6th
International Science, Social Sciences, Engineering and Energy Conference
17-19 December, 2014, Prajaktra Design Hotel, Udon Thani, Thailand
I-SEEC 2014
http//iseec2014.udru.ac.th
Solution of the Diophantine equation  2m    2n   z 2
x
y
Oam Sthityanak*, Nangnouy Songkampol, Usaporn Swekwi
Department of Mathematics, Faculty of Science and Technology,
Rajamangala University of Technology Thanyaburi,
Patumthani, Thailand
*
oam_st@hotmail.com
Abstract
In this paper, we give all non-negative integer solutions  x, y, z  of the exponential Diophantine equation
2   2 
m
x
n
y
 z where m and n are two given positive integers. In addition, we provide some useful
2
symbols and criteria for solving the Diophantine equation of form a  b  z .
x
y
2
Keyword: exponential Diophantine equation
1. Introduction
The Diophantine equation a  b  c has been studied by many mathematicians for long time.
This type of equation is known as the exponential Diophantine equation. It is also called the exponential
x
y
z
Diophantine equation for the equation a  b  z . In recent years, many papers have been published to
solve this type of equation in various values of a and b .
x
y
2
In 2007, Acu [1] solved the Diophantine equation 2  5  z . He gave that the equation has
exactly two non-negative integer solutions. Then, in 2011, A. Suvarnnamani, A. Singta and S.
x
y
2
Chotchaisthit [8] studied the Diophantine equation 4  7  z and 4  11  z . Since that, many
mathematicians such as S. Chotchaisthit [3], B. Sroysang [2], B. Peker and S. Cenberci I. [6], and many
other people investigated this type of equation.
x
y
2
x
y
2
2
The inspiration of this work is from the proceeding of A. Suvarnnamani [7], in 2011, that solved
for the non-negative integer solution of 2  2  z . He gave all solutions of this equation in general
form as stated following.
x
y
2
2. Preliminaries
A  {(2k  1, 2k  1, 2 ) | k is a positive integer.} ,
k
Theorem 2.1. [7] Let
B  {(2k  3, 2k , 3·2 ) | k is a non  negative integer.}
k
C  {(2k , 2k  3, 3·2 ) | k is a non  negative integer.}
k
and
The solution of the Diophantine equation 2  2  z is  x, y, z   A  B  C
x
y
2
Here, the Catalan’s conjecture proved by P. Mihailescu [6] is an important tool in our proofs. It
stated as follows.
Theorem 2.2. (Catalan’s conjecture [6])  3, 2, 2, 3 is a unique solution  a, b, x, y  for the Diophantine
x
y
equation a  b  1 where a , b, x and y are integer with min a, b, x, y  1 .
For convenient, we would like to use some special notations in this paper. We note here that
 1, 2, 3,  is the set of natural numbers.
For two given positive integers a and b greater than 1 , we write S  a, b  to denote the set of all
positive integer solutions  x, y, z  of the Diophantine equation a  b  z .
x
That is S  a, b  :
 x, y, z  
3

y
2
|a b  z .
x
y
2
 0, y, z  0  | 1  b  z  , and
 1  z  , and for the case x  0 and y  0
Moreover we define for the case x  0 as S 1, b  :
for the case y  0 as S  a,1 :
as S 1,1 :
 x, 0, z  
0, 0, z  0 0
 0 
|11  z
2
|a
x
y
2
 which is easy to prove that S 1,1   .
Proposition 2.3. Let a , b, m and n be positive integers and a and b are greater than 1.
(1)
S  b, a    y, x, z  |  x, y, z   S  a, b  . It is also valid for a or b is equal to 1.
(2)
S  2,1   3, 0, 3 and S 1, 2    0, 3, 3 .
(3) For a  2 and k  1 , if a  k  1 then S  a,1  1, 0, k , otherwise S  a,1  .
2

(4) If  x, y, z   S a , b
m
n
 then  mx, ny, z   S  a, b  .
x y
(5) If  x, y, z   S  a, b  then  , , z   S a m , bn
m n 

(6)

m
S a ,b
n


t u 
, ,v
m n 
3

 t , u, v   S  a , b 
(7) If a  b  1  mod 3 then S  a, b    .
 if and only if
.
x
m
and
y
n
are integer.
2
3

(8) If S  a, b    then S a , b
m
 .
n
(9) The set of all non-negative solutions of the Diophantine equation a  b  z
S  a, b   S 1, b   S  a,1
x
y
2
is
Proof.
(1) The proof is easy.
(2) The equation 2  1  z can be rewritten as z  2  1 . So, by Catalan’s conjecture, we derive
that  x, z    3, 3 is the only positive solution of the equation. Thus S  2,1   3, 0, 3 . Similarly,
x
2
2
x
S 1, 2   0, 3, 3
(3) Let a  2 and k  1 . Suppose that a  k  1 . Then we consider the equation a  1  z . That is
x
2
 z, a, x    3, 2, 3 where x  1 .
z  a  1 . By Catalan’s conjecture, we get that
2
x
2
Since a  2 , the
equation has no solution. When x  1 , we have a  1  z . That is k  1  1  z . So z  k . Therefore
2
2
2
 x, z   1, k  is the only positive solution of the equation a  1  z . Hence we get
S  a,1  1, 0, k  as desired. Assume that  x, 0, z   S  a,1 . So a  1  z . As above, this equation
x
2
x
2
will have a solution only when x  1 . That is a  z  1 . Since a  2 , z  k for some k  1 . Hence
2
a  k 1.
2

