Approach of the value of a rent when non-central moments of the capitalization factor are known: an R application with interest rates following normal and beta distributions


Ratio Mathematica                                         Volume 41, 2021, pp. 206-222 

 

              
 

206 

 

 

 

On the solutions of Pellian equation                        

𝑼𝟐 − 𝑫𝑽𝟐 = 𝒌𝟐𝑵 
 

Bilkis M. Madni * 

Devbhadra V. Shah † 
 

 

 

Abstract  

In this paper we consider a class of Pell’s equation 𝑈2 − 𝐷𝑉2 =
𝑘2𝑁, where 𝐷, 𝑁 are positive integers, 𝐷 is non-square and 𝑘 is any 

integer. When (𝑢, 𝑣) satisfy this equation, we define 
𝑢+𝑣√𝐷

𝑘
 to be its 

solution. We first introduce the class of solutions of this equation 

and call 
𝑢+𝑣√𝐷

𝑘
 to be the fundamental solution of the class, if 𝑣 is the 

smallest positive value which occurs in the solutions of that class. 

We first derive the necessary and sufficient condition for any two 

solutions of proposed equation to belong to the same class and the 

bounds for the values of 𝑢, 𝑣 occurring in the fundamental solution. 
We also derive an explicit formula which gives all the solutions of 

this equation. We further present some interesting recurrence 

relations connecting the values of 𝑢, 𝑣. Finally, we obtain the results 
for total number of positive solutions of proposed equation not 

exceeding any given positive real number 𝑍. 
Keywords: Pell’s equation, Solutions of Pell’s equation,                    

Recurrence relations. 

2010 AMS subject classification‡: 11D09, 11D45. 
 

 
 

 
* Bilkis Madni (Department of Mathematics, Veer Narmad South Gujarat University, Surat, 

India); bilkis1110@gmail.com 
† Devbhadra Shah (Department of Mathematics, Veer Narmad South Gujarat University, Surat, 

India); drdvshah@yahoo.com 
‡Received on July 15, 2021. Accepted on December 25, 2021. Published on December 31, 2021. 

doi: 10.23755/rm.v41i0.628. ISSN: 1592-7415. eISSN: 2282-8214. ©The Authors. This paper 

is published under the CC-BY licence agreement. 



On the solutions of Pellian equation 𝑈2 − 𝐷𝑉2 = 𝑘2𝑁 
 

207 

 

1. Introduction  
 

In number theory, Pell’s equation falls in the category of Diophantine 

equations and it is considered to be one of the oldest Diophantine equations. 

Specifically, the term Pell’s equation is used to refer any Diophantine equation 

of the form 

𝑢2 − 𝐷𝑣 2 = 𝑁,                                          (1.1) 
 where 𝐷 and 𝑁 are fixed non-zero integers and we wish to find integers 𝑢 and 
𝑣 that satisfy the equation. Throughout, we consider the integer 𝐷 to be positive 
and non-square. This condition is helpful because only it leaves open the 

possibility of infinitely many solutions in positive integers 𝑢, 𝑣. 
Pellian equation   

𝑥2 − 𝐷𝑦2 = 1,                                           (1.2) 
where 𝐷 is fixed positive non-square integer, attracted attention of early 
mathematicians. It is known that (1.2) always has infinitude of solutions. We 

refer to Stolt [11, 12] who nicely studied the equation 𝑥2 − 𝐷𝑦2 = ±4𝑁. In this 
paper we use the notions proposed by Stolt. Many authors considered some 

specific Pell equations and discussed about their solutions. For completeness 

we recall that there are many papers which considered different types of Pell’s 

equation. For extensive resources on Pell’s equation, one can refer [1] – [14]. 

In the present paper we consider the Pellian equation 

𝑈2 − 𝐷𝑉 2 = 𝑘2𝑁,                                       (1.3) 
where 𝐷, 𝑁 are positive integers, 𝐷 is non-square and 𝑘 is any integer. 
Throughout we assume that (1.3) is solvable. If (𝑢, 𝑣) satisfy this equation,             

we define 
𝑢+𝑣√𝐷

𝑘
 to be the solution of (1.3). If 

𝑢+𝑣√𝐷

𝑘
 is any solution of (1.3), 

𝑥 + 𝑦√𝐷 is any solution of (1.2) and 𝑥1 + 𝑦1√𝐷 is its smallest positive solution, 

then it is easy to observe that 
𝑢+𝑣√𝐷

𝑘
× (𝑥 + 𝑦√𝐷) is also a solution of (1.3).            

This solution is said to be associated with the solution 
𝑢+𝑣√𝐷

𝑘
 . Here we note 

that if two positive solutions 
𝑢𝛼+𝑣𝛼√𝐷

𝑘
 and 

𝑢𝛽+𝑣𝛽√𝐷

𝑘
 are associated then there 

exists an integer 𝑡 such that 

 
𝑢𝛽+𝑣𝛽√𝐷

𝑘
=

𝑢𝛼+𝑣𝛼√𝐷

𝑘
× (𝑥1 + 𝑦1√𝐷)

𝑡
; 𝑡 = 0, ±1, ± 2, … .        (1.4)      

The set of all solutions associated with other forms a class of solutions of 

(1.3). Among all the solutions 
𝑢+𝑣√𝐷

𝑘
 in any given class 𝐾 we now choose a 

solution 
𝑢1+𝑣1√𝐷

𝑘
 in the following way: Let 𝑣1 be the least non-negative value of 

𝑣 which occurs in 𝐾. Then in this case, 𝑢1 is also uniquely determined.                         

The solution 
𝑢1+𝑣1√𝐷

𝑘
 defined in this way is said to be fundamental solution of 



Bilkis M. Madni, Devbhadra V. Shah 

208 

 

the class. It can be observed that 
𝑢1+𝑣1√𝐷

𝑘
 is a fundamental solution of class 𝐾 if 

𝑢1+𝑣1√𝐷

𝑘
× (𝑥1 − 𝑦1√𝐷) is not a positive solution of (1.3). Thus if 

𝑢𝛼+𝑣𝛼√𝐷

𝑘
 is 

any fixed fundamental solution then all the positive solutions given by (1.4) are 

associated with each other. Moreover, we observe that if (1.3) is solvable, then 

it has only finite number of classes of solutions.  

 

2. The equation 𝑼𝟐 − 𝑫𝑽𝟐 = 𝒌𝟐𝑵 
 

 It is easy to observe that if 
𝑢+𝑣√𝐷

𝑘
 and 

𝑢′+𝑣′√𝐷

𝑘
 are any two integer solutions 

of (1.3) such that 𝑢 ≡ 𝑢′ and 𝑣 ≡ 𝑣′(𝑚𝑜𝑑 𝑁), then (
𝑢𝑢′−𝐷𝑣𝑣′

𝑘𝑁
,

𝑢𝑣′−𝑢′𝑣

𝑘𝑁
) is the 

positive solution of the equation 𝑅2 − 𝐷𝑆2 = 𝑘2. We first present the necessary 
and sufficient condition for any two solutions of (1.3) to be associated with each 

other. 

 

Theorem 2.1. If 
𝑢+𝑣√𝐷

𝑘
 and 

𝑢′+𝑣′√𝐷

𝑘
 are any two solutions of (1.3) then the 

necessary and sufficient condition for these two solutions to be associated with 

each other is that 
𝑢𝑣′ −𝑢′𝑣

𝑘𝑁
 is an integer. 

Proof. Since 
𝑢+𝑣√𝐷

𝑘
 and 

𝑢′+𝑣′√𝐷

𝑘
 are solutions of (1.3), we get 𝑢2 − 𝐷𝑣2 =

𝑘2𝑁 and 𝑢′
2

− 𝐷𝑣′
2

= 𝑘2𝑁. Multiplying these two equations we get 

(
𝑢𝑢′−𝐷𝑣𝑣′

𝑘𝑁
)

2

− 𝐷 (
𝑢𝑣′−𝑢′𝑣

𝑘𝑁
)

2

= 𝑘2. We first show that the solutions 
𝑢+𝑣√𝐷

𝑘
 and 

𝑢′+𝑣′√𝐷

𝑘
 of the Pellian equation (1.3) are associated with each other if and only 

if both 
𝑢𝑢′−𝐷𝑣𝑣′

𝑘𝑁
  and 

𝑢𝑣′−𝑢′𝑣

𝑘𝑁
 are integers. 

 If two numbers 
𝑢𝑢′ −𝐷𝑣𝑣′

𝑘𝑁
 and 

𝑣𝑢′−𝑢𝑣′

𝑘𝑁
 are integers, then  

(
𝑢𝑢′−𝐷𝑣𝑣′

𝑘𝑁
)

2

− 𝐷 (
𝑢𝑣′−𝑢′𝑣

𝑘𝑁
)

2

= 𝑘2. Thus 
𝑢𝑢′−𝐷𝑣𝑣′

𝑘𝑁
 + 

𝑢𝑣′−𝑢′𝑣

𝑘𝑁
√𝐷

𝑘
 is a solution of 

Pellian equation 𝑅2 − 𝐷𝑆2 = 𝑘2. Now  
𝑢+𝑣√𝐷

𝑘

𝑢′+𝑣′√𝐷

𝑘

=
𝑢+𝑣√𝐷

𝑢′+𝑣′√𝐷
= (𝑢 + 𝑣√𝐷) ×

𝑢′−𝑣′√𝐷

±𝑘2𝑁
=

𝑢𝑢′−𝐷𝑣𝑣′

𝑘2𝑁
−

𝑢𝑣′−𝑢′𝑣

𝑘2𝑁
√𝐷. 

