Edelweiss Applied Science and Technology 
ISSN: 2576-8484 
Vol. 8, No. 6, 4408-4414 
2024 
Publisher: Learning Gate 
DOI: 10.55214/25768484.v8i6.2965 
© 2024 by the authors; licensee Learning Gate 

© 2024 by the authors; licensee Learning Gate 
* Correspondence:  amarachandoul@yahoo.fr 

 
 
 
 

 

Solutions of the equation 𝒙𝟐 − (𝒑𝟐𝒒𝟐  ±  𝐚𝐩)𝒚𝟐 = 𝒌𝒕 

 
Amara Chandoul1*, Alanod Sibih2 
1Department of Computer Science 𝖠Multimedia, Higher Institute of Informatics 𝖠Multimedia of Sfax, Tunis Road, Km 10, Al 

−Ons, B. P. 242, 3021, Tunisia; amarachandoul@yahoo.fr (A.C.) 
2Department of Mathematics, Jamoum University College, Umm Al −Qura University, Holly Makkah 21955, Saudi Arabia; 
amsibih@uqu.edu.sa (A.S.). 

 

 

Abstract: In this note, the Diophantine equation 𝑥2 − (𝑝2𝑞2  ±  ap)𝑦2 = 𝑘𝑡 has been solved, and its 
positive integer solutions have been stated in terms of generalized Fibonacci, generalized Lucas, 

generalized Pell, and generalized Pell-Lucas sequences. We have discovered units for 𝑍[𝐷] in terms of 
the above sequences of numbers. 
Keywords: Diophantine equations, Generalized Fibonacci numbers, Generalized Lucas numbers, Generalized Pell 
numbers, Generalized Pell–Lucas numbers, Pell equations. 
 

1. Introduction  
Since ancient times, numerous mathematicians have studied number sequences. Particular 

attention has been paid to recurrent sequences like the Fibonacci, Lucas, Pell, or Pell-Lucas 
sequences, which are described below: 
 

Fibonacci number sequence 

𝐹𝑛+2 = 𝐹𝑛+1 + 𝐹𝑛 for 𝑛 ≥ 0 with  𝐹0 = 0 and 𝐹1 = 1.  
Lucas number sequence 

𝐿𝑛+2 = 𝐿𝑛+1 + 𝐿𝑛 for 𝑛 ≥ 0 with  𝐿0 = 2 and 𝐿1 = 1.  
Pell number sequence 

Φ𝑛+2 =  2Φ𝑛+1 + Φ𝑛 for 𝑛 ≥ 0 with  Φ0 = 0 and Φ1 = 1. 
Pell-Lucas number sequence 

Ψ𝑛+2 =  2Ψ𝑛+1 + Ψ𝑛 for 𝑛 ≥ 0 with  Ψ0 = 2 and Ψ1 = 2. 
In recent years, these sequences of numbers have significant applications in the fields 

including statistics, music, coding theory, cryptography and communication systems. [2], [3], [4]. 
As seen in [11,12,13], number sequences are generalized in several ways. One can see some 

generalizations of Pell and Pell–Lucas numbers. In this paper we will pay attention to the following 
generalization: 

Let 𝑠 and 𝑡 be two non-zero integers satisfying 𝑠2 + 4𝑡 > 0, The generalized Fibonacci and 
generalized Lucas sequences are, respectively, defined as: 
 

𝐹𝑛+2(𝑠, 𝑡) = 𝑠𝐹𝑛+1(𝑠, 𝑡) + 𝑡𝐹𝑛(𝑠, 𝑡) for 𝑛 ≥ 0 with  𝐹0(𝑠, 𝑡) = 0,  𝐹1(𝑠, 𝑡) = 1, and 
 

𝐿𝑛+2(𝑠, 𝑡) = 𝑠𝐿𝑛+1(𝑠, 𝑡) + 𝑡𝐿𝑛(𝑠, 𝑡) for 𝑛 ≥ 0 with  𝐿0(𝑠, 𝑡) = 2,  𝐿1(𝑠, 𝑡) = 1, 
 
Binet’s formulae for these sequences are:  



4409 

 
 

