Abdelmejid Bayad
email: abayad@maths.univ-evry.fr

Appolinaire Dossavi-Yovo
email: appolinairedossaviyovo@yahoo.fr

Alan Filipin
email: filipin@grad.hr

Alain Togbé
email: atogbe@pnc.edu

On the extensibility of the D(4)-triple {k -2, k + 2, 4k} over Gaussian integers

Keywords: Diophantine m-tuple, Pell equation, Linear forms in logarithms AMS Classification: 11D09, 11D45, 11B37, 11J86

In this paper, we prove that if {k -2, k + 2, 4k, d}, where k ∈ Z[i], k = 0, ±2, is a D(4)-quadruple in the ring of Gaussian integers, then d = 4k 3 -4k.

Introduction

Let n be a non-zero integer. A set of m distinct positive integers such that the product of any two of its elements increased by n is a perfect square is called D(n)-m-tuple. The most studied case is n = 1 in which we call D(1)-m-tuple simply Diophantine m-tuple. There is the folklore conjecture that there does not exist a Diophantine quintuple. This conjecture is intensively studied in recent years. To see all references the reader should consult the webpage [START_REF] Dujella | Diophantine m-tuples[END_REF]. The most important contribution was done by Dujella [START_REF] Dujella | There are only finitely many Diophantine quintuples[END_REF] in 2004 when he proved that there does not exist a Diophantine sextuple and that there are only finitely many Diophantine quintuples. Moreover, he also gave an upper bound for the number of Diophantine quintuples, which was recently significantly improved by various authors. The best known bound can be found in [START_REF] Cipu | Searching for Diophantine quintuples[END_REF]. But that upper bound is still too large such that it is impossible to completely solve the problem using today's computers. Therefore, the conjecture is still open. Moreover, in the case n = 1, there is a stronger version of the main conjecture which says that every D(1)-triple can be extended to a quadruple with a larger element in a unique way. More precisely, the conjecture says that if {a, b, c, d} is a Diophantine quadruple such that a < b < c < d, then d = a + b + c + 2(abc + rst), where r, s, and t are positive integers defined by ab + 1 = r 2 , ac + 1 = s 2 and bc + 1 = t 2 . Such Diophantine quadruples are called regular. So the conjecture implies that all Diophantine quadruples are regular. Similar results and conjectures exist when n = -1 and n = 4. In the case n = -1, there is a conjecture that there does not exist D(-1)-quadruple, while in the case n = 4, we have an analogous conjecture (as in the case n = 1) that there does not exist D(4)-quintuple. Those conjectures are also extensively studied recently, and for details the reader could again consult [START_REF] Dujella | Diophantine m-tuples[END_REF].

There are various generalizations of the above problems. One of them consists in taking elements of m-tuples in a commutative ring with unity. Let R be a commutative ring with unity 1. One can consider m different non-zero elements in R such that the product of any two distinct elements increased by some element of R is a perfect square in R. Recently, there are also results on the extendibility of Diophantine m-tuples in other rings such as the ring of Gaussian integers and rings of integers of quadratic fields. For example, in 2008, Franušić [START_REF] Franušić | On the extensibility of Diophantine triples {k-1, k+1, 4k} for Gaussian integers[END_REF] proved that the Diophantine triple {k -1, k +1, 4k} extends uniquely to a Diophantine quadruple in the ring Z [i]. The first, third and the fourth author [START_REF] Bayad | Extension of a parametric family of Diophantine triples in Gaussian integers[END_REF] proved the same result for the triple {k, 4k + 4, 9k + 6}. In this paper, we will consider a D(4)-tuple in the ring of Gaussian integers. Precisely, we study the D(4)-extensions of the triple {k -2, k + 2, 4k} in Z[i] and our main result is the following theorem.

Theorem 1.1. Let k ∈ Z[i] and k = 0, ±2. If {k -2, k + 2, 4k, d} is a D(4) quadruple in Z[i], then d = 4k 3 -4k.

That result was proven in the ring of integers by Fujita [START_REF] Fujita | The unique representation d = 4k(k 2 -1) in D(4)-quadruples {k -2, k + 2, 4k, d}[END_REF]. This is the first time that a D(4)tuple is studied in the ring of Gaussian integers. The organization of this paper is as follows. In Section 2, we determine a system of Pell equations and the sequences v m , w n and w n corresponding to our problem of extendibility. Moreover, we prove some useful results that we will need to prove Theorem 1.1. We find the relation between indices m and n in the equations v m = w n and v m = w n and the lower bound on |x| in terms of |k| in Section 3. To prove the main result for |k| ≥ 155402, we devote Section 4 to the use of a generalization of Bennett's result (see [START_REF] Jadrijević | A system of relative Pellian equations and related family of relative Thue equations[END_REF]), that consists in simultaneous rational approximations of algebraic numbers that are close to 1. The study of the remaining cases is done in two steps. First, by the means of the Baker-Davenport reduction method, we complete the proof of Theorem 1.1 for 5 < |k| < 155402 in Section 5. In the section 6, we deal with the second step that consists in looking the problem for 2 < |k| ≤ 5. In the final section, we study the cases |k| ≤ 2. This is really a laborious work. Our proof follows mostly the methods from [START_REF] Franušić | On the extensibility of Diophantine triples {k-1, k+1, 4k} for Gaussian integers[END_REF].

Solving a system of Pell equations

Let k ∈ Z[i], k = 0, ±2. The aim of this paper is to determine all D(4)-quadruples of the form {k -2, k + 2, 4k, d} in Z[i]. Thus, we have to solve the system

(k -2)d + 4 = x 2 , (k + 2)d + 4 = y 2 , (4k)d + 4 = z 2 , ( 1
)
in d, x, y, z ∈ Z[i]. Eliminating d in (1)
, we obtain the following system of diophantine equations

(k + 2)x 2 -(k -2)y 2 = 16 (2)
(4k)x 2 -(k -2)z 2 = 12k + 8. ( 3
) Lemma 2.1. Let k ∈ Z[i] with |k| > 2.
Then there exist x 0 and y 0 ∈ Z[i] such that

(i) (x 0 , y 0 ) is a solution of (2) (ii) the estimates |x 0 | 2 ≤ 8 |k -2| |k| -2 , ( 4
)
|y 0 | 2 ≤ 8 |k + 2| |k| -2 + 16 |k -2| . (5)
(iii) For each solution (x, y) of equation (2) there exists m ∈ Z such that

x √ k + 2 + y √ k -2 = (x 0 √ k + 2 + y 0 √ k -2) k + √ k 2 -4 2 m . (6)
Proof. If (x, y) is a solution of (2), then (x m , y m ) is obtained by

x m √ k + 2 + y m √ k -2 = (x √ k + 2 + y √ k -2) k + √ k 2 -4 2 m (7)
is also a solution of ( 2), for all m ∈ Z. Let (x 0 , y 0 ) be an element of the sequence (x m , y m ) m∈Z (defined by [START_REF] Dujella | There are only finitely many Diophantine quintuples[END_REF]) such that the absolute value |x 0 | is a minimal. We put

x √ k + 2 + y √ k -2 = (x 0 √ k + 2 + y 0 √ k -2) k + √ k 2 -4 2 x √ k + 2 + y √ k -2 = (x 0 √ k + 2 + y 0 √ k -2) k + √ k 2 -4 2 -1 = (x 0 √ k + 2 + y 0 √ k -2) k - √ k 2 -4 2 .
Due to minimality of |x 0 |, we have

|x 0 | ≤ |x | = x 0 k + y 0 (k -2) 2 |x 0 | ≤ |x"| = x 0 k -y 0 (k -2) 2 .
At least one of the expressions x 0 k+y 0 (k-2) 2 and x 0 k-y 0 (k-2) 2 must be greater or equal to x 0 k 2 , since x 0 k+y 0 (k-2) 2

+ x 0 k-y 0 (k-2) 2 ≥ 2 x 0 k 2 . Let us assume that x 0 k+y 0 (k-2) 2 ≥ x 0 k 2 . Hence, we get x 2 0 k 2 ≤ (x 0 k) 2 -(y 0 (k -2)) 2 4 ,
and

x 2 0 k 2 ≤ 4x 2 0 + 16(k -2) 4 .
Immediately, we obtain the following estimate for

|x 0 | |x 0 | 2 ≤ 8 |k -2| |k| -2 .
This implies that the estimate for

|y 0 | is (k -2)y 2 0 = (k + 2)x 2 0 -16 ≤ |k + 2| 8 |k -2| |k| -2 + 16.
It is obvious that there exist only finitely many x 0 an y 0 such that above estimates are fulfilled. Thus, the definition of x 0 implies there exist m 0 ∈ Z such that

x 0 √ k + 2 + y 0 √ k -2 = (x √ k + 2 + y √ k -2) k + √ k 2 -4 2 m 0
.

