On Second-Order Linear Sequences of Composite Numbers

Dan Ismailescu Department of Mathematics

Hofstra University Hempstead, NY 11549

USA

dan.p.ismailescu@hofstra.edu

Adrienne Ko

Ethical Culture Fieldston School

Bronx, NY 10471

USA

19ako@ecfs.org

Celine Lee Chinese International School

Hong Kong SAR

China

celinl@student.cis.edu.hk

Jae Yong Park The Lawrenceville School Lawrenceville, NJ 08648

USA

jpark19@lawrenceville.org

Abstract

We present a new proof of the following result of Somer:

Let (a, b) ∈ ℤ^{2} and let (x_{n})_{n≥0} be the sequence defined by some initial values x_{0} and x_{1} and the second-order linear recurrence

x_{n+1} = ax_{n} + bx_{n−1}

for n ≥ 1. Suppose that b ≠ 0 and (a, b) ≠ (2, −1), (−2, −1). Then there exist two relatively prime positive integers x_{0}, x_{1} such that |x_{n}| is a composite integer for all n ∈ ℕ.

The above theorem extends a result of Graham, who solved the problem when (a, b) = (1, 1).

1 Introduction

We give a new proof of the following result of Somer:

Theorem 1. [12, 3] Let (a, b) ∈ ℤ^{2} and let (x_{n})_{n≥0} be the sequence defined by some initial values x_{0} and x_{1} and the second-order linear recurrence

x_{n+1} = ax_{n} + bx_{n−1}  (1)

for n ≥ 1. Suppose that b ≠ 0 and (a, b) ≠ (2, −1), (−2, −1). Then there exist two relatively prime positive integers x_{0}, x_{1} such that |x_{n}| is a composite integer for all n ∈ ℕ.

Throughout the paper we will use the following convention: a nonnegative integer n is said to be composite if n ≠ 0, 1, and n is not a prime number.

Graham [5] considered the problem above in the particular case (a, b) = (1, 1). He found two relatively prime positive integers x_{0} and x_{1} such that the sequence x_{n+1} = x_{n} + x_{n−1}, consists of composite numbers only. Graham’s starting pair is

(x_{0}, x_{1}) = (331635635998274737472200656430763, 1510028911088401971189590305498785).

Graham’s technique was successively refined by Knuth [6], Wilf [16], and Nicol [10], who all found smaller pairs (x_{0}, x_{1}). The current record is due to Vsemirnov [15]:

(x_{0}, x_{1}) = (106276436867, 35256392432).  (2)

The above results are based on the fact that the Fibonacci sequence is a divisibility sequence, that is, F_{n} | F_{m} whenever n | m, and on finding a finite covering system of congruences r_{i} (mod m_{i}), 1 ≤ i ≤ t, such that there exist distinct primes p_{1}, p_{2}, … p_{t}, so that p_{i} | F_{m_{i}} for all i = 1, 2, …, t.

Somer’s proof of Theorem 1 is relatively short since it relies on several results of Bilu, Hanrot and Voutier [1], Choi [2], and Parnami and Shorey [11].

A few years later, unaware of Somer’s article, Dubickas, Novikas and S̆iurys [3] published a new solution that, although somewhat lengthier, is essentially self-contained.

In this paper, we present another free-standing proof of this theorem that, while comparable to the one in [3], differs from it in several important ways.

We summarize our plan for proving Theorem 1 as follows.

In Section 2 we prove three easy lemmata, which will be useful later on. We believe that Lemma 2 is of independent interest. In Section 3 we study two simple cases: (i) a = 0 and (ii) a^{2} + 4b = 0. In this section we also show that the condition (a, b) ≠ (±2, −1) is necessary. Section 4 deals with the case |b| ≥ 2. Following [3], we choose x_{1} ≡ 0 (mod |b|), since then (1) implies that x_{n} ≡ 0 (mod |b|) for all n ≥ 2. The main difficulty is to show that x_{0} and x_{1} can be chosen such that x_{n} ≠ −b, 0, b for every n ≥ 0. In this case, Dubickas et al. present a mainly existential proof which relies on a series of six lemmata. In contrast, our proof is constructive, as we provide explicit expressions for x_{0} and x_{1} as polynomials in a and b.

In Section 5 we consider the case |b| = 1. Dubickas, Novikas and S̆iurys prove that except for finitely many values of |a|, one can take x_{0} and x_{1} so that each x_{n} is divisible by one of five distinct appropriately chosen prime numbers. We prove that, with the exception of finitely many values of |a|, four primes suffice. The proof in the case |b| = 1 relies on the divisibility properties of the Lucas sequence of the first kind (u_{n})_{n≥0}, defined as

u_{0} = 0, u_{1} = 1 and u_{n+1} = au_{n} + bu_{n−1}, for n ≥ 1.  (3)

Finally, in Section 6 we prove that if |a| ≥ 3 and b = −1, then u_{n} is composite for all n ≥ 3. The interesting fact is that in this case it seems likely that there is no finite set of prime numbers p_{1}, p_{2}, … p_{t} such that each u_{n} is divisible by some p_{i}, i = 1, 2, … t.

2 Three useful lemmata

Lemma 2. Consider the sequence (x_{n})_{n≥0} given by (1). Then

x_{n+1}^{2} − ax_{n}x_{n+1} − bx_{n}^{2} = (−b)^{n}(x_{1}^{2} − ax_{0}x_{1} − bx_{0}^{2}).  (4)

Proof.

