On the solutions of the equation 𝒑 = 𝒙 𝟐 + 𝒚 𝟐 + 𝟏 in Lucas sequences

— In 1970, Motohashi proved that there are an infinite number of primes having the form 𝑥 2 + 𝑦 2 + 1 for some nonzero integers 𝑥 and 𝑦 . In this research, we present a technique for studying the solutions of the equation 𝑝 = 𝑥 2 + 𝑦 2 + 1 , where the unknowns are derived from some Lucas sequences of the first kind {𝑈 𝑛 (𝑃, 𝑄)} or the second kind {𝑉 𝑛 (𝑃, 𝑄)} with 𝑃 and 𝑄 are certain nonzero relatively primes integers. As applications to this technique, we apply our procedure in case of (𝑥, 𝑦, 𝑝) = (𝑈 𝑖 (𝑃, 𝑄), 𝑈 𝑗 (𝑃, 𝑄), 𝑈 𝑘 (𝑃, 𝑄)) or (𝑉 𝑖 (𝑃, 𝑄) , 𝑉 𝑗 (𝑃, 𝑄) , 𝑉 𝑘 (𝑃, 𝑄)) with 𝑖, 𝑗, 𝑘 ≥ 1,−2 ≤ 𝑃 ≤ 3 and 𝑄 = ±1 .


Introduction and preliminaries
Several scientists are interested in the study of prime numbers due to its usage and applications in numerous scientific domains such as mathematics and computer science. In reality, there exists an infinite number of prime numbers with many wellknown forms. For example, Edmund Landau [4] conjectured that there are an infinite number of primes of the type = 2 + 1. In addition, Shanks [9,8] conjectured that there exist an infinite number of prime numbers of the forms = 4 + 1 and = 1 2 ( 2 + 1) for some integers . Furthermore, Motohashi [6] proved that there are an infinite number of primes having the form = 2 + 2 + 1. On the other hand, it is also known that the Lucas sequences of the first kind { ( , )} (simply { }) or the second kind { ( , )} (simply { }) which are defined by the following relations provide infinitely many prime numbers (see e.g. [5] and [10]): where and are nonzero integers with ( , ) = 1. Additionally, the first and second types of Lucas sequences are associated in the identity 2 ( , ) = 2 ( , ) + 4 , where = 2 − 4 . Consequently, these sequences can be correspondingly expressed by the following formulas which are known as Binet's formulas: and ( , ) = + for ≥ 0, where is called the golden ratio and = −1 . Note that and are the roots of these sequences' characteristic polynomial that is defined by Thus, if / is not a root of unity then these sequences are said to be nondegenerate, and degenerate otherwise. As a result, they are degenerate only with ( , ) ∈ {(±1, 1), (±2, 1)}, for more details see e.g. [7]. Considering these sequences, these are several well-known sequences obtained with certain values of and . For instance, if ( , ) = (1, −1) we get the so called Fibonacci and Lucas sequences that are denoted by { } and { }, respectively.
Here, we present some results concerning of the Fibonacci numbers, Lucas numbers, Pell numbers and Pell-Lucas numbers (that we use later in the proofs of our main results) for which they respectively satisfy the following inequalities: Furthermore, Hashim, Szalay, and Tengely [3] proved that if ≥ 2 and − − 1 ≤ ≤ − 1 then Lucas sequences of the first and second kind satisfy the following inequalities: Considering that these sequences with the above mentioned forms of prime numbers, one may be interested in knowing whether or not there are infinitely many prime numbers of the forms = 2 + 1, 4 + 1, In [1,2], we studied such special solutions of the equations: = 2 + 1, = 4 + 1 and = 1 2 ( 2 + 1). In fact, we found out that these equations have a finite number of solutions, and that leads to having a finite number of such primes.