Therefore, we obtain

x √ k + 2 + y √ k -2 = (x 0 √ k + 2 + y 0 √ k -2) k + √ k 2 -4 2 -m 0
Analogously, all solutions of (3) are given by the following lemma.

Lemma 2.2. Let k ∈ Z[i] with |k| > 2.
Then, there exist x 1 and z 1 ∈ Z such that

(i) (x 1 , z 1 ) is a solution of (3) (ii) the estimates |x 1 | 2 ≤ |3k + 2| |k -2| |k -1| -1 , ( 8
)
|z 1 | 2 ≤ |4k| |3k + 2| |k -1| -1 + |12k + 8| |k -2| . (9)
(iii) For each solutions (x, z) of equation (3) there exist n ∈ Z such that

x √ 4k + z √ k -2 = (x 1 √ 4k + z 1 √ k -2) k -1 + √ k 2 -2k n . (10)
Now, we define the following sequences

v 0 = x 0 , v 1 = 1 2 (x 0 k + y 0 (k -2)) , v m+2 = kv m+1 -v m , (11)
v 0 = x 0 , v 1 = 1 2 (x 0 k -y 0 (k -2)) , v m+2 = kv m+1 -v m , (12)
for all m ∈ N. If x is a solution of (2), then there exists a nonnegative integer

m such that x = v m or x = v m .
Similarly, if x is a solution of (3), then there exists a nonnegative integer n such that x = w n or x = w n , where

w 0 = x 1 , w 1 = 1 2 (2x 1 (k -1) + z 1 (k -2)) , w n+2 = 2(k -1)w n+1 -w n , (13)
w 0 = x 1 , w 1 = 1 2 (2x 1 (k -1) -z 1 (k -2)) , w n+2 = 2(k -1)w n+1 -w n . (14)
The next result will specify the values of x 0 and y 0 .

Lemma 2.3. Let k ∈ Z[i], k = ±6 and |k| > 5.
Then, x 0 = ±2 and y 0 = ±2 are the only fundamental solutions of (2) and all solutions are represented by the sequences (v m ) m and (-v m ) m defined by

v 0 = 2, v 1 = 2k -2, v m+2 = kv m+1 -v m , m ∈ Z. (15)
Proof. Suppose (x 0 , y 0 ) is a fundamental solution of (2). Then we have

x 2 0 (k + 2) -y 2 0 (k -2) = 16.
Thus, we have

x 2 0 -y 2 0 = 16-4x 2 0 k-2 . Relation (1) implies x 2 0 -y 2 0 = 4d 0 , x 2 0 ≡ 4 (mod k - 2), with d 0 ∈ Z[i]
. Thus, we have

|d 0 | = |4 -x 2 0 | |k -2| ≤ 4 + |x 2 0 | |k -2| ≤ 12 |k| -2 . ( 16
)
If |k| > 14, then we have |d 0 | < 1, so d 0 = 0. Thus, we have x 2 0 -y 2 0 = 0. So we deduced x 0 = ±2 and y 0 = ±2. If 5 < |k| ≤ 14 and k = ±6, using Maple, the following relations

x 0 ∈ Z[i], y 0 ∈ Z[i], x 2 0 = 4 + d 0 (k -2), y 2 0 = 4 + d 0 (k + 2
) and ( 16) are simultaneous satisfied if and only if x 0 = ±2 and y 0 = ±2.

Let us mention that from this lemma, we have that for k = ±6 and |k| > 5, all possible d's are of the form d = 2d for some d ∈ Z[i]. We can conclude that, because those fundamental solutions will generate all possible elements that extend D(4)-pair {k -2, k + 2}, and it is easy to see that all of them will be of that form 2d . It furthermore implies, using the above mentioned result from [START_REF] Franušić | On the extensibility of Diophantine triples {k-1, k+1, 4k} for Gaussian integers[END_REF], that for such k and k = 2t for some t ∈ Z[i], the statement of the main Theorem is valid.

Before proceeding further, let us recapitulate our results. For |k| > 5 and k = ±6 the problem of solving (2) and ( 3) is reduced to looking for the intersections of recursive sequences, i.e. to solving the equations

v m = ±w n , v m = ±w n , m, n ≥ 0. ( 17
)
3 The congruence method

In this section, we will determine all fundamental solutions of equation ( 3) under the assumption that one of the equations in (17).

Lemma 3.1. Let k ∈ Z[i], |k| > 5 and k = ±6. If (x 1 , z 1
) is a fundamental solution of (3), then

x 1 (mod k -1) ∈ {±2, 0} or 1 2 (k -2)z 1 (mod k -1) ∈ {±2, 0}.
Proof. We have

(v m (mod k -1)) m≥0 = (2, 0, -2, -2, 0, 2, 2, 0, -2, -2, 0, 2, . . . ), (w n (mod k -1)) n≥0 = (x 1 , 1 2 (k -2)z 1 , -x 1 , - 1 2 (k -2)z 1 , x 1 , 1 2 (k -2)z 1 . . . ), (w n (mod k -1)) n≥0 = (x 1 , - 1 2 (k -2)z 1 , -x 1 , 1 2 (k -2)z 1 , x 1 , - 1 2 (k -2)z 1 , . . . ).
These congruence relations are obtained by induction from (15), ( 13), ( 14), respectively. The rest follows immediately from (17).

In what follows, we will discuss all the possibilities given in Lemma 3.1.

• x 1 ≡ ±2 (mod k -1): In this case, if x 1 = ±2, we have: |k -1| -2 ≤ |x 1 | .
From this relation and ( 8), we have

(|k| -3) 2 ≤ |x 1 | 2 ≤ |3k + 2| |k -2| |k -1| -1 .
Therefore, we obtain

|k| 3 -11 |k| 2 + 17 |k| -22 ≤ 0. If |k| ≥ 10, then |k| 3 -11 |k| 2 + 17 |k| -22 > 0. So, if |k| ≥ 10 then x 1 = ±2. If 5 < |k| < 10, then we put x 1 = ±2 + l 0 (k -1). Suppose that |l 0 | ≥ 2. We have |x 1 | ≥ 2 |k -1| -2.
This relation and ( 8) give

4 |k| 3 -27 |k| 2 + 44 |k| -36 ≤ 0. If |k| > 5, then we get 4 |k| 3 -27 |k| 2 + 44 |k| -36 > 0. So, if 5 < |k| < 10 then |l 0 | ≤ √ 2. In this case, l 0 = ±1 or l 0 = ±i or l 0 = ±(1 ± i). l 0 = ±1 implies x 1 = ±(k + 1) or x 1 = ±(k -3).
In this case, the following relation x 2 1 ≡ 4 (mod (k -2)), [START_REF] Franušić | On the extensibility of Diophantine triples {k-1, k+1, 4k} for Gaussian integers[END_REF] are not simultaneously satisfied.