 
Edelweiss Applied Science and Technology 
ISSN: 2576-8484 

Vol. 8, No. 6: 4408-4414, 2024 
DOI: 10.55214/25768484.v8i6.2965 
© 2024 by the authors; licensee Learning Gate 

 

𝐹𝑛(𝑠, 𝑡) =  
𝛼𝑛− 𝛽𝑛

𝛼−𝛽
, 𝐿𝑛(𝑠, 𝑡) =  𝛼

𝑛 + 𝛽𝑛, where 𝛼 and 𝛽 are the roots of equation 𝑥2 − 𝑠𝑥 − 𝑡 = 0.  𝛼 

and 𝛽 verify 𝛼 + 𝛽 = 𝑠, 𝛼 − 𝛽 = 2√𝑠2 + 4𝑡,  𝛼𝛽 = −𝑡.    
For further details of these sequences, see [5], [6], [7],[17], [18], [19], [20],[21].  

Let, now, 𝑘 and ℎ be two non-zero integers satisfying 𝑘2 + ℎ > 0, The generalized Pell and 
generalized Pell-Lucas sequences are, respectively, defined as: 

Φ𝑛+2(𝑘, ℎ) =  2𝑘Φ𝑛+1(𝑘, ℎ) + ℎΦ𝑛(𝑘, ℎ)  for 𝑛 ≥ 0 with  Φ0(𝑘, ℎ) = 0 and Φ1(𝑘, ℎ) = 1,  and 

Ψ𝑛+2(𝑘, ℎ) =  2𝑘Ψ𝑛+1(𝑘, ℎ) + ℎΨ𝑛(𝑘, ℎ)  for 𝑛 ≥ 0 with  Ψ0(𝑘, ℎ) = 2 and Ψ1(𝑘, ℎ) = 2𝑘 

Binet’s formulae for these sequences are: 

Φ𝑛(𝑘, ℎ) =  2𝑘
𝛼𝑛− 𝛽𝑛

𝛼−𝛽
, Ψ𝑛(𝑘, ℎ) =  𝛼

𝑛 + 𝛽𝑛, where 𝛼 and 𝛽 are the roots of equation 𝑥2 − 2𝑘𝑥 − ℎ =

0.  𝛼 and 𝛽 verify 𝛼 + 𝛽 = 2𝑘, 𝛼 − 𝛽 = 2√𝑘2 + ℎ,  𝛼𝛽 = −ℎ.  

In the plethora of integer sequences, the Fibonacci and Lucas sequences stand out as the two 
brightest. They have captivated both amateur and expert mathematicians for ages, and they never cease 
to enchant us with their beauty, myriad practical uses, and omnipresent propensity to appear in 
completely unexpected and unrelated contexts. They continue to be a fertile ground for creative 
amateurs and mathematicians alike.   

In literature, Fibonacci and Lucas numbers are used to resolve many Diophantine equations. 

The quadratic Diophantine equation of the form 𝑥2 − 𝐷𝑦2 = 𝑁, where 𝐷 is square free, generally 

known as Pell’s equation. In this case, if 𝐷 is square free, the continued fraction expansion of 𝐷 is 

periodic and it is given by 𝐷 =  𝑎0;  𝑎1 , 𝑎2 ,⋯ 𝑎𝑛−1, 𝑎𝑛̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅, where 𝑎𝑛 = 2𝑎0 and 𝑛 is the length of period. 

As it is known, we denote by 
𝑝1

𝑞1
= [𝑎0;  𝑎1 , 𝑎2 ,⋯ 𝑎𝑙] the 𝑙𝑡ℎ convergent of 𝐷, for 𝑙 ≥ 0. In (5), we have 

obtained some formulas for the integer solutions of the Pell equation 𝑥2 − 𝑦2 = ±𝑘2 for all 𝑘 ≥ 1. 
In [9], Bala and Mishra, considered the solutions of two Diophantine equations 𝑥2 − (𝑝2𝑞2  ±

 3p)𝑦2 = 𝑘𝑡 and 𝑥2 − (𝑝2𝑞2  ±  5p)𝑦2 = 𝑘𝑡. They generalized a previous result of Guney [10], who 
found all positive integer solutions of the equations  
 

𝑥2 − (𝑎2𝑏2 + 2𝑏)𝑦2 = 𝑁 when 𝑁 ∈ {±1,±4} in terms of generalized Fibonacci and Lucas  sequences. 
 
