rational diophantine quadruples.

The necessary literature survey that paved us a way to write this paper is given as follows: A collection {x1, x2 · · · xm} of m distinct non-zero rationals is called a rational Diophantine m-tuple if xixj + 1 is a perfect square for all 1 ≤ i < j ≤ m [1].The greek Mathematician Diophantus of Alexandria first analysed the problem of searching four numbers such that the product of any two of them increased by unity is a perfect square.He spotted a set of four positive rationals , , , with this property.Followed by Diophantus many Mathematicians [2–4] searched the triples or quadruples that can be extended to the next consecutive

tuples.Recently Gibbs found the first rational Diophantine sextuple , , , , , . Meanwhile Dujella, Kazalicki, Mikic and Szikszai proved that there are infinitely many rational Diophantine sextuples.Though these legends discovered the Diophantine m-tuples that are extendible,there are some cases where the rational Diophantine triples cannot be extended to a rational Diophantine quadruple [5, 6]. In this paper we have focused on such kind of numbers say rational polynomials,polygonal numbers, Centered pyramidal numbers and so on.We prove their non-extendability by showing that any two numbers of those quadruples does not meet the requirement of Diophantine m-tuples for the particular property. To acquire deep knowledge in Diophantine m-tuples and their generalizations, the readers are diverted to [7].

Definition 1. A set of m nonzero rationals {a1, a2 · · · am} is called a Regular Diophantine m-tuple if aiaj + q is a perfect square for all i, j = 1, 2, · · · m.If the condition is not satisfied for any one pair of (ai, aj), then it is known as an irregular Diophantine m-tuple.

quadruples with respect to Rational Numbers

Problem 1: Consider the 2-tuple (RQ1, RQ2) = , satisfying the 4W n W property D(n W + 1) ∀ n ≥ 3. To search for its extendability, let RQ3 be any other rational number with the same property. Here we can see that

RQ1.RQ3 + (n W + 1) = ψ (1) RQ2.RQ3 + (n W + 1) = Ω (2)

Eliminating RQ3 between (1) and (2), we get

(n W + 1)(RQ2 − RQ1) = ψ RQ2 − Ω RQ1 (3)

The choice ψ = j + RQ1H; Ω = j + RQ2H (4)

Substitute equation (4) in equation (3) which yields the succeeding equation 1 + 2n W (n W + 1) = j − RQ1.RQ2H . For the choice H = 1,the values of j, ψ are found to be j = , 2nW

2 2 2 2 3

4n W + n + 2 (n + 2) + W (8n ) ψ = . The values of ψ can be employed to derive the rational number RQ3 = . 4nW 4n W Hence (RQ1, RQ2, RQ3) is a Regular rational Diophantine triple with property D(n W + 1), ∀n ≥ 3.

Theorem 0.1. Let RQ1 = , RQ2 = , RQ3 = and

RQ4 = .Then (RQ1, RQ2, RQ3, RQ4) is an Irregular rational Diophantine quadruple with the property D(n W + 1) ,∀ n ≥ 3.

2 2 1 1 2 2 1 + 2n W RQ1RQ2 + (n W + 1) = + (n W + 1) = 4W n W 2nW 2 2 (n + 2) + 4n W 4nW 2 2 (n + 4) + (16n )W + 16n W + 16n W

2 2 (n + 2) + 2n W 2 2 1 (n + 4) + 16n W 2 2 RQ2RQ4 + (n W + 1) = + (n W + 1) n W 4n W

2 2 (n + 2) + W (8n ) (n + 4) + 16n W 4n W 4n W + (n W + 1)

(n + 6n + 8) + 12n W

For all the combinations of the triples satisfying D(n W + 1) ,∀ n ≥ 3, except RQ1RQ4 + (n W + 1) are perfect Squares. Hence (RQ1, RQ2, RQ3, RQ4) is an Irregular rational Diophantine quadruple.

Corollary 0.1. Let RQ1 = , RQ2 = , RQ3 = and

RQ4 = .Then (RQ1, RQ2, RQ3, RQ4) is a D(n W −1) ∀ n ≥ 3, Irregular rational Diophantine quadruple.

Problem 2: Examine the 2-tuple (RQ5, RQ6) = , fulfilling the property D(4 W1 + 2).To determine whether it can be extended, let RQ7 be a different rational number, achieving the same property. Here we can see that

n1 2 2

RQ5RQ7 + (4 W1 + 2) = ψ (5)

n1 2 2

RQ6RQ7 + (4 W1 + 2) = Ω (6)

Applying the same procedure described in the previous problem, it is obtained the Regular rational Diophantine

triple, , ,

satisfying D(4 W1 + 2).

