Download Solutions of the diophantine equation 2x + py = z2

Int. J. of Mathematical Sciences and Applications,
Vol. 1, No. 3, September 2011
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Solutions of the diophantine equation 2x + py = z 2
Alongkot Suvarnamani
Department of Mathematics, Faculty of Science and Technology,
Rajamangala University of Technology Thanyaburi (RMUTT),
Thanyaburi, Pathum Thani, 12110, Thailand.
E-mail: [email protected] or [email protected]
Abstract
In this paper, we study the diophantine equation 2x + py = z 2 where where p is
a prime number and x, y and z are non-negative integers.
1
Introduction
In 2002, J. Sandor studied two diophantine equations 3x + 3y = 6z and 4x + 18y = 22z .
After that D. Acu (2007) studied the diophantine equation of form 2x + 5y = z 2 . He
found that this equation has exactly two solutions in non-negative integer (x, y, z) ∈
{(3, 0, 3), (2, 1, 3)}. Then A. Suvarnamani, A. Singta and S. Chotchaisthit (2011) found
solutions of two diophantine equations 4x + 7y = z 2 and 4x + 11y = z 2 .
Now, we study the diophantine equation of form
2x + py = z 2
(1),
where p is a prime number and x, y and z are non-negative integers.
2
Main Theorem
From the diophantine equation (1), we have
2x + 2y = z 2
(2),
where p = 2. From the diophantine equation (2), we consider in 3 cases.
2010 Mathematics Subject Classification: 11D61.
Key words and phrases: diophantine equations, exponential equations.
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A. Suvarnamani
case 1: x = y. The diophantine equation (2) becomes
2x+1 = z 2 .
So, z = 2k where k is a non-negative integer. That is x = 2k − 1.
But it is impossible for k = 0. Hence, the solution of the diophantine
equation (2) is (x, y, z) = (2k − 1, 2k − 1, 2k ) where k is a positive integer.
case 2: x > y. The diophantine equation (2) becomes
2y (2x−y + 1) = z 2 .
That is z = a · 2k where k is a non-negative integer and a2 = 2x−y + 1.
So, y = 2k.
Since a2 = 2x−y + 1, we get 2x−y = a2 − 1 = (a − 1)(a + 1) = 2v · 2x−y−v , where
a − 1 = 2v and a + 1 = 2x−y−v , x − y > 2v and v is a non-negative integer. Then we
obtain 2v · (2x−y−2v − 1) = 2.
For v = 0, we get 2x−y − 1 = 2. It is impossible.
For v = 1, we have 2x−y−2 − 1 = 1. So, 2x−y−2 = 2. That is x − y − 2 = 1. Then
we get x = 2k + 3 and a = 3. Hence, the solution of the diophantine equation (1) is
(x, y, z) = (2k + 3, 2k, 3 · 2k ) where k is a non-negative integer.
case 3: x < y. It is similarly with case 2.
Hence, the solution of the diophantine equation (2) is
(x, y, z) = (2k, 2k + 3, 3 · 2k )
where k is a non-negative integer.
In conclusion, we have
Main Theorem 2.1. Let A = {(2k − 1, 2k − 1, 2k )|k is a positive integer.},
B = {(2k + 3, 2k, 3 · 2k )|k is a non-negative integer.} and
C = {(2k, 2k + 3, 3 · 2k )|k is a non-negative integer.}.
The solution of the diophantine equation (2) is (x, y, z) ∈ A ∪ B ∪ C.
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Solutions of the diophantine equation 2x + py = z 2
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Main Theorem 2.2. Consider the diophantine equation (1) where p is a prime number
which is more than 2.
(i) For each prime number p, the diophantine equation (1) has a solution (x, y, z) =
(3, 0, 3).
(ii) For p = 3, the diophantine equation (1) has a solution
(x, y, z) = (4, 2, 5).
(iii) For p = 1 + 2k+1 where k is non-negative integer, the diophantine equation (1) has
a solution (x, y, z) = (2k, 1, 1 + 2k ).
Proof. From the diophantine equation (1), we consider in 2 case.
