Abstract

In the present paper, we consider some D(s) Diophantine triples for a prime integer s with its negative case/value although there exist infinitely many Diophantine triples. We give several properties of such Diophantine triples and prove that they are non – extendability to D(s) Diophantine quadruple using algebraic and elementary number theory structures.

2010 Mathematics Subject Classification: 11A07, 11D09, 11A15.

1. Introduction and Preliminaries

Number Theory includes lots of questions on whole types of numbers. These questions on numbers also related with sets, equations and so on… The Diophantine n- tuple was started to work by Diophantus of Alexandria who was one of the crucial mathematicians in the algebra. Diophantine problems can not be solved easily. They contain different kind of methods and some of them related with the integer solutions of Pell equations.

There are lots of papers on the Pell equations and Diophantine n- tuple in the literature. In ¹ Beardon & his coauthor and Dujella et al. in ² gave some special and significant results on some of Diophantine triples such as strong Diophantine triples, regular Diophantine triples etc… Gopalan ^{3, 4} worked on both dio special diophantine triples and pellian equation.

The author Özer ^{5, 6, 7, 8, 9} has prepared her papers on various types of Diophantine triple with different algebraic methods. Some of the basic informations (Definitions, Examples, Theorems, Lemmas etc…) such as quadratic residues, quadratic reciprocity law, legendre & Jacobi symbols are given for algebraic and elementary number theory in ¹⁰, ¹¹ and ¹². For further results and informations, we refer to ^{13, 14, 15, 16, 17, 18, 19, 20, 21} books cited herein for readers.

In the current paper we choose s= 29 and consider some of the D(29) Diophantine triples. We prove that they aren’t extendable to D(29) Diophantine quadruple. Then, we demonstrate some properties useful to determine of the such sets. For the proof of theorems, we use the factorization method in set of integers and integer solutions of Pell (Pellian) equations or Pell like equations as well as quadratic resıdue, quadratc reciprocity theorem, legendre symbol, etc...

Now, we give some basic notations useful for proving our theorems as follow:

Definition 1.1. ^{1, 13} A Diophantine n-tuple with the property D(s) (it sometimes representatives as with n-tuple) for an integer s is a n-tuple of different positive integers such that is always a square of an integer for every distinct i, j.

As a special case, If n =3 then it is called D(s) - Diophantine triple.

Definition 1.2. ² If - Diophantine triple satisfies the condition

(1.1)

Then, it is called - Regular Diophantine Triple.

Definition 1.3. ^{10, 12, 21} (Quadratic Residue) Let be an odd prime, We call that is a quadratic residue if a nonzero number is a square and is abbreviated QR. A number which isn’t congruent to a square (mod ) is called a quadratic nonresidue (mod ) and is shortened NR. If a number that is congruent to 0 modulo p is neither a residue nor a nonresidue.

Lemma 1.1. ¹⁰ Let where is an odd prime. Then is a quadratic residue if and only if

Definition 1.4. ⁴ Legendre Symbol) Legendre symbol is introduced the following notation for prime

(1.2)

Note from above that

Lemma 1.2. ^{12, 17} Let be a Legendre Symbol and be a prime number. Then, followings are satisfied.

(1.3)

(1.4)

Note. The legendre symbol can be naturally generalized to the case when a and b are odd and coprime numbers.

Definition 1.5. ^{11, 20} (Jacobi Symbol) Suppose Then the Jacobi symbol, also represented as is defined as follows:

(1.5)

The Jacobi symbol satisfies some multiplicative properties as follows:

(1.6)

As usual with empty products, we set

Theorem 1.1. ^{14, 15, 16} Quadratic Reciprocity Theorem) If p, q are distinct odd primes, then following equation is hold.

(1.7)

Theorem 1.2. ^{12, 21} (Quadratic Reciprocity Law) If u, v are odd numbers such that (u, v) = 1, then followings can be obtained by Theorem 1.1.

(1.8)

(1.9)

2. Theorems and Results

Theorem 2.1. A set is a non-extendible regular Diophantine triple.

Proof. Let us consider Definition 1.2 and regularity condition (1.1). Then, we can see that is a regular triple. Assuming that can be extended for any positive integer and is a Diophantine quadruple. Then, there exist integers such that;

(2.1)

(2.2)

(2.3)

Dropping between (2.1) and (2.3), we have

(2.4)

the both side of (2.4) can be factorized in the set of integers. So, we have following table:

Elimination of between (2.1) and (2.2), we get

(2.5)

Considering above solutions, we have and respectively. If we substitute these values into the (2.5) we obtain and This shows that any values of is not integer solution of (2.5).

So, there is not any such and the can not be extended to Diophantine quadruple.

Theorem 2.2. and are regular Diophantine triple but they can not extendible to Diophantine quadruple.

