1. Introduction
The Diophantine problem endeavors a vast area for research in number theory because of its diversity as well as its characteristic property of having immense ways to find the solutions. Thus, Diophantine problems attract number theorists all the time. Some recent studies on Diophantine problems especially on the generalization of Pell equations to higher degrees and the relationship of Diophantine equations with the ring of algebraic integers can be seen in [
1,
2,
3,
4]. Another remarkable work in the field of Diophantine equations is by Shang [
5] in which a necessary condition of solvability of Diophantine equation
over
has been derived. The Prouhet Tarry Escott problem of size
n and degree
k (
-PTE problem) focuses on determining two disjoint sets of integers, say,
and
where these two sets satisfy the Diophantine system
;
. If
, it is called the ideal solutions and otherwise called non-ideal. It was Bastein who proved the impossibility of
where
do not form a permutation of
. He applied a result from the elementary symmetric function which states that the two distinct sets of roots of a polynomial equation of degree
n have the same elementary symmetric functions [
6]. Later, Tarry proved that the first
integers can be split up into two equivalent classes each consists of
integers where the sum of the
powers in one class will be equal to that of the second class for
. It is to be noted that the system of equations
is equivalent to the system
,
,
. Thus, the PTE problem can be reformulated as the problem of detecting two polynomial equations of same degree such that both the equations have the same integral roots and the first
n coefficients are equal to each other [
6].
Several works exists in evaluating the solutions of the PTE problem [
7,
8,
9,
10,
11,
12,
13,
14,
15,
16,
17,
18,
19]. Choudhary [
20,
21] studied PTE problem with the additional condition of equal product of integers and then established the complete ideal solutions for the fourth degree case. Dickson [
22] established a method for finding all integral solutions of the
and
-PTE problem. Later, Gopalan and Srikanth [
23] found a general form of parametric soluitons of non-ideal
-PTE problem. The different approaches to the 2nd degree problem and its related problems over finite field can be seen in [
24,
25]. Bolker et al. [
26] first observed the relation between the PTE problem and the Prouhet-Thue-Morse (PTM) sequence. Later, Nguyen [
27] has derived the solutions of general PTE problem by using the product generating formula for PTM sequence. Recently, Srikanth and Veena [
28] performed a detailed survey on the PTE problem and addressed the difficulties as well as future directions of the problem systematically.
The PTE problem is the most general case of easier waring problem which concerns the integral solutions of the equation
. Ramanujan [
29] provided the integral solutions of the equation
for the given natural numbers
. Rabin and Shallit [
30] constructed a randomized polynomial-time algorithm for finding one representation of the given integer
n as
. Elia [
31] proved that the prime numbers can be proclaimed as the sum of four squares. Recently, Borkovich and Jagy [
32] discovered a new design for intgers as the sum of three squares.
In the present study, the authors aim to develop some new parametric forms of solutions of the
,
and
PTE problems and also to study the Fibonacci like pattern in the solutions of non-ideal PTE problem. In
Section 2, a new proof for the non-existence of solutions of
-PTE problem is presented.
2. On the Ideal PTE Problem
Theorem 1. The system of Diophantine equations and has no integer solutions.
Proof. Legendre [
6] showed that for any positive integer
, the set of all positive integers can be exemplified as
for some integers
if and only if
where
and
b is any integer. Therefore, the equation
has no integral solutions. □
Lemma 1. A new parametric form of solutions of the -PTE problem is given by , , , , ; where , the set of all integers for , and .
Proof. Let
. Take
,
where
and
are any integers. Thus
. Similarly, if
and
, then
. Now, applying these values in second degree equation it provides
Now
is of the form
which is equal to
. Thus
. Hence
This proves the Lemma 1. □
Lemma 2. All integral solutions of the relation satisfy the following conditions
- 1.
.
- 2.
,
where is an integer, , is any divisor of and t is an integer.
Proof. Consider the form . Let and where and are integers.
It is clear that
if and only if
if and only if
, where
,
is any divisor of
and
and
w are any integers. Now, take
, where
t is any integer,
is any divisor of
t and
. i.e.,
Thus, we obtain . Hence the proof. □
Replacing x by , y by , z by , w by in Lemma 2 and combining the results of both Lemma 1 and Lemma 2, we obtain Theorem 2.
Theorem 2. The parametric form , , , , and , where
- 1.
satisfy the relation
- 2.
- 3.
- 4.
- 5.
- 6.
