Advances in Pure Mathematics, 2011, 1, 155-159
doi:10.4236/apm.2011.14028 Published Online July 2011 (http://www.SciRP.org/journal/apm)
155
On Polynomials Solutions of Quadratic
Diophantine Equations
Amara Chandoul
Institut Supérieure d’Informatique et de Multimedia de Sfax, Sfax, Tunisia
Received April 11, 2011; revised May 14, 2011; accepted May 25, 2011
Abstract
Let be a polynomial in

:PPt




\0,1.X In this paper, we consider the number of polynomial
solutions of Diophantine equation






2
:4244=0EXP PYPXPPY
22 2
 . We also obtain some
formulas and recurrence relations on the polynomial solution


,
nn
X
Y.E
=1ax by
22
bxyydx ey
22
=
of
Keywords: Polynomial Solutions, Pell’s Equation, Diophantine Equation
1. Introduction
A Diophantine equation is an indeterminate polynomial
equation that allows the variables to be integers only.
Diophantine problems have fewer equations than un-
known variables and involve finding integers that work
correctly for all equations. In more technical language,
they define an algebraic curve, algebraic surface or more
general object, and ask about the lattice points on it. The
word Diophantine refers to the Hellenistic mathematician
of the 3rd century, Diophantus of Alexandria, who made
a study of such equations and was one of the first
mathematicians to introduce symbolism into algebra. The
mathematical study of Diophantine problems Diophantus
initiated is now called Diophantine analysis. A linear
Diophantine equation is an equation between two sums
of monomials of degree zero or one. While individual
equations present a kind of puzzle and have been
considered th rough ou t h istory, th e f ormulat ion of g en eral
theories of Diophantine equations was an achievement of
the twentieth century. For example, the equation
is known the linear Diophantine equation. In
general, the Diophantine equation is the equation given
by
=0fax c
The equation
x
DyND, with given integers
and and unknowns
N
x
and , is called Pell’s
equation If is negative, it can have only a finite
number of solutions. If is a perfect square, say
, the equation reduces to

y

=
D
2
=Da D
x
ayx ayN
1D
and again there is only a finite number of solutions. The
most interesting case of the equation arises when

be a positive non-square.
Although J. Pell contributed very little to the analysis
of the equation, it bears his name because of a mistake
by Euler.
Pell’s equation was solved by Lagrange
in terms of simple continued fractions. Lagrange was the
first to prove that has infinitly many
solutions in integers if is a fixed positive
non-square integer. If the lenght of the periode of
22
=1xDy
22
=1xDy1DD
l21
=vk xP is , all positive solutions are given by

and
21vk
=yQ

if k is odd, and by 1vk
x
=P=yQ
 and 1vk

if is even, where and
k=1,2,v nn
denotes
the nth convergent of the continued fraction expansion of
PQ
.D=xP (2 1)(1)
=,
vk
yQ

: =1,2,yv
Incidentally, (2 1)(1)vk and
v 22
= , are the positive solutions of xDy

1


l22
=1xDy
,:=1,2,
vv
xyv
provided that is odd.
There is no solution of other than

 given by

11
=
v
vv
x
Dyx Dy
,
,
where 11
x
y is the least positive solution called the
fundamental solution, which there are different method
for finding it. The reader can find many references in the
subject in th e book [1].
We recall that there are many papers in which are
considered different types of Pell's equation. Many
authors such as Tekcan [2], Kaplan and Williams [3],
Matthews [4], Mollin, Poorten and Williams [5],
Stevenhagen [6] and the others consider eome specific
Pell equations and their integer solutions. In [2,7], It

156
22
=4xDy
A. CHANDOUL
considered the equation


xD
22
and the equa-
tion and he obtained some formulas for
its integer solutions. He mentioned two conjecture which
was proved by A. S. Shabani [8]. In [9], we extend the
work in [2,7] by considering the Pell equation
22
=9,y
2
=
x
Dy
kkD1 
22
=1PDQ
DP Q
DD

PQ

1,

Dt
1
[ ,...,]
when be a positive non-square
and , we obtain some formulas for its integer
solutions.
2
In [10], A. Dubickas and J. Steuding are interested in
the polynomial solutions of the Pell equation given by
where is a fixed polynomial, and are
polynomials in the same variables as and with
coefficients in the same field or ring as those of . The
solutions and are called trivial.
All other solutions are called non-trivial. The main
difficulty in solving polynomial Pell equations is to
determine whether non-trivial solutions exist or not. In
case, if there is at least one non-trivial solution, all
solutions are obtained as powers of the smallest non-trivial
solution. They prove this for polynomials in one
variable with coefficients in The proof is purely
algebraic and extends without change to arbitrary
polynomials in several variables