In this paper, we generalize the results of Bala and Mishra [9], by solving 

𝑥2 ± (𝑝2𝑞2 − 𝑎𝑝)𝑦2 = 𝑘𝑡 
Then we express its positive integer solutions in the current study. Using generalized Lucas, 

generalized Pell, and generalized Pell-Lucas sequences.  
To do, we need the following theorems (1,2,3,2.1): 
 
Theorem 1, 

There is no positive integer solution to the equation 𝑥2 − 𝐷𝑦2 = −1 if the length of the period of 

√𝐷 's continued fraction expansion is even, while the fundamental solution of equation 𝑥2 − 𝐷𝑦2 = 1 is 
𝑝𝑛−1

𝑞𝑛−1
. 

 
 
 
 



4410 

 
 

 
Edelweiss Applied Science and Technology 
ISSN: 2576-8484 

Vol. 8, No. 6: 4408-4414, 2024 
DOI: 10.55214/25768484.v8i6.2965 
© 2024 by the authors; licensee Learning Gate 

 

Theorem 2, 

There are infinitely many solutions to the Pell equation 𝑥2 − 𝐷𝑦2 = 1 if 𝐷 is a natural number that 
is not a perfect square. All positive solutions can be obtained by the formula 

𝑥𝑛 + 𝑦𝑛√𝐷 = (𝑥1 + 𝑦1√𝐷)
𝑛, for all 𝑛 > 1, 

where (𝑥1, 𝑦1) is fundamental solution of equation 𝑥2 − 𝐷𝑦2 = 1. 
 
Theorem 3, 

Let 𝑁 be a positive integer and (𝑢1, 𝑣1) be the fundamental solution of equation 𝑥2 − 𝐷𝑦2 = 𝑁. 
Then all positive solutions of 𝑥2 − 𝐷𝑦2 = 𝑁 can be obtained by the formula 

𝑥𝑛 + 𝑦𝑛√𝐷 = (𝑢1 + 𝑣1√𝐷)(𝑥1 + 𝑦1√𝐷)
𝑛, for all 𝑛 > 1, where (𝑥1, 𝑦1) is fundamental solution of 

equation 𝑥2 − 𝐷𝑦2 = 1.  
 
We are now ready to present our main results. 
 
2. Main Results 
2.1. Theorem 

Let 𝐷 = 𝑝2𝑞2 − 𝑎𝑝,  with 𝑞 > 𝑎 being a multiple of 𝑎 and 𝑝, 𝑞 being positive integers chosen in 

such a way that 𝐷 is not a perfect square, then 

1. √𝐷 = √𝑝2𝑞2 − 𝑎𝑝 = [𝑝𝑞 − 1; 1,
2(𝑞−𝑎)

𝑎
, 1, 29𝑝𝑞 − 1)

̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅
]. 

2. The fundamental solution of the Pell equation 𝑥2 − 𝐷𝑦2 = 1 is (𝑥1, 𝑦1) = (
2𝑝𝑞2−𝑎

𝑎
 ,
2𝑞

𝑎
). 

3. All positive integer solution (𝑥𝑛, 𝑦𝑛) of the Pell equation 𝑥2 − 𝐷𝑦2 = 1 are provided by 
 

𝑥𝑛 = 
𝛼𝑛+𝛽2

2
=

1

2
𝑙𝑛(𝑠, 𝑡) =

1

2
 𝑙𝑛  (2

2𝑝𝑞2−𝑎

𝑎
, −1),  and 𝑦𝑛 = 

𝛼𝑛+𝛽2

2√𝐷
=

2𝑞

𝑎
𝐹𝑛(𝑠, 𝑡) =

2𝑞

𝑎
𝐹𝑛 (2

2𝑝𝑞2−𝑎

𝑎
, −1) =

𝑞

2𝑝𝑞2−𝑎
Φ𝑛 (2

2𝑝𝑞2−𝑎

𝑎
, −1) , for all 𝑛 ≥ 1. 

