# On Polynomial Solutions of Pell’s Equation

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

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## Abstract

Polynomial Pell’s equation isx2−Dy2=±1, where D is a quadratic polynomial with integer coefficients and the solutions X,Y must be quadratic polynomials with integer coefficients. Let D=a2x2+a1x+a0 be a polynomial in Zx. In this paper, some quadratic polynomial solutions are given for the equation x2−Dy2=±1 which are significant from computational point of view.


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## Body

## 1. Introduction

LetD be a quadratic polynomial with integer coefficient. We consider polynomial Pell’s equation(1)x2−Dy2=±1,where X,Y are quadratic polynomials with integer coefficients. Many fruitful research studies have been conducted in this direction.In 1976, Nathanson [1] proved that when D=x2+d, equation (1) has a nontrivial solution if and only if d=±1,±2.In 2001, Mollin et al. [2] described the background to the study of continued fraction expansions of D where D is a polynomial, especially quadratic, which we study herein.


## 2. Notation and Preliminaries

LetD be a given square-free natural number. It is well known that Pell’s equation(2)x2−Dy2=1,always has an infinite number of integer solutions (see e.g. Nagell [3]). All the solutions of (2) with positive x and y are obtained by the formula An+BnD=a+bDn where n=1,2,3,…, and a+bD is the fundamental solution of (2).Pell’s equation(3)x2−Dy2=−1,may not always have integer solutions. When this equation is solvable in integers for some given D, if L+MD is its fundamental solution, we have the relation(4)a+bD=L+MD2,where a+bD is the fundamental solution of (2).For a given value ofD, in order to determine the fundamental solution of (2) or (3) (when it exists), one may use the method of continued fraction expansion of D. The penultimate convergent of this expansion provides the fundamental solution.From computational point of view, it would be advantageous to group the values ofD into several classes such that the values of D in a specific class are associated with a similar pattern of continued fraction expansion. It is found that polynomial expressions for D come in handy for this purpose.As usuala0;a1,…,an;2a0¯ will denote the simple infinite periodic continued fraction(5)a0+1a1+1/an−1+⋱/an+1/2a0+1/a1+⋱.Ai/Bi will denote the i′th convergent:(6)a0+1a1+1/a2+⋯1/ai.Use will also be made of the following:(7)Ai=aiAi−1+Ai−2,Bi=aiBi−1+Bi−2,A−1=B−2=1,A−2=B−1=0,whereAiBi−1−Ai−1Bi=−1i−1,each of these being valid for i=0,1,2,….Definition 1.
(see [4]). A polynomial ft∈ℤt is called a Fermat−Pell polynomial in one variable if there exists polynomial ct,ht∈ℤt such that(8)ct2−ftht2=1,for allt∈ℤ.
A triple of polynomialsct,ht,ft satisfying equation (8) constitutes a polynomial solution to Pell’s equation. A Fermat−Pell polynomial is said to have a polynomial continued fraction expansion if there exists an integer T and a positive integer n such that(9)ft=a0t;a1t,…,ant¯,for all integral t≥T, where ait∈ℤt. For a fixed Fermat−Pell polynomialft, a pair of polynomials ct,ht is said to be the fundamental polynomial solution to equation (9) if ct,ht constitutes the smallest solution in positive integers to equation (8) for all non-negative values of t≥T.Definition 2.
(see [5]). For real x, let the symbol x denote the greatest integer ≥x.Theorem 1 (see [5]).
SupposeD=4t2+12t+5, where t is any natural number. Then, a+bD is the fundamental solutions of (2) where a=4t3+18t2+24t+9, b=2t2+6t+4.Theorem 2 (see [5]).
IfD=4t2+4t+5 where t is any natural number, then L+MD is the fundamental solution of (3) where L=4t3+6t2+6t+2, M=2t2+2t+1.Theorem 3 (see [6]).
LetD≥1 be a non-square integer; then:(10)D=a0;2a0¯⇔D=a02+1,where D is a positive integer.


## 3. Main Results

LetD=a2x2+a1x+a0 be a polynomial in ℤx. In this section, we will prove our main results. As a start, we record the following propositions.Proposition1.
LetD=Dt, where t is any natural number. If D=a0t;2a0t¯, then:(1)
L+MD is the fundamental solution for the equation x2−Dy2=−1, where L=a0t,M=1.(2)
a+bD is the fundamental solution for the equation x2−Dy2=1, where a=2a0t2+1,b=2a0t.Proof.
Let us assumeD=a0t;2a0t¯. Then:(1)
The penultimate convergent is(11)a0t1.Thus, we get the relation stated above, and by (10),(12)L2−DM2=a0t2−a0t2+112=−1.(2)
The penultimate convergent is(13)a0t+12a0t=2a0t2+12a0t.Thus, we get the relation stated above, and by (10),(14)a2−Db2=2a0t2+12−a02+12a0t2=1.Theorem 4.
Lett,m,p be any natural numbers, p be a prime, and a+bD be the fundamental solution of (2); then:(1)
IfD=m2t2+m, then a=2mt2+m,b=2t.(2)
IfD=p2t2+2p, then a=pt2+1,b=t.(3)
IfD=m2t2+2mt, then a=mt+1,b=1.(4)
IfD=4t2+4m+4t+m2+m, then a=4t+2m+1,b=2.(5)
IfD=16m+12t2+8m+12t+m2+m, then a=32m+1t2+16m+1t+2m+1,b=8m+2.Proof.
(1) We have(15)mt<D<mt+1,and so D=mt. Now, expanding D into a continued fraction expansion, we obtain(16)D=mt+m2t2+m−mt=mt+1m2t2+m+mt/m2t2+m−m2t2=mt+12t+m2t2+m−mt/m=mt+12t+1/mm2t2+m+mt/m2t2+m−m2t2=mt+12t+1/2mt+m2t2+m−mt=mt;2t,2mt¯.
The penultimate convergent is(17)mt+12t=2mt2+12t.
Thus, we get the relation(18)2mt2+12−m2t2−m2t2=1.
Similarly, (2)–(4) and (7) can all be proven.Theorem 5.
SupposeD=16k2t2+8k2t+k2+k, where t,k are any natural numbers. Then:(1)
a+bD is the fundamental solution of (2) where a=2k4t+12+1,b=24t+1.(2)
Ifk=1, then L+MD is the fundamental solution of (3) where L=4t+1,M=1.Proof.
(1) ExpandingD into a continued fraction expansion, we obtain(19)D=4kt+k;24t+k,24kt+k¯.
The penultimate convergent is(20)4kt+k+124t+k=2k4t+12+124t+1.
Thus, we get the relation(21)2k4t+12+12−16k2t2+8k2t+k2+k24t+1=1.
(2) Fork=1, D=16t2+8t+2. Now, expanding D into a continued fraction expansion, we obtain(22)D=4t+1;24t+1¯.
By Proposition1, L=4t+1,M=1; then, 4t+1+16t2+8t+2 is the fundamental solution of (3).Theorem 6.
SupposeD=t2+2ct+c2+1, where t,c are any natural numbers. Then:(1)
L+MD is the fundamental solution of (3) where L=t+c,M=1.(2)
a+bD is the fundamental solution of (2) where a=2t+c2+1,b=2t+c.Example 1.
LetD=4t2+2; then, 4t2+1+2t4t2+2 is the fundamental solution of (2), by (1) in Theorem 3.Example 2.
LetD=t2+4t+5; then, t+2+t2+4t+5 is the fundamental solution of (3) and 2t+22+1+2t+2t2+4t+5 is the fundamental solution of (2), by Theorem 5.



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