,=1,

0

0

s
Dt over
every field of characteristic t


2
4 4 =0tty

2
16 0t

:PPt[ ]\{0,1}.X
 
42
=0
Pt X

,
n
2.
In [11,12], the number of integer solutions of Dio-
phantine equation and
Diophantine equation over
is considered, where


22 2
42xttyt x 


22 2
16 4xttyt x 
2.t16 =ty
2. Main Results
Let be a polynomial in In this
paper, we consider the number of polynomial solutions
of Diophantine equation
 

 

2
22
2
:
44
E XPtPtY
PtPt Y

 (1)
We also obtain some formulas and recurrence relations
on the polynomial solution n
X
Y.E
E
E
E
=
=
of
Note that the resolution of in its present form is
difficult, that is, we can not determine how many
solutions has and what they are. So, we have to
transform into an appropriate Diophantine equation
which can be easily solved. To get this let
:
X
UH
YVK
T



 (2)
be a translation for some
H
and
By applying the transformation to we get T,E
 



   


22 2
2
::
424 4
=0
TEE UHPtPtVK
PtUHPtPtV K
 
 


(3)

In (3), we obtain

224UH Pt
 
and




22
2244VKPt KPtPt Pt So we get


=2 1HPt and Consequently for =2.K



=2 1XU Pt

= 2,YV
 
and we have the Dio-
phantine equation


2
22
:1EUPtPt V
.
K


 (4)
which is a Pell equation.

Now, we try to find all polynomial so lutions

,UV
nn
of


TE and then we can retransfer all results from


TE E.T
E

E

to by using the inverse of
Theorem 2.1: Let be the Diophantin e equation in
(3), then
1) The fundamental solution of is




11
,=2 1,2UV Pt

2) Define the sequence

,
nn
UV



1
1
1
21
1
21
2
212 2,2
221
n
n
n
UPt
V
UU
PtPt Ptn
VV
Pt










  
by

 
 





(5)


,UV E

Then nn
3) The solutions
is a solution of


,
nn
UV





2
11
11
=2 122
=22 1
nn n
nn n
UPtUPtPtV
VUPtV
satisfy the recurrence
relations




 




2n
(6)
for


4) The solutions ,
nn
UV

satisfy the recurrence
relations

 
  
12 3
12 3
=4 3
=4 3
nnnn
nnnn
UPtUUU
VPt VVU
 
 





4nn
(7)
for


,
nn
UV can be given by 5) The -th solution

A. CHANDOUL
157
 
1
=1;2,22, ,2,2
n
nn times
UPt PtPt
V


 2,2 ,1
n






 
2 1,2Pt
E



2
2=1Pt
=1n
 
1,2
,n


2
=1.
n
Pt V
1.n







1
1
1
1
1
2
1
1
2
1
2
1
2
1
n
n
n
n
nn
nn
U
Pt
V
U
Pt
V
U
V
Pt V
V



































2
2
2
1
1
n
nn
Pt V
Pt V







11
,
nn
UV
 E


(8)
Proof.
1) It is easily seen that is the
fundamental polynomial solution of since

11
,=UV
  
22
21Pt Pt
2) We prove it using the method of mathematical
induction. Let , by (5) we get

,=UV

2Pt
11 which is the fundamental solu-
tion and so is a so lution of . Now, we assume that the
Diophantine equation (4) is satisfied for that is
E



2
2
:
n
EU Pt
 We try to show that this
equation is also satisfied for Applying (5), we
find that
 

 

 

 

  

2
1
1
2
2
2
2
212
22
212 2
22
212
22
212
=22
212
=22
n
n
UPt Pt
VPt
PtPt Pt
Pt
Pt Pt
Pt
PtPt Pt
Pt
Pt UPt
UPt


 

 















(9)
Hence, we conclude that
 




 

 

2
22
11
2
2
2
22
=21 2
22
==
nn
n
nn
UPtPtV
PtU Pt
Pt Pt U
UPtPtV



 

So is also solution of
3) Using (9), we find that




 11
2n tPtV


2n
=4,n
 
2
11
=2 12
=22 1
nn
nn n
UPtU P
VUPtV






for
4) We prove it using the method of mathematical
induction. For we get


 
 
1
2
2
22
3
=2 1
=881
=324818 1
UPt
UPt Pt
UPtPt Pt



 
432
4=128256160321UPtPtPtPt
and


 
 
 
 
 
 


 

432
4
32
22 2
32 1
128256160321
43 324010
21
=4 3
3248181 881
21
=4 3
UPtPt PtPt
PtPtPt Pt
Pt
Pt
PtPtPtPtPt
Pt
PtUUU






Hence










 
12 3
=4 3
nnnn
UPtUUU
So



=4.n
,n
is satisfied
for Let us assume that this relation is satisfied for
that is,