Theorem 0.2. Let RQ5 = , RQ6 = , RQ7 =

D(4 W1 + 2) Irregular rational Diophantine quadruple.

n 2 1 1 n 2 RQ5RQ6 + (4 W1 + 2) = + (4 W1 + 2)

n 2 2

1 + 4 W1

n n 2 2

(1 + 2 ) + 4 W1

n 2 1 (1 + 2 ) + 2(8 )W1 n 2 RQ5RQ7 + (4 W1 + 2) = + (4 W1 + 2)

n n 2 2

(1 + 2 ) + 4 W1

n1 2 n1 2 2 n1 2 n1

n 2 (2 + 2 ) + (4 W1 ) + 2(4 )W1 (2(2 ) + 1) RQ5RQ8 + (4 W1 + 2) = (2 W1) n 2 1 (1 + 2 ) + 2(8 )W1 n 2 RQ6RQ7 + (4 W1 + 2) = + (4 W1 + 2) 4 W1 4 W1

n n 2 2

(1 + 2 ) + 8 W1

n 2 1 (2 + 2 ) + 4(8 )W1 n 2 RQ6RQ8 + (4 W1 + 2) = + (4 W1 + 2) 4 W1 4 W1

n n 2 2

n 2 (2 + 2 ) + 8 W1 RQ6RQ8 + (4 W1 + 2) =

n1 2 n1 2 n1 2 n1 2

n 2 (1 + 2 ) + 2(8 )W1 (2 + 2 ) + 4(8 )W1 RQ7RQ8 + (4 W1 + 2) = 4 W1 4 W1 + (4 W1 + 2)

n n n 2 2

(2 + 3(2 ) + 4 ) + 3(8 )W1

Since RQ5.RQ8 + (4 W1 + 2) is not perfect square, (RQ5, RQ6, RQ7, RQ8) is an irregular rational Diophantine quadruple.

Corollary 0.2. Let RQ5 = , RQ6 = , RQ7 =

D(4 W1 − 2) Irregular rational Diophantine quadruple.

for the Polygonal Numbers

Problem:

(2W + 1)n − (2W − 1)n (3W + 1)n − (3W − 1)n Consider the 2-tuple GP n = , GP n = rewarding

the propertyD .To ascertain whether it can be extended,let GP n be a rational number such that n (3W − 5W − 1) + n (12W − 2) − n (6W − 5W + 1) 2 GP n GP n + = θ (7)

n (3W − 5W − 1) + n (12W − 2) − n (6W − 5W + 1) 2 GP n GP n + = τ (8) Eliminating GP n between (7) and (8)we get

(GP n − GP n )

= θ GP n − τ GP n (9) Consider θ = j + GP n H; τ = j + GP n H (10) The application of the equation (11) in equation(9),we get new equation

= j − GP n GP n H 3W n (5W + 1)n − (2W − 1)n Consider H = 1.We get j = , deputy j and H in θ, θ = The rational number can be determined using the value of θ n (22W + 15W + 2) − n (32W − 6W − 4) + n (10W − 9W + 2) GP n = 2[(2W + 1)n − (2W − 1)n] Hence(GP n , GP n , GP n ) is a Regular rational Diophantine triple with the Property

(2W + 1)n − (2W − 1)n Theorem 0.3. Let GP n = , (3W + 1)n − (3W − 1)n GP n = , n (22W + 15W + 2) − n (32W − 6W − 4) + n (10W − 9W + 2) GP n = and 2[(2W + 1)n − (2W − 1)n] n (78W + 41W + 5) + n (−120W + 12W + 10) + n (42W − 29W + 5) GP n = .

2[(3W + 1)n − (3W − 1)n] Then (GP n , GP n , GP n , GP n ) is a

D irregular rational Diophantine quadruple.

Proof : For all the combinations of the quadruple satisfying

D ,except

GP n .GP n + are perfect Squares. Hence

(GP n , GP n , GP n , GP n )is an Irregular rational Diophantine quadruple. □

Consider the general expression for Heptagonal and Nonagonal numbers, we get the Regular rational Diophantine quadruple triple

2 2 4 3 2 4 3 2

5n − 3n 7n − 5n 60n − 56n + 12n n + 46n − 15n , , with the property D and irregular rational 2 2 5n − 3n 4 Diophantine quadruple 5n − 3n 7n − 5n 60n − 56n + 12n 399n − 446n + 115n , , , with the same 2 2 5n − 3n 2(7n − 5) property.