 

1. The Fundamental solution of Pell equation 𝑥2 − 𝐷𝑦2 = 𝑘𝑡 is 
 

(𝑥1, 𝑦1) =   (
2𝑝𝑞2 − 𝑎

𝑎
𝑘
𝑡

2,
2𝑞

𝑎
𝑘
𝑡

2). 

 

2. All positive solutions of Pell equation 𝑥2 − 𝐷𝑦2 = 𝑘𝑡is given by 

𝑥𝑛+1 = 
2𝑝𝑞2−𝑎

𝑎
𝑘
𝑡

2𝑥𝑛 + 2𝑞 
𝑝2𝑞2−𝑎𝑝

𝑎
𝑘
𝑡

2𝑦𝑛 and 𝑦𝑛+1 = 
2𝑞

𝑎
𝑘
𝑡

2𝑥𝑛 + 
2𝑝𝑞2−𝑎

𝑎
𝑘
𝑡

2𝑦𝑛 for all 𝑛 ≥ 1 

Proof: 1. 

√D = √p2q2 − ap 

 

= (pq − 1) + √p2q2 − ap − (pq − 1) 

 



4411 

 
 

 
Edelweiss Applied Science and Technology 
ISSN: 2576-8484 

Vol. 8, No. 6: 4408-4414, 2024 
DOI: 10.55214/25768484.v8i6.2965 
© 2024 by the authors; licensee Learning Gate 

 

= (pq − 1) +
1

√p2q2−ap+(pq−1)

2pq−ap−1

 

= (pq − 1) +
1

1 +
√p2q2−ap−(pq−ap)

2pq−ap−1

 

= (pq − 1) +
1

1 +
1

√p2q2−ap+(pq−ap)

ap

 

= (pq − 1) +
1

1 +
1

2(q−a)

a
+
√p2q2−ap−(pq−ap)

ap

 

= (pq − 1) +
1

1 +
1

2(q−a)

a
+

1

1+ 
1

2(pq−1)+√p2q2−ap−(pq−1)

 

= [pq − 1; 1,
2(q − a)

a
, 1,2(pq − 1)

̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅
] 

2. The period length of √𝐷 's continued fraction expansion is 4, making it even. Then, according to 

Theorem 1.1, the fundamental solution of equation 𝑥2 − 𝐷𝑦2 = 1 is given by 

𝑝3
𝑎
𝑞3
𝑎

.  

Let 
𝑝3

𝑞3
= [𝑝𝑞 − 1; 1,

2(𝑞−𝑎)

𝑎
, 1] = (𝑝𝑞 − 1) +

1

1+
2(𝑞−𝑎)

𝑎
+1
=

2𝑝𝑞2−𝑎

2𝑞
. 

 

Then, (𝑥1, 𝑦1) = (
2𝑝𝑞2−𝑎

𝑎
,
2𝑞

𝑎
). 

 

3. According to Theorem 1.1, all of the positive integral solutions to the equation 𝑥2 − 𝐷𝑦2 = 1 is 
provided by 

𝑥𝑛 + 𝑦𝑛√𝐷 = (
2𝑝𝑞2−𝑎

𝑎
+
2𝑞

𝑎
√𝐷)

𝑛

,  and 

𝑥𝑛 − 𝑦𝑛√𝐷 = (
2𝑝𝑞2−𝑎

𝑎
−
2𝑞

𝑎
√𝐷)

𝑛

  

 

Set 𝛼 = 
2𝑝𝑞2−𝑎

𝑎
+
2𝑞

𝑎
√𝐷  and 𝛽 =

2𝑝𝑞2−𝑎

𝑎
−
2𝑞

𝑎
√𝐷, then 𝛼 + 𝛽 = 2(

2𝑝𝑞2−𝑎

𝑎
) and 𝑡 = −𝛼𝛽 =

−1. Thus, for all 𝑛 ≥ 1. 