[\begin{matrix}x_{n+2} & x_{n+1} \\ x_{n+1} & x_{n} \end{matrix}] = [\begin{matrix}ax_{n+1} + bx_{n} & ax_{n} + bx_{n−1} \\ x_{n+1} & x_{n} \end{matrix}] = [\begin{matrix}a & b \\ 1 & 0 \end{matrix}] · [\begin{matrix}x_{n+1} & x_{n} \\ x_{n} & x_{n−1} \end{matrix}] =

= [\begin{matrix}a & b \\ 1 & 0 \end{matrix}] · [\begin{matrix}a & b \\ 1 & 0 \end{matrix}] · [\begin{matrix}x_{n} & x_{n−1} \\ x_{n−1} & x_{n−2} \end{matrix}] = [\begin{matrix}a & b \\ 1 & 0 \end{matrix}]^{2} · [\begin{matrix}x_{n} & x_{n−1} \\ x_{n−1} & x_{n−2} \end{matrix}] = ⋯ = [\begin{matrix}a & b \\ 1 & 0 \end{matrix}]^{n} · [\begin{matrix}x_{2} & x_{1} \\ x_{1} & x_{0} \end{matrix}] .

Taking determinants on both sides we obtain

det [\begin{matrix}x_{n+2} & x_{n+1} \\ x_{n+1} & x_{n} \end{matrix}] = det [\begin{matrix}a & b \\ 1 & 0 \end{matrix}]^{n} · det [\begin{matrix}x_{2} & x_{1} \\ x_{1} & x_{0} \end{matrix}] ⟹ x_{n+2} · x_{n} − x_{n+1}^{2} = (−b)^{n} · (x_{2}x_{0} − x_{1}^{2})

⟹ (ax_{n+1} + bx_{n})x_{n} − x_{n+1}^{2} = (−b)^{n} · ((ax_{1} + bx_{0})x_{0} − x_{1}^{2}), which after expanding becomes x_{n+1}^{2} − ax_{n}x_{n+1} − bx_{n}^{2} = (−b)^{n}(x_{1}^{2} − ax_{0}x_{1} − bx_{0}^{2}), as claimed.

Lemma 3. Consider the sequence (x_{n})_{n≥0} given by (1). Suppose that 1 ≤ |x_{0}| < |x_{1}| and |a| > |b| ≥ 1. Then the sequence (|x_{n}|)_{n≥0} is strictly increasing.

Proof. We use induction on n. By hypothesis, the statement is true for n = 0. Suppose that |x_{n−1}| < |x_{n}| for some n ≥ 1. We intend to prove that |x_{n}| < |x_{n+1}|. Indeed

|x_{n+1}| = |ax_{n} + bx_{n−1}| ≥ |ax_{n}| − |bx_{n−1}| = |a||x_{n}| − |b||x_{n−1}| >
> |a||x_{n}| − |b||x_{n}| = (|a| − |b|)|x_{n}|, by the induction hypothesis.

Using |a| − |b| ≥ 1, we obtain |x_{n+1}| > |x_{n}|, which completes the induction.

Lemma 4. Let n_{1}, n_{2} and n_{3} be three positive integers such that no prime number p divides all of them. Then there exists an integer k ≥ 2 such that n_{1} and n_{2} + kn_{3} are relatively prime.

Proof. Let d := gcd(n_{2}, n_{3}). Note that gcd(n_{1}, d) = 1, otherwise d divides n_{1}, n_{2} and n_{3}.

By Dirichlet’s theorem on arithmetic progressions there exists a k such that n_{2}/d+kn_{3}/d is a prime number greater than n_{1}. Then d (n_{2}/d + k · n_{3}/d) = n_{2} + k n_{3} is both greater and relatively prime to n_{1}.

3 Two simple special cases: (i) a = 0 and (ii) a^{2} + 4b = 0

Case (i): Since a = 0, it can be easily proved that x_{2n} = b^{n}x_{0} and x_{2n+1} = b^{n}x_{1}. It suffices to take x_{0} = 4 and x_{1} = 9 to obtain that x_{n} is composite for all n ≥ 0.

Case (ii): If a^{2} + 4b = 0, then a must be even; a = 2c and therefore b = −a^{2}/4 = −c^{2}. We divide the proof into two cases: |b| ≥ 2 and b = −1.

If |b| ≥ 2, we have |c| ≥ 2. Let us now take x_{0} = 4c^{2} − 1 and x_{1} = 2c^{3}. Then x_{0} and x_{1} are relative prime positive composite integers. One immediately obtains x_{2} = c^{2}; thus x_{2} is also composite. Also, it can be easily proved that x_{n} = c^{n} ((n − 1) − (2n − 4)c^{2}) for all n ≥ 3.

Note that for any n ≥ 3 one cannot have x_{n} = 0; otherwise, 4 ≤ c^{2} = (n−1)/(2n−4) ≤ 1, which is impossible. It follows that |x_{n}| is composite for all n ≥ 3.

If b = −1, then |a| = 2. If a = 2, then a simple induction shows that

x_{n+1} = (n + 1)x_{1} − nx_{0} = x_{1} + n(x_{1} − x_{0}) for all n ≥ 0,

that is, (x_{n})_{n≥0} is an arithmetic sequence whose first term and common difference are relatively prime. By Dirichlet’s theorem on primes in arithmetic progressions, it follows that |x_{n}| is a prime number for infinitely many values of n. If a = −2, then one can show that x_{n+1} = (−1)^{n}(x_{1} +n(x_{0} +x_{1})). In this case, (x_{n})_{n≥0} is the union of two arithmetic sequences, both of which have the first term and the common difference relatively prime. Again, for any choice of x_{0} and x_{1} relatively prime, |x_{n}| is a prime for infinitely many n. This proves the necessity of the condition (a, b) ≠ (±2, −1).

4 The case |b| ≥ 2

Based on the results from the previous section, from now on we can assume that

|a| ≥ 1 and a^{2} + 4b ≠ 0.  (5)

We divide the proof into three separate cases:

Case I: |a| > |b|, