# The Polynomial Solutions of Quadratic Diophantine EquationX2−ptY2+2KtX+2ptLtY = 0

Authors: Hasan Sankari; Ahmad Abdo
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this study, we consider the number of polynomial solutions of the Pell equationx2−pty2=2 is formulated for a nonsquare polynomial pt using the polynomial solutions of the Pell equation x2−pty2=1. Moreover, a recurrence relation on the polynomial solutions of the Pell equation x2−pty2=2. Then, we consider the number of polynomial solutions of Diophantine equation E:X2−ptY2+2KtX+2ptLtY=0. We also obtain some formulas and recurrence relations on the polynomial solution Xn,Yn of E.


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## Body

## 1. 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. The equationax+by=1 is known as the liner Diophantine equation. In general, the Diophantine equation is the equation given by(1)ax2+bxy+cy2+dx+ey+f=0.The equationx2−Dy2=N, with given integers D and N and unknowns x and y, is called Pell’s equation. The most interesting case of the equation arises when D≠1 be a positive nonsquare. Pell’s equation x2−Dy2=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 x2−Dy2=2 was considered, and some formulas of its integer solutions were obtained. In [6, 7], the number of integer solutions of Diophantine equation x2−t2−1y2−4t−2x+4t2−4ty=0 and Diophantine equation x2−t2−1y2−16t−4x+16t2−16ty=0 over ℤ is considered, where t≥2. In [3, 8], the number of polynomial solutions of Diophantine equation(2)x2−pt2−pty2−4pt−2x+4pt2−4pty=0and Diophantine equation(3)x2−pty2−2p′tx+4pty+p′t2−4pt−1=0over ℤ was considered, where pt be a polynomial in ℤt/0,1.


## 2. Preliminaries

In this section, we introduce the objects we need later and collect some important facts about them.In [4], Li proved that the Pell equation x2−Dy2=1 has infinitely positive solutions. If x1,y1 is the fundamental solution, then for n=2,3,…, xn+ynD=x1+y1D2. The pairs xn,yn are all the positive solutions of the Pell equation. The xn ’s and yn ’s are strictly increasing to infinity and satisfy the recurrence relations:(4)xn+2=2x1xn+1−xn,yn+2=2x1yn+1−yn.Theorem 1.
(Tekcan (see [5])). Let X1,Y1=K,L be the fundamental solution of the Pell equation x2−Dy2=2 and x1,y1=a,b be the fundamental solution of the Pell equation x2−Dy2=1. Then, the other solutions of the Pell equation x2−Dy2=2 are Xn,Yn:(1)
Forn≥2,(5)XnYn=KLDLKxn−1yn−1.(2)
Forn≥1,(6)Xn+1Yn+1=aXn+bDYnbXn+aYn.(3)
Forn≥4,(7)Xn+1=2K2+1Xn−1−Xn−2+Xn−3,Yn+1=2K2+1Yn−1−Yn−2+Yn−3.


