EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 16, No. 4, 2023, 2693-2702
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Solutions of some quadratic Diophantine equations

Alanod M. Sibih

Department of Mathematics, Jamoum University College, Umm Al-Qura University, Holly
Makkah 21955, Saudi Arabia

Abstract. Let P (t)±i = t2k±itm be a non square polynomial and Q(t)±i = 4k2t4k−2+i2m2t2m−2±
4imkt2k+m−2 − 4t2k ∓ 4itm − 1 be a polynomial, such that k ≥ 2m and i ∈ {1, 2}. In this paper,
we consider the number of integer solutions of Diophantine equation

E : x2 − P (t)±i y
2 − 2P ′(t)±i x+ 4P (t)±i y +Q(t)±i = 0.

We extend a previous results given by A. Tekcan and A. Chandoul et al. We also derive some
recurrence relations on the integer solutions of a Pell equation.

2020 Mathematics Subject Classifications: 11D45, 11Y65

Key Words and Phrases: Diophantine equation, Pell’s equation, Continued fraction, Quadratic
residue

1. Introduction

Let f(x1, x2, . . . , xn) be a polynomial with integer coefficients in one or more variables.
A Diophantine equation is an algebraic equation

f(x1, x2, . . . , xn) = 0

for which integer solutions are sought.
The problem to be solved is to determine whether or not a given Diophantine equation
has solutions in the domain of integer numbers.
In the case where the Diophantine equation is solvable, there are some natural questions:
∗ ) Is the number of solutions finite or infinite ?
∗∗) Is it possible to determine all solutions ?
In 1900, Hilbert [4] asked for general algorithm to determine, in a finite number of steps,
the solvability of any given Diophantine equation. In other words, he asked if there are
any universal method of solving all Diophantine equations.
Unfortunately, it was proven by Matyasevich, in 1970, that this problem is unsolvable [3].

DOI: https://doi.org/10.29020/nybg.ejpam.v16i4.4940

Email address: amsibih@uqu.edu.sa (Alanod M. Sibih)

https://www.ejpam.com 2693 © 2023 EJPAM All rights reserved.



A. M. Sibih / Eur. J. Pure Appl. Math, 16 (4) (2023), 2693-2702 2694

The absence of a general algorithm was not by itself obstacle to involve more than technique
in solving Diophantine equations. In fact, Diophantine equations can be very creative and
mathematiciens usually have to exhibit creativity to solve these questions.
One of the best-known techniques is that one based on reduction of the Diophantine
equation of arbitrary size with many arbitrary unknowns to another equation having a
fixed degree and fixed number of unknowns.
Another one of the most common techniques used to examine Diophantine equations
problem is that based on considering residues by checking certain common modulos on
each term of the equation, one can either arrive at a contradiction to prove that there’s no
solution, or to find the unique solutions that satisfy the equation. This technique assumes
basic knowledge of modular arithmetic as well as important notions and theorem like the
quadratic residues modulo a prime number p and Euler theorem.
Recently, there are a number of paper have been written and published by A. Tekcan using
the techniques mentioned above.
This paper offers an extension of one of the results given by A. Tekcan [2] and A. Chandoul
et al. [1].
In [2], Tekcan consider the number of integer solutions of Diophantine equation E : x2 −
(t2 − t)y2 − (4t − 2)x + (4t2 − 4t)y = 0 over Z, where t ≥ 2. Then, we assume that the
Diophantine equation E can be extended to the form

E : x2 − P (t)y2 − 2P ′(t)x+ 4P (t)y + (P ′(t))2 − 4P (t)− 1 = 0

where P (t) be a non-square polynomial.
Few later years, Chandoul et al. [1] considered the number of integer solutions of Dio-
phantine equation E1 : x2 − P (t)y2 − 2P ′(t)x + 4P (t)y + P ′(t)2 − 4P (t) − 1 = 0. They
derived some recurrence relations on the integer solutions (xn, yn) of E1 and giving a nice
generaliations of previous results given by Tekcan [2]. These extensions allows us to solve
many types of such equations. We also derive some recurrence relations on the integer
solutions of a Pell equation.
Another advantage of our results is that the procedure can be implemented by computer,
which allows us to obtain all the solutions after the insertion of the coefficients and the
verification of the conditions of the method.