𝑥𝑛 =
𝛼𝑛+𝛽𝑛

2
=

1

2
𝐿𝑛(𝑠, 𝑡) =

1

2
𝐿𝑛 (2

2𝑝𝑞2−𝑎

𝑎
, −1) =

1

2
Ψ𝑛 (

2𝑝𝑞2−𝑎

𝑎
, −1), 



4412 

 
 

 
Edelweiss Applied Science and Technology 
ISSN: 2576-8484 

Vol. 8, No. 6: 4408-4414, 2024 
DOI: 10.55214/25768484.v8i6.2965 
© 2024 by the authors; licensee Learning Gate 

 

 
and 

𝑦𝑛 =
𝛼𝑛−𝛽𝑛

2√𝐷
=

2𝑞

𝑎
𝐹𝑛(𝑠, 𝑡) =

2𝑞

𝑎
𝐹𝑛 (2

2𝑝𝑞2−𝑎

𝑎
, −1) =

𝑞

2𝑝𝑞2−𝑎
Ψ𝑛 (

2𝑝𝑞2−𝑎

𝑎
, −1), 

For all 𝑛 ≥ 1. 
 

4. It is easy to verify (4) from (1). 
5. The fifth part can be proved using the theorem. 

 
2.2. Corollary  

Let 𝐷 = 𝑝2𝑞2 − 𝑎𝑝, with 𝑞 > 𝑎 being a multiple of 𝑎 and 𝑝, 𝑞 being positive integers chosen in such 

a way that 𝐷 is not a perfect square, then, the equation 𝑥2 − 𝐷𝑦2 = 1  has no positive integer solution. 
 
Proof: 
It is simple to conclude using the theorem..; 

 
2.3. Theorem 

1. Let 𝐷 = 𝑝2𝑞2 + 𝑎𝑝, with 𝑞 > 𝑎 being a multiple of 𝑎 and 𝑝, 𝑞 being positive integers chosen in 

such a way that 𝐷 is not a perfect square, then, √𝐷 = √𝑝2𝑞2 + 𝑎𝑝 = [𝑝𝑞;
2𝑞

𝑎
, 2𝑝𝑞

̅̅ ̅̅ ̅̅ ̅̅ ̅
].  

2. The fundamental solution of the Pell equation 𝑥2 − 𝐷𝑦2 = 1  is  

(𝑥1, 𝑦1) = (
2𝑝𝑞2+𝑎

𝑎
,
2𝑞

𝑎
).  

3. All positive integer solution (𝑥𝑛, 𝑦𝑛) of the Pell equation 𝑥2 − 𝐷𝑦2 = 1 are provided by 

𝑥𝑛 =
𝛼𝑛+𝛽𝑛

2
=

1

2
𝐿𝑛(𝑠, 𝑡) =

1

2
𝐿𝑛 (2

2𝑝𝑞2−𝑎

𝑎
, −1) =

1

2
Ψ𝑛 (

2𝑝𝑞2−𝑎

𝑎
, −1), 

and 

𝑦𝑛 =
𝛼𝑛−𝛽𝑛

2√𝐷
=

2𝑞

𝑎
𝐹𝑛(𝑠, 𝑡) =

2𝑞

𝑎
𝐹𝑛 (2

2𝑝𝑞2+𝑎

𝑎
, −1) =

𝑞

2𝑝𝑞2−𝑎
Φ𝑛 (2

2𝑝𝑞2+𝑎

𝑎
, −1), 

for all 𝑛 ≥ 1. 
4.  The Fundamental solution of Pell equation 𝑥2 − 𝐷𝑦2 = 𝑘𝑡 is  

(𝑥1, 𝑦1) = (
2𝑝𝑞2+𝑎

𝑎
𝑘
𝑡

2,
2𝑞

𝑎
𝑘
𝑡

2).  

 

5. All positive solutions of Pell equation 𝑥2 − 𝐷𝑦2 = 𝑘𝑡  are given by 

𝑥𝑛+1 =
2𝑝𝑞2 + 𝑎

𝑎
𝑘
𝑡

2𝑥𝑛 + 2𝑞
𝑝2𝑞2 + 𝑎𝑝

𝑎
𝑘
𝑡

2𝑦𝑛 

and 

𝑦𝑛+1 =
2𝑞

𝑎
𝑘
𝑡

2𝑥𝑛 +
2𝑝𝑞2 + 𝑎

𝑎
𝑘
𝑡

2𝑦𝑛 

 

for all 𝑛 ≥ 1. 
Proof: 
  The proof is similar to theorem 1. 
 