Clearly this is a solution of (1.2). Thus, both the solutions 
𝑢𝑢′−𝐷𝑣𝑣′

𝑘𝑁
 and 

𝑣𝑢′−𝑢𝑣′

𝑘𝑁
 

of (1.3) are associated.  

 Conversely let two solutions 
𝑢+𝑣√𝐷

𝑘
 and 

𝑢′+𝑣′√𝐷

𝑘
 of (1.3) be associated with 

each other. Then clearly, they should lie in the same solution class of (1.3). Then 



On the solutions of Pellian equation 𝑈2 − 𝐷𝑉2 = 𝑘2𝑁 
 

209 

 

we have 

𝑢+𝑣√𝐷

𝑘

𝑢′+𝑣′√𝐷

𝑘

=
𝑢𝑢′−𝐷𝑣𝑣′

𝑘2𝑁
−

𝑢𝑣′−𝑢′𝑣

𝑘2𝑁
√𝐷, which should be some solution of 

(1.2). In this case we have 𝑢2 − 𝐷𝑣2 = 𝑘2𝑁 and 𝑢′
2

− 𝐷𝑣′
2

= 𝑘2𝑁. This gives  

(
𝑢𝑢′−𝐷𝑣𝑣′

𝑘𝑁
)

2

− 𝐷 (
𝑣𝑢′−𝑢𝑣′

𝑘𝑁
)

2

= 𝑘2. Thus,  

𝑘 × (
𝑢𝑢′−𝐷𝑣𝑣′

𝑘2𝑁
,

𝑢𝑣′−𝑢′𝑣

𝑘2𝑁
) = (

𝑢𝑢′−𝐷𝑣𝑣′

𝑘𝑁
,

𝑢𝑣′−𝑢′𝑣

𝑘𝑁
)  

is an integer solution of 𝑅2 − 𝐷𝑆2 = 𝑘2. Hence 
𝑢𝑢′−𝐷𝑣𝑣′

𝑘𝑁
 and 

𝑢𝑣′−𝑢′𝑣

𝑘𝑁
 are 

integers.  

 Thus, it is now sufficient to show that 
𝑢𝑢′ −𝐷𝑣𝑣′

𝑘𝑁
 is an integer when 

𝑢𝑣′−𝑢′𝑣

𝑘𝑁
 

is an integer; and 
𝑢𝑣′−𝑢′𝑣

𝑘𝑁
 is not an integer when 

𝑢𝑢′−𝐷𝑣𝑣′

𝑘𝑁
 is not an integer. Since 

𝑢+𝑣√𝐷

𝑘
 and 

𝑢′+𝑣′√𝐷

𝑘
 are solutions of (1.3), we have 𝑢2 − 𝐷𝑣2 = ±𝑘2𝑁 , 𝑢′

2
−

𝐷𝑣′
2

= ±𝑘2𝑁. Multiplying these two equations we get 
(𝑢𝑢′ − 𝑣𝑣′𝐷)2 − 𝐷(𝑢𝑣′ − 𝑣𝑢′)2 = 𝑘2(𝑘𝑁)2.                     (2.1) 

It is obvious from (2.1) that 𝑢𝑢′ − 𝐷𝑣𝑣 ′ is divisible by 𝑘𝑁 when 𝑢𝑣′ − 𝑢′𝑣 is 

divisible by 𝑘𝑁. That is, 
𝑢𝑢′−𝐷𝑣𝑣′

𝑘𝑁
 is an integer when 

𝑢𝑣′−𝑢′𝑣

𝑘𝑁
 is an integer. 

Conversely, if 
𝑢𝑢′−𝐷𝑣𝑣′

𝑘𝑁
 is not an integer, then there exists an integer 𝑑 such that 

𝑑 | 𝑘𝑁 but 𝑑 ∤ (𝑢𝑢′ − 𝐷𝑣𝑣′). Now from (2.1) it is seen that if 𝑑 | (𝑢𝑣′ − 𝑢′𝑣), 
then 𝑑 | (𝑢𝑢′ − 𝑣𝑣′𝐷) too, which is contrary to the assumption. Thus                        

𝑑 ∤ (𝑢𝑣′ − 𝑢′𝑣), that is 
𝑢𝑣′−𝑢′𝑣

𝑘𝑁
 is not an integer, as required. This proves the 

required result.  

 

 We now derive the bounds for the values of 𝑢, 𝑣 occurring in the 
fundamental solutions.  

 

Theorem 2.2. If 
𝑢+𝑣√𝐷

𝑘
 is the smallest fundamental solution of the class 𝐾 of 

the Pellian equation (1.3) and if 𝑥1 + 𝑦1√𝐷 is the fundamental solution of (1.2), 

then 0 < |𝑢| ≤ 𝑘√
(𝑥1+1)𝑁

2
 and 0 ≤ 𝑣 ≤ 𝑘𝑦1√

𝑁

2(𝑥1+1)
 . 

Proof. If both the inequalities to be proved are true for a class 𝐾, then they are 
also true for the conjugate class �̅�. Thus, we can assume that 𝑢 is positive.              

Note that 
𝑢𝑥1−𝐷𝑣𝑦1

𝑘
=

𝑢𝑥1

𝑘
− (𝑦1√𝐷) (

𝑣√𝐷

𝑘
) =

𝑢𝑥1

𝑘
− √𝑥1

2 − 1 √
𝑢2−𝑘2𝑁

𝑘2
 

                      =
𝑢𝑥1

𝑘2
− √(𝑥1

2 − 1) (
𝑢2

𝑘2
− 𝑁) > 0.  



Bilkis M. Madni, Devbhadra V. Shah 

210 

 

 Now consider the solution (
𝑢+𝑣√𝐷

𝑘
) (𝑥1 − 𝑦1√𝐷) =

𝑢𝑥1−𝐷𝑣𝑦1+(𝑥1𝑣−𝑦1𝑢)√𝐷 

𝑘
 

of (1.3) which belongs to the same class as 
𝑢+𝑣√𝐷

𝑘
 . But 

𝑢+𝑣√𝐷

𝑘
 is the fundamental 

solution of the class and by above 
𝑢𝑥1−𝐷𝑣𝑦1

𝑘
 is positive. Thus, we must have 

𝑢𝑥1−𝐷𝑣𝑦1

𝑘
≥

𝑢

𝑘
 , as 

𝑢

𝑘
 occurs in fundamental solution of (1.3). From this inequality 

it now follows that 𝑢𝑥1 − 𝐷𝑣𝑦1 ≥ 𝑢. This gives 𝑢(𝑥1 − 1) ≥ 𝐷𝑣𝑦1, which 
gives 𝑢2(𝑥1 − 1)

2 ≥ 𝐷2𝑣2𝑦1
2 = (𝑢2 − 𝑘2𝑁)(𝑥1

2 − 1). This can be written as 

𝑢2
𝑥1−1

𝑥1+1
≥ 𝑢2 − 𝑘2𝑁, which eventually gives 𝑢2 ≤

(𝑥1+1)𝑘
2𝑁

2
 . This proves the 

first of required inequality. 

 Again by (1.3) we have 𝐷𝑣2 = 𝑢2 − 𝑘2𝑁. Using above inequality, we get 

𝐷𝑣2 ≤
(𝑥1+1)𝑘

2𝑁

2
− 𝑘2𝑁 =

(𝑥1−1)𝑘
2𝑁

2
 . This gives  

𝑣 ≤ 𝑘√
(𝑥1−1)𝑁

2𝐷
= 𝑘√

(𝑥1
2−1)𝑁

2𝐷(𝑥1+1)
 . 

Since 𝑥1 + 𝑦1√𝐷 is the solution of (1.2), we thus get 0 ≤ 𝑣 ≤ 𝑘𝑦1√
𝑁

2(𝑥1+1)
 , as 

required. 

 

 We now present an illustration to demonstrate the results of this theorem. 

 

Illustration. Consider the Pellian equation 𝑈2 − 7𝑉2 = 18. Then clearly,           
𝐷 = 7, 𝑘 = 3 and 𝑁 = 2. If we consider the equation 𝑥2 − 7𝑦2 = 1, then it is 

easy to see that 8 + 3√7 is its fundamental solution. This gives 𝑥1 = 8, 𝑦1 = 3. 

Now if 
𝑢+𝑣√7

3
 is the smallest fundamental solution of equation 𝑈2 − 7𝑉2 = 18, 

then by above theorem we should have  

0 < |𝑢| ≤ 3 × √
9×2

2
 and 0 ≤ 𝑣 ≤ 3 × 3√

2

2×9
 . 

This gives 0 < |𝑢| ≤ 9 and 0 ≤ 𝑣 ≤ 3. These are indeed true as 
5+√7

3
 is the 

smallest fundamental solution of the equation 𝑈2 − 7𝑉2 = 18. 
 

 We now prove a very important result which produces all the fundamental 

solutions of (1.3). 

