ON THE X –COORDINATES OF PELL EQUATIONS THAT ARE PRODUCTS OF TWO PADOVAN NUMBERS

Let { P n } n ≥ 0 be the sequence of Padovan numbers deﬁned by P 0 = 0, P 1 = P 2 = 1, and P n +3 = P n +1 + P n for all n ≥ 0. In this paper, we ﬁnd all positive square-free integers d ≥ 2 such that the Pell equations x 2 − dy 2 = (cid:96) , where (cid:96) ∈ {± 1 , ± 4 } , have at least two positive integer solutions ( x, y ) and ( x (cid:48) , y (cid:48) ) such that each of x and x (cid:48) is a product of two Padovan numbers.

Let d ≥ 2 be a positive integer which is not a square.It is well known that the Pell equations where ∈ {±1, ±4}, have infinitely many positive integer solutions (x, y).By putting (x 1 , y 1 ) for the smallest positive solution to (1.1), all solutions are of the form (x k , y k ) for some positive integer k, where for all k ≥ 1, and = ±1 and for all k ≥ 1, and = ±4.
Furthermore, the sequence {x k } k≥1 in both cases ∈ {±1, ±4} is binary recurrent.In fact, the following formulas , for = ±1, and hold for all positive integers k.
Recently, Kafle et al. [14] studied the Diophantine equation where x l are the x-coordinates of the solutions of the Pell equation (1.1) (in the case = ±1) for some positive integer l and {F n } n≥0 is the sequence of Fibonacci numbers given by F 0 = 0, F 1 = 1, and F n+2 = F n+1 + F n for all n ≥ 0. They proved that for each square free integer d ≥ 2, there is at most one positive integer l such that x l admits the representation (1.2) for some nonnegative integers 0 ≤ m ≤ n, except for d ∈ {2, 3, 5}.Furthermore, they explicitly stated all the solutions for these exceptional cases.
In the same spirit, Rihane et al. [22] studied the Diophantine equation where x n are the x-coordinates of the solutions of the Pell equations (1.1), for some positive integers n and {P m } m≥0 is the sequence of Padovan numbers.They proved that for each square free integer d ≥ 2, there is at most one positive integer x participating in the Pell equations (1.1), that is a Padovan number with a few exceptions of d that can be effectively computed.Furthermore, the exceptional cases in (1.3) were d ∈ {2, 3, 5, 6} (for the case = ±1) and d ∈ {5} (for the case = ±4).Several other related problems have been studied where x l belongs to some interesting positive integer sequences.For example, see [2,3,6,7,8,9,10,12,15,16,17,18,19].

Main Results
In this paper, we study a problem related to that of Kafle et al. [14] but with the Padovan sequence instead of the Fibonacci sequence.We also extend the results from the Pell equation (1.1) in the case = ±1 to the case = ±4.In both cases we find that there are only finitely many solutions that we effectively compute.
Since P 1 = P 2 = P 3 = 1, we discard the situations when n = 1 and n = 2 and just count the solutions for n = 3.Similarly, P 4 = P 5 = 2, we discard the situation when n = 4 and just count the solutions for n = 5.The main aim of this paper is to prove the following results.
Theorem 2.1.For each integer d ≥ 2 which is square-free, there is at most one positive integer k such that Theorem 2.2.For each integer d ≥ 2 which is square-free, there is at most one positive integer k such that except when d ∈ {3, 5, 6, 77} in the +4 case and d ∈ {2, 5, 13, 29, 65, 257} in the −4 case.
For the exceptional values of d listed in Theorem 2.1 and Theorem 2.2, all solutions (k, n, m) are listed at the end of the proof of each result.The main tools used in this paper are the lower bounds for linear forms in logarithms of algebraic numbers and the Baker-Davenport reduction procedure, as well as the elementary properties of Padovan numbers and solutions to Pell equations.Computations are done with the help of a computer program in Mathematica.

