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 ℤ by the ring M2(ℤ) 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(ℤ). The Tarry–Escott problem over Mm(R) for a given ring R can be stated as follows: given k,m,n∈ℕ and a ring R, two different multisets

A={A1,A2,…,An} and B={B1,B2,…,Bn},

where Ai,Bi∈Mm(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(ℤ) 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(ℤ) 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(ℤ) 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(ℤ).

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

αAn+βBn=αCn+βDn,

where α,β∈ℚ with α+β≠0 and n∈{1,2}.

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

Lemma 2.1. Let m and n be positive integers,

let Ai,Bi∈Mm(ℚ) (i=1,…,n), and let αi∈ℚ. If

∑ i=1nαiAi=∑ i=1nαiBi and ∑ 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 C∈Mm(ℚ).

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} and B={B1,B2,…,Bn}

be subsets of Mm(ℚ). If A=2B, then for any matrix C∈Mm(ℚ) 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(ℚ). We say that A=kB and X=kY are equivalent if there exist M and N in Mm(ℚ) such that for all i,

Xi=MAi+N and Yi=MBi+N.

Definition 2.4. Let k, m and n be positive integers. Let A={A1,…,An},B={B1,…,Bn} be subsets of Mm(ℚ). 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(ℚ) is equivalent to a reduced solution.

Proof. Let A={A1,…,An} and B={B1,…,Bn} be two subsets of Mm(ℚ) such that A=2B. Now let X={X1,X2,…,Xn} and Y={Y1,Y2,…,Yn}, where Xi=Ai−S, Yi=Bi−S 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(ℚ); 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∈ℂ. Suppose that X and Y are two elements in M2(ℂ) such that XY≠YX. 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(ℚ) and X≠Y. Then X2=Y2 and XY=YX if and only if there exist nonzero matrices A,B∈M2(ℚ) such that

AB=BA=0_,X=A+B2 and Y=A−B2.

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=(A−B)/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(ℚ). Then {X,−X}=2{Y,−Y} if and only if X,Y belong to one of the following two classes.

• 1. XY≠YX, X2=Y2, X=abc−a and Y=wxy−w where

  a,b,c,w,x,y are rationals such that a2+bc=w2+xy and (bx≠cx or ax≠bw or ay≠cw).

• 2. XY=YX, X2=Y2, and there exist nonzero matrices A,B∈M2(ℚ) such that

  AB=BA=0_,X=A+B2 and Y=A−B2.

  where A and B are of the following forms:

  • (a) A=amacmc and B=wx−wm−xm where acmw≠0 and aw+xc=0,

  • (b) A=00cd and B=w0−cwd0 where cdw≠0,

  • (c) A=0b0d and B=w−bwd00 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: XY≠YX. Then by Theorem 2.6, tr(X)=tr(Y)=0 and detX+detiY=0. Let X=abc−a,Y=wxy−w where a,b,c,w,x,y∈ℚ. Since detX+detiY=0, detX=detY. This implies that a2+bc=w2+xy as desired. Now note that XY=aw+byax−bwcw−aycx+aw and YX=aw+cxbw−axay−cwby+aw. Thus we have by≠cx or ax≠bw or ay≠cw.

Case 2: XY=YX. By Lemma 2.7, there exist A,B∈M2(ℚ) such that AB=BA=0_. Suppose A=abcd and B=wxyz. Thus we have the following system of equations:

(2.1)aw+by=aw+cx=0
(2.2)ax+bz=bw+xd=0
(2.3)cw+dy=ay+cz=0
(2.4)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: abcd≠0. 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=wx−wm−xm 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 cd≠0 then cw+dy=0. Then y=−cw/d. Thus we obtain solutions of the form A=00c0 and B=w0−cwdd where cdw≠0.

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

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

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): bd≠0. Then bw+dx=0 and x=−bw/d. Thus we obtain solutions of the form A=0b0d and B=w−bwd00 where bdw≠0.

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

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

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=12−1−2,B=44−2−2. It is easy to see that AB=BA=0_. Next, we let X=(A+B)/2 and Y=(A−B)/2. Then

X=5/23−3/2−2,Y=−3/2−11/20.

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(ℚ). We need two more auxiliary results.

Lemma 2.10. Let m be a positve integer. Let A1,A2,B1,B2 be matrices in Mm(ℚ). Let α1 and α2 be nonzero rational numbers such that α1+α2≠0. If

α1A1n+α2A2n=α1B1n+α2B2n

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

α1A 1′+α2A 2′=0_=α1B 1′+α2B 2′ and α1A 1′2+α2A 2′2=α1B 1′2+α2B 2′2.

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′=Ai−Sα1+α2,Bi′=Bi−Sα1+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 1′2+α2A 2′2=α1B 1′2+α2B 2′2.

Lemma 2.11. Let m be a positive integer. Let α1 and α2 be nonzero rational numbers such that α1+α2≠0. There exist A1,A2,B1,B2∈Mm(ℚ) 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+α2≠0. Then all solutions of the diophantine equation

α1X1n+α2X2n=α1Y1n+α2Y2n

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

We end this paper with an example.

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

2X1n+3X2n=2Y1n+3Y2n  (1≤n≤2)

over M2(ℚ).

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

2A1+3A2=2B1+3B2=0_ and 2A12+3A22=2B12+3B22.

By Theorem 2.8, it suffices to find A1,B1∈M2(ℚ) such that A12=B12. We work out some solutions for two different cases.

Case 1: A1B1=B1A1. Let A1=5/23−3/2−2 and B1=−3/2−11/20. Then we have

A2=−23A1=−5/3−214/3 and B2=−23B112/3−1/30.

Case 2: A1B1≠B1A1. Let A1=21−4−2 and B1=31−9−3. Then

A2=−23A1=−4/3−2/38/34/3 and B2=−23B1=−2−2/362.

Therefore,

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

for n=1, 2 where C∈M2(ℚ).

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