 

Theorem 2.3. If 𝑢𝑖 + 𝑣𝑖 √𝐷 runs through all the fundamental solutions of (1.1) 

and 
𝑟𝑗+𝑠𝑗√𝐷

𝑘
 runs through all the fundamental solutions of  𝑅2 − 𝐷𝑆2 = 𝑘2, then 

all the fundamental solutions of 𝑈2 − 𝐷𝑉2 = 𝑘2𝑁 are covered by 
𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
= (

𝑟𝑗+𝑠𝑗√𝐷

𝑘
) (𝑢𝑖 ± 𝑣𝑖 √𝐷).                            (2.2)   



On the solutions of Pellian equation 𝑈2 − 𝐷𝑉2 = 𝑘2𝑁 
 

211 

 

Proof. If we multiply the surd conjugate of (2.2) with (2.2) then we get 
𝐴𝑖𝑗

2 −𝐷𝐵𝑖𝑗
2

𝑘2
= (

𝑟𝑗
2−𝐷𝑠𝑗

2

𝑘2
) (𝑢𝑖

2 − 𝐷𝑣𝑖
2). Since 𝑢𝑖

2 − 𝐷𝑣𝑖
2 = 𝑁 and 𝑟𝑗

2 − 𝐷𝑠𝑗
2 = 𝑘2,  

we get 𝐴𝑖𝑗
2 − 𝐷𝐵𝑖𝑗

2 = 𝑘2𝑁. This shows that 
𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
 defined by (2.2) are the 

solutions of (1.3). 

  We next show that the solutions 
𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
 defined by (2.2) covers all the 

fundamental solutions of (1.3). On the contrary assume that there exists some 

positive solution, say 
𝑋+𝑌√𝐷

𝑘
 of (1.3) which is not covered by (2.2). Then this 

solution will lie between any two successive solutions of (1.3) of some fixed 

class generated by 
𝑟𝑗+𝑠𝑗√𝐷

𝑘
 . This means for some fixed 𝑗, we have 

(
𝑟𝑗+𝑠𝑗√𝐷

𝑘
) (𝑢𝑖 ± 𝑣𝑖 √𝐷) ≤

𝑋+𝑌√𝐷

𝑘
< (

𝑟𝑗+𝑠𝑗√𝐷

𝑘
) (𝑢𝑖+1 ± 𝑣𝑖+1√𝐷). 

Then 𝑢𝑖 ± 𝑣𝑖 √𝐷 ≤ (
𝑋+𝑌√𝐷

𝑘
) (

𝑘

𝑟𝑗+𝑠𝑗√𝐷
) < 𝑢𝑖+1 ± 𝑣𝑖+1√𝐷. This gives 

𝑢𝑖 ± 𝑣𝑖 √𝐷 ≤ (
𝑋+𝑌√𝐷

𝑘
) (

𝑟𝑗−𝑠𝑗√𝐷

𝑘
) < 𝑢𝑖+1 ± 𝑣𝑖+1√𝐷. 

We denote  

𝜖 + 𝛿√𝐷 = (
𝑋+𝑌√𝐷

𝑘
) (

𝑟𝑗−𝑠𝑗√𝐷

𝑘
).                               (2.3) 

Then 

𝑢𝑖 ± 𝑣𝑖 √𝐷 ≤ 𝜖 + 𝛿√𝐷 < 𝑢𝑖+1 ± 𝑣𝑖+1√𝐷.                     (2.4) 
 To prove the required result, it is sufficient to prove that 

(i) 𝜖 + 𝛿√𝐷 is a solution of (1.1), and 

(ii) 𝜖 > 0 and 𝛿 > 0 or 𝛿 < 0 depending on the sign of 𝑢𝑖 ± 𝑣𝑖 √𝐷. 
This will produce one positive solution of (1.1) between two consecutive 

fundamental solutions of (1.1) for any fixed class 𝑗, which is a contradiction. 
 To prove (i), we multiply surd conjugate of (2.3) with (2.3). This gives              

𝜖 2 − 𝐷𝛿 2 = (
𝑋2−𝐷𝑌2

𝑘2
) (

𝑟𝑗
2−𝐷𝑠𝑗

2

𝑘2
). Since 𝑋2 − 𝐷𝑌2 = 𝑘2𝑁 and 𝑟𝑗

2 − 𝐷𝑠𝑗
2 = 𝑘2, 

we get 𝜖 2 − 𝐷𝛿 2 = 𝑁, which proves the first part.  
 Next, we show that 𝜖 and 𝛿 defined by (2.3) are positive. Now 𝑟𝑗

2 − 𝐷𝑠𝑗
2 =

𝑘2 implies (
𝑟𝑗+𝑠𝑗√𝐷

𝑘
) (

𝑟𝑗−𝑠𝑗√𝐷

𝑘
) = 1 and 

𝑋+𝑌√𝐷

𝑘
> 1. Since 0 <

𝑟𝑗+𝑠𝑗√𝐷

𝑘
< ∞, 

clearly 0 <
𝑟𝑗−𝑠𝑗√𝐷

𝑘
< ∞. Since 𝜖 + 𝛿√𝐷 = (

𝑋+𝑌√𝐷

𝑘
) (

𝑟𝑗−𝑠𝑗√𝐷

𝑘
), we get                      

0 < 𝜖 + 𝛿√𝐷 < ∞. Also, since 𝜖 2 − 𝐷𝛿 2 = 𝑁, we get 0 <
𝑁

𝜖−𝛿√𝐷
< ∞, that is               

0 <
𝜖−𝛿√𝐷

𝑁
< ∞. This gives 0 < 𝜖 − 𝛿√𝐷 < ∞. Adding this result with                     

0 < 𝜖 + 𝛿√𝐷 < ∞ proves that 𝜖 > 0. 



Bilkis M. Madni, Devbhadra V. Shah 

212 

 

 We further observe by (2.4) that 

1 < 𝑢𝑖 ± 𝑣𝑖 √𝐷 ≤ 𝜖 + 𝛿√𝐷 < 𝑢𝑖+1 ± 𝑣𝑖+1√𝐷 < ∞. 

By considering the ‘+’ sign, we get 0 <
1

𝜖+𝛿√𝐷
≤

1

𝑢𝑖+𝑣𝑖√𝐷
< 1.                                       

Then 0 <
𝜖−𝛿√𝐷

𝑁
≤

𝑢𝑖−𝑣𝑖√𝐷

𝑁
, which gives 0 < 𝜖 − 𝛿√𝐷 ≤ 𝑢𝑖 − 𝑣𝑖 √𝐷. 

Subsequently we get 

2𝛿√𝐷 = (𝜖 + 𝛿√𝐷) − (𝜖 − 𝛿√𝐷) ≥ (𝑢𝑖 + 𝑣𝑖 √𝐷) − (𝑢𝑖 − 𝑣𝑖 √𝐷) = 2𝑣𝑖 √𝐷. 

This gives 𝛿 ≥ 𝑣𝑖 > 0.  

Also, if we select ‘–’ sign in 𝑢𝑖 ± 𝑣𝑖 √𝐷, then we have 

𝜖 + 𝛿√𝐷 < 𝑢𝑖+1 − 𝑣𝑖+1√𝐷 < ∞, 

which implies that 0 <
1

𝑢𝑖+1−𝑣𝑖+1√𝐷
<

1

𝜖+𝛿√𝐷
. Then 0 <

𝑢𝑖+1+𝑣𝑖+1√𝐷

𝑁
<  

𝜖−𝛿√𝐷

𝑁
. 

This gives 𝜖 − 𝛿√𝐷 > 𝑢𝑖+1 + 𝑣𝑖+1√𝐷. Thus, we get 

2𝛿√𝐷 = (𝜖 + 𝛿√𝐷) − (𝜖 − 𝛿√𝐷) < (𝑢𝑖+1 − 𝑣𝑖+1√𝐷) − (𝑢𝑖+1 + 𝑣𝑖+1√𝐷) 

             = −2𝑣𝑖+1√𝐷.  
This gives 𝛿 < −𝑣𝑖+1 < 0. Hence 𝛿 > 0 or 𝛿 < 0 depends on the sign of                 

𝑢𝑖 ± 𝑣𝑖 √𝐷, which proves (ii). Hence all the fundamental solutions 
𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
 of 

(1.3) are covered by (2.2). 

 

 We illustrate this by an example which justifies the meaning of “… covered 

by (2.2)”. 

 

Illustration. As earlier we once again consider the equation 𝑈2 − 7𝑉2 = 18. 
Then we have 𝐷 = 7, 𝑘 = 3 and 𝑁 = 2. Next, we consider the Pellian equation 

𝑅2 − 7𝑆2 = 9 and if 
𝑟𝑗+𝑠𝑗√7

3
 runs through all of its fundamental solutions, then 

it can be observed that 
𝑟1+𝑠1√7

3
=

4+√7

3
, 

𝑟2+𝑠2√7

3
=  

11+4√7

3
, 

𝑟3+𝑠3√7

3
=  

24+9√7

3
. 

Also 𝑢𝑖 + 𝑣𝑖 √7 runs through all the fundamental solutions of 𝑢
2 − 7𝑣2 = 2, 

then we have 𝑢1 ± 𝑣1√7 = 3 ± √7. Then above theorem claims that all the 
fundamental solutions of 𝑈2 − 7𝑉2 = 18 are covered by 

 
𝐴1𝑗+𝐵1𝑗√2

3
= (

𝑟𝑗+𝑠𝑗√𝐷

3
) (3 ± √7); 𝑗 = 1, 2, 3. 