In the rest of this section, we present our procedure for investigating the solutions ( , , ) = ( , , ) with , , ≥ 1 of equation (14), where denotes a generalized Lucas number of the first or second kind, namely = or . We indeed apply the technique of this procedure in case of −2 ≤ ≤ 3 and = ±1, where { } and { } are nondegenerate sequences. It is clear that the equation (14) has such special solutions only if < and < . Hence, in order to find all the solutions ( , , ) = ( , , ) with , , ≥ 1, we fix the condition that ≤ < . Then after determining the corresponding solutions, we permute the first two components of the these solutions. In the following, we summarize the general steps of the procedures under the condition 1 ≤ ≤ < for finding the solutions ( , , ) = ( , , ) of equation (14), with , , ≥ 1. In other words, we consider the equation with 1 ≤ ≤ < and ( , , ) = ( , , ) or ( , , ). Then we proceed follow the following stapes: ▪ We first divide equation (15) by with using Binet's formulas presented by (4) or (5) and the inequalities presented by (8)-(13). Then after some simplifications, we obtain an upper bound for , say ≤ .
▪ By dividing the inequality (21) by 1 3 +2 , we get that which may also be written as Therefore, ≤ 3.
In equation (19), we substitute the values of such that ∈ {2, 3} to obtain the corresponding values of and . This can be done in the following steps: − For = 2, we obtain the equation − For = 3, we have Lastly, we have to acquire the values of and corresponding to each values of = 1 and 2. ▪ Firstly, we combine equation (23)  Proof: To begin the proof, we must first determine the upper bound for in the equation 134 such that , , ≥ 1 and ≤ < . Initially, we divide equation (25) by to obtain Subsequently, as ≤ . Thus, by replacing the inequality (9) and (identity (6)  in the later inequality, we conclude that This implies that So, by dividing the last inequality by 1 3 −1 , we discover that  (14), then it has no such solution. Proof:. We start by determining an upper bound for in the equation where , , ≥ 1 under the condition that ≤ < . By following the argument applied the proofs of Theorems 1 and 2, we first find the solutions of (27) under the condition that 1 ≤ ≤ < . Dividing the latter equation by with using the fact of inequality (10) and Binet's formula of Pell sequence presented in (7) imply that ▪ If = 1, we get that
for , , ≥ 1. Following the same strategy as in the proofs of the previous theorems with the assumption of 1 ≤ ≤ < , we ultimately conclude that   1, 3). Proof: To begin the proof, we first determine an upper bound for in the equation where , , ≥ 1 such that ≤ < . Next, we divide the above equation by with using ≤ to get Consequently, we conclude that by substituting inequality (12) and identity (4) into inequality (31) (with ( , ) = ((3 + √5)/2, (3 − √5)/2)). By replacing the assumption 1 ≤ ≤ < with the fact = −1 in the last inequality, we arrive to the inequality Hence, by dividing the last inequality by 3 +2 we get that Therefore, where , , ≥ 1. Using the same technique used in the proofs of the preceding theorems with the condition 1 ≤ ≤ < , we finally conclude that where , , ≥ 1 and ≤ < . Below is a summary of the necessary actions to achieve this bound: ▪ First, we divide equation (35) by and then use ≤ to obtain ▪ Therefore, we get that by combining inequality (12) and (identity (4) with ( , ) = ((3 + √13)/2, (3 − √13)/2)) into inequality (36). By replacing the assumption ≤ < with the fact = −1 in the latter inequality, we find that ▪ Dividing inequality (37) by 3 +2 , we find that With some simplification, we get that < 15.151 which implies that < 2.275. Hence, ≤ 2. We plug the values of in equation (35) to obtain the values of and corresponding to = 1 and 2.
Theorem 8. Suppose that = , = and = with , , ≥ 1, then equation (14) is not solvable over the integers , and . Proof: We begin by getting an upper bound for (with 1 ≤ ≤ < ) in the equation Similarly, by using the argument applied the proofs of the previous theorems, we get ≤ 2.
To get the values of and that correspond to the values of ∈ {1, 2}, we substitute the values of into equation (38).