  
12 3
=4 3
nnnn
UPtUUU
 
 (10)
Then using (9) and (10), we conclude that





112
=4 3
nnnn
UPtUUU



completing the proof.
Similarly, we prove that


 
12 3
=4 3,4
nnnn
VtVVVn
 


=1n
5) We prove it using the method of mathematical
induction. For , we have





1
1
21221 1
==1
22 2
=1;2
Pt Pt
UPt
V
Pt





E

which is the fundamental solution of . Let us assume
that the n-th solution


,
nn
UV is given by
 
1
=1;2,2 2,,2,2 2,2
n
nn times
UPt PtPt
V




 





11
,.
nn
Uy


and we show that it hold s for
U
sing (6) , we have

A. CHANDOUL
C APM
158
   




     


 


2
1
1
212 221211
22 122 1
1
11
21
nn
nn n
n
nnn nn
n
n
PtUPtPt VPtU UPtPtV
U
VUPtV UPtV
Pt
U
Pt V


  

 



1
n
Pt V 
as
opyright © 2011 SciRes.
 

11
1
1
211
22
2
2()2
1
1
22() 2
n
Pt Pt Pt
V
Pt
Pt
Pt
 









1
111
2(
) 22
n
U
Pt 
we get




 
1
1
11
21
22 1
222
=1;2,2 2,,2,2 2
n
nt imes
Pt
V
Pt
Pt
Pt Pt Pt








 

1
111
22
2
,2
n
U
Pt






E
E

.
could be transformed into the Diophantine equation
via the transformation T Also, we showed that


=2 1XU Pt

=2.
E
E.T
D
E
and YV So, we can retransfer
all results from to by applying the inverse of
Thus, we can give the following mai n theorem:
Theorem 2.2: Let be the Diophantine equation in
(1). Then
1) The fundamental (minimal) solution of is
completing the proof.
As we reported above, the Diophantine equation




11
,=4 2,4XYPt

2) Define the sequence







1
,=21,2
nn nn
n
XYU PtV

where


 
,
nn
Y

,
nn
defined in (5). Then
X
X
Y
.E
is a
solution of So it has infinitely many integer
solutions


,.
nn
XY 

3) The solutions


,
nn
X
Y satisfy the recurrence
relations (see (11))
4) The solutions


,
nn
X
Y satisfy the recurrence
relations (see (12))


=2 1,Pt t Then Example 2.3: Let




11
,=41,2UV t is the fundamental solution of


22 2
:42=1EUt tV

2
2
2
2
2
4132161
418 4
=2
241 12864 5
Uttt
ttt
Vttt

 
  


 






232
2
3
2
3
4 125619236 1
418 4
2
241 12864 5
Utttt
ttt
Vttt

 
  



 









 
 
22
11
11
=2 1228102
=228 6
nn n
nn n
XPtXPtPtYPtPt
YXPY Pt


 

 2
    

 
2
12 3
123
=4 316248
=4316 16
nnnn
nnnn
XPtX XXPtPt
YPYYY Pt
 
 





4
and some other solutions are


1t




for n (11)

t for n (12)

A. CHANDOUL
159
2
41
260 1
7
tt
t









2
5292 1
9
tt
t






3
2
4
4
43
32
41
8 4
2
241
20482048 63
1024768 156
Ut
ttt
Vt
tt
tt
  

 







3
2
5
5
543
43 2
41
418 4
2
241
16192204808896 15
8096 8192 2656 304
Ut
ttt
Vt
ttt
ttt
 
  

 

 



Further

1
1
=2
U
V
41
;2= 2
t
t

2
3216 1
16 4
tt
Vt


2
2
=2 ;2,4 ,2=
Utt

32
19236 1
64 5
ttt
tt
32
3
256
=2;2,4 ,2,4 ,2=128
Uttt
V




4
4
43
32
=2 ;2,4 ,2,4 ,2,4 ,2
2048 2048
=1024 768
tttt
V
tt
tt


2
63260 1
1567
U
tt
t



5
2
2
55292 1
304 9
U
tt
tt





22 22
:42 42168=0EXttYtXt tY 
2
232202
16 6
Xtt
t

 



32
25619240 2
64 7
ttt
tt





432
04863264 2
156 9
tttt
t





32
2
8961552962
2656304 11
ttt
tt





22
xD
22
xm
5
54 3
43
=2;2,4 ,2,4 ,2,4 ,2,4 ,2
16192204808896 1
=8096 8192 2656
ttttt
U
ttt
tt


It can be concluded now, that the fundamental solution
of
is Some other solutions are

82,4.t
2
Y


3
X



2
3128
Y

42048 2
X



32
41024 768
Ytt


54
5
43
1619220480 8
8096 8192
Xtt
Xtt
 

 
5 
3. Acknowledgements
We would like to thank Saäd Chandoul and Massöuda
Loörayed for helpful discussions and many remarks.
4. References
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22
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
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




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


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