l 0 = ±i implies x 2 1 ≡ 3 ± 4i (mod (k -2)
). This relation and ( 1) imply 1 ± 4i ≡ 0 (mod (k -2)). We deduce |k| ≤ √ 17 + 2. In this case, relation ( 8) is not satisfied.

l 0 = 1 ± i and x 1 = 2 + l 0 (k -2) imply |x 1 | ≥ √ 2 |k| - √ 2.
These relations and (8) imply

2 |k| 3 -11 |k| 2 + 6 |k| -8 ≤ 0. This inequality is not satisfied if 5 < |k| < 10. l 0 = -1 ± i and x 1 = 2 + l 0 (k -2) imply 2x 3 -(7 + 4 √ 5)x 2 + (8 √ 5 + 6)x -24 ≤ 0. This inequality is not satisfied if |k| ≥ √ 41. If 5 < |k| < √ 41, using Maple in the various cases the relation (8) is not satisfied. If x 1 = -2 + l 0 (k -2), we have similar results. So, if x 1 ≡ ±2 (mod k -1) and |k| > 5 , then x 1 = ±2. • x 1 ≡ 0 (mod (k -1)): In this case, if x 1 = 0, we have |k -1| ≤ |x 1 |
. This relation and ( 8) give

(|k| -1) 2 ≤ |x 1 | 2 ≤ |3k + 2| |k -2| |k -1| -1 .
Therefore, we obtain

|k| 3 -7 |k| 2 + |k| -6 ≤ 0. If |k| ≥ 7, then |k| 3 -7 |k| 2 + |k| -6 > 0.
So we deduce that x 1 = 0. Suppose that

x 1 = l 1 (k -1). If 5 < |k| < 7 then |l 1 | ≤ 1. If |l 1 | = 1
, thus relations (1) or ( 8) are not satisfied. This implies that l 1 = 0. Consequently, if x 1 ≡ 0 (mod (k -1)) then x 1 = 0.

If

x 1 = 0, then relation (3) implies -(k -2)z 2 1 = 12k + 8. In this case, using Maple, the equation -(k -2)z 2 1 = 12k + 8 has no solution in Z[i] for all k ∈ Z[i] with 5 < |k| < 7. If x 1 = ±2, then relation (3) implies z 1 = ±2. • k = 6:
In this case, the fundamental solutions of ( 2) are (x 0 , y 0 ) = (±2, ±2), (x 0 , y 0 ) = (0, ±2i) and the fundamental solutions of ( 3

) are (x 1 , z 1 ) = (±2, ±2), (x 1 , z 1 ) = (±(2 + 2i), ±(4 + 6i)), (x 1 , z 1 ) = (±(2 -2i), ±(4 -6i)).
In Section 2, the solutions of (2) are given by recurrence sequences (11) and (12). In these cases, we have

q 0 = 0, q 1 = 4i, q m+2 = 6q m+1 -q m , m ∈ N, q 0 = 0, q 1 = -4i, q m+2 = 6q m+1 -q m , m ∈ N, v 0 = 2, v 1 = 10, v m+2 = 6v m+1 -v m , m ∈ N.
From ( 13) and ( 14), all solutions of (3) are given by

x 0 = 2 + 2i, x 1 = 18 + 22i, x n+2 = 10x n+1 -x n , n ∈ N, x 0 = 2 + 2i, x 1 = 2 -2i, x n+2 = 10x n+1 -x n , n ∈ N, u 0 = 2 -2i, u 1 = 18 -22i, u n+2 = 10u n+1 -u n , n ∈ N, u 0 = 2 -2i, u 1 = 2 + 2i, u n+2 = 10u n+1 -u n , n ∈ N, w 0 = 2, w 1 = 14, w n+2 = 10w n+1 -w n , n ∈ N, w 0 = 2, w 1 = 6, w n+2 = 10w n+1 -w n , n ∈ N.
So, we have the following relations:

a) q m = ±x n or q m = ±x n or q m = ±w n or q m = ±w n or q m = u n or q m = u n , b) q m = ±x n or q m = ±x n or q m = ±w n or q m = ±w n or q m = u n or q m = u n , c) v m = ±x n or v m = ±x n or v m = ±w n or v m = ±w n or v m = u m or v m = u n .
We see that (q m (mod (10) = (0, 4i, 4i, 0, -4i, -4i, 0, 4i, 4i, ...), (q m (mod 10) = (0, -4i, -4i, 0, 4i, 4i, 0, -4i, -4i, ...), By these relations, we deduce that the only following equality are possible:

(v m (mod 10) = (2, 0, -2, -2, 0, 2, 2, 0, -2, ...), (x n (mod 10) = (2 + 2i, -2 + 2i, -2 -2i, 2 -2i, 2 + 2i, -2 + 2i, ...), (x n (mod 10) = (2 + 2i, 2 -2i, -2 -2i, -2 + 2i, 2 + 2i, 2 -2i, ...), (u n (mod 10) = (2 -2i, -2 -2i, -2 + 2i, 2 + 2i, 2 -2i, -2 -2i, -2 + 2i, ...), (u n (mod 10) = (2 -2i, 2 + 2i, -2 + 2i, -2 -2i,
v m = ±w n , or v m = ±w n .
• The case k = -6 is similar to the case k = 6.

The above results can be summarized in the following lemma.

w 0 = 2, w 1 = 3k -4, w n+2 = (2k -2)w n+1 -w n , (18)
w 0 = 2, w 1 = k, w n+2 = (2k -2)w n+1 -w n , (19)
for n ∈ N.

We now prove the following results:

Lemma 3.3. Let k ∈ Z[i] and |k| > 5.
• The equation v m = w 2n+1 has no solution.

• The equation v m = w 2n+1 has no solution.

Proof. The result comes from the following facts:

(v m (mod k -1)) m≥0 = (2, 0, -2, -2, 0, 2, 2, 0, -2, -2, 0, 2, . . . ), (w n (mod k -1)) m≥0 = (2, -1, -2, 1, 2, -1, -2, 1, 2, -1, -2, . . . ), (w n (mod k -1)) m≥0 = (2, 1, -2, -1, 2, 1, -2, -1, 2, 1, -2, -1, . . . ).
Lemma 3.4. The sequences (v m ) m , (w n ) n and (w n ) n defined by (15), ( 18) and (19) respectively, satisfy the following congruences

(v m (mod k)) m≥0 = (2, -2, -2, 2, 2, -2, -2, 2, 2, -2, ...), (w n (mod k)) n≥0 = (-1) n (2n + 2), (w n (mod k)) n≥0 = (-1) n+1 (2n -2).
Proof. This can be verified by induction.

Lemma 3.4 implies the following lemma.

x = v m = w n or x = v m = -w n or x = v m = w n or x = v m = -w n ,
where (v m ) m , (w n ) n and (w n ) n are defined by (15), ( 18) and (19), respectively. Now, observe that v 0 = w 0 = w 0 = 2 and v 2 = w 2 = 2k 2 -2k -2. So, x = ±2 and x = ±(2k 2 -2k -2) are solutions of the system of equations ( 2) and (3). The solutions x = ±2 are not interesting, because they correspond to d = 0. The solutions x = ±(2k 2 -2k -2) correspond to d = 4k 3 -4k. Since we intend to prove that this is the unique nontrivial extension of the D(4)-triple {k -2, k + 2, 4k} in Z[i]. We have to show that the system of equations ( 2), (3) has no other solution. Our next step is to determine an upper bound for all solutions of (2) and (3) that are different from the previous ones.

