On the Tarry-Escott and Related Problems for matrices over
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 e-mail : firstname.lastname@example.org and email@example.com
Department of Mathematics, Faculty of Science, Kasetsart University, Bangkok 10900, Thailand e-mail : firstname.lastname@example.org
Received: October 31, 2022; Revised: July 2, 2022; Accepted: July 5, 2023
Reduced solutions of size 2 and degree n of the Tarry-Escott problem over are determined. As an application, the diophantine equation , where are rational numbers satisfying and , is completely solved for .
Keywords: Diophantine Equation, Matrix Equation, Tarry-Escott Problem
A Diophantine equation is an equation, usually with integral or rational coefficients, in which the sought-after unknowns are also integers. In 1989, Vaserstein  suggested solving classical problems of number theory substituting the ring by the ring of 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 and such that
for We call n the size of the solution and k the degree. We abbreviate the above system by writing
Solutions with k=n-1 are called ideal solutions. The Tarry-Escott problem has been extensively investigated in the literature; see for instance ,  and also .
In 2006, Choudhry  introduced a matrix analog of the Tarry-Escott problem by considering the problem over . The Tarry–Escott problem over for a given ring R can be stated as follows: given and a ring R, two different multisets
where , constitute a non-trivial solution of the Tarry–Escott problem of size n and degree k over if
abbreviated as . Choudhry  obtained, in parametric terms, two distinct pairs of matrices A1,A2 and B1,B2 in such that 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 such that
which also holds simultaneously for all integral values of n, whether positive or negative. Further, he obtained a parametric solution over of the equation
for all integral values of n. This solution leads to another arbitrarily long multigrade chain of matrices in .
In the present work, we present a different approach to obtain solutions of the Tarry-Escott problem over ; our approach also provides additional solutions different from those of Choudhry. As an application of our main result, general solutions, over , are determined for the diophantine equation
First, we prove an auxiliary result which will be used later.
Lemma 2.1. Let m and n be positive integers,
let , and let . If
for any .
Proof. Since and , it is easy to see that
Immediate from Lemma 2.1 is
Corollary 2.2. Let m and n be positive integers and let
be subsets of . If , then for any matrix we have
We next define equivalent solutions.
Definition 2.3. Let k, m and n be positive integers. Let
be subsets of . We say that and are equivalent if there exist M and N in such that for all i,
Definition 2.4. Let k, m and n be positive integers. Let be subsets of . Then a solution is called a reduced solution if
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 is equivalent to a reduced solution.
Proof. Let and be two subsets of such that . Now let and , where , for and It is easy to see that
Thus is a reduced solution. Since and , by Lemma 2.1, is equivalent to a reduced solution .
We now consider the so-called symmetric solutions of the Tarry-Escott Problem over ; these are integral matrices X and Y satisfying
for all positive integers n. It suffices to show that . We first recall a result from .
Theorem 2.6. Let . Suppose that X and Y are two elements in such that . If where I is the identity matrix, then and .
We prove now another auxiliary result.
Lemma 2.7. Suppose X and Y are nonzero elements in and . Then if and only if there exist nonzero matrices such that
Proof. Suppose and XY=YX. Next, we let A = X+Y and B=X-Y. Then the results follows easily. For the converse, we suppose that and where . Then it is easy to see that XY=YX and . Hence the converse holds as desired.
From Lemma 2.7, in order to find commutative solutions of for all positive integer n, it suffices to solve for matrices A and B such that , and this leads us to our first main result.
Theorem 2.8. Suppose X and Y are nonzero elements in . Then if and only if X,Y belong to one of the following two classes.
1. , where
a,b,c,w,x,y are rationals such that and ( or or ).
2. , and there exist nonzero matrices such that
where A and B are of the following forms:
(a) and where and aw+xc=0,
(b) and where ,
(c) and where bdw≠ 0,
(d) and where dw≠ 0,
(e) and where cy≠ 0,
(f) and where bx≠ 0.
Proof. Suppose .
Case 1: . Then by Theorem 2.6, and . Let where . Since , . This implies that as desired. Now note that . Thus we have or or .
Case 2: XY=YX. By Lemma 2.7, there exist such that . Suppose and . Thus we have the following system of equations:
Since , this implies that or . We may assume that . Thus ad-bc=0.
Case 2.1: . Since ad-bc=0, . Let . By (2.1), . By (2.2), . Thus and as desired.
Case 2.2: ad = bc =0. Thus there are 4 cases to consider.