Proof. From the (1.1) condition in Definition 1.2, it is easily seen that both and are regular triples. Let’s suppose that there exists a positive integer such that is a Diophantine quadruple. Then the following equations have integral solutions for

(2.6)

(2.7)

(2.8)

Let consider (2.6) and (2.8) and eliminate Then, we obtain

(2.9)

Factorization of the left side of (2.9), we get

(2.10)

If we search the solutions of the (3.13), we have

We obtain the equation

(2.11)

from (2.6) and (2.7). If we use Table 2, we get or By putting these values into the (2.11), we have or respectively. This is a contradiction since they are not integer solutions of (2.11).

Let us look at the and assume that it can be extendible to Diophantine quadruple. In a similar way, we get

(2.12)

(2.13)

(2.14)

for where positive integer. From (2.12) and (2.14), we obtain

(2.15)

If we put the solutions of (we get Table 2 for solutions if we consider (2.12) and (2.13)) from the Table 2 into the (2.15), then we have which satisfying the equation (2.15).

Therefore, it is a contradiction and is nonextendible.

Theorem 2.3. and are not only regular Diophantine triple but also non-extendible.

Proof. It is seen that is a regular Diophantine triple by applying (1.1) condition from Definition 1.2. Suggesting that is a Diophantine quadruple where is positive integer. So, there are such that

(2.16)

(2.17)

(2.18)

By simplifying between (2.16) and (2.18), we obtain

(2.19)

and likewise a (2.16) and (2.17) we get

(2.20)

By factorizing (2.19), we obtain following table for solutions.

By substituting or into the (2.20), we get or respectively. It shows that this is a contradiction since As a result, there isn’t any such and also the can not be extended to Diophantine quadruple.

Given that be a Diophantine quadruple for Using Definition 1.1, we obtain

(2.21)

(2.22)

(2.23)

for By eliminating from (2.21) and (2.22), we get an equation similar to (2.19). So, we can use results of Table 3 for instead of From (2.21) and (2.23), we also have

(2.24)

Putting or into the (2.24), we get or respectively. This is a contradiction because is not an integer solution of (2.24). So, is nonextendable.

Theorem 2.4. diophantine triples are regular and nonextendible to Diophantine quadruple.

Proof. Let us start with the regularity of and Both of them satisfy the (1.1) condition in the Definition 1.2. So, they are regular.

By a contraction method, assume that is Diophantine quadruple for From the Definition of 1.1, we get following equations

(2.25)

(2.26)

(2.27)

for Simplification of (2.26) and (2.27), following equation is got;

(2.28)

And a similar way, we obtain

(2.29)

from (2.25) and (2.26). Since (2.28) can be factorized, we get following tables for soutions in the set of integers.

Using the values from Table 4 and substitute into the (2.29), we have or show that is not integer solution of (2.29). This is a contradiction. So, is a regular nonextendable Diophantine triple.

In the same vein, ıf we suppose that is Diophantine quadruple for positive integer then we obtain

(2.30)

(2.31)

(2.32)

for From (2.30) and (2.31), we obtain similar equation of (2.28) and we get same solutions of Table 4 for We have

(2.33)

by dropping from (2.30) and ( 2.32). If we put or into the (2.33), or are obtained respectively. It is easy to seen that is not integer solution for (2.33) whıch is a contradiction. Thus, can not be extendible.

Remark 2.1. There are lots of various regular Diophantine triples and one may determine others using our method.

Theorem 2.5. There is no involves elements satisfy either of the states of affairs as follows:

(a) is divided by 3 or multiplies of 3.

(b) is divided by 8 or multiplies of 8.

(c) is divided by 11 or multiplies of 11.

(d) is divided by 17 or multiplies of 17.

(e) is divided by 19 or multiplies of 19.

Proof. (a) Let us assume that and are elements of set for We have

(2.34)

for an integer If we apply (mod 3) on (2.34), then we obtain

(2.35)

and from (1.3) of Lemma 1.2,

(2.36)

is got. It is a contradiction. Thus, there isn’t any contains the types of elements in (a).

(b) Given that and are elements of By the Definition 1.1, we have

(2.37)

for integer If we apply on (2.37), then follwing quadratic equivalent is had.

(2.38)

It is clear that is satisfied from Lemma 1.1, Definition 1.3 and residue classes of modulo 8 is satified. It shows that (2.38) can not solved implies that it is acontradiction. Therefore, 5 is non-quadratic residue of (mod 8). So, there is no involves the types of elements in (b).

(c) Supposing that and, is divided by 11 or multiplies of 11 in Then,

(2.39)

satisfies for integer Using modulo 11, we have

(2.40)

From (1.7) in Theorem 1.1, we obtain

(2.41)

Applying properties of Legendre symbols on and using (1.6) from Definition 1.5, we have

(2.42)

Then, we obtain It is a contradiction. Consequently, can not be an element of Diophantine set.

(d) Assuming that is an element of set If is divided by 17 or multiplies of 17 is an element of set so we obtain

(2.43)

for the integer In a similar manner, we get

(2.44)

Using Definition 1.5, we have

(2.45)

From Theorem 1.1 (Quadratic Reciprocity Theorem), following equation is got.