- 7.
and are integers such that where t is any interger
provides a new parametric form of the integral solutions of the -PTE problem.
Proof. By Lemma 1, the parametric form of all integral solutions of -PTE problem , , , , ; where and . Thus, the proof completes immediately if we replace x, y, z and w by , , and respectively in Lemma 2. □
Example 1. Let , and . Then be any divisor of 2. Take . Then we have , , and . So, , , , , , and . Thus we obtain solution sets as and .
Corollary 1. If we take or in Theorem 2, then the Diophantine system and does not posess any integral solutions.
Proof. Assume in Theorem 2. Then, we get and . Thus the solutions becomes , , , , and where and are any integers with . i.e., we obtain solution sets as and where X and Y are not distinct as one is a permutation of other. □
Theorems 3 and 4 also provide different forms of parametric solutions of -PTE problem in such a way that Theorem 3 provides two parametric solutions and Theorem 4 provides three parametric solutions.
Theorem 3. A two parametric form of integral solutions of the Diophantine system and is given by , , , , and .
Proof. Let
,
,
,
,
and
. Then,
Thus we obtain
. Now, consider
and
Thus, we obtain . Hence the proof. □
Example 2. Let and in Theorem 3. Then we have , , , , and where these and satisfy and .
Theorem 4. A three parametric form of integral solutions of the Diophantine system and is given by , , , , and provided .
Proof. Let
. Take
. Then
Let
and
be two constants such that
Then
and
. Multiply (
2) by
. Then, we obtain
(
1) and (
3) gives
. i.e.,
Similarly take
and
. Then as in the previous case we obtain
,
and
. Now, apply these general values in the second degree equation and simplifying we obtain
Put
. Then we have
Hence , , , , and . □
Example 3. Let , and . Then . So we obtain , , , , , where and .
According to Frolov [
10], if
, then
. Thus by the repeated application of this, infinite number of solutions can be generated.
Theorem 5. Let t, , are any integers and be any divisor of . Then, the integer can be written as the sum of three perfect squares in two disparate ways.
Proof. As per the assumptions in Theorem 2, choose the integers and k.
Let , , and . Consider the equation .
We know that
where
. Thus
Similarly, we know that
where
. Thus
and thus
Comparing the values of
and
we get,
Hence the proof. □
3. On the Non-Ideal PTE Problem
Some new parametric forms of solutions of the
-PTE problem and
-PTE problem have been discussed in
Section 3.
Lemma 3. A parametric form of integral solutions of the -PTE problem is given by , , , , , , , and , where .
Proof. Consider the equation . Let . Then we have . Let , and . Then . Similarly, if , and , then . Substituting these values in , we obtain .
i.e.,
.
From this,
k can be written as
i.e.,
Hence the proof. □
Lemma 4. The integral solutions of the relation satisfy the following conditions:
- (I)
- (II)
- (III)
where is any integer, such that are any integers and .
Proof. Let
,
and
. Then,
and
Thus, if and only if if and only if ; where .
If the parameters , , , , and in Lemma 4 are replaced by and then the simultaneous applications of Lemmas 3 and 4, we obtain Theorem 6.
Theorem 6. The parametric form , , , , , . and where
satisfy
and are any integers such that is any divisor of
provide integral solutions of the non-ideal -PTE problem.
Proof. By Lemma 3, the parametric form of the integral solutions -PTE problem is , , , , , , and where with . If we replace by , by , by , by , by , by in Lemma 4 we will have the theorem. □
Example 4. Let , , and . Then . Take . Then , , , and . So , , , , , , , .
Theorem 7 also provides another parametric solution of -PTE problem.
Theorem 7. A four parametric form of integral solutions of -PTE problem is given by , , , , , , and ; provided .
Proof. Let
,
. Then we obtain
Let
and
are two integers such that
. Then
and
. i.e.,
Multiplying this with
, we obtain
Thus,
and
. Similarly, we obtain
and
. Applying the values of
s and
s in the second degree equation, we obtain
After elementary simplification, we obtain
. Now take
and
and after simplifying further we obtain
So, by substituting the value of k, we obtain , , and . This proves the theorem. □
Example 5. Let , , and . Then . . Thus , , and .
Theorem 8. Any integer of the form , where with can be disclosed as the sum of four perfect squares in two distinct ways.