## 3. New Results

Our principal result is the following.Theorem 2.
Letx1,y1=at,bt be the fundamental solution of the Pell equation:(8)x2−pty2=1.
Then, the other solutions of the Pell equationx2−pty2=1 are xn,yn, where(1)
Forn≥1,(9)xnyn=atbtptbtatn10.(2)
Forn≥2,(10)xn=atxn−1+btptyn−1,yn=btxn−1+atyn−1.(3)
Forn≥4,(11)xn=2at−1xn−1+xn−2−xn−3,yn=2at−1yn−1+yn−2−yn−3.Proof.
(1)
We prove it using the method of mathematical induction. Letn=1, and we get x1,y1=at,bt, which is the fundamental solution of equation (8). Now, we assume that (9) is satisfied for n, that is,(12)xnyn=atbtptbtatn10.We try to show that this equation is also satisfied forn+1. Applying (9), we find that(13)xn+1yn+1=atbtptbtatn+110=atbtptbtatatbtptbtatn10=atbtptbtatxnyn=atxn+btptynbtxn+atyn.Hence, we conclude that(14)xn+12−ptyn+12=atxn+btptyn2−ptbtxn+atyn2=xn2−ptyn2=1.So,xn+1,yn+1 is also solution of equation (8).(2)
Using (13), we find that(15)xn=atxn−1+btptyn−1,yn=btxn−1+atyn−1,Forn≥2.(3)
We prove it using the method of mathematical induction. Forn=4, we get(16)x1=at,x2=2at2−1,x3=4at3−3at,x4=8at4−8at2+1.
Hence,(17)2at−1x3+x2−x1=2at−14at3−3at+2at2−1−at=8at4−8at2+1=x4.
So,xn=2at−1xn−1+xn−2−xn−3 is satisfied for n=4. Let us assume that this relation is satisfied for n, that is,(18)xn=2at−1xn−1+xn−2−xn−3,then, using (13) and (18), we conclude that(19)xn+1=2at−1xn+xn−1−xn−2,completing the proof.
Similarly, we prove that(20)yn=2at−1yn−1+yn−2−yn−3,∀n≥4.
Now, we give a relation betweenKt,Lt and at,bt.Theorem 3.
IfKt,Lt be the fundamental solution of the equation (21)x2−pty2=2,then Kt2−1,KtLt be a solution of the equation (8) and(22)Kt2+ptLt22,KtLt=Kt2−1,KtLt.Proof.
Hence, it is easily seen that(23)Kt2−12−ptKtLt2=Kt2Kt2−ptLt2−2Kt2+1=1,Kt2+ptLt22,KtLt=Kt2+Kt2−22,KtLt=Kt2−1,KtLt,since Kt,Lt is the fundamental solution of the equation x2−pty2=2, i.e, Kt2−ptLt2=2.Theorem 4.
LetU1,V1=Kt,Lt be the fundamental solution of the equation (21) and x1,y1=at,bt be the fundamental solution of the equation (8). Then,(1)
(24)UnVn=KtLtptLtKtxn−1yn−1,∀n≥2.(2)
(25)Un+1=atUn+btptVn,Vn+1=btUn+atVn,∀n≥1.(3)
The solutionUn,Vn satisfies the recurrence relations(26)Un=2atUn−1−Un−2,Vn=2atVn−1−Vn−2,∀n≥3.(4)
The solutionUn,Vn satisfies the recurrence relations(27)Un=2Kt2−1Un−1−Un−2+Un−3,Vn=2Kt2−1Vn−1−Vn−2+Vn−3,∀n≥4.Proof.
(1)
From (24), we get(28)UnVn=Ktxn−1+Ltptyn−1Ltxn−1+Ktyn−1.Hence, it is easily seen that(29)Un2−ptVn2=Ktxn−1+Ltptyn−12−ptLtxn−1+Ktyn−12=Kt2xn−12−ptyn−12−ptLt2xn−12−ptyn−12=Kt2−ptLt2=2.Sincexn−12−ptyn−12=1, and Kt2−ptLt2=2.(2)
From (24), we get(30)UnVn=KtLtptLtKtxn−1yn−1.Then,(31)xn−1yn−1=KtLtptLtKt−1UnVn=KtLtUn+2−Kt2Vn2Lt−LtUn+KtVn2.Then,(32)xn−1=KtLtUn+2−Kt2Vn2Lt,yn−1=−LtUn+KtVn2.On the other hand, by using (10) and (32), we get(33)Un+1Vn+1=KtLtptLtKtxnyn=KtLtptLtKtatxn−1+btptyn−1btxn−1+atyn−1=xn−1Ktat+ptLtbt+yn−1Ktbt+ptLtatxn−1Ltat+btKt+yn−1btLtpt+atKt.Applying (32) and (33), we find(34)Un+1=xn−1Ktat+ptLtbt+yn−1Ktbt+ptLtat=KtLtUn+2Kt2+1Vn2LtatKt+btLtpt+−LtUn+KtVn2btLtpt+atKt=2atUn+2btptLtVn2Lt=atUn+btptvn.Similarly, we prove that(35)Vn+1=btUn+atVn,∀n≥3.(3)