Lemma 3.6. Let k ∈ Z[i] and |k| > 5. If x ∈ Z[i]\{±2, ±(2k 2 -2k -2)
} is a solution of the system of equations (2) and (3), then 2n ≥ |k| -4.

Proof. If m ≥ 2, then |w 1 | < |v m | . Otherwise, |w 1 -v 1 | = 0, |w 1 -v 3 | = 0, |w 1 -v 4 | = 0. We have |v 0 | < |v 1 | . Suppose |v m | < |v m+1 |. We have |v m+2 | ≥ |kv m+1 | -|v n | = (|k| -1) |v m+1 | + |v m+1 | -|v m | > |v m+1 | . |w 1 | < |v 4 | < |v 5 | < .... |v 1 -w 2 | = 0, |v 3 -w 2 | = 0, |v 4 -w 2 | = 0, |v 5 -w 2 | = 0, |w 2 | < |v 6 | < |v 7 | < ...
So, by Lemma 3.5 if x is a solution of the system of equations ( 2) and ( 3), then we have 2n So, we conclude that

-2 ≥ 2 + |k| or 2n -2 ≥ -2 + |k| or 2n + 2 ≥ 2 + |k| or 2n + 2 ≥ -2 + |k|. Lemma 3.7. Let k ∈ Z[i] and |k| > 5. If x ∈ Z[i]\{±2, ±(2k 2 -2k -2)} is a solution of the system of equations (2) and (3) then |x| ≥ |k| (2 |k| -3) |k|-5 . Proof. If x ∈ Z[i]\{±2, ±(2k 2 -2k -2)}, then n ≥ 2. We have |w 2 | ≤ |w 3 | . Now, assume that |w n | ≤ |w n+1 | . From (18), we have that |w n+2 | ≥ (2 |k| -3) |w n+1 | + |w n+1 | -|w n | ≥ (2 |k| -3) |w n+1 | .
|w n+1 | ≥ |k| (2 |k| -3) n . Analogously, if n ≥ 2, we obtain |w n | ≥ |k| (2 |k| -3) n-1 . From Lemma 3.6, If x ∈ Z[i]\{±2, ±(2k 2 -2k -2)
} is a solution of the system of equations ( 2) and (3) then

|x| ≥ |k| (2 |k| -3) |k|-5 .

An application of the theorem on simultaneous approximations

In this section, we prove that if the parameter |k| is large enough, then x = ±2 and x = ±(2k 2 -2k -2) give all solutions of the system of equations ( 2), [START_REF] Bennett | On the number of solutions of simultaneous Pell equations[END_REF]. For that reason, we apply the following generalization of Bennett's theorem [START_REF] Bennett | On the number of solutions of simultaneous Pell equations[END_REF] on simultaneous rational approximations of square roots which are close to one.

Theorem 4.1. ( [START_REF] Jadrijević | A system of relative Pellian equations and related family of relative Thue equations[END_REF]) Let θ i = 1 + a i T , i = 1, 2, with a 1 and a 2 pairwise distinct quadratic integers in the imaginary quadratic field K and let T be an algebraic integer of K. Further, let

M = max{|a 1 | , |a 2 |}, |T | > M and l = 27 |T | 64 |T | -M , L = 27 16 |a 1 | 2 |a 2 | 2 |a 1 -a 2 | 2 (|T | -M ) 2 > 1, p = 2 |T | + 3M 2 |T | -2M , P = 16 |a 1 | 2 |a 2 | 2 |a 1 -a 2 | 2 min{|a 1 | , |a 2 | , |a 1 -a 2 |} 3 (2 |T | + 3M ). Then max θ 1 - p 1 q , θ 2 - p 2 q > c |q| -λ ,
for all algebraic integers p 1 , p 2 , q ∈ K, where

λ = 1 + log P log L , c -1 = 4pP (max{1, 2l}) λ-1 .
First, let us show the following technical lemma.

1 = ± 1 + 4 k -2 , θ (1)
1 = -θ (2)
2 = ± 1 + 2 k -2 , θ (1) 1 , θ (1)
2 = -θ (2)
where the signs are chosen such that

θ (1) 1 - y x ≤ θ (2) 1 - y x , θ (1)
2 -

z 2x ≤ θ (2) 2 - z 2x .
Then, we obtain

θ (1) 1 - y x ≤ 16 |k 2 -4| |x| 2 , θ (1)
2 -

z 2x ≤ |3k + 2| k(k -2) |x| 2 .
Proof. We have

θ (1) 1 - y x = (θ (1) 1 ) 2 - y 2 x 2 • θ (1) 1 + y x -1 = 16 |k -2| |x| 2 θ (2) 1 - y x -1
.

Because of the assumptions on θ

(1)

1 and θ (2) 1 , we get θ (2) 1 - y x ≥ 1 2 θ (2) 1 - y x + θ (1) 1 - y x ≥ 1 2 θ (1) 1 -θ (2) 1 = k + 2 k -2 .
Hence, we see that

θ (1) 1 - y x ≤ 16 |k 2 -4| |x| 2 .
Similarly, we get

θ (1) 2 - z 2x = (θ (1)
2

) 2 - z 2 (2x) 2 • θ (1) 2 + z 2x -1 = |3k + 2| |k -2| |x| 2 θ (2) 2 - z 2x -1 , and θ (2) 2 - z 2x ≥ k k -2 ,
the other estimate is obtained. Now, we apply Theorem 4.1 on θ

(1)

1 and θ

2 . In our case, we have

a 1 = 4, a 2 = 2, T = k -2, M = 4 and l = 27 |k -2| 64 |k -2| -4 , L = 27 4096 (|k -2| -4) 2 , p = 2 |k -2| + 12 2 |k -2| -8 = |k -2| + 6 |k -2| -4 , P = 1024(|k -2| + 6).
If |k| ≥ 155402, then we have L > 1. Thus, the condition L > 1 of Theorem 4.1 is satisfied. So, we conclude that max θ

(1)

1 - y x , θ (1)
2 -

z 2x > c |2x| -λ , (20)
where

λ = 1 + log P log L , c -1 = 4pP (max{1, 2l}) λ-1 .
If we assume that |k| ≥ 155402, then max{1, 2l} = 1 and c -1 = 4pP. Furthermore, according to Lemma 4.1, we have max θ

(1)

1 - y x , θ (1)
2 -

z 2x ≤ 3.1 |x| 2 ,
and inequality (20) implies

1 4 |k -2| -4 |k -2| + 6 • 1 1024(|k -2| + 6) |2x| -λ < 3.1 |x| 2 .
Hence,

|2x| 2-λ ≤
= ±2, y = ±2, z = ±2 and x = ±(2k 2 -2k -2), y = ±(2k 2 + 2k -2), z = ±(4k 2 -2).

Linear forms in three logarithms

In this section, we will use the already well-known method in solving the equations of the form v m = w n or v n = w n , that consists in searching for the intersection of binary recurrence sequences. This will be done using Baker's theory on linear forms in logarithms. It gives the upper bound of m (and n) which will later be reduced using the reduction method. For more details, one can refer to [START_REF] Dujella | An absolute bound for the size of Diophantine m-tuples[END_REF] (for the integer case) and [START_REF] Franušić | On the extensibility of Diophantine triples {k-1, k+1, 4k} for Gaussian integers[END_REF] (for the case of Gaussian integers). We assume 5 < |k| < 155402 and we will use it at several places. Let v m = w n or v m = w n , for n > 2 and |k| > 5. Then we have

v m = √ k + 2 + √ k -2 √ k + 2 k + √ k 2 -4 2 m + √ k + 2 - √ k -2 √ k + 2 k - √ k 2 -4 2 m , w n = √ 4k + √ k -2 √ 4k k -1 + √ k 2 -2k n + √ 4k - √ k -2 √ 4k k -1 - √ k 2 -2k n , w n = √ 4k - √ k -2 √ 4k k -1 + √ k 2 -2k n + √ 4k + √ k -2 √ 4k k -1 - √ k 2 -2k n .