Proof. Let
,
,
,
,
,
,
and
; provided
. Then,
Comparing these two we get
Hence the proof. □
Theorem 9 is a particular case of the result by Nguyen [
11] which illustrates a particular solution of the non-ideal PTE problem
;
by using PTM sequence.
Theorem 9. The first eight non-negative integers can be partitioned into two sets of equal size such that the elements in each set satisfy the non-ideal -PTE problem.
Proof. Consider the Prouhet-Thue-Morse sequence defined by
where
is the base-2 expansion of the integer
n. Now, consider the first eight non-negative integers, say
. We define two disjoint sets
and
in such a way that
Then, we obtain
and
such that
Hence the proof. □
Remark 1. In Theorem 9, if we consider the generalized Prouhet-Thue-Morse sequence under modulo 3, then we can partition the first 27 non-negative integers into three distinct sets of equal size satisfying the relations ; r=1,2.
Remark 2. Consider the assignment as in Theorem 9. Now, define ; for and . Let be a vector consisting of two arbitrary complex values such that . Define to be a polynomial of degree 7 whose coefficients belong to A and repeat according to . i.e., So, we obtain . Put and define . Then, Let denotes the derivative of for . i.e., However, both and can also be written as and . Since and , we get and . From the choice of and , we get and where and .
Theorem 10. The parametric form of all integral solutions of the non-ideal PTE problem is given by , , , , , , , , and .
Proof. Let
,
,
,
,
and
.Then we get
and
Let
and
be two integers such that
. Then
and
. So
. Multiplying by
, we get
Adding and subtracting
we obtain
Thus,
and
. Similarly, we get
and
. Now, substitute the values of
’s and
’s in (7), we get
and
. Take
,
and
. After performing elementary calculations we obtain
provided
. Thus,
,
,
and
. Hence the proof. □
Example 6. Let , , , and . Then . Thus, , , and . Hence .
Theorem 11. If , , , and are any integers with , then the integer of the form can be represented as the sum of five perfect squares in two distinct ways.
Proof. Let
,
,
,
,
,
,
,
,
and
.Then,
Comparing these two we get
Hence the proof. □
4. On Fibonacci Like Pattern in PTE Problem
Because of the uncertainty in nature, the Fibonacci numbers are always become fishy to the mathematicians. One can expound the Fibonacci sequence of numbers using the looping as
.
Some remarkable studies on Fibonacci numbers and their applications can be seen in [
33,
34,
35]. In
Section 4, the Fibonacci like pattern appearing in the solutions of the
-PTE problem is analyzed by impossing an additonal condition
for
to the problem.
Note 1. Consider the Diophantine equationwhere and . If we replace and in (12) by and , we obtainwhich is the -PTE problem. Thus, any solution of will provide solution to (12). Note 2. It does not guarantee that the solutions of ideal PTE problem satisfy the Fibonacci like pattern. This is because, in the -PTE problem, if and , then after some algebraic operations the system will reduce to -PTE problem which has no solutions.
Theorem 12. A two parameter family of infinitely many integral solutions of the system of equations and where for , is given by , , , , , , and , provided .
Proof. Consider the following equations
Put
,
,
and
. Then, we have
Thus (
13) and (14) becomes
Take
and
. Then, we have
and
. Substituting these values, we get
Thus the solutions of the problem is given by , , , , , , and provided . □
Example 7. Let and . Then . , , , , , , and .
Corollary 2. The primes 2 and 3 divides .
Proof. Let
. As per the assumptions in Theorem 12, we have
,
,
,
,
,
,
and
. Then
So, the primes 2 and 3 divides C. The other prime divisors of C will obtain accordingly as the number since and can take any integers such that . □
If denote the least positive integer such that the Diophantine system ; and for , possess nontrivial integer solutions, then from Note 1 and Theorem 12 we obtain and . Thus, we arrive at the Theorem 13.
Theorem 13. .
Proof. Then there are
elements in
A. Let
. Then we can take
,
as
and
for some integers
and
;
. We define an equivalence relation on
A by
if and only if
is a permutation of
. For example, if
is an element in
A, then all its six permutations like
are equivalent to
. The reason for defining the equivalence relation like this is that generally in PTE problems if
satisfies the left hand side equality then the right hand side solution should not be the permutation of
. Since
has atmost
distinct permutations, there are
distinct classes in
. Define
for
. Note that
. So there are atmost
distinct sets
. Choose
. Then, we have
since
. So the number of possible
is less than the number of distinct
. Thus, the two distinct sets
and
form a solution of degree
k. □