From (25), we get(36)Un=atUn−1+btptbtUn−2+atVn−2=atUn−1+bt2ptUn−2+atbtvn−2=atUn−1+at2−1Un−2+atbtVn−2=atUn−1+atatUn−2+btptvn−2−Un−2=2atUn−1−Un−2.Similarly, we prove that(37)Vn=2atVn−1−Vn−2,∀n≥3.(4)
From (22) and (26), we get(38)Un=Un−Un−1+Un−1=2atUn−1−Un−2−2atUn−2+Un−3+Un−1=2at+1Un−1−Un−2+Un−3=2Kt2−1Un−1−Un−2+Un−3.
Similarly, we prove that(39)Vn=2Kt2−1Vn−1−Vn−2+Vn−3,∀n≥4,and we consider the number of polynomial solutions of Diophantine equation. Now,(40)E:x2−pty2−2Ktx+2ptLty=0,where Kt,Lt is the fundamental solution of equation (21).
We have to transformE into an appropriate Diophantine equation which can be easily solved. To get this, let(41)T:x=U+M,y=V+N.be a translation for some M and N.
By applying the transformationT to E, we get(42)TE=U+M2−ptV+N2−2KtU+M+2ptLtV+N=0.
In (42), we obtain U2M−2Kt and V−2ptN+2ptLt. So we get M=Kt and N=Lt. Consequently, for x=U+Kt,y=V+Lt, we have the Diophantine equation(43)E˜:U2−ptV2=2,which is Pell equation.
Now, we try to find all polynomial solutionsUn,Vn of TE, and then, we can retransfer all results from TE to E by using the inverse of T.Theorem 5.
LetE be the Diophantine equation in (40), where Kt,Lt is the fundamental solution of equation (43); then,(1)
The fundamental (minimal) solution ofE is(44)X1,Y1=2Kt,2Lt.(2)
Define the sequence(45)xn,ynn≥1=Un+Kt,Vn+Lt,where Un,Vn is defined in (24). Then, Xn,Yn is a solution of E. So, it has infinitely many integer solutions Xn,Yn∈ℤ×ℤ.(3)
The solutionsXn,Yn satisfy the recurrence relations(46)Xn+1=atXn+btptYn+1−atKt−btptLt,Yn+1=btXn+atYn−btKt+1−atLt,Forn≥1.(4)
The solutionsXn,Yn satisfy the recurrence relations(47)Xn=2atXn−1−Xn−2+21−atKt,Yn=2atYn−1−Yn−2+21−atLt,Forn≥3.(5)
The solutionsXn,Yn satisfy the recurrence relations(48)Xn=2Kt2−1Xn−1−Xn−2+Xn−3,Yn=2Kt2−1Yn−1−Yn−2+Yn−3.Forn≥4.Example 1.
Letpt=t2−2; then, U1,V1=t,1 is the fundamental solution of(49)E˜:U2−t2−2V2=2.
By Theorem3, x1,y1=t2−1,t is the fundamental solution of(50)x2−t2−2y2=1.
By Theorem4, it is easily seen that some other solutions of equation E˜ are(51)U2V2=2t3−3t2t2−1,U3V3=4t5−10t3+5t4t4−6t2+1,U4V4=8t7−30t5+29t3−9t8t6−20t4+12t2−1,U5V5=16t9−76t7+112t5−69t3+11t16t8−56t6+60t4−20t2+1.
It can be concluded now that the fundamental solution of(52)E:x2−t2−2y2−2tx+2t2−4y=0is 2t,2. By Theorem 5, it is easily seen that some other solutions of equation E are (53)X2Y2=2t3−2t2t2,X3Y3=4t5−10t3+6t4t4−6t2+2,X4Y4=8t7−30t5+29t3−8t8t6−20t4+12t2,X5Y5=16t9−76t7+112t5−69t3+12t16t8−56t6+60t4−20t2+2.



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