Now, let us define

P = √ k + 2 + √ k -2 √ k + 2 k + √ k 2 -4 2 m , Q = √ 4k ± √ k -2 √ 4k k -1 + √ k 2 -2k 2n , where k -2 k + 2 , ± k -2 4k , 1 - 4 k 2 , 1 - 1 (k -1) 2 are chosen such that Re k -2 k + 2 ≥ 0, Re ± k -2 4k ≥ 0, Re 1 - 4 k 2 ≥ 0, Re 1 - 1 (k -1) 2 ≥ 0.
The equation v m = w 2n or v m = w 2n implies that

P + 4 k + 2 P -1 = Q + 3k + 2 4k Q -1 . ( 23
)
We have

|P | = 1 + k -2 k + 2 k 2 m 1 + 1 - 4 k 2 m > k 2 2 > 2.5 2 > 6. |Q| = 1 ± k -2 4k |k -1| 2n 1 + 1 - 1 (k -1) 2 2n > |k -1| 2n > 4 4 ≥ 256.
Furthermore, from (23) we have

||P | -|Q|| ≤ |P -Q| ≤ 3k + 2 4k |Q| -1 + 4 k + 2 |P | -1 < 0.23.
P -Q P ≤ 3k + 2 4k |Q| -1 |P | -1 + 4 k + 2 |P | -2 < 2.19 |P | -2 < 0.061.
Finally, we get

log |P | |Q| = log 1 - |P | -|Q| |P | < 2.19 |P | -2 + (2.19 |P | -2 ) 2 < 0.061• 2.19• |P | -2 < k 2 -2m < 6 -m .
The above expression can be written as a linear form in three logarithms:

m log k + √ k 2 -4 2 -2n log k -1 + √ k 2 -2k + log √ 4k( √ k + 2 + √ k -2) √ k + 2( √ 4k ± √ k -2) < 6 -m . ( 24
) Lemma 5.1. If v m = w 2n , then |P | = |Q| for k ∈ Z[i], k = 0, ±2.
α = k -2 k + 2 , β = k -2 k .
Then, we have

P = a + bα Q = c + dβ, where a, b, c, d ∈ Q[i]. Furthermore, the assumption v m = w n implies that a = c because v m = a + bα + a -bα = 2a, w n = c + dβ + c -dβ = 2c.
Moreover, we have

|P | 2 = p + uα + uα + q|α| 2 , |Q| 2 = r + vβ + vβ + s|β| 2 .
where p, q, r, s

∈ Q and u, v ∈ Q[i].
In the exactly same way as in [START_REF] Franušić | On the extensibility of Diophantine triples {k-1, k+1, 4k} for Gaussian integers[END_REF] we can prove several important facts:

• The complex numbers α and β are algebraic numbers of degree 2, for k ∈ Z[i], k = 0, ±2.

• The basis for Q[i](α, α), considered as a vector space over

Q[i] is B α = {1, α, α, |α| 2 },
and the basis for

Q[i](β, β) is B β = {1, β, β, |β| 2 }. • The set B = {1, α, α, |α| 2 , β, β, |β| 2 } is linearly independent. Now we have |P | 2 ∈ Q[i](α, α) and |Q| 2 ∈ Q[i](β, β
) and they are uniquely represented. Thus, the assumption |P | 2 = |Q| 2 implies p = q = r = s = 0 because B is linearly independent set. So we have P = a and Q = c which is not possible because we showed that a = c.

Put Λ = m log k + √ k 2 -4 2 -2n log k -1 + √ k 2 -2k + log √ 4k( √ k + 2 + √ k -2) √ k + 2( √ 4k ± √ k -2) .
From (24) and 5.1, if |k| > 5, we have 0

< |Λ| < 6 -m . Lemma 5.2. Let k ∈ Z[i] such that |k| > 5. If v m = ±w n or v m = ±w n then n ≤ m < 1.33n.
Proof. The statement is trivially satisfied if m = n = 0, which is the only possibility if one of the index is equal to 0. Moreover, it can be proved by induction that

(2 |k| -2)(|k| -1) m-1 ≤ |v m | ≤ (2 |k| + 2)(|k| + 1) m-1 , (3 |k| -4)(2 |k| -3) n-1 ≤ |w n | ≤ (3 |k| + 4)(2 |k| + 3) n-1 , |k| (2 |k| -3) n-1 ≤ |w n | ≤ |k| (2 |k| + 3) n-1 , for n, m ≥ 1. So v m = w n or v m = w n implies |k| (2 |k| -3) n-1 ≤ (2 |k| + 2)(|k| + 1) m-1 ,
which yields n < m. On the other hand, v m = w n or v m = w n also implies

2(|k| -1) m ≤ (3 |k| + 4)(2 |k| + 3) n-1 < 2(2 |k| + 3) n , if |k| > 5.
Now, if we take the logarithm of both sides of the inequality and use the condition |k| > 5, we easily get m < 1.33n.

We use the following theorem of Baker and Wüstholz (see [START_REF] Baker | Logarithmic forms and group varieties[END_REF], p.20) to obtain a upper bound for m.

Theorem 5.1. Let Λ be a nonzero linear form in logarithms of l algebraic numbers α 1 , . . . , α l with rational integer coefficients b 1 , . . . , b l . Then

log Λ ≥ -18(l + 1)!l l+1 (32d) l+2 h (α 1 ) • • • h (α l ) log(2ld) log B,
where B = max(|b 1 | , . . . , |b l |) and d is the degree of the number field generated by α 1 , . . . , α l over the rationals.

Here

h (α) = max(h(α), 1 d |log α| , 1 d ),
where h(α) denotes the standard logarithmic Weil height of α.

In our case, we have

log |m log α 1 -2n log α 2 + log α 3 | ≥ -18• 4!3 4 (32d) 5 h (α 1 ) • • • h (α l ) log(2ld) log B,
where

α 1 = k + √ k 2 -4 2 , α 2 = k -1 + √ k 2 -2k , α 3 = √ 4k( √ k + 2 + √ k -2) √ k + 2( √ 4k ± √ k -2) .
The minimal polynomials of α 1 , α 2 are respectively

p 1 (x) = (2x 4 -|k| 2 x 2 + 2) 2 -k 2 -4 2 x 4 , (25)
p 2 (x) = (x 4 -2 |k -1| 2 + 1) 2 -4 k 2 -2k 2 x 4 . (26)
The roots of p 1 are

x 1 , x 2 = ± k + √ k 2 -4 2 = ±α 1 , x 3 , x 4 = ± k - √ k 2 -4 2 , x 5 , x 6 = ± |k| 2 -|k 2 -4| + (|k| 2 -|k 2 -4|) 2 -4 2 , x 7 , x 8 = ± |k| 2 -|k 2 -4| -(|k| 2 -|k 2 -4|) 2 -4 2 . If i = 1, 2 then |x i | > 1 and if i = 3, 4, 5, 6, 7, 8 then |x i | ≤ 1. Thus, if 5 < |k| < 155402 then we have h(α 1 ) = 1 4 log(2 k + √ k 2 -4 ) < 3.34.
The roots of p 2 are

x 1 , x 2 = ± k -1 + √ k 2 -2k = ±α 2 , x 3 , x 4 = ± k -1 - √ k 2 -2k , x 5 , x 6 = ± |k -1| 2 -|k 2 -2k| + (|k -1| 2 -|k 2 -2k|) 2 -1, x 7 , x 8 = ± |k -1| 2 -|k 2 -2k| -(|k -1| 2 -|k 2 -2k|) 2 -1. If i = 1, 2, then |x i | > 1 and if i = 3, 4, 5, 6, 7, 8, thus |x i | ≤ 1. So, if 5 < |k| < 155402 then we have h(α 2 ) = 1 4 log k -1 + √ k 2 -2k < 3.17.
For the estimate of h(α 3 ), we compute the minimal polynomial using Maple. However, because we do not need this estimate to be so accurate (we will significantly improve the bound on m using the reduction method), we only estimate the leading coefficient of the minimal polynomial and the conjugates of α 3 . For that we use |k| > 5. We have Lemma 5.3. Let M be a positive integer, let p/q be a convergent of the continued fraction of the irrational γ such that q > 6M , and let A, B, µ be some real numbers with A > 0 and B > 1.

