Journal of Indian Acad. Math.

ISSN: 0970-5120

Vol. 47, No. 1 (2025) pp. 191-204

A. Deepshika¹

INVESTIGATING THE EXISTENCE OF

THE EXPONENTIAL DIOPHANTINE

RECTANGLES OVER STAR AND

PRONIC NUMBERS

and

J. Kannan²

Abstract: This paper is focused on collecting a different sort of rectangle

called Exponential Diophantine Rectangles over Star and Pronic Numbers.

We demonstrated that there is only one Exponential Diophantine Rectangle

over the Star numbers and no Exponential Diophantine rectangles over the

Pronic numbers. Python programming is provided for the existence of such

rectangles.

Keywords: Binomial Expansion, Catalan’s Conjecture, Diophantine

Equation, Exponential Diophantine Equation, Star numbers,

Pronic numbers, Rectangles.

Mathematics Subject Classification (2020) No.: 11A07, 11D61, 11D72.

1. Introduction

An exponential Diophantine equation is a special type of Diophantine

equation where the variables exist in exponents. Many authors solved the different

forms of exponential Diophantine equations. In particular, William Sobredo Gayo, Jr.

and Jerico Bravo Bacani [12] solved the exponential Diophantine equation of the

form M_p^x (M_q 1)^y z²and Mahalakshmi, M. et al. [5], [6], and [7] solved

various Diophantine equations to collect various geometrical shapes, including peble

triangles and almost equilateral triangles.

We define and collect the exponential Diophantine rectangle over Special

numbers (ED Rectangles over Special numbers), inspired by the above. In this paper

we deal only with two types of special numbers especially star and Pronic numbers.

After the introduction basis preliminaries provided. In section (3), the definition

192

A. DEEPSHIKA AND J. KANNAN

of ED Rectangles over Star numbers (S_m) and the lemmas needed for the main

theorem, while subsections establish theorems for the existence of solutions to the

Exponential Diophantie equations. Python code is displayed for the existence of such

rectangles. Section (3) provides the definition theorems and Python programming for

the existence of exponential Diopahantine Rectangles over Pronic numbers.

2. Preliminaries

This section contains basic definitions and lemmas required for this article.

Lemma 2.1 (Catalan’s Conjecture): (3, 2, 2, 3) is the unique solution for

the exponential Diophantine equation a^x b^y 1, where a,b, x, y   such

that min{a,b, x, y}  2

.

Definition 2.1 (Binomial Expansion): For x   and n   , the

n(n 1)

2!

expansion

for

(1  x)ⁿ

is

1  nx 

x² 

and

for

n(n 1)

2 !

(1  x)ⁿis 1  nx 

x² .

n

In general,

(x  y)ⁿ



nC_kx^{n k k}

y



nC_kx^ky^{n k}

.



k 0

Definition

2.2: The m^th

Star number

(S_m)

is given by

S_m 6m² 6m  1, for m  

.

Example 1: S₁ 1 S₁₀ 541.

,

Definition 2.3: The m^thPronic numbers (P_m) are of the form

P_m m² m , for all m  

:

Definition 2.4: An Exponential Diophantine rectangle is defined as a

rectangle with the length (l) and breadth (b) as (l,b)  (rp  (r  1)(x  y)

,

q(r  1)  (r  2)(x  y)) where p,q, r   and x;y are non-negative

integers such that

DIOPHANTINE RECTANGLES OVER STAR AND PRONIC NUMBERS 193

p^x q^x r^y

(1)

3. Exponential Diophantine Rectangles over Star Numbers

This section defines Exponential Diophantine Rectangles over S_m, provides

some lemmas for solving exponential Diophantine equations, and is divided into two

subsections that examine some theorems. The existence of Exponential Diophantine

Rectangles over S_mis proved by python programming with certain limits.