2. Main results

Let P (t) = t2k±itm and Q(t) = 4k2t4k−2+i2m2t2m−2±4imkt2k+m−2−4t2k∓4itm−1,
where k ≥ 2m and i ∈ {1, 2}. We consider the equation

E : x2 − P (t)y2 − 2P ′(t)x+ 4P (t)y +Q(t) = 0 (1)

Theorem 1. Let P (t)±i = t2k ± itm, then the continued fraction of
√
P (t)±i is given as

follow ;



A. M. Sibih / Eur. J. Pure Appl. Math, 16 (4) (2023), 2693-2702 2695

1)
√
P (t)+1 =


[
tk; 2

]
, if t = 1[

tk; 2tk−m, 2tk
]
, if t ≥ 2

2)
√

P (t)−1 =
[
tk; 1, 2tk−m − 2, 1, 2tk − 2

]
, t ≥ 2

3)
√
P (t)+2 =

[
tk; tk−m, 2tk

]
4)

√
P (t)+2 =

[
tk − 1; 1, tk−1 − 2, 1, 2tk−1 − 2

]
, if t ≥ 3

Proof. We have ;

√
P (t)+1 =

√
t2k + tm = tk +

√
t2k + tm − tk

= tk +
1√

t2k + tm + tk

tm

= tk +
1

2tk−m +

√
t2k + tm − tk

tm

= tk +
1

2tk−m +
1√

t2k + tm + tk

= tk +
1

2tk−m +
1

2tk +
√
t2k + tm − tk

Hence,
√
P (t)+1 =


[
tk; 2

]
, if t = 1[

tk; 2tk−m, 2tk
]
, if t ≥ 2

Similarly, one can find the requered form of the continued fractions.

Theorem 2. Let P (t) = t2k ± itm, where k ≥ 2m ̸= 0 and i ∈ {1, 2} be a non square
polynomial and let the Diophantine equation

E : x2 − P (t)y2 − 2P ′(t)x+ 4P (t)y + (P ′(t))2 − 4P (t)− 1 = 0.

Then
(1) The fundamental (minimal) solution of E is (x1, y1) = (u1+2kt2k−1± imtm−1, v1+2)
(2) Define the sequence {(xn, yn)}n≥1 = {(un + 2kt2k−1 ± imtm−1, vn + 2)}, then (xn, yn)
is a solution of E. So it has infinitely many integer solutions (xn, yn) ∈ Z× Z.



A. M. Sibih / Eur. J. Pure Appl. Math, 16 (4) (2023), 2693-2702 2696

(3) The solutions (xn, yn) satisfy the recurrence relations
xk = u1xk−1 + (a0u1 + α)yn−1 − u1(2a0 + 2kt2k−1 ± imtm−1)− 2α+ 2kt2k−1 ± imtm−1

yk = v1xk−1 + (a0v1 + β)yn−1 − v1(2a0 + 2kt2k−1 ± imtm−1)− 2β + 2

for k ≥ 2.

Theorem 3. Let P (t) = t2k + i, where k ̸= 0 and i ∈ {−2,−1, 1, 2} be a non square
polynomial and let the Diophantine equation

E : x2 − P (t)y2 − 2P ′(t)x+ 4P (t)y + (P ′(t))2 − 4P (t)− 1 = 0.

Then
(1) The fundamental (minimal) solution of E is (x1, y1) = (u1+2kt2k−1± imtm−1, v1+2)
(2) Define the sequence {(xn, yn)}n≥1 = {(un + 2kt2k−1 ± imtm−1, vn + 2)}, then (xn, yn)
is a solution of E. So it has infinitely many integer solutions (xn, yn) ∈ Z× Z.
(3) The solutions (xn, yn) satisfy the recurrence relations

xk = u1xk−1 + (a0u1 + α)yn−1 − u1(2a0 + 2kt2k−1 ± imtm−1)− 2α+ 2kt2k−1 ± imtm−1

yk = v1xk−1 + (a0v1 + β)yn−1 − v1(2a0 + 2kt2k−1 ± imtm−1)− 2β + 2

if i ̸= 0 for k ≥ 2.