3.1.
The Padovan sequence.Here, we recall some important properties of the Padovan sequence {P n } n≥0 .The characteristic equation and Furthermore, the Binet formula is given by where Numerically, the following estimates hold: From (3.1), (3.2), and (3.5), it is easy to see that the contribution the complex conjugate roots β and γ, to the right-hand side of equation (3.3), is very small.In particular, setting e(n) holds for all n ≥ 1.Furthermore, by induction, one can prove that α n−2 ≤ P n ≤ α n−1 holds for all n ≥ 4. (3.7) 3.2.Linear forms in logarithms.Let η be an algebraic number of degree d with minimal primitive polynomial over the integers where the leading coefficient a 0 is positive and the η (i) 's are the conjugates of η.Then the logarithmic height of η is given by In particular, if η = p/q is a rational number with gcd(p, q) = 1 and q > 0, then h(η) = log max{|p|, q}.The following are some of the properties of the logarithmic height function h(•), which will be used in the next sections of this paper without a reference: Here, we recall the result of Bugeaud, Mignotte, and Siksek ( [4], Theorem 9.4, pp.989), which is a modified version of the result of Matveev [20].This result is one of our main tools in this paper.Theorem 3.1 (Matveev according to Bugeaud, Mignotte, Siksek).Let η 1 , . . ., η t be positive real algebraic numbers in a real algebraic number field K ⊂ R of degree D, b 1 , . . ., b t be nonzero integers, and assume that 3.3.Reduction procedure.During the calculations, we get upper bounds on our variables which are too large, thus we need to reduce them.To do so, we use some results from the theory of continued fractions.
For the treatment of linear forms homogeneous in two integer variables, we use the well-known classical result in the theory of Diophantine approximation.It is called the Legendre criterion.Lemma 3.1 (Legendre).Let τ be an irrational number, p 0 q 0 , p 1 q 1 , p 2 q 2 , . . .be all the convergents of the continued fraction of τ and M be a positive integer.Let N be a nonnegative integer such that q N > M .Then putting a(M ) := max{a i : i = 0, 1, 2, . . ., N }, the inequality holds for all pairs (r, s) of positive integers with 0 < s < M .
For a nonhomogeneous linear form in two integer variables, we use a slight variation of a result due to Dujella and Pethő (see [11], Lemma 5a).For a real number X, we write ||X|| := min{|X − n| : n ∈ Z} for the distance from X to the nearest integer.Lemma 3.2 (Dujella, Pethő).Let M be a positive integer, p q be a convergent of the continued fraction of the irrational number τ such that q > 6M , and A, B, µ be some real numbers with A > 0 and B > 1.At various occasions, we need to find a lower bound for linear forms in logarithms with bounded integer coefficients in three variables.In this case we use the LLL algorithm that we describe below.Let τ 1 , τ 2 , . . .τ t ∈ R and the linear form We put X := max{X i }, C > (tX) t and consider the integer lattice Ω generated by where C is a sufficiently large positive constant.
Lemma 3.3 (LLL algortthm).Let X 1 , X 2 , . . ., X t be positive integers such that X := max{X i } and C > (tX) t be a fixed sufficiently large constant.With the above notation on the lattice Ω, we consider a reduced base {b i } to Ω and its associated Gram-Schmidt orthogonalization base {b * i }.We set For the proof and further details, we refer the reader to the book of Cohen, (Proposition 2.3.20 in [5], pp.58-63).Finally, the following Lemma is also useful.It is Lemma 7 in [13].

Proof of Theorem 2.1
Let (x 1 , y 1 ) be the smallest positive integer solution to the Pell quation (1.1) in the case = ±1.We Put From which we get that Then We assume that (k 1 , n 1 , m 1 ) and (k 2 , n 2 , m 2 ) are triples of integers such that We asuume that 1 ≤ k 1 < k 2 .We also assume that 3 ≤ m i < n i for i = 1, 2. We set (k, n, m) := (k i , n i , m i ), for i = 1, 2. Using the inequalities (3.7) and (4.4), we get from (4.5) that The above inequalities give Dividing through by log α and setting c 2 := 1/ log α, we get that and since α 3 > 2, we get To fix ideas, we assume that We also put Inequality (4.7) together with the fact that δ > α 2 ( so, c 2 log δ > 2), tells us that Besides, given that k 1 < k 2 , we have by (3.7) and (4.5) that Thus, we get that and by (3.6), we have . Thus, we have We assume that n + m ≥ 10.
for n + m ≥ 10 (because 2/α 5 < 1/2), since the inequality |y| < 2|e y − 1| holds for all y ∈ − 1 2 , 1 2 , it follows that e |Γ 1 | < 2 and so Thus, we get that (4.12) We apply Theorem 3.1 on the left-hand side of (4.11) with the data: Furthermore, we take the number field K := Q( √ d, α) which has degree D := 6.Since max{1, k, n+ m} ≤ 2n, we take B := 2n.First, we note that the left-hand side of (4.11) is non-zero, since otherwise, The left-hand side belongs to the quadratic field Q( √ d), while the right-hand side belongs to the cubic field Q(α).These fields only intersect when both sides are rational numbers.Since δ k is a positive algebraic integer and a unit, we get that to δ k = 1.Hence, k = 0, which is a contradiction.Thus, Λ 1 = 0. Now, we can apply Theorem 3.1.
We record what we have proved so far.4.2.Absolute bounds.We recall that (k, n, m) = (k i , n i , m i ), where 3 ≤ m i < n i , for i = 1, 2 and 1 ≤ k 1 < k 2 .Further, n i ≥ 4 for i = 1, 2. We return to (4.12) and rewrite We do a suitable cross product between Γ 1 and k 1 , k 2 to eliminate the term involving log δ in the above linear forms in logarithms: where λ := min We need to find an upper bound for λ.If 8n 4 /α λ > 1/2, we then get We apply Theorem 3.1 with the data: We take the number field K := Q(α) and D := 3. We begin by checking that e Γ 3 − 1 = 0 (so Γ 3 = 0).This is true because α and 2a 2 are multiplicatively independent, since α is a unit in the ring of integers Q(α) while the norm of 2a 2 is 8/529.We note that |k 1 − k 2 | < k 2 < n 4 .Further, from (4.18), we have given that m 3 ≥ 1.So, we can take B := 25n 4 .By Theorem 3.1, with the same A 1 := log 529 + log(2a 2 ) and A 2 := log α, we have that We set {i, j} = {1, 2} and return to (4.12) to replace (k, n, m) = (k i , n i , m i ): and also return to (4.16), replacing with (k, n, m) = (k j , n j , m j ): We perform a cross product on (4.22) and (4.23) in order to eliminate the term on log δ: where ν := min 1≤i,j≤2 As before, we need to find an upper bound on ν.If 20n 2 /α ν > 1/2, then we get Otherwise, |Γ 4 | < 1/2, so we have In order to apply Theorem 3.1, first we check if e Γ 4 = 1, we obtain (2a Since α is a unit, the right-hand side in above is an algebraic integer.This is a contradiction because k 1 < k 2 so k i − k j = 0, and neither (2a) nor (2a) −1 are algebraic intgers.Hence e Γ 4 = 1.By assuming that ν ≥ 100, we apply Theorem 3.1 with the data: t := 2, and the inequalities (4.21) and (4.26).We get The above inequality also holds when ν < 100.Further, it also holds when the inequality (4.25) holds.So the above inequality holds in all cases.Note that the case {i, j} = {2, 1} leads to or the minimum is n j and from the inequality (4.21) we get that By the inequality (4.7), By substituting this into Lemma 4.1, we get n 4 < 9.77 × 10 79 (log n 4 ) 6 .Also, by Lemma 3.4, with the data r := 6, H := 9.77 × 10 79 , and L := n 4 , we get that n 2 ≤ n 4 < 2.44 × 10 95 .This immediately gives that n 1 ≤ n 3 < 6.56 × 10 29 and m 1 ≤ m 3 < 6.50 × 10 29 .
We record what we have proved.
Lemma 4.2.Let (k i , n i , m i ) be a solution to