Definition 3.1 (Exponential Diophantine Rectangles overS_m): An

Exponential Diophantine rectangle over star numbers (S_m) is defined as a

rectangle

(l,b)  ((m  1)S_m1 m(x  y)

x, y are non-negative integers such that

with

the

length

(l)

and

breadth

(b)

as

,

mS_m (m  1)(x  y)) where m   and

S_m^x_1 S_m^x (m  1)^y

(2)

(3)

or

S_m^x_1 S_m^x (m  1)^y

Notation: Exponential Diophantine Rectangles over star numbers - ED

Rectangles over S_m

.

Lemma 3.1: The inequality (1  n)^x 2 holds for all n, x  1

.

Proof: Let us show this by using induction hypothesis on

n

. Now, for

n  2, the inequality becomes 3^x 2 . This is true for x  1. Now, assume that

the inequality (1  n)^y 2 holds for n  k . That is, (k  1)^y 2 for k  1

.

We have to show that the given inequality holds for n  k  1. We know that

(k  2)  (k  1) implies (k  2)^y 2, for all k, y  1

.



Lemma 3.2: The equation b⁴ 6b² 4b³ 4b  1  0 has no positive

integer solution for b  1

.

194

A. DEEPSHIKA AND J. KANNAN

Proof: The equation b⁴ 6b² 4b³ 4b  1  2, 3 (mod 4), which is an

absurd one.



Lemma 3.3: If y  2, the inequality

(1  n)^y 4n

holds for all

n  1

.

Lemma 3.4: The equation

m¹² 12m¹¹ 66m¹⁰ 220m⁹ 495m⁸

 792m⁷ 924m⁶ 792m⁵ 495m⁴ 220m³ 66m² 1  0

has

no

solution m  

.

Proof:

The

equation

m¹² 12m¹¹ 66m¹⁰ 220m⁹ 495m⁸

 792m⁷ 924m⁶ 792m⁵ 495m⁴ 220m³ 66m² 1

0 (mod 10)

Hence it has no solution.



3.1 The exponential Diophantine equation S_m^x_1 S_m^x (m  1)^y

:

The existence of the solution to the equation S_m^x_1 S_m^x (m  1)^yis

discussed here.

Theorem

3.1:

The

exponential

Diophantine

equation

for all

S_m^x_1 S_m^x (m  1)^yhas only one solution (x, y, m)  (0, 1, 1)

x, y  ^ {0} and m  

.

Proof:

Consider

the

exponential

Diophantine

equation

S_m^x_1 S_m^x (m  1)^yand solve this in various possibilities.

Possibility 1: For x  0 and y  0, the equation (2) has no solution.

Possibility 2: If x  0 and y  1, then we obtain m  1 from the exponential

Diophantine equation.

DIOPHANTINE RECTANGLES OVER STAR AND PRONIC NUMBERS 195

Possibility 3: If x  1 and y  0, the equation becomes 12m² 1 which

has no solution as m  

.

Possibility 4: For x  y  1 it

,

reduces

to

a

quadratic

equation

8

24

12m² m  1  0. On solving this we obtain m 

  .

Possibility 5: When x  0 and y  1, the equation becomes (m  1)^y 2

.

This is not possible by lemma (3.3).

Possibility 6: For x  1 and y  0, the equation becomes (6m² 6m  1)^x

 (6m² 6m  1)^x 1. Here S_m^x S_x^m_1 0 (mod 2) which

is an absurd one.

Possibility 7: For x  1 and y  1, we get 12m² 2  (m  1)^y. By using

definition (2.1) and equating the coefficient of

m

², we get the

value y  4 . Putting the value of

y

in 12m² 2  (m  1)^y

implies (m  1)⁴ 12m² 2. By lemma (3.2), it has no

solutions.

Possibility 8: Now

x  1

and

y  1

,

the

equation

changes

into

S_m^x S_m^x_1 m  1  0. For m  1, it reduces to 13^x 1

,

which is an impossible one as x  1. Now, m  1 for,

S_m S_m1 m  1  12m² m  1  0

.

This

is

not

possible as m  1

.

Possibility 9: Here x, y  1 and m  1, the equation has no solution by lemma

(2.1). Now for x, y, m  1, this possibility fails by using

definition (2.1).