Thus, 
𝐴11+𝐵11√2

3
= (

4+√7

3
) (3 ± √7) =

19+7√7

3
 ,

5−√7

3
 ;  

𝐴12+𝐵12√2

3
= (

11+4√7

3
) (3 ± √7) =

61+23√7

3
,

𝟓+√𝟕

𝟑
 and  

𝐴13+𝐵13√2

3
= (

24+9√7

3
) (3 ± √7) =

135+51√7

3
,

𝟗+𝟑√𝟕

𝟑
 . 



On the solutions of Pellian equation 𝑈2 − 𝐷𝑉2 = 𝑘2𝑁 
 

213 

 

It can be observed that 
5+√7

3
,

9+3√7

3
 and 

19+7√7

3
 are the only three fundamental 

solutions of the equation 𝑈2 − 7𝑉2 = 18. Thus 
𝐴1𝑗+𝐵1𝑗√2

3
; 𝑗 = 1,2,3 covers all 

the fundamental solutions of  𝑈2 − 7𝑉2 = 18. 

 Here the smallest fundamental solution of 𝑈2 − 7𝑉2 = 18 is 
5+√7

3
 and                 

𝑥1 + 𝑦1√𝐷 = 8 + 3√7 is the fundamental solution of 𝑥
2 − 7𝑦2 = 1. Then 

every fundamental solution of 𝑈2 − 7𝑉2 = 18 should be smaller than 
5+√7

3
× (8 + 3√7) =

61+23√7

3
. This is indeed true as the only fundamental 

solutions of 𝑈2 − 7𝑉2 = 18 are 
𝐴11+𝐵11√2

3
=

19+7√7

3
,

𝐴12+𝐵12√2

3
=

5+√7

3
 and 

𝐴13+𝐵13√2

3
=

9+3√7

3
 . 

 

 We next derive an explicit formula which produces all the positive 

solutions of (1.3). 

 

Theorem 2.4. If 𝑥1 + 𝑦1√𝐷 is the smallest positive solution of (1.2) and 
𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
 defined by (2.2) runs through all the fundamentals solutions of (1.3), 

then all the integer solutions 
𝑢𝑖𝑗,𝑛+𝑣𝑖𝑗,𝑛√𝐷

𝑘
 of 𝑈2 − 𝐷𝑉2 = 𝑘2𝑁 are given by  

𝑢𝑖𝑗,𝑛+𝑣𝑖𝑗,𝑛√𝐷

𝑘
= (

𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
) (𝑥1 + 𝑦1√𝐷)

𝑛
; 𝑛 ≥ 0.              (2.5) 

Proof. If we consider the surd conjugate of (2.5) and multiply with (2.5),                 

we get 
𝑢𝑖𝑗,𝑛

2 −𝐷𝑣𝑖𝑗,𝑛
2

𝑘2
= (

𝐴𝑖𝑗
2 −𝐷𝐵𝑖𝑗

2

𝑘2
) (𝑥1

2 − 𝐷𝑦1
2)𝑛. Since 𝐴𝑖𝑗

2 − 𝐷𝐵𝑖𝑗
2 = 𝑘2𝑁 and 

𝑥1
2 − 𝐷𝑦1

2 = 1, we get 𝑢𝑖𝑗,𝑛
2 − 𝐷𝑣𝑖𝑗,𝑛

2 = 𝑘2𝑁. Thus 
𝑢𝑖𝑗,𝑛+𝑣𝑖𝑗,𝑛√𝐷

𝑘
 defined by (2.5) 

are the solutions of (1.3). 

 We next show that the solutions 
𝑢𝑖𝑗,𝑛+𝑣𝑖𝑗,𝑛√𝐷

𝑘
 defined by (2.5) gives all the 

positive solutions of (1.3). On the contrary assume that there exists some 

positive solution, say 
𝑋+𝑌√𝐷

𝑘
 of (1.3) which is not covered by (2.5). Then this 

solution will lie between any two successive solutions of (1.3) of some classes 

generated by 
𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
 (for a fixed 𝑖, 𝑗). This means for some fixed 𝑖, 𝑗 and for 

some fixed 𝑚, we have 

(
𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
) (𝑥1 + 𝑦1√𝐷)

𝑚
≤

𝑋+𝑌√𝐷

𝑘
< (

𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
) (𝑥1 + 𝑦1√𝐷)

𝑚+1
. 

Then (
𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
) ≤ (

𝑋+𝑌√𝐷

𝑘
) (𝑥1 − 𝑦1√𝐷)

𝑚
< (

𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
) (𝑥1 + 𝑦1√𝐷). 

We denote  



Bilkis M. Madni, Devbhadra V. Shah 

214 

 

𝜖+𝛿√𝐷

𝑘
= (

𝑋+𝑌√𝐷

𝑘
) (𝑥1 − 𝑦1√𝐷)

𝑚
.                              (2.6) 

Thus, 

(
𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
) ≤

𝜖+𝛿√𝐷

𝑘
< (

𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
) (𝑥1 + 𝑦1√𝐷).                  (2.7) 

 To prove the required result, it is sufficient to prove that 

 (i) 
𝜖+𝛿√𝐷

𝑘
 is a solution of (1.3) 

 (ii) 𝜖 > 0, 𝛿 > 0. 

 (iii) 
𝜖+𝛿√𝐷

𝑘
 is the solution of (1.3) smaller than all its fundamental solutions for 

the fixed classes 𝑖, 𝑗. This will contradict the fact that 
𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
 runs through 

every fundamental solution of (1.3) for some fixed value of 𝑖, 𝑗.  
  To prove (i), we take surd conjugate of (2.6) and multiply it with (2.6).               

This gives 
𝜖2−𝐷𝛿2

𝑘2
= (

𝑋2−𝐷𝑌2

𝑘2
) (𝑥1

2 − 𝐷𝑦1
2)𝑚. Since 𝑋2 − 𝐷𝑌2 = 𝑘2𝑁 and 

𝑥1
2 − 𝐷𝑦1

2 = 1, we get 𝜖 2 − 𝐷𝛿 2 = 𝑘2𝑁, as required. 

 Again, since 1 < (𝑥1 + 𝑦1√𝐷)
𝑚

< ∞ and 
𝑋+𝑌√𝐷

𝑘
> 1, we have                               

0 < (𝑥1 − 𝑦1√𝐷)
𝑚

< 1. Hence, we get 0 <
𝜖+𝛿√𝐷

𝑘
< ∞, as 

𝜖+𝛿√𝐷

𝑘
=

(
𝑋+𝑌√𝐷

𝑘
) (𝑥1 − 𝑦1√𝐷)

𝑚
. This gives 0 <

𝜖−𝛿√𝐷

𝑘𝑁
< ∞, that is 0 <

𝜖−𝛿√𝐷

𝑘
< ∞, as 

𝜖 2 − 𝐷𝛿 2 = 𝑘2𝑁. Adding this result with 0 <
𝜖+𝛿√𝐷

𝑘
< ∞ proves that 𝜖 > 0. 

 We further observe by (2.7) that 1 <
𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
≤

𝜖+𝛿√𝐷

𝑘
< ∞. Taking 

reciprocal, we get 0 <
𝑘

𝜖+𝛿√𝐷
≤

𝑘

𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷
< 1. Then        0 <

𝜖−𝛿√𝐷

𝑘𝑁
≤

𝐴𝑖𝑗−𝐵𝑖𝑗√𝐷

𝑘𝑁
. 

This gives 0 <
𝜖−𝛿√𝐷

𝑘
≤

𝐴𝑖𝑗−𝐵𝑖𝑗√𝐷

𝑘
. Subsequently we get 

2𝛿√𝐷

𝑘
= (

𝜖+𝛿√𝐷

𝑘
) − (

𝜖−𝛿√𝐷

𝑘
) ≥ (

𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
) − (

𝐴𝑖𝑗−𝐵𝑖𝑗√𝐷

𝑘
) =

2𝐵𝑖𝑗√𝐷

𝑘
 . 

 This now gives 𝛿 ≥ 𝐵𝑖𝑗 > 0. Hence both 𝜖 > 0, 𝛿 > 0, which proves (ii).  

  Finally, we prove that 
𝜖+𝛿√𝐷

𝑘
 is the smallest solution of (1.3) for the fixed 

classes 𝑖, 𝑗. On the contrary suppose 
𝜖+𝛿√𝐷

𝑘
 is positive solution of (1.3) but not 

the smallest solution of any fixed classes 𝑖, 𝑗. In this case 
𝜖+𝛿√𝐷

𝑘
(𝑥1 − 𝑦1√𝐷) 

will be the positive solution of (1.3). Then by (2.7) we have 
𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
≤

𝜖+𝛿√𝐷

𝑘
< (

𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
) (𝑥1 + 𝑦1√𝐷), 

which gives  

(
𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
) (𝑥1 − 𝑦1√𝐷) ≤

𝜖+𝛿√𝐷

𝑘
(𝑥1 − 𝑦1√𝐷) <

𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
. 



On the solutions of Pellian equation 𝑈2 − 𝐷𝑉2 = 𝑘2𝑁 
 

215 

 

Here 
𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
 runs through every fundamental solution of (1.3). Thus, we can 

now say that 
𝜖+𝛿√𝐷

𝑘
(𝑥1 − 𝑦1√𝐷) is a positive solution of (1.3) smaller than all 

the fundamental solutions 
𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
 of (1.3), which is a contradiction. This 

contradiction finally assures that there cannot exist any solution of (1.3) which 

is not covered by (2.5), as required.  
 

