The Polynomial Solutions of Quadratic Diophantine Equation X 2 − p ( t ) Y 2 + 2 K ( t ) X + 2 p ( t ) L ( t ) Y = 0

In this study, we consider the number of polynomial solutions of the Pell equation x 2 − p ( t ) y 2 � 2 is formulated for a nonsquare polynomial p ( t ) using the polynomial solutions of the Pell equation x 2 − p ( t ) y 2 � 1. Moreover, a recurrence relation on the polynomial solutions of the Pell equation x 2 − p ( t ) y 2 � 2. Then, we consider the number of polynomial solutions of Diophantine equation E : X 2 − p ( t ) Y 2 + 2 K ( t ) X + 2 p ( t ) L ( t ) Y � 0. We also obtain some formulas and recurrence relations on the polynomial solution ( X n , Y n


Introduction
A Diophantine equation is an indeterminate polynomial equation that allows the variables to be integers only. Diophantine problems have fewer equations with unknown variables and involve finding integers that work correctly for all equations. e equation ax + by � 1 is known as the liner Diophantine equation. In general, the Diophantine equation is the equation given by ax 2 + bxy + cy 2 + dx + ey + f � 0. (1) e equation x 2 − Dy 2 � N, with given integers D and N and unknowns x and y, is called Pell's equation. e most interesting case of the equation arises when D ≠ 1 be a positive nonsquare. Pell's equation x 2 − Dy 2 � 1 was solved by Lagrange in terms of simple continued fractions. We recall that there are many studies in which there are different types of Pell's equation. Many authors such as Tekcan [1], Matthews [2], Chandoul [3], and Li [4] have researched. In [5], the equation x 2 − Dy 2 � 2 was considered, and some formulas of its integer solutions were obtained. In [6,7], the number of integer solutions of Diophantine equation [3,8], the number of polynomial solutions of Diophantine equation and Diophantine equation

Preliminaries
In this section, we introduce the objects we need later and collect some important facts about them.

New Results
Our principal result is the following.

Proof
(1) We prove it using the method of mathematical induction. Let n � 1, and we get (x 1 , y 1 ) � (a(t), b(t)), which is the fundamental solution of equation (8). Now, we assume that (9) is satisfied for n, that is, We try to show that this equation is also satisfied for n + 1. Applying (9), we find that Hence, we conclude that So, (x n+1 , y n+1 ) is also solution of equation (8).
(2) Using (13), we find that For n ≥ 2. (3) We prove it using the method of mathematical induction. For n � 4, we get 2

Journal of Mathematics
Hence, Let us assume that this relation is satisfied for n, that is, then, using (13) and (18), we conclude that completing the proof. Similarly, we prove that Now, we give a relation between (K(t), L(t)) and (a(t), b(t)). □ Theorem 3. If (K(t), L(t)) be the fundamental solution of the equation then (K(t) 2 − 1, K(t)L(t)) be a solution of the equation (8) and Proof. Hence, it is easily seen that since (K(t), L(t)) is the fundamental solution of the equation , L(t)) be the fundamental solution of the equation (21) and (x 1 , y 1 ) � (a(t), b(t)) be the fundamental solution of the equation (8). en, (3) e solution (U n , V n ) satisfies the recurrence relations (4) e solution (U n , V n ) satisfies the recurrence relations

Journal of Mathematics 3
Hence, it is easily seen that Since en, en, On the other hand, by using (10) and (32), we get

)a(t) + p(t)L(t)b(t)) + y n− 1 (K(t)b(t) + p(t)L(t)a(t))
x n− 1 (L(t)a(t) + b(t)K(t)) + y n− 1 (b(t)L(t)p(t) + a(t)K(t)) . (33) Applying (32) and (33), we find Similarly, we prove that (3) From (25), we get Similarly, we prove that (4) From (22) and (26), we get Similarly, we prove that and we consider the number of polynomial solutions of Diophantine equation. Now, where (K(t), L(t)) is the fundamental solution of equation (21). We have to transform E into an appropriate Diophantine equation which can be easily solved. To get this, let T: x � U + M, be a translation for some M and N. By applying the transformation T to E, we get In (42), we obtain U(2M − 2K(t)) and V(− 2p(t)N + 2p(t)L(t)). So we get M � K(t) and N � L(t). Consequently, for x � U + K(t), y � V + L(t), we have the Diophantine equation which is Pell equation. Now, we try to find all polynomial solutions (U n , V n ) of T(E), and then, we can retransfer all results from T(E) to E by using the inverse of T.
(2) Define the sequence where (U n , V n ) is defined in (24). en, (X n , Y n ) is a solution of E. So, it has infinitely many integer so- For n ≥ 1. (4) e solutions (X n , Y n ) satisfy the recurrence relations

Journal of Mathematics
For n ≥ 3. (5) e solutions (X n , Y n ) satisfy the recurrence relations For n ≥ 4.

Data Availability
No data were used to support this study.

Conflicts of Interest
e authors declare that they have no conflicts of interest.