Hence, there is only one solution for the Diophantine equation

S_m^x_1 S_m^x (m  1)^y

(

i.e.,(x, y, m)  (0,1, 1))

.



196

A. DEEPSHIKA AND J. KANNAN

3.2 The exponential Diophantine equation S_m^x_1 S_m^x (m  1)^y

:

This subsection provides the theorem for determining the solution for the equation

S_m^x_1 S_m^x (m  1)^y

.

Theorem 3.2: No integral solution exists for the exponential

Diophantine equation S_m^x_1 S_m^x (m  1)^yx, y     {0} and m  

.

Proof:

Consider

the

exponential

Diophantine

equation

S_m^x_1 S_m^x (m  1)^yand solve this in various possibilities.

Possibility 1: For x  0 and y  0, this possibility fails.

Possibility 2: x  0 and y  1, the equation becomes m  1, which

contradicts.

Possibility 3: Here x  y  1, then on solving the above equation, the value of

1

11

m

obtained as

  .

Possibility 4: For x  1 and y  0, we obtain m  

.

Possibility 5: Now x  0 and y  1, then from the equation we obtain

(m  1)^y 0, as m  1. This possibility fails.

Possibility 6: When x  1 and y  0, the equation becomes S_m^x_1 S_m^x_1

.

But S_m^x_1 S_m^x 1(mod 2). This is not possible.

Possibility 7: For x  1 and y  1, now by the definition (2.1) we get

y  12.If y  12, then this possibility fails by lemma (3.4).

Possibility 8: When

y  1

and

x  1

,

the equation reduces into

m  1  S_m1 S_mwhich implies 11m  1

.

Possibility 9: x  1 and

y  1, This possibility also fails by Binomial

Expansion.

DIOPHANTINE RECTANGLES OVER STAR AND PRONIC NUMBERS 197

Therefore the given equation has no solution.



Theorem 3.3: (27, 1) is the only one ED Rectangle over S_m

.

Proof: By the theorem (3.1) and (3.2), there exists only one (x, y, z) and so

there only one ED Rectangle exists over S_m

.



3.3 Python Programing for Existence of ED Rectangles over S_m: In this

section, we provided the python programming for the existence and non existence of

the ED Rectangles over S_m

1 # ED rectangle over Star number

2 import math

3 def rectangle ():

4 print (’x\ty\tn\tSm\tSn\t(l,b)’)

5

for x in range (0, m + 1) :

6

for y in range (0, m + 1) :

7

for n in range (1, m + 1) :

8

Sm = 6* n **2 + 6* n + 1

9

Sn = 6* n **2 - 6* n + 1

10

11

12

13

l = Sm *(n + 1) + n*(x + y)

b = n*Sn + (n -1) *(x + y)

if (Sm)**x + ( Sn)**x == ( n + 1) **y :

print (x,’\t’, y,’\t’, n,’\t’, Sm ,’\t’, Sn ,’\t’ ,(l, b))

14 m = int ( input (" Enter the maximum range :"))

15 # m is the maximum range

16 rectangle ()

Coding 1: Calculating the solution for S_m^x S_m^x_1 (m  1)^y

Figure 1: Output: Coding 1

198

A. DEEPSHIKA AND J. KANNAN

18 # ED rectangle over Star number

19 import math

20 def rectangle ():

21 print (’x\ty\tn\tSm\tSn\t(l,b)’)

22

23

24

25

26

27

28

29

30

for x in range (0,m + 1) :

for y in range (0,m + 1) :

for n in range (1,m + 1) :

Sm = 6* n **2 + 6* n + 1

Sn = 6* n **2 - 6* n + 1

l = Sm *(n + 1) + n*(x + y)

b = n*Sn + (n -1) *(x + y)

if (Sm)**x - ( Sn)**x == ( n + 1) **y :

print (x,’\t’,y,’\t’,n,’\t’, Sm ,’\t’,Sn ,’\t’ ,(l,b))

31 m = int ( input (" Enter the maximum range :"))

32 #m is the maximum range

33 rectangle ()

Coding 2: Calculating the solution for S_m^x S_m^x_1 (m  1)^y

Figure 2: Output: Coding 2

4. Exponential Diophantine Rectangles over Pronic Numbers

This section includes defintion of Exponential Diophantine Rectangles over

P_mand also contains three subsections. First two subsections provide the theorems

for solving the exponential Diopahntine equations. In the final subsection the python

programming is provided for the existence of Exponential Diophantine Rectangles

over P_mwithin a specific limit.