 We illustrate this theorem by the following example: 

 

Illustration. Once again we consider the Pellian equation 𝑈2 − 7𝑉2 = 18 

whose all the fundamental solutions are given by 
𝐴11+𝐵11√2

3
=

19+7√7

3
;  

𝐴12+𝐵12√2

3
=  

5+√7

3
 and 

𝐴13+𝐵13√2

3
=  

9+3√7

3
 and 𝑥1 + 𝑦1√𝐷 = 8 + 3√7 

is the fundamental solution of 𝑥2 − 7𝑦2 = 1. Then the above theorem asserts 

that all the integer solutions 
𝑢1𝑗,𝑛+𝑣1𝑗,𝑛√7

3
 of Pellian equation 𝑈2 − 7𝑉2 = 18 are 

covered by 
𝑢1𝑗,𝑛+𝑣1𝑗,𝑛√7

3
= (

𝐴1𝑗+𝐵1𝑗√7

3
) (𝑥1 + 𝑦1√𝐷)

𝑛
; 𝑗 = 1,2,3 ;  𝑛 ≥ 0. This 

gives 
𝑢11,𝑛+𝑣11,𝑛√7

3
= (

19+7√7

3
) (8 + 3√7)

𝑛
 ;  

𝑢12,𝑛+𝑣12,𝑛√7

3
= (

5+√7

3
) (8 + 3√7)

𝑛
 and 

𝑢13,𝑛+𝑣13,𝑛√7

3
= (

9+3√7

3
) (8 + 3√7)

𝑛
. 

  

 We now prove some recurrence relations for the values of 𝑢𝑖𝑗,𝑛 and 𝑣𝑖𝑗,𝑛. 

We assume that 𝑥1 + 𝑦1√𝐷 is the fundamental solution of (1.2), 𝑢𝑖 + 𝑣𝑖 √𝐷 runs 

through all the fundamental solutions of (1.1), 
𝑟𝑗+𝑠𝑗√𝐷

𝑘
 runs through all the 

fundamental solutions of 𝑅2 − 𝐷𝑆2 = 𝑘2, 𝑟 ≥ 1 is a fixed integer and 𝑛 ≥ 1. 

 

Theorem 2.5. a) 𝑢𝑖𝑗,𝑛+𝑟 =
𝑥𝑟𝑢𝑖𝑗,𝑛+𝐷𝑦𝑟𝑣𝑖𝑗,𝑛

𝑘
, 𝑣𝑖𝑗,𝑛+𝑟 =

𝑦𝑟𝑢𝑖𝑗,𝑛+𝑥𝑟𝑣𝑖𝑗,𝑛

𝑘
 . 

                b) 𝑢𝑖𝑗,𝑛+𝑟 =
𝑥1𝑢𝑖𝑗,𝑛+𝑟−1+𝐷𝑦1𝑣𝑖𝑗,𝑛+𝑟−1

𝑘
, 𝑣𝑖𝑗,𝑛+𝑟 =

𝑦1𝑢𝑖𝑗,𝑛+𝑟−1+𝑥1𝑣𝑖𝑗,𝑛+𝑟−1

𝑘
. 

                c) 𝑢𝑖𝑗,𝑛+2𝑟 =
2𝑘𝑥𝑟𝑢𝑖𝑗,𝑛+𝑟−𝑢𝑖𝑗,𝑛

𝑘2
 , 𝑣𝑖𝑗,𝑛+2𝑟 =

2𝑘𝑥𝑟𝑣𝑖𝑗,𝑛+𝑟−𝑣𝑖𝑗,𝑛

𝑘2
.        

Proof. (a) By (2.5), we have  
𝑢𝑖𝑗,𝑛+𝑟+𝑣𝑖𝑗,𝑛+𝑟√𝐷

𝑘
= (

𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
) (𝑥1 + 𝑦1√𝐷)

𝑛+𝑟
= (

𝑢𝑖𝑗,𝑛+𝑣𝑖𝑗,𝑛√𝐷

𝑘
) (𝑥𝑟 + 𝑦𝑟 √𝐷)   

                         =
(𝑥𝑟𝑢𝑖𝑗,𝑛+𝐷𝑦𝑟𝑣𝑖𝑗,𝑛)+(𝑦𝑟𝑢𝑖𝑗,𝑛+𝑥𝑟𝑣𝑖𝑗,𝑛)√𝐷

𝑘
 . 

Hence 𝑢𝑖𝑗,𝑛+𝑟 =
𝑥𝑟𝑢𝑖𝑗,𝑛+𝐷𝑦𝑟𝑣𝑖𝑗,𝑛

𝑘
, 𝑣𝑖𝑗,𝑛+𝑟 =

𝑦𝑟𝑢𝑖𝑗,𝑛+𝑥𝑟𝑣𝑖𝑗,𝑛

𝑘
 . 

(b) To prove the second result, we first consider 𝑟 = 1 and replace 𝑛 by 𝑛 − 1 
in the above result. We thus get 



Bilkis M. Madni, Devbhadra V. Shah 

216 

 

𝑢𝑖𝑗,𝑛 =
𝑥1𝑢𝑖𝑗,𝑛−1+𝐷𝑦1𝑣𝑖𝑗,𝑛−1

𝑘
, 𝑣𝑖𝑗,𝑛 =

𝑦1𝑢𝑖𝑗,𝑛−1+𝑥1𝑣𝑖𝑗,𝑛−1

𝑘
. 

Then 𝑢𝑖𝑗,𝑛+𝑟 =
𝑥𝑟𝑢𝑖𝑗,𝑛+𝐷𝑦𝑟𝑣𝑖𝑗,𝑛

𝑘
 

                     =
1

𝑘
{𝑥𝑟 (

𝑥1𝑢𝑖𝑗,𝑛−1+𝐷𝑦1𝑣𝑖𝑗,𝑛−1

𝑘
) + 𝐷𝑦𝑟 (

𝑦1𝑢𝑖𝑗,𝑛−1+𝑥1𝑣𝑖𝑗,𝑛−1

𝑘
)} 

                     =
1

𝑘
{𝑥1 (

𝑥𝑟𝑢𝑖𝑗,𝑛−1+𝐷𝑦𝑟𝑣𝑖𝑗,𝑛−1

𝑘
) + 𝐷𝑦1 (

𝑥𝑟𝑣𝑖𝑗,𝑛−1+𝑦𝑟𝑢𝑖𝑗,𝑛−1

𝑘
)} 

                    =
𝑥1𝑢𝑖𝑗,𝑛+𝑟−1+𝐷𝑦1𝑣𝑖𝑗,𝑛+𝑟−1

𝑘
  

Value of 𝑣𝑖𝑗,𝑛+𝑟 can be obtained accordingly. 

 

(c) To prove the final part, we replace 𝑛 by 𝑛 + 𝑟 in the first result and using 
that in (a) above, we obtain 

𝑢𝑖𝑗,𝑛+2𝑟 =
1

𝑘
{𝑥𝑟 𝑢𝑖𝑗,𝑛+𝑟 + 𝐷𝑦𝑟 (

𝑦𝑟𝑢𝑖𝑗,𝑛+𝑥𝑟𝑣𝑖𝑗,𝑛

𝑘
)}  

              =
1

𝑘
{𝑥𝑟 𝑢𝑖𝑗,𝑛+𝑟 +

𝐷𝑦𝑟
2𝑢𝑖𝑗,𝑛

𝑘
+ 𝑥𝑟 (

𝐷𝑦𝑟𝑣𝑖𝑗,𝑛

𝑘
)}  

              =
1

𝑘
{𝑥𝑟 𝑢𝑖𝑗,𝑛+𝑟 +

𝐷𝑦𝑟
2𝑢𝑖𝑗,𝑛

𝑘
+ 𝑥𝑟 (𝑢𝑖𝑗,𝑛+𝑟 −

𝑥𝑟𝑢𝑖𝑗,𝑛

𝑘
)} 

              =
1

𝑘
{2𝑥𝑟 𝑢𝑖𝑗,𝑛+𝑟 −

𝑢𝑖𝑗,𝑛

𝑘
(𝑥𝑟

2 − 𝐷𝑦𝑟
2)} 

Since 𝑥𝑟 + 𝑦𝑟 √𝐷 is a solution of (1.2), we have 𝑥𝑟
2 − 𝐷𝑦𝑟

2 = 1. Thus, we obtain 

𝑢𝑖𝑗,𝑛+2𝑟 =
2𝑘𝑥𝑟𝑢𝑖𝑗,𝑛+𝑟−𝑢𝑖𝑗,𝑛

𝑘2
 . 

Value of 𝑣𝑖𝑗,𝑛+2𝑟 can also be obtained accordingly. 

 

       We further derive some more interesting properties related with the value 

of 𝑢𝑖𝑗,𝑛 and 𝑣𝑖𝑗,𝑛. The following interesting recursive formula connects three 

𝑢𝑖𝑗,𝑛’s as well as 𝑣𝑖𝑗,𝑛’s when the suffixes are in arithmetic progression.  