h(α 3 ) ≤ 1 32 log(|a 0 | |α | 32 ) < 14.
Let ε := µq -M γq , where • denotes the distance from the nearest integer. If ε > 0, then there is no solution to the inequality

0 < nγ -m + µ < AB -n , in positive integers m, n with log(Aq/ε) log B ≤ n ≤ M. • x 1 = ±(1 -2i) , z 1 = ±4i for k = -1 -2i, • x 1 = ±(2 + 3i) , z 1 = ±(10 + 6i) and x 1 = ±(2 + i) , z 1 = ±(6 + 2i) for k = 2 + i, • x 1 = ±(2 -3i) , z 1 = ±(10 -6i) and x 1 = ±(2 -i) , z 1 = ±(6 -2i) for k = 2 + i, • x 1 = ±(2 -i) , z 1 = ±(2 -2i) for k = -2 + i, • x 1 = ±(2 + i) , z 1 = ±(2 -2i) for k = -2 -i, • x 1 = ±(2 + 2i) , z 1 = ±(6 + 4i) and x 1 = 0, z 1 = ±(2 + 4i) for k = 2 + 2i, • x 1 = ±(2 -2i) , z 1 = ±(6 -4i) and x 1 = 0, z 1 = ±(2 -4i) for k = 2 -2i, • x 1 = ±(2 -2i) , z 1 = ±(-2 + 4i) for k = -2 + 2i, • x 1 = ±(2 + 2i) , z 1 = ±(-2 -4i) for k = -2 -2i, • x 1 = ±3 , z 1 = ±4 for k = -3, • x 1 = ±3i , z 1 = ±10i and x 1 = ±i , z 1 = ±6i for k = 4, • x 1 = ±1 , z 1 = ±2i for k = -4.
Each of the above cases will be treated separately.

• k = 4i
In Section 2, the solutions of ( 2) are given by recurrence sequences (11) and ( 12). In this cases we have q 0 = 1 -i, q 1 = -1 + 3i, q m+2 = 4iq m+1 -q m , m ∈ N,

q 0 = 1 -i, q 1 = 5 + i, q m+2 = 4iq m+1 -q m , m ∈ N, u 0 = 1 -3i, u 1 = -1 + i, u m+2 = 4iu m+1 -u m , m ∈ N, u 0 = 1 -3i, u 1 = 13 + 3i, u m+2 = 4iu m+1 -u m , m ∈ N, v 0 = 2, v 1 = -2 + 8i, v m+2 = 4iv m+1 -v m , m ∈ N.
The relations (13) and ( 14) giving all solutions of (3) become

x 0 = 4i, x 1 = -34 -8i, x n+2 = 2(4i -1)x n+1 -x n , n ∈ N, x 0 = 4i, x 1 = 2, x n+2 = 2(4i -1)x n+1 -x n , n ∈ N, w 0 = 2, w 1 = -4 + 12i, w n+2 = 2(4i -1)w n+1 -w n , n ∈ N, w 0 = 2, w 1 = 4i, w n+2 = 2(4i -1)w n+1 -w n , n ∈ N.
So, we have the following relations: a): q m = ±x n or q m = ±x n or q m = ±w n or q m = ±w n , b): q m = ±x n or q m = ±x n or q m = ±w n or q m = ±w

(q m (mod 2) = (1 -i, -1 + i, -1 + i, 1 -i, 1 -i, -1 + i, ...), (q m (mod 2) = (1 -i, 1 + i, -1 + i, -1 -i, 1 -i, 1 + i, ...), (u m (mod 2) = (1 -i, -1 + i, -1 + i, 1 -i, 1 -i, -1 + i, -1 + i, ...), (u m (mod 2) = (1 -i, 1 + i, -1 + i, -1 -i, 1 -i, 1 + i, -1 + i, ...),
(v m (mod 2) = (0, 0, 0, 0, 0, 0, 0, 0, ...), (x n (mod 2) = (0, 0, 0, 0, 0, 0, 0, 0, 0, ...), (x n (mod 2) = (0, 0, 0, 0, 0, 0, 0, 0, 0, ..., ), (w n (mod 2) = (0, 0, 0, 0, 0, 0, 0, 0, 0, ...), (w n (mod 2) = (0, 0, 0, 0, 0, 0, 0, 0, 0, ...).

By these relations, we deduce that the only possible equations are:

v m = ±w n , or v m = ±w n , or v m = ±x n , or v m = ±x n .
The equations v m = ±w n and v m = ±w n can be solved similarly as in the previous sections. Moreover, x n = w n+1 and x n+1 = w n .

• k = 1 + 2i In Section 2, the solutions of ( 2) are given by recurrence sequences (11) and ( 12). In these cases, we have q 0 = 4i, q 1 = -10 + 4i, q m+2 = (1 + 2i)q m+1 -q m , m ∈ N, q 0 = 4i, q 1 = 2, q m+2 = (1 + 2i)q m+1 -q m , m ∈ N,

v 0 = 2, v 1 = 4i, v m+2 = (1 + 2i)v m+1 -v m , m ∈ N.
The relations (13) and ( 14) that give all solutions of (3) become

x 0 = 1 + 2i, x 1 = -10 + 4i, x n+2 = 2(2i)x n+1 -x n , n ∈ N, x 0 = 1 + 2i, x 1 = 2, x n+2 = 2(2i)x n+1 -x n , n ∈ N, w 0 = 2, w 1 = -1 + 6i, w n+2 = 2(2i)w n+1 -w n , n ∈ N, w 0 = 2, w 1 = 1 + 2i, w n+2 = 2(2i)w n+1 -w n , n ∈ N.
Thus, the following cases should be analyzed: a): q m = ±x n or q m = ±x n or q m = ±w n or q m = ±w n , b): q m = ±x n or q m = ±x n or q m = ±w n or q m = ±w n , c):

v m = ±x n or v m = ±x n or v m = ±w n or v m = ±w n . Moreover, q m = v m+1 , q m+2 = v m , x n = w n+1 , x n+1 = w n . So, the cases a), b), c) can be reduced to v m = ±w n or v m = ±w n .
This case can be solved similarly as in the previous sections.