Definition 4.1: An Exponential Diophantine rectangle over Pronic

numbers P_mis defined as a rectangle with the length (l) and breadth (b) as

(l,b)  (P_m1(m  1)  m(x  y)

,

mP_m (m  1)(x  y))

where m  

and x, y are non-negative integers such that

DIOPHANTINE RECTANGLES OVER STAR AND PRONIC NUMBERS 199

P_m^x_1 P_m^x (m  1)^y

(4)

(5)

or

P_m^x_1 P_m^x (m  1)^y

Notation - Exponential Diophantine Rectangles over Pronic numbers- ED

Rectangles over P_m

.

4.1 The exponential Diophantine equation P_m^x_1 P_m^x (m  1)^y: This

subsections contains the theorem for finding the solution for the exponential

Diophantine equation P_m^x_1 P_m^x (m  1)^y

.

Theorem 4.1: The only solution for the Exponential Diophantine

equation P_m^x_1 P_m^x (m  1)^yare (x, y, m)  {(0,1, 1)} with m   and

x, y  ^ 0}

.

Proof:

Consider

the

exponential

Diophantine

equation

P_m^x_1 P_m^x (m  1)^yand solve this in various possibilities.

Possibility 1: Wheneverx  y  0, there is no possibility.

Possibility 2: Here x  0 and y  1, we obtain m  1

.

Possibility 3: Suppose x  1 and y  0, the equation 4 reduced to 12m² 1

which is an impossible one.

Possibility 4: Now x  y  1, the equation (4) reduces to a quadratic equation

2m² 3m  1  0 and it is not solvabe over



.

Possibility 5: x  0 and y  1 the equation (4) reduces(m  1)^y 2. For

m  1, it becomes y  1 which is contradiction to our

assumption. For m  1, (m  1)^yis always greater than 2. This

possibility fails.

200

Possibility 6: For

A. DEEPSHIKA AND J. KANNAN

y  0 and x  1 the equation (4) reduces to

,

P_m^x_1 P_m^x 1. Thus, P_m^x_1 P_m^x 0 (mod 2). We get a

contradiction.

Possibility 7: Whenever x  1 and y  1, the equation reduced into

2m² 4m  2  (m  1)^y. By using Binomial Expansion we

get y  2. Now for y  2, the equation reduces into a quadratic

equation 2m  1  0 and it is not solvable.

Possibility 8: Here y  1 and x  1, we have m  1  P_{m 1} P_mimplies

1  m² 3m  2. This is an impossible one.

Possibility 9: When x, y  1 Now for m  1, we obtain an impossible one. If

m  1 and by using the definition 2.1, the possibility fails.



The given equation has only one solution.



4.2 The exponential Diophantine equation P_m^x_1 P_m^x (m  1)^y: This

subsection discusses about the solution for the equation P_m^x_1 P_m^x (m  1)^y

.

Theorem 4.2: There is no solution exists for the Exponential

Diophantine

equation

P_m^x_1 P_m^x (m  1)^ywith

m  

and

x, y  ^ {0}

.

Proof: Consider the equation P_m^x_1 P_m^x (m  1)^yand deal with

different possibilities.

Possibility 1: Wheneverx  y  0, this is one fails.

Possibility 2: Suppose x  0 and y  1, we obtainm  1. This is a

contradiction.

Possibility 3: If x  1 and y  0, then m  

.

DIOPHANTINE RECTANGLES OVER STAR AND PRONIC NUMBERS 201

Possibility 4: Forx  y  1, we obtainm  1

.

Possibility 5: When x  0 and y  1, the equation becomes (m  1)^y 0

.

This is impossible.