 

Corollary 2.6. (a) 𝑢𝑖𝑗,𝑛𝑢𝑖𝑗,𝑛+2𝑟 − 𝑢𝑖𝑗,𝑛+𝑟
2 = 𝑘2𝐷𝑁𝑦𝑟

2.                           

                        (b) 𝑣𝑖𝑗,𝑛𝑣𝑖𝑗,𝑛+2𝑟 − 𝑣𝑖𝑗,𝑛+𝑟
2 = −𝑘2𝑁𝑦𝑟

2. 

Proof. By (2.5) we have 
𝑢𝑖𝑗,𝑛+𝑣𝑖𝑗,𝑛√𝐷

𝑘
= (

𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
) (𝑥1 + 𝑦1√𝐷)

𝑛
. 

Taking its surd conjugate, we get 
𝑢𝑖𝑗,𝑛−𝑣𝑖𝑗,𝑛√𝐷

𝑘
= (

𝐴𝑖𝑗−𝐵𝑖𝑗√𝐷

𝑘
) (𝑥1 − 𝑦1√𝐷)

𝑛
. 

For convenience we write 𝛾 = 𝑥1 + 𝑦1√𝐷, �̅� = 𝑥1 − 𝑦1√𝐷 and 𝜇𝑖𝑗 =
𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
, 𝜇𝑖𝑗̅̅̅̅ =

𝐴𝑖𝑗−𝐵𝑖𝑗√𝐷

𝑘
. Then we have 

𝑢𝑖𝑗,𝑛+𝑣𝑖𝑗,𝑛√𝐷

𝑘
= 𝜇𝑖𝑗 𝛾

𝑛 and 

𝑢𝑖𝑗,𝑛−𝑣𝑖𝑗,𝑛√𝐷

𝑘
= 𝜇𝑖𝑗̅̅̅̅ �̅�

𝑛. 

Adding and subtracting these two relations, we get 

𝑢𝑖𝑗,𝑛 =
𝑘

2
{𝜇𝑖𝑗 𝛾

𝑛 + 𝜇𝑖𝑗̅̅̅̅ �̅�
𝑛} and 𝑣𝑖𝑗,𝑛 =

𝑘

2√𝐷
{𝜇𝑖𝑗 𝛾

𝑛 − 𝜇𝑖𝑗̅̅̅̅ �̅�
𝑛}. 



On the solutions of Pellian equation 𝑈2 − 𝐷𝑉2 = 𝑘2𝑁 
 

217 

 

It can be easily observed that 𝛾�̅� = 𝑥1
2 − 𝐷𝑦1

2 = 1 and 𝜇𝑖𝑗 𝜇𝑖𝑗̅̅̅̅ =
𝐴𝑖𝑗

2 −𝐷𝐵𝑖𝑗
2

𝑘2
= 𝑁. 

Then 

 𝑢𝑖𝑗,𝑛𝑢𝑖𝑗,𝑛+2𝑟 − 𝑢𝑖𝑗,𝑛+𝑟
2  

         =
𝑘2

4
{
(𝜇𝑖𝑗 𝛾

𝑛 + 𝜇𝑖𝑗̅̅̅̅ �̅�
𝑛)(𝜇𝑖𝑗 𝛾

𝑛+2𝑟 + 𝜇𝑖𝑗̅̅̅̅ �̅�
(𝑛+2𝑟))

−(𝜇𝑖𝑗 𝛾
𝑛+𝑟 + 𝜇𝑖𝑗̅̅̅̅ �̅�

(𝑛+𝑟))
2 } 

         =
𝑘2

4
{

𝜇𝑖𝑗 𝜇𝑖𝑗̅̅̅̅ (𝛾
𝑛+2𝑟 �̅�  𝑛 + 𝛾 𝑛�̅�  (𝑛+2𝑟))

−2𝜇𝑖𝑗 𝜇𝑖𝑗̅̅̅̅ 𝛾
𝑛+𝑟 �̅� (𝑛+𝑟)

} =
𝑘2

4
{𝑁(𝛾 2𝑟 + �̅�  2𝑟) − 2𝑁}       

         =
𝑘2𝑁

4
(𝛾 𝑟 − �̅�  𝑟 )2. 

Since 𝑥𝑟 + 𝑦𝑟 √𝐷 = (𝑥1 + 𝑦1√𝐷)
𝑟

= 𝛾 𝑟 and 𝑥𝑟 − 𝑦𝑟 √𝐷 = (𝑥1 − 𝑦1√𝐷)
𝑟

=

�̅�  𝑟, we get 𝑢𝑖𝑗,𝑛𝑢𝑖𝑗,𝑛+2𝑟 − 𝑢𝑖𝑗,𝑛+𝑟
2 = 𝑘2𝐷𝑁𝑦𝑟

2. 

Second result can be proved accordingly. 

 

 Following result gives some more recurrence relations in the form of a 

determinant. 

 

Theorem 2.7. a)  |
𝑢𝑖𝑗,𝑛 𝑢𝑖𝑗,𝑛+𝑟
𝑣𝑖𝑗,𝑛 𝑣𝑖𝑗,𝑛+𝑟

| = 𝑘𝑦𝑟 𝑁 

                       b) |
𝑢𝑖𝑗,𝑛+𝑟−1 𝑢𝑖𝑗,𝑛+𝑟
𝑣𝑖𝑗,𝑛+𝑟−1 𝑣𝑖𝑗,𝑛+𝑟

| = 𝑘𝑦1𝑁 

           c) |

1 1 1
𝑢𝑖𝑗,𝑛−𝑟 𝑢𝑖𝑗,𝑛 𝑢𝑖𝑗,𝑛+𝑟
𝑣𝑖𝑗,𝑛−𝑟 𝑣𝑖𝑗,𝑛 𝑣𝑖𝑗,𝑛+𝑟

| = −2𝑘𝑁𝑦𝑟 (𝑥𝑟 − 1). 

Proof. We only prove (c), since first two results follow easily through theorem 

2.5. Now 

|

1 1 1
𝑢𝑖𝑗,𝑛−𝑟 𝑢𝑖𝑗,𝑛 𝑢𝑖𝑗,𝑛+𝑟
𝑣𝑖𝑗,𝑛−𝑟 𝑣𝑖𝑗,𝑛 𝑣𝑖𝑗,𝑛+𝑟

|

= |
𝑢𝑖𝑗,𝑛 𝑢𝑖𝑗,𝑛+𝑟
𝑣𝑖𝑗,𝑛 𝑣𝑖𝑗,𝑛+𝑟

| − |
𝑢𝑖𝑗,𝑛−𝑟 𝑢𝑖𝑗,𝑛+𝑟
𝑣𝑖𝑗,𝑛−𝑟 𝑣𝑖𝑗,𝑛+𝑟

| + |
𝑢𝑖𝑗,𝑛−𝑟 𝑢𝑖𝑗,𝑛
𝑣𝑖𝑗,𝑛−𝑟 𝑣𝑖𝑗,𝑛

| 

                             = 𝑘𝑦𝑟 𝑁 − 𝑘𝑦2𝑟 𝑁 + 𝑘𝑦𝑟 𝑁 = 𝑘𝑁(2𝑦𝑟 − 𝑦2𝑟 ). 

Now (𝑥1 + 𝑦1√𝐷)
2𝑟

= 𝑥2𝑟 + 𝑦2𝑟 √𝐷 = (𝑥𝑟 + 𝑦𝑟 √𝐷)
2
. This gives 𝑦2𝑟 =

2𝑥𝑟 𝑦𝑟  . Thus 

|

1 1 1
𝑢𝑖𝑗,𝑛−𝑟 𝑢𝑖𝑗,𝑛 𝑢𝑖𝑗,𝑛+𝑟
𝑣𝑖𝑗,𝑛−𝑟 𝑣𝑖𝑗,𝑛 𝑣𝑖𝑗,𝑛+𝑟

| = 𝑘𝑁(2𝑦𝑟 − 2𝑥𝑟 𝑦𝑟 ) = −2𝑘𝑁𝑦𝑟 (𝑥𝑟 − 1). 

 



Bilkis M. Madni, Devbhadra V. Shah 

218 

 

3. Number of solutions up to a desired limit 
 

 In this final section, we define and obtain the values of the sums 𝑅(𝑍) =
∑ 1𝑢𝑖𝑗,𝑛+𝑣𝑖𝑗,𝑛√𝐷

𝑘
 ≤ 𝑍

𝑈2−𝐷𝑉2=𝑘2𝑁

, 𝑆(𝑍) = ∑ 1𝑢𝑖𝑗,𝑛 ≤ 𝑍
𝑈2−𝐷𝑉2=𝑘2𝑁

 and 𝑇(𝑍) = ∑ 1𝑣𝑖𝑗,𝑛≤ 𝑍
𝑈2−𝐷𝑉2=𝑘2𝑁

, the 

total number of positive solutions 
𝑢𝑖𝑗,𝑛+𝑣𝑖𝑗,𝑛√𝐷

𝑘
, 𝑢𝑖𝑗,𝑛 and 𝑣𝑖𝑗,𝑛 respectively of 

(1.3) that do not exceed any given large positive real number 𝑍.                                      

For convenience, we denote 𝛿 =
1

log 𝛾
, 𝛾 = 𝑥1 + 𝑦1√𝐷 and let 𝒜𝑖𝑗 =

𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
 

runs through all the fundamental solutions of (1.3) for any fixed class 𝑖. We also 
assume that (1.1) has 𝛽 fundamental solutions and the equation 𝑅2 − 𝐷𝑆2 = 𝑘2 
has 𝜂 fundamental solutions.  
 Thus, throughout we have 1 ≤ 𝑖 ≤ 𝛽 and 1 ≤ 𝑗 ≤ 𝜂. We first obtain the 
value of 𝑅(𝑍) which gives the number of all the solutions of (1.3) not exceeding 
any fixed given positive real number 𝑍.    
 

