+α2A22=α1B12+α2B22.

Combining the results of Lemmas 2.10 and 2.11, we arrive at our final main result.

Theorem 2.12. Let α1 and α2 be nonzero rational numbers such that α1+α20. Then all solutions of the diophantine equation

α1X1n+α2X2n=α1Y1n+α2Y2n

for n=1 and 2 over M2() are the form X1=A1+C, X2=(α1A1)/α2+C, Y1=B1+C, Y2=(α1B1)/α2+C where A12=B12 and CM2().

We end this paper with an example.

Example 2.13. We show how to solve the following system of equations

2X1n+3X2n=2Y1n+3Y2n  (1n2)

over M2().

By Lemma 2.10, we can restrict our attention to find solutions A1,A2,B1,B2 in M2() such that

2A1+3A2=2B1+3B2=0_and2A12+3A22=2B12+3B22.

By Theorem 2.8, it suffices to find A1,B1M2() such that A12=B12. We work out some solutions for two different cases.

Case 1: A1B1=B1A1. Let A1=5/233/22 and B1=3/211/20. Then we have

A2=23A1=5/3214/3 and B2=23B112/31/30.

Case 2: A1B1B1A1. Let A1=2142 and B1=3193. Then

A2=23A1=4/32/38/34/3 and B2=23B1=22/362.

Therefore,

2(A1+C)n+3(A2+C)n=2(B1+C)n+3(B2+C)n

for n=1, 2 where CM2().

### Acknowledgements.

We would like to express our gratitude to the referee(s) for comments and suggestions.

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