• k = -2 + i In Section 2, the solutions of (2) are given by recurrence sequences (11) and (12). In these cases, we have q 0 = 6 -2i, q 1 = -8 + 10i, q m+2 = (-2 + i)q m+1 -q m , m ∈ N,

q 0 = 6 -2i, q 1 = -2, q m+2 = (-2 + i)q m+1 -q m , m ∈ N, v 0 = 2, v 1 = -6 + 2i, v m+2 = (-2 + i)v m+1 -v m , m ∈ N.
The relations (13) and ( 14) that all solutions of (3) are

x 0 = 2 -i, x 1 = -8 + 10i, x n+2 = 2(-3 + i)x n+1 -x n , n ∈ N, x 0 = 2 -i, x 1 = -2, x n+2 = 2(-3 + i)x n+1 -x n , n ∈ N, w 0 = 2, w 1 = -10 + 3i, w n+2 = 2(-3 + i)w n+1 -w n , n ∈ N, w 0 = 2, w 1 = -2 + i, w n+2 = 2(-3i)w n+1 -w n , n ∈ N.
So, the following cases should be analyzed: a): q m = ±x n or q m = ±x n or q m = ±w n or q m = ±w n , b):

q m = ±x n or q m = ±x n or q m = ±w n or q m = ±w n , c): v m = ±x n or v m = ±x n or v m = ±w n or v m = ±w n . Moreover, q m = -v m+1 , q m+2 = v m , x n = -w n+1 , x n+1 = -w n . So, the cases a), b), c) are reduced to v m = ±w n or v m = w n .
These equations can be solved similarly as in the previous sections.

• k = -2 + 2i
In the Section 2, the solutions of (2) are given by recurrence sequences (11) and (12). In this cases we have q 0 = 2 -2i, q 1 = 4i, q m+2 = (-2 + 2i)q m+1 -q m , m ∈ N, q 0 = 4i, q 1 = -10 -6i, q m+2 = (-2 + 2i)q m+1 -q m , m ∈ N,

q 0 = 4i, q 1 = 2 -2i, q m+2 = (-2 + 2i)q m+1 -q m , m ∈ N, v 0 = 2, v 1 = -6 + 4i, v m+2 = (-2 + 2i)v m+1 -v m , m ∈ N.
The relations (13) and ( 14) that give all solutions of (3) are

x 0 = 2 -2i, x 1 = -2, x n+2 = 2(-3 + 2i)x n+1 -x n , n ∈ N, x 0 = 2 -2i, x 1 = -2 + 20i, x n+2 = 2(-3 + 2i)x n+1 -x n , n ∈ N, w 0 = 2, w 1 = -10 + 3i, w n+2 = 2(-3 + i)w n+1 -w n , n ∈ N, w 0 = 2, w 1 = -2 + i, w n+2 = 2(-3i)w n+1 -w n , n ∈ N.
Hence, the following cases should be analyzed: a): q m = ±x n or q m = ±x n or q m = ±w n or q m = ±w n , b): q m = ±x n or q m = ±x n or q m = ±w n or q m = ±w n , c): q m = ±x n or q m = ±x n or q m = ±w n or q m = ±w n , d): v m = ±x n or v m = ±x n or v m = ±w n or v m = ±w n . Moreover, we have q m+1 = q m , q m = q m+1 , w n = -x n+1 , w n+1 = -x n . So, the cases a), b), c), d) are reduced to v m = ±w n or v m = w n or q m = ±w n or q m = ±w n .

These equations v m = ±w n or v m = w n can be solved similarly as in previous sections. Moreover, by the congruence method, the equations w n = ±q m have no solution. Now, we study the equations q m = ±w n . One can check that for the small values of m, n, i.e. m, n ≤ 2 and one gets the only solutions: q 0 = -w 1 = 2 -2i, which gives d = 2i. Thus, we have to solve q m = ±w n , for n > 2. The same way as in Lemma 5.2, we get n < m. Moreover, using Baker's theory on linear forms in logarithms we get

0 < |m log α 1 -n log α 2 + log α 3 | < 1.42 -m ,
where

α 1 = -1 + i + √ -2i -1 , α 2 = -3 + 2i + √ 4 -12i , α 3 = (1 -i) √ -8 + 8i √ -8 + 8i - √ -4 + 2i .
Now, we have to combine this bound with that obtained by Baker-Wüstholz theorem. Similarly as for general case, we get h(α 1 ) < 0.27, h(α 2 ) < 0.5, h(α 3 ) < 3.61.

This yields m log m < 5.54• 10 29 and then m < 4.1• 10 31 . The previous reduction method confirms the result.

• k = -3 In Section 2, the solutions of (2) are given by recurrence sequences (11) and (12). In this cases we have q 0 = 6i, q 1 = -14i, q m+2 = (-3)q m+1 -q m , m ∈ N,

q 0 = 6i, q 1 = -4i, q m+2 = (-3)q m+1 -q m , m ∈ N, q 0 = 4i, q 1 = -6i, q m+2 = (-3)q m+1 -q m , m ∈ N, v 0 = 2, v 1 = -8, v m+2 = (-3)v m+1 -v m , m ∈ N.
The relations (13) and ( 14) of all solutions of (3) are given by x 0 = 3, x 1 = -22, x n+2 = -8x n+1 -x n , n ∈ N,

x 0 = 3, x 1 = -2, x n+2 = -8x n+1 -x n , n ∈ N, w 0 = 2, w 1 = -13, w n+2 = -8w n+1 -w n , n ∈ N, w 0 = 2, w 1 = -3, w n+2 = -8w n+1 -w n , n ∈ N.

Therefore, the following cases will be analyzed: a): q m = ±x n or q m = ±x n or q m = ±w n or q m = ±w n , b): q m = ±x n or q m = ±x n or q m = ±w n or q m = ±w n , c): q m = ±x n or q m = ±x n or q m = ±w n or q m = ±w n , d): v m = ±x n or v m = ±x n or v m = ±w n or v m = ±w n . We have q m ∈ iZ, q m ∈ iZ, q m ∈ iZ, x n ∈ Z, x n ∈ Z, w n ∈ Z, w n ∈ Z. So in the cases a), b), c) have no solution. Moreover, x n = -w n+1 and x n+1 = -w n . Consequently, the case d) can be summed to v m = ±w n or v m = w n . This case can be solved similarly as in previous sections.

• k = 4

In Section 2, the solutions of (2) are given by recurrence sequences (11) and (12). In these cases, we have q 0 = 1 -i, q 1 = 3 -5i, q m+2 = 4q m+1 -q m , m ∈ N, q 0 = 1 -i, q 1 = 1 + i, q m+2 = 4q m+1 -q m , m ∈ N,

v 0 = 2, v 1 = 6, v m+2 = 4v m+1 -v m , m ∈ N.
The relations (13) and ( 14) of all solutions of (3) become So, we have the following relations: a): q m = ±x n or q m = ±x n or q m = ±w n or q m = ±w n or q m = u n or q m = u n , b): q m = ±x n or q m = ±x n or q m = ±w n or q m = ±w n or q m = u n or q m = u n , c): v m = ±x n or v m = ±x n or v m = ±w n or v m = ±w n or v m = u n or v m = u n .

x 0 = 3i, x 1 = 19i, x n+2 = -6x n+1 -x n , n ∈ N, x 0 = 3i, x 1 = -i, x n+2 = -6x n+1 -x n , n ∈ N,
We have:

(q m (mod 4)) = (1 -i, -1 -i, -1 + i, 1 + i, 1 -i, ...), (q m (mod 4)) = (1 -i, 1 + i, -1 + i, -1 -i, 1 -i, ...), (v m (mod 4)) = (2, 2, -2, -2, 2, 2, -2, -2, ...), (x n (mod 4)) = (-i, -i, -i, -i, -i, -i, -i, ..., ), (x n (mod 4)) = (-i, -i, -i, -i, -i, -i, -i, ..., ), (u n (mod 4)) = (i, i, i, i, i, i, i, i, i, , ..., ), (u n (mod 4)) = (i, i, i, i, i, i, i, i, i, ..., ), (w n (mod 4)) = (2, 0, -2, 0, 2, 0, -2, 0, 2, ...).

nontrivial extension of the D(4)-triple {i -2, i + 2, 4i} is obtained for d = -8i, i.e. 4k 3 -4k, for k = i.