Possibility 6: Now

y  0 and x  1, we deal with two possibilities. For

m  1, we obtain a contradiction. For m  1, then by lemma

(2.1) P_m1and P_mare obtained as 3 and 2 respectively, which

contradicts.

Possibility 7: For x  1 and y  1, by using Definition (2.1) we obtain y  2

. This is an absurd one.

Possibility 8: Here x  1 and y  1, the equation becomes m  1  2m  2

(not possible).

Possibility 9: x, y  1, When m  1, we get an impossible one and m  1

this possibility fails (by Definition 2.1)

Hence, the Diophantine equation P^x

 P^x (m  1)^yhas no solution.



m

m1

Theorem 4.3: There is no ED Rectangles over Pronic numbers exists.

Proof: By the above two theorems (4.1) and (4.2), we get the side of ED

rectangle over P_mis (13, 0), this is not possible and conclude that there is no ED

rectangles over Pronic numbers.

4.3 Python Programing for Existence of ED Rectangles over P_m: The

Python programming for the existence and nonexistence of the exponential

Diophantine equation solution is provided in this part; however, the ED rectangle

over P_m, does not exist.

³⁴#ED rectangle over Pm

³⁵import math

³⁶def rectangle ():

³⁷print (’x\ty\tm\tPm\tPn\t(l,b)’)

38 for x in range (0, n + 1) :

39 for y in range (0, n + 1) :

202

A. DEEPSHIKA AND J. KANNAN

40 for m in range (1, n + 1) :

41 Pm = (m + 1) *(m + 2)

42 Pn = m*(m - 1)

43 l = Pm *(m + 1) +m*(x + y)

44 b = m*Pn + (m - 1) *(x + y)

45 if (Pm)**x + ( Pn)**x ==( m + 1) **y :

46

print (x,’\t’,y,’\t’,m,’\t’, Pm ,’\t’,Pn ,’\t’ ,(l,b))

⁴⁷n = int ( input (" Enter the maximum range :"))

⁴⁸#n is the maximum range

⁴⁹rectangle ()

Coding 3: Calculating the solution for P_m^x_1 P_m^x (m  1)^y

Figure 3: Output: Coding 3

⁵⁰#ED rectangle over Pm

⁵¹import math

⁵²def rectangle ():

⁵³print (’x\ty\tm\tPm\tPn\t(l,b)’)

54 for x in range (0,n +1) :

55

56

57

58

59

60

61

62

for y in range (0,n +1) :

for m in range (1,n +1) :

Pm =(m +1) *(m +2)

Pn=m*(m -1)

l=Pm *(m +1) +m*(x+y)

b=m*Pn +(m -1) *(x+y)

if (Pm)**x -( Pn)**x == ( m +1) **y :

print (x,’\t’,y,’\t’,m,’\t’, Pm ,’\t’,Pn ,’\t’ ,(l,b))

⁶³n = int ( input (" Enter the maximum range :"))

⁶⁴#n is the maximum range

⁶⁵rectangle ()

Coding 4: Calculating the solution for P_m^x_1 P_m^x (m  1)^y

Figure 4: Output: Coding 4

DIOPHANTINE RECTANGLES OVER STAR AND PRONIC NUMBERS 203

Since the sides of the rectangles are positive, but we have l  13 and b  0 and

therefore does not exists ED rectangle overP_m

.

5. Conclusion

Finally we infer that there exists only one ED Rectangle over Star numbers

and there is no ED Rectangles over Pronic numbers. In the future, this could be

employed in cryptographic concepts like it helps to develop efficient algorithms.

Additionally, it can be applied to furnishings design (making tables, chairs, etc.),

building construction, graphic design (such as creating logos), etc. One can also work

on these topics using various types of equations.

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1, 2. Department of Mathematics,

Ayya Nadar Janaki Ammal College,

(Autonomous, affiliated to Madurai

Kamaraj University, Madurai),

(Received, January 8, 2025)

(Revised, January 20, 2025)

Sivakasi- 626 124, Tamil Nadu, India.

1. E-Mail: deepshi20mar@gmail.com

2. E-Mail: jayram.kannan@gmail.com