Theorem 3.1. 𝑅(𝑍) = 𝛿 {𝛽𝜂 log(𝑍) − log(∏ ∏ 𝒜𝑖𝑗
𝜂
𝑗=1

𝛽
𝑖=1 )} + 𝐶, where 𝐶 is 

the effective constant such that 0 ≤ 𝐶 < 𝛽𝜂.  
Proof. To find the value of 𝑅(𝑍) = ∑ 1𝑢𝑖𝑗,𝑛+𝑣𝑖𝑗,𝑛√𝐷

𝑘
 ≤ 𝑍

𝑈2−𝐷𝑉2=𝑘2𝑁

, we first find the number 

of positive solutions 
𝑢𝑖𝑗,𝑛+𝑣𝑖𝑗,𝑛√𝐷

𝑘
 of (1.3) that do not exceed 𝑍 for some fixed 

class 𝑖 = 𝛼 (1 ≤ 𝑗 ≤ 𝜂). Since 𝛾 and 𝒜𝑖𝑗  are solutions of (1.2) and (1.3) 

respectively, (2.5) can be written as 
𝑢𝛼𝑗,𝑛+𝑣𝛼𝑗𝑛√𝐷

𝑘
= 𝒜𝛼𝑗 𝛾

𝑛. Now, for any given 

𝑍, it is clear that for some fixed class 𝑖 = 𝛼, there exists some n such that 
𝑢𝛼𝑗,𝑛+𝑣𝛼𝑗,𝑛√𝐷

𝑘
≤ 𝑍 <

𝑢𝛼𝑗,𝑛+1+𝑣𝛼𝑗,𝑛+1√𝐷

𝑘
 . Then we get 𝒜𝛼𝑗 𝛾

𝑛 ≤ 𝑍 < 𝒜𝛼𝑗 𝛾
𝑛+1. 

This implies 𝑛 <
log 𝑍−log 𝒜𝛼𝑗

log 𝛾
< 𝑛 + 1. Since n is an integer, we get                        

𝑛 = [
log 𝑍−log 𝒜𝛼𝑗

log 𝛾
], where [𝑥] denotes the integer part of 𝑥.  

 Now since [𝑥] = 𝑥 − {𝑥}, where {𝑥} is the fractional part of 𝑥 and as                    
0 ≤ {𝑥} < 1, we have   

𝑅(𝑍) = ∑ ∑ [
log 𝑍−log 𝒜𝑖𝑗

log 𝛾
]

𝜂
𝑗=1

𝛽
𝑖=1 = ∑ ∑ (

log 𝑍−log 𝒜𝑖𝑗

log 𝛾
+ 𝑐′)

𝜂
𝑗=1

𝛽
𝑖=1 , 

where 0 ≤ c′ < 1. Thus 𝑅(𝑍) =
1

𝑙𝑜𝑔 𝛾
∑ ∑ (log 𝑍 − log 𝒜𝑖𝑗 )

𝜂
𝑗=1

𝛽
𝑖=1 + 𝛽𝜂𝑐

′.                    

If we write 𝐶 = 𝛽𝜂𝑐′, then we get 0 ≤ 𝐶 < 𝛽𝜂 and 

𝑅(𝑍) = 𝛿 {𝛽𝜂 log(𝑍) − log(∏ ∏ 𝒜𝑖𝑗
𝜂
𝑗=1

𝛽
𝑖=1 )} + 𝐶. 

 



On the solutions of Pellian equation 𝑈2 − 𝐷𝑉2 = 𝑘2𝑁 
 

219 

 

 We next find the value of 𝑆(𝑍). 
 

Theorem 3.2. 𝑆(𝑍) = 𝛿 {𝛽𝜂 log(2𝑍/𝑘) − log(∏ ∏ 𝒜𝑖𝑗
𝜂
𝑗=1

𝛽
𝑖=1 )} + 𝐶, where 𝐶 

is the effective constant such that −𝛽𝜂 ≤ 𝐶 < 𝛽𝜂. 
Proof. To find the value of 𝑆(𝑍), we first find the number of positive solutions 
of (1.3) where the values 𝑢𝑖𝑗,𝑛 of 𝑈 do not exceed 𝑍 for some fixed class 𝑖 = 𝛼 

(1 ≤ 𝑗 ≤ 𝜂). Now (2.5) can be written as 
𝑢𝛼𝑗,𝑛+𝑣𝛼𝑗𝑛√𝐷

𝑘
= 𝒜𝛼𝑗 𝛾

𝑛. Then for any 

given 𝑍, it is clear that for some fixed class 𝑖 = 𝛼, there exists some n such that 

𝑢𝛼𝑗,𝑛 ≤ 𝑍 < 𝑢𝛼𝑗,𝑛+1. Since 𝒜𝑖𝑗 =
𝐴𝑖𝑗+𝐵𝑖𝑗√𝐷

𝑘
 , we write 𝒜𝑖𝑗̅̅ ̅̅ ̅ =

𝐴𝑖𝑗−𝐵𝑖𝑗√𝐷

𝑘
.                     

We also have 𝛾 −1 = 𝑥1 − 𝑦1√𝐷. Since 𝛾 and 𝒜𝑖𝑗 , 𝒜𝑖𝑗̅̅ ̅̅ ̅ are the solutions of (1.2) 

and (1.3) respectively, we have  𝛾𝛾 −1 = 1 and  𝒜𝛼𝑗  𝒜𝛼𝑗̅̅ ̅̅ ̅ = 𝑁. Then (2.5) can 

be written as  
𝑢𝛼𝑗,𝑛+𝑣𝛼𝑗,𝑛√𝐷

𝑘
= 𝒜𝛼𝑗 𝛾

𝑛.                                    (3.1) 

Now taking surd-conjugate of (3.1) we get 
𝑢𝛼𝑗,𝑛−𝑣𝛼𝑗,𝑛√𝐷

𝑘
= 𝒜𝛼𝑗̅̅ ̅̅ ̅𝛾

−𝑛. Adding this 

with (3.1) we now have  

𝑢𝛼𝑗,𝑛 =
𝑘

2
{𝒜𝛼𝑗 𝛾

𝑛 + 𝒜𝛼𝑗̅̅ ̅̅ ̅𝛾
−𝑛} =

𝑘

2
{𝒜𝛼𝑗 𝛾

𝑛 +
𝑁

𝒜𝛼𝑗
𝛾 −𝑛}. 

Since 𝑢𝛼𝑗,𝑛 ≤ 𝑍 < 𝑢𝛼𝑗,𝑛+1, for some n, we get 

𝒜𝛼𝑗 𝛾
𝑛 +

𝑁

𝒜𝛼𝑗
𝛾 −𝑛 ≤

2𝑍

𝑘
< 𝒜𝛼𝑗 𝛾

𝑛+1 +
𝑁

𝒜𝛼𝑗
𝛾 −𝑛−1. 

Also, since 
𝑁

𝒜𝛼𝑗
𝛾 −𝑛 > 0, 𝒜𝛼𝑗̅̅ ̅̅ ̅ =

𝑁

𝒜𝛼𝑗
< 𝒜𝛼𝑗  and 𝛾

−1 < 𝛾, we have                        

𝒜𝛼𝑗 𝛾
𝑛 <

2𝑍

𝑘
< 2𝒜𝛼𝑗 𝛾

𝑛+1. Now 𝛾 = 𝑥1 + 𝑦1√𝐷 > 2. Then we have       

𝒜𝛼𝑗 𝛾
𝑛 <

2𝑍

𝑘
< 𝒜𝛼𝑗 𝛾

𝑛+2. This implies  𝑛 <
log 2𝑍−log 𝒜𝛼𝑗−log 𝑘

log 𝛾
< 𝑛 + 2. Since 

n is an integer, we get 𝑛 ≤ [
log 2𝑍−log 𝒜𝛼𝑗−log 𝑘

log 𝛾
] ≤ 𝑛 + 1. This implies  

𝑛 = [
log 2𝑍−log 𝒜𝛼𝑗 −log 𝑘

log 𝛾
] or 𝑛 = [

log 2𝑍−log 𝒜𝛼𝑗 −log 𝑘

log 𝛾
] − 1. 

Thus 𝑆(𝑍) = ∑ ∑ [
log 2𝑍−log 𝒜𝛼𝑗−log 𝑘

log 𝛾
]

𝜂
𝑗=1

𝛽
𝑖=1 + 𝑐, where 𝑐 = 0  or −1. 