• k = 2i. In this case, the fundamental solution of ( 2) is (x 0 , y 0 ) = (2, 2) and the fundamental solution of ( 3) is (x 1 , z 1 ) = [START_REF] Bayad | Extension of a parametric family of Diophantine triples in Gaussian integers[END_REF][START_REF] Bayad | Extension of a parametric family of Diophantine triples in Gaussian integers[END_REF]. By repeating the procedure described in Section 5, we conclude that the above equations have two solutions v 0 = w 0 = w 0 = 2 and v 2 = w 2 = -10 -4i. So, the nontrivial extension of the D(4)-triple {2i -2, 2i + 2, 8i} is obtained for d = -40i, i.e. 4k 3 -4k, for k = 2i.

• k = 1 + i.

In this case, the fundamental solution of ( 2) is (x 0 , y 0 ) = (2, 2) and the fundamental solution of ( 3) is (x 1 , z 1 ) = (2, 2). Thus, we get two solutionsv 0 = w 0 = w 0 = 2 and v 2 = w 2 = -4 + 2i. Therefore, the nontrivial extension of the D(4)-triple {-1 + i, 3 + i, 4 + 4i} is obtained for d = -12 + 4i, i.e. 4k 3 -4k, for k = 1 + i.

• k = 1 -i, k = -2i, k = i. By conjugating, these cases become the same as the previous cases.

• k = -1 -i, k = -1 + i, k = -1. Using Lemma 6.1, we make similar conclusion for these cases.

  2 -2i, 2 + 2i, ...), (w n (mod 10) = (2, 4, -2, -4, 2, 4, -2, -4, ...), (w n (mod 10) = (2, -4, -2, 4, 2, -4, -2, 4, ...).

Lemma 3 . 2 .

 32 Let k ∈ Z[i] and |k| > 5. If at least one of the equations in (17) is solvable, then all fundamental solutions of equation (3) are x 1 = ±2, z 1 = ±2 and related sequences (w n ) and (w n ) are given by

Lemma 3 . 5 .

 35 Let k ∈ Z[i], |k| > 5 and let x ∈ Z[i] be a solution of the system of equations (2) and (3). Then, there exist m, n ∈ N, (2n ± 2) ≡ ±2 (mod (k)), such that

  So, we get |w n+2 | ≥ |w n+1 | . One can see that |w 2 | ≥ |k| (2 |k| -3). Assume that |w n | ≥ |k| (2 |k| -3) n-1 . From (18), we have that |w n+1 | ≥ |k| (2 |k| -3) |w n | + |w n | -|w n-1 | ≥ |k| (2 |k| -3) |w n | .

Lemma 4 . 1 .

 41 Let k ∈ Z[i] and |k| > 5 and let (x, y, z) ∈ Z[i] 3 be a solution of the system of (2) and (3). Furthermore, let θ

Hence,

  |P | ≤ |Q| + 0.23 ≤ 1.001 |Q| , which yields |Q| -1 ≤ 1.001 |P | -1 and

Proof.

  The proof is analogous as [8, Lemma 5.2]. Assume that |P | = |Q|. If P = Q, then (23) implies 3k 2 -8k + 4 = 0. The only solution in Z[i] of this equation is k = 2, so we conclude that P = Q. Let us denote

  n , c): u m = ±x n or u m = ±x n or u m = ±w n or u m = ±w n , d): u m = ±x n or u m = ±x n or u m = ±w n or u m = ±w n , e): v m = ±x n or v m = ±x n or v m = ±w n or v m = ±w n .Thus, we get:

u 0 =

 0 i, u 1 = 9i, u n+2 = -6u n+1 -x n , n ∈ N, u 0 = i, u 1 = -3i, u n+2 = -6u n+1 -u n , n ∈ N, w 0 = 2, w 1 = 8, w n+2 = -6w n+1 -w n , n ∈ N, w 0 = 2, w 1 = 4, w n+2 = -6w m+1 -w m , n ∈ N.

  If |k| ≥ 155402, then we have a contradiction. Therefore, we just prove the following statement.Theorem 4.2. Let k ∈ Z[i] and |k| ≥ 155402. Then, all solution of the system of equations (2) and (3) are given by x

	12697.6	|k -2| + 6 |k -2| -4	(|k -2| + 6).	(21)
	If |k| ≥ 155402, then 2 -λ ≥ 0.000014. Thus, using the estimate for x, |x| ≥ |k| (2 |k| -3) |k|-5
	(from Lemma 3.7), and after taking the logarithm of (21), we obtain
	(2 -λ) (log 2 + log |k| + (|k| -5) log(2 |k| -3)) < log 12697.6	|k -2| + 6 |k -2| -4	(|k -2| + 6)
					(22)

  m < 2.91• 10 33 . Therefore, we have just proved that for 5 < |k| < 155402, equation v m = w n or v m = w n implies m < 2.91• 10 33 . As this bound for m is very large, we have to reduce it.

	|α | <	4 |k|(2 |k| + 2) |k| -2( 4 |k| -|k| + 2)	< 5.27.
	Applying Baker-Wüstholz theorem with l = 3 and d ≤ 8• 32• 8 = 2048, we get -m log 6 >
	-34992• 1.3• 10 24 • 3.34• 3.17• 14.74• log 12288• log 2m, which yields
		m log 2m	< 3.74• 10 31

74, as |a 0 | ≤ ( |k| + 2( 4 |k| + |k| + 2)) 32 < 4.51• 10 181 , and

Acknowledgments

The authors express their gratitude to the anonymous referee for constructive suggestions to improve an earlier draft of this paper. The third author was supported by Croatian Science Foundation under the project no. 6422. The last author is partially supported by Purdue University Northwest.

As 0 < m log α 1 -2n log α 2 + log α 3 < 6 -m , we apply Lemma 5.3 with γ = log α 1 2 log α 2 , µ = log α 3 2 log α 2 , A = 1 2 log α , B = 6, M = 2.91 • 10 33 .

The program was developed in Mathematica running with 500 digits. For the computations, if the first convergent such that q > 6M does not satisfy the condition ε > 0, then we use the next convergent until we find the one that satisfies the conditions. In a few days all the computations were done. After the first run, we obtained M ≤ 67. For the second run, we set M = 67 to obtain another bound M = 16. The third run of the reduction method yields n ≤ m ≤ 14. In this range, we check the equations v m = w n and v m = w n and see that they confirm the result in the main theorem.

The case 2 < |k| ≤ 5

There are some extra fundamental solutions of ( 2) and ( 3) for certain values of parameter k.

Precisely, these fundamental solutions of (2) also appear (besides x = ±2):

, and for (3), we have:

By these relations, we deduce that the only possible equations are:

The equations v m = ±w n and v m = ±w n can be solved similarly as in previous sections.

By conjugating, this cases becomes the same as the previous one.

This relation is equivalent to

Using Lemma 6.1, the above cases become the same as the previous cases.

7

The case 1 ≤ |k| ≤ 2

• k = 1.

In this case, system (1) is equivalent to the following system of Pell equations

We have z 2 ≡ 0 (mod 4), so there exists z ∈ Z[i] such that z = 2z . Equations ( 27) and ( 28) are equivalent to 3x 2 + y 2 = 16 (29) 28) is (x, y, z) = (2, 2, 2). We deduce that d = 0.

• k = i.

In this case, the fundamental solution of ( 2) is (x 0 , y 0 ) = (2, 2) and the fundamental solution of ( 3) is (x 1 , z 1 ) = [START_REF] Bayad | Extension of a parametric family of Diophantine triples in Gaussian integers[END_REF][START_REF] Bayad | Extension of a parametric family of Diophantine triples in Gaussian integers[END_REF]. By repeating the procedure described in Section 5, we conclude that the above equations have two solutions v 0 = w 0 = w 0 = 2 and v 2 = w 2 = -4 -2i. So, the