 
 
 
 
 



4413 

 
 

 
Edelweiss Applied Science and Technology 
ISSN: 2576-8484 

Vol. 8, No. 6: 4408-4414, 2024 
DOI: 10.55214/25768484.v8i6.2965 
© 2024 by the authors; licensee Learning Gate 

 

3. Applications 

One application of our results is to find the units of ℤ[√𝐷]. The unity, in𝑐, is 1and units are those 

invertible elements. In the following, we will find units of ℤ[√𝐷], where 𝐷 = 𝑝2𝑞2 ± 𝑎𝑝, with 𝑞 > 𝑎 

being a multiple of 𝑎 and 𝑝, 𝑞 being positive integers chosen in such a way that 𝐷 is not a perfect square. 

Let 𝑢 + 𝑣√𝐷, where 𝑢, 𝑣 ∈ ℤ be a unit element in ℤ[√𝐷], then there exists some 𝑐 + 𝑑√𝐷 ∈ ℤ[√𝐷]  
such that 

(𝑢 + 𝑣√𝐷)(𝑐 + 𝑑√𝐷) = 1. 

So, 𝑐 + 𝑑√𝐷 =
1

𝑢+𝑣√𝐷
=

𝑢−𝑣√𝐷

𝑢2−𝑣2𝐷
, then if we consider the quantity 𝑁(𝑢 + 𝑣√𝐷) =  𝑢2 − 𝑣2𝐷, one 

can verify that 𝑁 is multiplicative, and then if 𝑢 + 𝑣√𝐷 is invertible, 𝑁 must be invertible too, so it is 

either 1 or −1. Then, we get 𝑢2 −𝐷𝑣2 = ±1. 
 

Further 𝑢2 −𝐷𝑣2 = −1 has no positive solution. To solve 𝑢2 − 𝐷𝑣2 = 1, we apply our theorems. 
 

Let for example, 𝑝 = 2, 𝑞 = 14, 𝑎 = 7, then 𝐷 = 22(14)2 − 7.14 = 686 which is, clearly, not a 
perfect square. 
 

Fundamental solution of  𝑥2 − 868𝑦2 = 1 is (
2.2(14)2−7

7
,
2.14

7
) = (195,4). 

All positive integer solutions of 𝑥2 − 868𝑦2 = 1 are 

{
 
 

 
 𝑥𝑛 =

1

2
𝐿𝑛(390,−1) =  

1

2
Ψ𝑛(195,−1)

𝑦𝑛 = 4𝐹𝑛(390,−1) =  
1

195
Φ𝑛(195,−1).

 

    We will then be required to find 𝐹𝑛 and 𝐿𝑛 for all 𝑛 > 1. Let  

{
 
 

 
 
𝐿𝑛(390,−1) = 390 L𝑛−1(390,−1) − L𝑛−2(390,−1),   

∀ 𝑛 > 2, 𝐿0(390,−1) = 2, 𝐿1(390,−1) = 390

 
𝐹𝑛(390,−1) = 390 F𝑛−1(390,−1) − 𝐹𝑛−2(390,−1)

∀ 𝑛 > 2, 𝐹0(390,−1) = 0, 𝐹1(390,−1) = 1

 

 

So, 195 + 4√686 $195+4 is a unit of ℤ[√686]. As 𝐿2(390,−1) =  152098 and 𝐹2(390,−1) =

 390, then 76049 + 1560√686 is another unit of ℤ[√686], and so on. We can clearly express units of 

ℤ[√686] in terms of generalized Fibonacci, generalized Lucas, generalized Pell, and generalized Pell-
Lucas numbers. 
 
4. Conclusion 

The Diophantine equation 𝑥2 − (𝑝2𝑞2 ± 𝑎𝑝)𝑦2 = 𝑘𝑡 has been solved, and its positive integer 
solutions have been stated in terms of generalized Fibonacci, generalized Lucas, generalized Pell, and 

generalized Pell-Lucas sequences. We have discovered units for ℤ[√𝐷] in terms of the above sequences 
of numbers. 