5.
Reducing the bounds for n 1 and n 2 In this section, we reduce the upper bounds for n 1 and n 2 given in Lemma 4.2 reasonably enough so that these can be treated computationally.For this, we return to the inequalities for Γ 3 , Γ 4 , and Γ 5 .
5.1.The first reduction.We divide both sides of the inequality (4.18) by (k 2 − k 1 ) log α.We get that log(2a We assume that λ ≥ 10.Below we apply Lemma 3.1.We put τ := log(2a Thus, by Lemma 3.1, we have that (5.2) Hence, combining the inequalities (5.1) and (5.2), we obtain given in the inequality (4.24), via the procedure described in Subsection 3.3 (LLL algorithm).We recall that Γ 4 = 0. We apply Lemma 3.3 with the data: t := 3, τ 1 := log(2a 2 ), τ 2 := log(2aP m j ), τ 3 := log α, We set X := 25 × 10 96 as an upper bound to |x i | < 25n 2 for all i = 1, 2, 3, and C := (5X) We note that the upper bound for n 2 represents a very good reduction of the bound given in Lemma 4.2.Hence, it is expected that if we start our reduction cycle with the new bound on n 2 , then we even get a better bound on n 1 .Indeed, returning to (5.1), we take M := 1.5 × 10 40 and computationally verify that q 77 > M > n 2 > k 2 − k 1 and a(M ) := max{a i : 0 ≤ i ≤ 77} = a 13 = 149, from which it follows that λ ≤ 666.We now return to (5.3), where putting X := 3.75 × 10 41 and C := (5X) 5 , we apply the LLL algorithm to λ ∈ [1,666] 5.2.The final reduction.Returning back to (4.1) and (4.3) and using the fact that (x 1 , y 1 ) is the smallest positive solution to the Pell equation (1.1), we obtain Thus, we return to the Diophantine equation x k 1 = P n 1 P m 1 and consider the equations with Besides the trivial case k 1 = 1, with the help of a computer search in Mathematica on the above equations in (5.4), we list the only nontrivial solutions in Table 1.
With the help of Mathematica, we obtain the results in Table 3.Thus, max{b t,n 2 −m 2 : t = 1, 2, . . ., 5 and m 2 = 1, 2, . . .b t } ≤ 351.Thus, by Lemma 3.2, we have that n 2 ≤ 351, for all t = 1, 2, . . ., 5, and by the inequality (4.10) we have that n 1 ≤ 2n 2 .From the fact that δ k ≤ α n+m , we can conclude that k 1 < k 2 ≤ 194.Collecting everything together, our problem is reduced to search for the solutions for (4.5) in the following range: After a computer search on the equation (4.5) on the above ranges, we obtained the following solutions, which are the only solutions for the exceptional d cases we have stated in Theorem 2.1.
This completes the proof of Theorem 2.1.

Proof of Theorem 2.2
The proof of Theorem 2.2 follows similar arguments as in the proof of Theorem 2.1.So, we do not give the details here.We leave it as an easy exersise to the reader.
Below, we give the exceptional d cases we have stated in Theorem 2.
By Lemma 3.4, we have that n 2 < 1.66 × 10 35 , a better bound than that obtained in the previous step of the reduction cycle.We record what we have proved.Lemma 5.1.Let (k i , n i , m i ) be a solution to x k i

Table 3 .
Final reduction computation results