Then 𝑆(𝑍) =
1

𝑙𝑜𝑔 𝛾
∑ ∑ (log 2𝑍 − log 𝒜𝑖𝑗 − log 𝑘 + 𝑐 + 𝑐

′)
𝜂
𝑗=1

𝛽
𝑖=1 , where                

0 ≤ c′ < 1. Now considering 𝑐 + 𝑐′ = 𝐶′, we have −1 ≤ 𝐶′ < 1. We can now 

write 𝑆(𝑍) =
𝛽𝜂

𝑙𝑜𝑔 𝛾
log(2𝑍/𝑘) −

1

𝑙𝑜𝑔 𝛾
∑ ∑ log 𝒜𝑖𝑗

𝜂
𝑗=1

𝛽
𝑖=1 + 𝛽𝜂𝐶

′. 

If we write 𝐶 = 𝛽𝜂𝐶′, then we get −𝛽𝜂 ≤ 𝐶 < 𝛽𝜂 and 

𝑆(𝑍) = 𝛿 {𝛽𝜂 log(2𝑍/𝑘) − log(∏ ∏ 𝒜𝑖𝑗
𝜂
𝑗=1

𝛽
𝑖=1 )} + 𝐶. 

 

 Finally, we find the value of 𝑇(𝑍). 



Bilkis M. Madni, Devbhadra V. Shah 

220 

 

 

Theorem 3.3. 𝑇(𝑍) = 𝛿 {𝛽𝜂 log(2√𝐷𝑍/𝑘) − log(∏ ∏ 𝒜𝑖𝑗
𝜂
𝑗=1

𝛽
𝑖=1 )} + 𝐶, 

where 𝐶 is the effective constant such that 0 ≤ 𝐶 < 2𝛽𝜂. 
Proof. To find the value of 𝑇(𝑍), we first find the number of positive solutions 
of (1.3) where the values 𝑣𝑖𝑗,𝑛 of 𝑉 do not exceed 𝑍 for some fixed class 𝑖 = 𝛼 

(1 ≤ 𝑗 ≤ 𝜂). Now by (2.5) since 
𝑢𝛼𝑗,𝑛+𝑣𝛼𝑗,𝑛√𝐷

𝑘
= 𝒜𝛼𝑗 𝛾

𝑛 and 
𝑢𝛼𝑗,𝑛−𝑣𝛼𝑗,𝑛√𝐷

𝑘
=

𝒜𝛼𝑗̅̅ ̅̅ ̅𝛾
−𝑛, on subtraction, we get  

𝑣𝛼𝑗,𝑛 =
𝑘

2√𝐷
{𝒜𝛼𝑗 𝛾

𝑛 − 𝒜𝛼𝑗̅̅ ̅̅ ̅𝛾
−𝑛} =

𝑘

2√𝐷
{𝒜𝛼𝑗 𝛾

𝑛 −
𝑁

𝒜𝛼𝑗
𝛾 −𝑛}. 

Since 𝑣𝛼𝑗,𝑛 ≤ 𝑍 < 𝑣𝛼𝑗,𝑛+1, for some n, we have 

𝒜𝛼𝑗 𝛾
𝑛 −

𝑁

𝒜𝛼𝑗
𝛾 −𝑛 ≤

2√𝐷𝑍

𝑘
< 𝒜𝛼𝑗 𝛾

𝑛+1 −
𝑁

𝒜𝛼𝑗
𝛾 −𝑛−1. 

Thus, we have 𝒜𝛼𝑗 𝛾
𝑛−1 ≤

2√𝐷𝑍

𝑘
< 𝒜𝛼𝑗 𝛾

𝑛+1. This implies 

𝑛 − 1 <
log 2√𝐷𝑍−log 𝒜𝛼𝑗 −log 𝑘

log 𝛾
< 𝑛 + 1. 

Since n is an integer, we have 

𝑛 = [
log 2√𝐷𝑍−log 𝒜𝛼𝑗− log 𝑘

log 𝛾
] or𝑛 = [

log 2√𝐷𝑍−log 𝒜𝛼𝑗−log 𝑘

log 𝛾
] + 1.  

Thus 𝑇(𝑍) = ∑ ∑ [
log 2√𝐷𝑍−log 𝒜𝛼𝑗−log 𝑘

log 𝛾
]

𝜂
𝑗=1

𝛽
𝑖=1 + 𝑐, where 𝑐 = 0  or 1. Then 

𝑇(𝑍) =
1

𝑙𝑜𝑔 𝛾
∑ ∑ (log 2√𝐷𝑍 − log 𝒜𝑖𝑗 − log 𝑘 + 𝑐 + 𝑐

′)
𝜂
𝑗=1

𝛽
𝑖=1 , 

where 0 ≤ c′ < 1. Considering 𝑐 + 𝑐′ = 𝐶′, we have 0 ≤ 𝐶′ < 2. Thus, we 

now write 𝑇(𝑍) =
𝛽𝜂

𝑙𝑜𝑔 𝛾
log(2√𝐷𝑍/𝑘) −

1

𝑙𝑜𝑔 𝛾
∑ ∑ log 𝒜𝑖𝑗

𝜂
𝑗=1

𝛽
𝑖=1 + 𝛽𝜂𝐶

′. 

If we write 𝐶 = 𝛽𝜂𝐶′, then we get 0 ≤ 𝐶 < 2𝛽𝜂 and 

𝑇(𝑍) = 𝛿 {𝛽𝜂 log(2√𝐷𝑍/𝑘) − log(∏ ∏ 𝒜𝑖𝑗
𝜂
𝑗=1

𝛽
𝑖=1 )} + 𝐶. 

 

 The Following interesting conclusions are now an easy consequence from 

these theorems. 

 

Corollary 3.4. 𝑇(𝑍) − 𝑆(𝑍) ≈ 𝛿𝛽𝜂 log √𝐷.     
  

Corollary 3.5. If the solutions 
𝑢𝑖𝑗,𝑛+𝑣𝑖𝑗,𝑛√𝐷

𝑘
 of 𝑈2 − 𝐷𝑉2 = 𝑘2𝑁 are considered 

as lattice points within the square [0, 𝑍] × [0, 𝑍], then density of these lattice 
points is zero.             

 This follows from the fact that lim
𝑛→∞

log 𝑍

𝑍
= 0. 

 



On the solutions of Pellian equation 𝑈2 − 𝐷𝑉2 = 𝑘2𝑁 
 

221 

 

4. Conclusions 

 
 In this paper, we derived the necessary and sufficient condition for any two 

solutions of 𝑈2 − 𝐷𝑉2 = 𝑘2𝑁 to belong to the same class and the bounds for 
the values of 𝑢, 𝑣 occurring in the fundamental solution. We also derived an 
explicit formula which gives all its positive solutions. We further obtained some 

interesting recurrence relations connecting the values of 𝑢, 𝑣. Finally, we 
obtained the results for total number of its positive solutions not exceeding any 

given positive real number 𝑍. 

 

 

References  
 

[1] Andreescu Titu, Andrica Dorin, Cucurezeanu Ion. An Introduction to 
Diophantine Equations. Birkhäuser Boston Inc, Secaucus, United States. 

2010. 

[2]   Burton D. M. Elementary Number Theory. Tata Mc Graw–Hill Pub. Co. 
Ltd., New Delhi. 2007. 

[3]   Kaplan P., Williams K. S. Pell’s Equation 𝑥2 − 𝐷𝑦2 = −1, −4 and 
continued fractions. Journal of Number Theory. 23, 169 – 182, 1986. 

[4] LeVeque William J. Topics in Number Theory, Vol. I. Addition-Wesley 
Pub. Co., Inc. 137 – 158, 1956. 

[5] Madni Bilkis M., Shah Devbhadra V. Alternate proofs for the infinite 
number of solutions of Pell’s equation. International Journal of 

Engineering, Science and Mathematics. 7 (4), 255 – 259, April 2018.  

[6]   Matthews K. The Diophantine Equation 𝑥2 − 𝐷𝑦2 = 𝑁, 𝐷 > 0. 
Expositiones Mathematicae.18, 323 – 331, 2000. 

[7]   Mollin R. A., Poorten A. J., Williams H. C. Halfway to a solution of 𝑥2 −
𝐷𝑦2 = −3. Journal de Theorie des Nombres Bordeaux. 6, 421 – 457, 1994. 

[8]   Shah Devbhadra V. On the solutions of Diophantine Equations 𝑥2 −
𝐷𝑦2 = ±4𝑁. Journal of V.N.S.G. University. 4 – B, 56 – 65, 2007. 

[9]   Steuding Jörn. Diophantine Analysis. Chapman & Hall/CRC, Taylor & 
Francis Group, Boca Raton. 2005.  

[10] Stevenhagen P. A density conjecture for the negative Pell equation. 
Computational Algebra and Number Theory, Mathematics and its 

Applications. 325, 1995. 

[11] Stolt Bengt. On the Diophantine equation 𝑢2 − 𝐷𝑣2 = ±4𝑁, Part I. Arkiv 
för Matematik. 2 (1), 1 – 23, 1951. 

[12] Stolt Bengt. On the Diophantine equation 𝑢2 − 𝐷𝑣2 = ±4𝑁, Part II. Arkiv 
för Matematik. 2 (10), 251 – 268, 1951. 



Bilkis M. Madni, Devbhadra V. Shah 

222 

 

[13]  Tekcan A. The Pell Equation 𝑥2 − 𝐷𝑦2 = ±4. Applied Mathematical 
Sciences. 1 (8), 363 – 369, 2007. 

[14]  Telang S. G. Number Theory. Tata Mc Graw–Hill Pub. Co. Ltd., New 
Delhi. 2004.