 
 
 



4414 

 
 

 
Edelweiss Applied Science and Technology 
ISSN: 2576-8484 

Vol. 8, No. 6: 4408-4414, 2024 
DOI: 10.55214/25768484.v8i6.2965 
© 2024 by the authors; licensee Learning Gate 

 

Copyright:  
© 2024 by the authors. This article is an open access article distributed under the terms and conditions 
of the Creative Commons Attribution (CC BY) license (https://creativecommons.org/licenses/by/4.0/). 
 
References 
[1] J.W. LeVeque, Topics in Number Theory 1, 2, Dover Publications, 2002. 
[2] T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1981. 
[3] D. Redmond, Number Theory: An Introduction to Pure and Applied Mathematics 201, CRC Press, 1996. 
[4] D. Redmond, Number Theory: An Introduction, Markel Dekker. Inc., New York,1996. 
[5] A. Chandoul, “The Pell Equation,” Research Journal of Pure Algebra, Vol. 1, No. 2, 2011, pp. 11-15. 

[6] Sankari, H., & Abdo, A, The Polynomial Solutions of Quadratic Diophantine Equation 𝑥2 − 𝑝(𝑡)𝑦2 + 2𝑘(𝑡)𝑥 +
2𝑝(𝑡)𝑙(𝑡)𝑦 = 0, Journal of Mathematics, 2021, 1-6.  

[7]  H Utz, W. R, Positive solutions of the Diophantine equation, International Journal of Mathematics and Mathematical 
Sciences, 5, 311-314 

[8] Akbik, S, On a class of diophantine equations. International Journal of Mathematics and Mathematical Sciences, 29, 545-
553. 

[9]  Bala, R., & Mishra, V.  Solutions of equations 𝑥2 − (𝑝2𝑞2 ± 3𝑝)𝑦2 = 𝑘𝑡. Examples and Counterexamples, 2. 

[10]  Guney, M. Solutions of the Pell equations 𝑥2 − (𝑎2𝑏2 + 2𝑏)𝑦2 = 𝑁 when 𝑁 ∈ {±1,±4}. Mathematica Aeterna, 2(7), 
629-638. 

[11]  Edson, M., & Yayenie, O. (2009). A New Generalization of Fibonacci Sequence & Extended Binet's Formula. 

[12]  S. Falcon, A. Plaza, The 𝑘-Fibonacci sequence and the Pascal 2-Triangle, Chaos, Solitons and Fractals 33(2007), 38-49. 
[13]  A. F. Horadam, A generalized Fibonacci sequence, Amer. Math. Monthly 68(1961), 455-459 
[14]  D. V. Jaiswal, On a generalized Fibonacci sequence, LabdevJ. Sci. Tech. Part A 7(1969), 67-71. 
[15]  A. T. Krassimir, A. C. Liliya, S. D. Dimitar, A new perspective to the generalization of the Fibonacci sequence, 

Fibonacci Quart. 23(1985), no. 1, 21-28 
[16]  L. J. Zai, L.J. Sheng, Some properties of the generalization of the Fibonacci sequence, Fibonacci Quart. 25(1987), no. 2, 

111-117. 
[17]  Kalman, D., Mena R., The Fibonacci Numbers exposed, Mathematics Magazine 76(2003), 167-181. 
[18] McDaniel, W.L., Diophantine Representation of Lucas Sequences, The Fibonacci Quarterly 33 (1995), 58-63. 
[19] Melham, R., Conics Which Characterize Certain Lucas Sequences, The Fibonacci Quarterly 35 (1997), 248-251. 
[20]  Ribenboim, P., My Numbers, My Friends, Springer-Verlag New York, Inc., 2000. 
[21]  Robinowitz, S., Algorithmic Manipulation of Fibonacci Identities, in: Application of Fibonacci Numbers, vol. 6, 

Kluwer Academic Pub., Dordrect, The Netherlands, 1996, pp. 389-408. 

 

https://creativecommons.org/licenses/by/4.0/