Here, we show that: if P is a non perfect square polynomial, then (1) has an infinitude
of integer solutions. In this case we find a closed expression (xn, yn), the general positive
integer solution, by an original method.
Note that the resolution of E in its present form is difficult, that is, we can not determine
how many solutions E has and what they are. So, we have to transform E into a Pell
equation which can be easily solved. To get this let

T :

{
x = u+ P ′(t) = u+ 2kt2k−1 ± imtm−1

y = v + 2
(2)

we get, T (E) := Ẽ, such that

Ẽ : (u+ 2kt2k−1 ± imtm−1)2 − (t2k ± itm)(v + 2)2 − (4kt2k−1 ± 2imtm−1)
(u+ 2kt2k−1 ± imtm−1) + (4t2k ± 4itm)(v + 2) + 4k2t4k−2 + i2m2t2m−2

±4imkt2k+m−2 − 4t2k ∓ 4itm − 1

Then, the equation (1) becomes

Ẽ : u2 − (t2k ± itm)v2 = 1 (3)



A. M. Sibih / Eur. J. Pure Appl. Math, 16 (4) (2023), 2693-2702 2697

which is a Pell equation.
It is known that the above Pell equation is always solvable. Its solutions are related to
the continued fraction expansion of

√
P (t).

We will be concerned with the continued fraction expansions of
√
P (t), where P (t) is

a non-square. In fact, this continued fractions have a very interesting form, which is
summarized in the next theorem.

Theorem 4. Let P (t) be a prime. Then
√
P (t) = [a0; a1, a2, · · · , al, 2a0], where the

repeating portion, excluding the last term, is symmetric upon reversal, and the central
term may appear either once or twice.

Theorem 5. Let
√

P (t) =
[
a0; a1, a2, · · · , al, 2a0

]
denote the continued fraction expansion

of period lenght l, where P (t) be a non-square polynomial.

Let
pn
qn

be the nth convergent of
√
P (t). Then

(1) The fundamental solution of the Pell equation Ẽ in (3) is (u1, v1), such that
u1 = pl−1

, if l is even,
v1 = ql−1

and


u1 = p2l−1

, if l is even
v1 = q2l−1

Set {(uk, vk)} = {(pkl−1, qkl−1)} where

pkl−1

qkl−1
=

a0; a1, a2, · · · , al,︸ ︷︷ ︸
l−1

2a0, a1, a2, · · · , al, 2a0, a1, a2, · · · , al︸ ︷︷ ︸
(k−1)l−1

, if l is even.

And
p2kl−1

q2kl−1
=

a0; a1, · · · , al, 2a0, a1, · · · , al︸ ︷︷ ︸
2l−1

, 2a0, a1, · · · , al, 2a0, · · · , a1, · · · , al,︸ ︷︷ ︸
(2k−2)l−1

 ,

if l is odd. Then (uk, vk) is a solution of Ẽ.

(2) The consecutive solutions (uk−1, vk−1) and (uk, vk) the Pell equation Ẽ in (3) sat-
isfy 

uk = u1uk−1 + (a0u1 + α)vk−1

, forall k ≥ 2, if l is even
vk = v1uk−1 + (a0v1 + β)vk−1

where α = xl−2 and β = xl−2.
and 

uk = u1uk−1 + (a0u1 + η)vk−1

, forall k ≥ 2, if l is odd
vk = v1uk−1 + (a0v1 + δ)vk−1



A. M. Sibih / Eur. J. Pure Appl. Math, 16 (4) (2023), 2693-2702 2698

where η = x2l−2 and δ = x2l−2.

To prove this theorem, we need the following Lemma

Lemma 1. Let
√

P (t) =
[
a0; a1, a2, · · · , al, 2a0

]
denote the continued fraction expansion

of period lenght l. Then {
a0xkl−1 + xkl−2 = P (t)ykl−1

a0ykl−1 + ykl−2 = xkl−1

for all k ≥ 2.

Proof. (Lemma 1)
We have

√
P (t) =

[
a0; a1, a2, · · · , a1, 2a0

]
.

Thus, we may write
√
P (t) =

[
a0; a1, a2, · · · , akl−1, a0 +

√
P (t)

]
, then

√
P (t) =

(a0 +
√

P (t))xkl−1 + xkl−2

(a0 +
√

P (t))ykl−1 + ykl−2

, which gives rise to the equation

P (t)ykl−1 +
√
P (t)(a0ykl−1 + ykl−2) = (a0xkl−1 + xkl−2) +

√
P (t)xkl−1.

Which yields, a0xkl−1 + xkl−2 = P (t)ykl−1 and a0ykl−1 + ykl−2 = xkl−1.