Case 1: if x is an odd number.
That is x = 2k + 1 where k is a non-negative integer.
The diophantine equation (1) becomes
z 2 − 22k+1 = py
or
1
1
(z − 2k+ 2 )(z + 2k+ 2 ) = py ,
1
1
where z − 2k+ 2 = pu and z + 2k+ 2 = py−u , y > 2u and u is a non-negative integer.
3
3
Then we obtain py−u − pu = 2k+ 2 or pu (py−2u − 1) = 2k+ 2 . Clearly, it is impossible if u > 0.
For u = 0, we get
3
py − 1 = 2k+ 2
or
3
py − 2k+ 2 = 1
(3).
The diophantine equation (3) is a diophantine equation by Catalan’s type ab − cd = 1.
So, it has in positive integer number (> 1) only the solutions p = 3, y = 2 and k + 32 = 3.
But it is impossible.
3
For y = 1, we get p − 2k+ 2 = 1. It is impossible, too.
For y = 0, we get z 2 − 22k+1 = 1 which it is no solution if z = 0 or z = 1. However, it is a diophantine equation by Catalan’s type ab − cd = 1. So, it has in positive
integer number(> 1) only the solutions z = 3 and 2k + 1 = 3. That is k = 1. Hence, a
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solution of the diophantine equation (1) is (x, y, z) = (3, 0, 3).
Case 2: if x is an even number.
That is x = 2k where k is a non-negative integer.
The diophantine equation (1) becomes
z 2 − 22k = py
or
(z − 2k )(z + 2k ) = py ,
where z − 2k = pv and z + 2k = py−v , y > 2v and v is non-negative integer.
Then we obtain py−v − pv = 2k+1 or pv (py−2v − 1) = 2k+1 . Clearly, it is impossible
if v > 0.
For v = 0, we get
py − 1 = 2k+1
or
py − 2k+1 = 1
(4).
The diophantine equation (4) is a diophantine equation by Catalan’s type ab − cd = 1.
So, it has in positive integer number (> 1) only the solutions p = 3, y = 2 and k + 1 = 3.
That is k = 2.
Hence, a solution of the diophantine equation (1) is (x, y, z) = (4, 2, 5) if p = 3.
For y = 1, we get p − 2k+1 = 1 or p = 1 + 2k+1 . Hence, a solution of the diophantine equation (1) is (x, y, z) = (2k, 1, 1 + 2k ) if p = 1 + 2k+1 .
For y = 0, we get z 2 − 22k = 1. It is impossible.
Acknowledgements
I would like to thank the referee(s) for his comments and suggestions on the manuscript.
This work was supported by the Faculty of Sciences and Technology, Rajamangala University of Technology Thanyaburi (RMUTT), Thailand.
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References
[1] D. Acu, On a diophantine equation 2x + 5y = z 2 , General Mathematics, Vol. 15,
No. 4 (2007), 145-148.
[2] M.B. David, Elementary Number Theory, 6th ed., McGraw-Hill, Singapore, 2007.
[3] H.R. Kenneth, Elementary Number Theory and its Application, 4th ed., Addison
Wesley Longman, Inc., 2000.
[4] L.J. Mordell, Diophantine Equations, Academic Press, New York, 1969.
[5] J. Sandor, On a diophantine equation 3x + 3y = 6z , Geometric theorems, Diophantine equations, and arithmetric functions, American Research Press Rehobot
4, 2002, 89-90.
[6] J. Sandor, On a diophantine equation 4x + 18y = 22z , Geometric theorems, Diophantine equations, and arithmetric functions, American Research Press Rehobot
4, 2002, 91-92.
[7] W. Sierpinski, Elementary Theory of Numbers , Warszawa, 1964.
[8] J.H. Silverman, A Friendly Introduction to Number Theory, 2nd ed., Prentice-Hall,
Inc., New Jersey, 2001.
[9] A. Suvarnamani, A. Singta and S. Chotchaisthit On two Diophantine equations
4x + 7y = z 2 and 4x + 11y = z 2 , Science and Technology RMUTT Journal, Vol.
1, No.1 (2011).
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