(4) Let  x, y, z   S a , b
m
n
 . Thus  a    b 
m
x
n
y
 z . Then we have a  b  z which means
mx
2
ny
2
 mx, ny, z   S  a, b  .
(5) Let
 x, y , z   S  a , b  .

m
definition of S a , b
n
To prove necessity, assume that 
x y 
m
n
, , z   S a , b . By the
m n 
 ,  x , y , z  
m n
sufficiency, we suppose that
x

and
y
y
and
2
y
m
  b    z
a
x
n
m
y
n
2

are integer as required. To prove
n
are integer. Then consider that a
m
a  b  z . Thus
x
x
. This means
m
n
m
 x, y , z   S  a , b  ,
3

  b   a
x
m
n
y
n
x
 b . Since
y
x y
. Therefore we have  , , z   S a m , bn
m n 


as desired.
(6) By (4) and (5).
(7) Let a  b  1  mod 3 . Assume that S  a, b   . So let
 x, y, z   S  a, b  . Then we have
a  b  1  mod 3 and a  b  z . That is a  b  2  mod 3 . Hence z  2  mod 3 . This leads a
contradiction since z  0  mod 3 or z  1  mod 3 for any positive integer z . Hence S  a, b    .
(8) We will prove by contrapositive. Suppose that S  a , b    . Thus let  x, y, z   S  a , b  .
By (4), we obtain  mx, ny, z   S  a, b  . This means S  a, b    . The proof is completed.
x
x
y
2
y
2
x
y
2
2
m
(9) The proof is obvious.
n
m
n

4
Then we can restate the Theorem 2.1 by using our notation as following proposition.
Proposition 2.4. S  2, 2   S1  S2  S3 where
S1  {(2k  1, 2 k  1, 2 ) | k  },
k
S2  {(2k  3, 2k , 3·2 ) | k  }
k
k
S3  {(2k , 2k  3, 3·2 ) | k  }.
We can easily see that S1 , S 2 and S 3 are disjoint sets so we will call S1 , S 2 and S 3 as classes of
S  2, 2  .
3. Main Results

m
Theorem 3.1. If m and n are positive even integers then S 2 , 2
n
.
Proof. Let m and n be positive even integers. Then we write m  2 s and n  2t for some positive
integers s and t . So 2m  2 2 s  4 s and 2n  22 t  4t . Since 4  1 mod 3 , by Proposition 2.3

m
(7), S 2 , 2
n
  S 4 , 4    .
s

t
Theorem 3.2. Let m be a positive odd integer, n be a positive even integer, and d  gcd  m, n  .
n m
r
( i 1)   
 n

nr  3d m
r

2 d
d
If d | 3 then S  2 , 2    (i  1) 
, (i  1)  , 3  2
 i   where  x1 , y1  is
dm
d
d
 d


m
n
a positive integer solution of mx  ny  d , and 3y1  qm  r , where q 

m
S 2 ,2
n
.

m
Proof. By Proposition 2.3 (6), we get S 2 , 2
n


t u 
, ,v
m n 
3
and 0  r  m . Otherwise

 t, u, v   S  2, 2 
. From
Proposition 2.4, we can see that the value u in the class S 1 and S 3 must be odd which is not divisible by

m
even integer n . So S 2 , 2
n


2k  3 2k
k 
, ,3 2  
m
n

2k  3
3
k

.
2k
. Hence we gain a linear equation mx  ny  3 .
m
n
This linear equation, see [4], will have a solution only when d | 3 , otherwise it has no integer solution,
First, we can have that x 

m
which makes S 2 , 2
n
and y 
.
So let d | 3 . We first solve the linear equation mx  ny  d by using Euclidean Algorithm. This
always give us a solution, called  x0 , y0  , and other solutions can be obtained by substituting integer
l
5


into the formula  x, y    x0 
n
d
l , y0 

d 
m
l  for l 
, see [4, p40]. Since x0 
n
d
l and y0 
m
l have
d
the same trend, if x 0 or y0 is not positive, the formula allows us to choose a new solution such that both
values are positive, say
x , y 
1
1
. Then we multiply the linear equation by
3
so we get
d
 3 x   n  3 y   3 Therefore x  n i  3 x and y  m i  3 y where i is an integer.
 1
1
1
1 
d
d
d
d
d  d 
m
Next it may occur that, when we set i  1 , we can find a positive integer s such that
3
m
3
(i  s )  x1 and
(i  s )  y1 are still positive integers, which means the index i is not suitable.
d
d
d
d
3x
3y
So we will find the largest value of such s . Thus we get s  1  1 and s  1  1 . That
n
m
3x
3y
means s  min 1  1 ,1  1
. Since s
must be the largest positive integer,
n
m
n