Proof. (Theorem 4)
(1) We prove the theorem only for even number l. It is easily seen that x2kl−1−P (t)y2kl−1 =

xkl−1ykl−1 − ykl−1xkl−2. Then x2kl−1 − P (t)y2kl−1 = (−1)kl. Thus, if l is even x2kl−1 −
P (t)y2kl−1 = 1 which yields (uk, vk) are solutions of Ẽ for all k ≥ 1 and (u1, v1) is the
fundamental solution.
We can also prove it using the method of mathematical induction. In fact, if l is even, we
have



A. M. Sibih / Eur. J. Pure Appl. Math, 16 (4) (2023), 2693-2702 2699

uk
vk

=
xkl−1

ykl−1
=

a0; a1, a2, · · · , a1,︸ ︷︷ ︸
l−1

2a0, a1, a2, · · · , a1, 2a0, · · · , a1, a2, · · · , a1︸ ︷︷ ︸
(k−1)l−1



=

a0; a1, a2, · · · , a1,︸ ︷︷ ︸
l−1

a0 + a0, a1, a2, · · · , a1, 2a0, · · · , a1, · · · , a1︸ ︷︷ ︸
(k−1)l−1



=

a0; a1, a2, · · · , a1,︸ ︷︷ ︸
l−1

a0 +
x(k−1)l−1

y(k−1)l−1



=

(
a0 +

x(k−1)l−1

y(k−1)l−1

)
xl−1 + xl−2(

a0 +
x(k−1)l−1

y(k−1)l−1

)
yl−1 + yl−2

=
a0y(k−1)l−1xl−1 + x(k−1)l−1xl−1 + y(k−1)l−1xl−2

a0y(k−1)l−1yl−1 + x(k−1)l−1yl−1 + y(k−1)l−1yl−2

.

Then

u2k − P (t)v2k = (a0y(k−1)l−1xl−1 + x(k−1)l−1xl−1 + y(k−1)l−1xl−2)
2

−P (t)(a0y(k−1)l−1yl−1 + x(k−1)l−1yl−1 + y(k−1)l−1yl−2)
2

= (u1uk−1 + (a0u1 + α)vk−1)
2 − P (t)(v1uk−1 + (a0v1 + β)vk−1)

2

= u21u
2
k−1 + 2u1(a0u1 + α)uk−1vk−1 + (a0u1 + α)2v2k−1

−P (t)v21u
2
k−1 − 2P (t)(a0v1 + β)v1uk−1vk−1 − P (t)(a0v1 + β)2v2k−1

= (u21 − P (t)v21)u
2
k−1 −

[
(P (t)(a0v1 + β)2 − (a0u1 + α)2

]
v2k−1

+2 [u1(a0u1 + α)− P (t)v1(a0v1 + β)]uk−1vk−1

Using the above lemma, we have
(P (t)(a0v1 + β)2 − (a0u1 + α)2 = P (t)u21 − P (t)2v21 = P (t)(u21 − P (t)v21) = P (t) and
u1(a0u1 + α)− P (t)v1(a0v1 + β) = 0. Hence, we conclude that

u2k − P (t)v2k = u2k−1 − P (t)v2k−1 = 1

So (uk, vk) is also solution of Ẽ. Completing the proof.
(2) This assertion is clear by the above.

As we reported above, the Diophantine equation E could be transformed into the
Diophantine equation Ẽ via the transformation T. Also, we showed that x = u + P ′(t)
and y = v + 2. So, we can retransfer all results from Ẽ to E by applying the inverse of T.
Thus, we can give the following main theorem

Theorem 6. Let D be the Diophantine equation in (1). Then
(1) The fundamental (minimal) solution of E is (x1, y1) = (u1 + P ′(t), v1 + 2)



A. M. Sibih / Eur. J. Pure Appl. Math, 16 (4) (2023), 2693-2702 2700

(2) Define the sequence {(xn, yn)}n≥1 = {(un + P ′(t), vn + 2)}, where {(xn, yn)} defined
in (3). Then (xn, yn) is a solution of E. So it has infinitely many integer solutions
(xn, yn) ∈ Z× Z.
(3) The solutions (xn, yn) satisfy the recurrence relations

xk = u1xk−1 + (a0u1 + α)yn−1 − u1(2a0 + P ′(t))− 2α+ P ′(t)
, if l is even

yk = v1xk−1 + (a0v1 + β)yn−1 − v1(2a0 + P ′(t))− 2β + 2

for k ≥ 2,and
xk = u1xk−1 + (a0u1 + η)yn−1 − u1(2a0 + P ′(t))− 2η + P ′(t)

, if l is odd
yk = v1xk−1 + (a0v1 + δ)yn−1 − v1(2a0 + P ′(t))− 2δ + 2

for k ≥ 2.