(2.46)

We write from the properties of Legendre symbol. Using (1.4) (Lemma 1.2) we have

(2.47)

Besides, holds since it satisfies (1.6) condition and

It is shown that there isn’t any integer satisfies (2.44). This is a contradiction. So, there is no set contains elements such that divided by 17 or multiplies of 17.

(e) Supposing that is divided by 19 or multiplies of 19 , is an element in We get

(2.48)

for any element of the set and it satisfies for integer

Similarly, we obtain

(2.49)

Using Definition 1.5 and condition (1.6), we have

(2.50)

Considering (1.7) (Quadratic Reciprocity Theorem), we have.

(2.51)

It is easily seen that from (1.4). So, we have

(2.52)

Additionally, satisfies due to (1.4). So, we obtain

Hence, there is no integer satisfies (2.49). It is a contradiction. As a result, is not an element of

Theorem 2.6. A set is regular but can not be extended to the Diophantine quadruple.

Proof. It is clear that is regular since it satisfies (1.1) condition in Definition 1.2. Let us suppose that can be extendible to Diophantine quadruple. Then, there is a positive integer such that So, there are integers and following equations are hold.

(2.53)

(2.54)

(2.55)

Dropping from (2.54) and (2.55), we have

(2.56)

If we use factorization method into the (2.56), we obtain solutions as following table:

Reducing from (2.53) and (2.55), then we have

(2.57)

By using Table 5, we calculate If we substitute these values into the (2.57), we have respectively. It shows that is not integer solution of (2.57) and it is contradiction. So, there is no positive integer and the set can be nonextended to Diophantine quadruple.

Theorem 2.7. A is both regular and nonextendible Diophantine triple.

Proof. Regularity of Diophantine triple is satisfied by (1.1) from Definition 1.2. In a similar way, let us supposing that be a Diophantine quadruple for positive integer We have such that

(2.58)

(2.59)

(2.60)

If we simplify from (2.58) and (2.59), we have an equation corresponds to (2.56) for We get same solutions of Table 5 for We obtain If we eliminate from (2.58) and (2.60), then following equation is had.

(2.61)

Substituting into the (2.61), we have respectively. It shows that is not integer solution for (2.61) and it is a contradiction.

Hence, there is no positive integer and can not be extended to Diophantine quadruple.

Remark 2.2. There are a great deal of various regular Diophantine triples and one may detect others with factorization method used in this paper.

Theorem 2.8. There isn’t any Diophantine set contains elements hold any of the following circs;

(i) is divided by 4 or multiplies of 4.

(ii) is divided by 7 or multiplies of 7.

(iii) is divided by 17 or multiplies of 17.

(iv) is divided by 23 or multiplies of 23.

Proof. (i) Supposing that both and also is divided by 4 or multiplies of 4, be elements in By the Definition 1.1, we have

(2.62)

for an integer Applying modulo 4 on the (2.62), we obtain

(2.63)

If is odd integer, then holds. Otherwise, satisfies. This implies that (2.63) doesn’t have solution and it is a contradiction.

Therefore, If such that is divided by 4 or multiplies of 4, then

(ii) In a same vein, assuming that is an element of and such that is divided by 7 or multiplies of 7. Then we have

(2.64)

holds for integer Considering (2.64) with modulo 7, following is got.

(2.65)

To see whether or not (2.65) a solution, we can search result of Legendre symbol,

(2.66)

From (1.4), we get and from (1.7) we obtain

(2.67)

By use of properties of Legendre Symbol, it is found This implies that So, we calculate and it is a contradiction.

Thus, if satisfies the condition (ii).

(iii) Analogously, given that and is divided by 17 or multiplies of 17, be elements of We can write,

(2.68)

for integer Applying modulo 17 on (2.68), we get an equivalent as follow;

(2.69)

Considering Quadratic Reciprocity Theorem (1.7) [from Theorem 1.1], we have

(2.70)

From (1.4) and properties of Legendre Symbol, holds. It means that if we substituting into the (2.70).

So, doesn’t include element holds is divided by 17 or multiplies of 17.

(iv) Given that and satisfies circ (iv) be elements of By the Definition 1.1, we have

(2.71)

for integer Using modulo 23, following equivalent is obtained.

(2.72)

From (1.7) of Theorem 1.1, then

(2.73)

is got. By use of Legendre Symbol’s properties, (1.4) and (1.7), we have

Hence, is divided by 23 or multiplies of 23, can not be element of

3. Conclusion

In the paper, we considered several D(29) Diophantine triples and proved that they can not be extended to D(29) Diophantine quadruple. Also, we demonstrated some properties of them by using factorization method in set of integers, integer solutions of Pell (Pellian) equations, quadratic residue, quadratic reciprocity theorem, legendre symbol so on. The results can be used to evaluate or estimate other results on them. Moreover, the conclusions would play significant role in the further study of Diophantiner D(n) sets.

References