.
n
m 
3 x1 3 y1 3d
3x
3y
3d
3  3 


 0 , 1 1  1 1 .
From m  x1   n  y1   3 , we have
. Since
n
m mn
mn
n
m
d  d 
3
y
3
y


 1.
This forces s  1  1  1 
 m 
 m 
s   min 1 
3 x1
3 y1
,1 
Let 3y1  qm  r , where q 
3y1  ms
that
r

d
d
and
d
3 x1  ns
d
3y1
m

nr  3d
dm
n

. Then we can easily show
m
. Now we have k 
d

r
q

nm
3 y1 
 (i  s ) 
 so
2 d
d 

 i   . And then it can be rewritten as
 d


n m
r
 n
nr  3d m
r
 (i 1) d   i   .
m
n
S  2 , 2    (i  1) 
, (i  1)  , 3  2 2 d



dm
d
d
 d



m
S 2 ,2
n

 n
m

and 0  r  m . So
 
i  s  
n m
3 x1 m
3y
, (i  s )  1 , 3  2 2
d d
d
(is )
d
3 y1
d
Theorem 3.3. Let m be a positive even integer , n be a positive odd integer, and d  gcd  m, n  .

m
If d | 3 then S 2 , 2
n
 n
   d (i  1)  d , d (i  1) 

r m
mr  3d

m n
,3 2 2
d
( i 1) 
r
d

dn
a positive integer solution of ny  mx  d , and 3x1  qn  r , where q 

m
S 2 ,2
n
.

 i   where  x1 , y1  is


and 0  r  n . Otherwise
6

Proof. By Proposition 2.3 (1) and Theorem 6.
Theorem 3.4. Let m and n be positive odd integers and d  gcd(m, n) .
 n
T1  
Let
 d
m
 2i  1 ,  2i  1 , 2
mn  2 i 1  d
2d
d


 | i  ,



 1r  3  r  
n  m 
r
r
2i 

 d 
1  3  nr  3d m 
1  3  r
2 d 
2
 n 





  ,3  2  
T2    2i 
,  2i 
 i




d
2
dm
d
2
d
 






where
x , y 
1
1





is a positive integer solution of mx  ny  d , and 3y1  qm  r , where q 
and
0r m,
and

 1r  3  r  
m  n 
r
r
2i 







 d 

1

3

1

3
2
d
2
 n
   r m     mr  3d


T3    2i 
 ,
2i 

,3  2  
 i




d
2
d
d
2
dn
 






where
x , y 
1
is a positive integer solution of ny  mx  d , and
1

0  r  n . Otherwise S 2 , 2

m
If d | 3 then S 2 , 2
n
m
n
.
  T T
1
2

 T3 , otherwise S 2 , 2
m
n
3x1  qn  r , where q 

m
n


t u 
, ,v
m n 
3
and
T .
1
Proof. By Proposition 2.3 (6) and Proposition 2.4,
S 2 ,2





 t , u, v   S
1

 S2  S3 .
Case 1.  t , u, v   S1 . With loss of generality, we assume that m  n  1 .
Then x 
2k  1
,y 
m
Then y 
mx
n


 t , u ,v


m n 
3
,z  k .
n
m
d x
n
d
is odd, l  2i  1 where i 
get
2k  1
. Since y must be integer and
. Therefore we obtain k 
 t , u, v   S

n
d
mn
d
|
m
d
n
l where l 

mn (2i  1)  d
2
 n
 
(2i  1),
m
and then we
2d
mn ( 2 i 1)  d
(2i  1), 2
. Since 2k  1
d
(2i  1)  1
d
d
that this result is also true for any positive odd integers m and n .
1
, x
2d


 i   . We can easily prove


Case 2.  t , u, v   S2 . By mean of the proof of Theorem 3.2, the evenness of n does not matter so that we
can use its result and then adjust the index for a fit value. Thus, let us consider
7

 t , u ,v


m n 
3

n m
r
( j 1)   
 n

nr  3d m
r

2 d
d
|  t , u , v   S 2   ( j  1) 
, ( j  1)  , 3  2
 j  .
dm
d
d
 d


Since  t , u, v   S2 and n is odd,
 1
r
u
must be even. Thus r and j  1 must have the same parity. Hence
n
3
where i  . Then we obtain T2 as desired.
2
Case 3.  t , u, v   S3 . The proof is similar to the case 2.
we set j  1  2i 

Then the conclusion comes easily.
Acknowledgements
I am very thankful to the Faculty of Sciences and Technology, RMUTT for financial support and
I would like to thank my colleagues who encourage me to do this work.
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