As an application, we can give the following examples:

Example 1. Let P (t) = t4 + 4t3 + 6t2 + 4t+ 2, Then√
P (t) =

[
t2 + 2t+ 1; 2t2 + 4t+ 2

]
.

So, (u1, v1) = (2t4 + 8t3 + 4t2 + 3, 2t2 + 4t+ 2) is the fundamental solution of

Ẽ1 : u2 − (t4 + 4t3 + 2t2 + 2)v2 = 1

and the other solutions are given by


uk = (2t4 + 8t3 + 4t2 + 3)uk−1 + (2t6 + 12t5 + 30t4 + 40t3 + 32t2 + 16t+ 4)vk−1

vk = (2t2 + 4t+ 2)uk−1 + (2t4 + 8t3 + 4t2 + 3)vk−1

For k ≥ 2.

Morover, let n = t2 + 2t+ 1, then P (t) become D(n) = n2 + 1. Then√
D(n) =

[
n; 2n

]
.

So, (u1, v1) = (2n2 + 1, 2n) is the fundamental solution of

Ẽ1 : u2 − (n2 + 1)v2 = 1

and the other solutions are given by



A. M. Sibih / Eur. J. Pure Appl. Math, 16 (4) (2023), 2693-2702 2701


uk = (2n2 + 1)uk−1 + (2n3 + 2n)vk−1

vk = 2nuk−1 + (2n2 + 1)vk−1

For k ≥ 2.
Then the fundamental solution of

E1 : x2 − (n2 + 1)y2 − 4nx+ (4n2 + 4)y − 2 = 0

is (x1, y1) = (2n2 + 2n+ 1, 2n+ 2) and the other solutions are given, for k ≥ 2, by
xk = (2n2 + 1)xk−1 + (2n3 + 2n)yk−1 − 8n3 − 4n

yk = 2nxk−1 + (2n2 + 1)yk−1 − 8n2 + 2n− 2.

Further, for t = 1, P (t) = 17. Then √
P (t) =

[
4; 8

]
.

So, (u1, v1) = (33, 8) is the fundamental solution of

Ẽ1 : u2 − 17v2 = 1

and the other solutions are given by
uk = 33uk−1 + 136vk−1

vk = 8uk−1 + 33vk−1

For k ≥ 2.

Then the fundamental solution of

E1 : x2 − 17y2 − 64x+ 68y + 955 = 0

is (x1, y1) = (65, 10) and the other solutions are given, for k ≥ 2, by
xk = 33xk−1 + 136yk−1 − 1296

yk = 8xk−1 + 33yk−1 − 320.

Example 2. In this example, we consider the number of integer solutions of the Diophan-
tine equation

E : x2 − (t2 + t)y2 − (4t+ 2)x+ (4t2 + 4t)y = 0

We have P (t) = t2+ t, thus P ′(t) = 2t+1 and the continued fraction expansion of
√

P (t)
is



REFERENCES 2702

√
P (t) =

[
t; 2, 2t

]
which yields,

u1
v1

= [t; 2] =
2t+ 1

2
. Then the fundamental solution of E is (x1, y1) =

(2t+ 1 + P ′(t), 2 + 2) = (4t+ 2, 4) and the other solutions are given by
xk = (2t+ 1)xk−1 + (2t2 + 2t)yk−1 − 8t2 − 6t

for, k ≥ 2
yk = 2xk−1 + (2t+ 1)yk−1 − 8t− 2

Acknowledgements

The authors would like to thank the Deanship of Scientific Research at Umm Al-Qura
University for supporting this work by Grant Code: (22UQU 4350388DSR01)

References

[1] Chandoul, A., Marques, D., Albrbar, S. S, The Quadratic Diophantine Equations
x2 − P (t)y2 − 2P ′(t)x + 4P (t)y + P ′(t)2 − 4P (t) − 1 = 0, Journal of Mathematics
Research, 11(2), (2019), 30-38 .

[2] A. Tekcan, Quadratic Diophantine Equation x2−(t2−t)y2−(4t−2)x+(4t2−4t)y = 0
, Bull. Malays. Math. Sci. Soc, (2)33 (2) (2010), 273-280.

[3] Y. V. Matiyasevich, Solution of the tenth problem of Hilbert, Mat. Lapok, 21: (1970)
83-87.

[4] David Hilbert, Mathematische Probleme. Vortrag, gehalten auf dem Internationalen
Mathematiker Kongress zu Paris 1900, Nachr. K. Ges. Wiss., G. Ottingen, Math.-
Phys.Kl, (1900), 253-297.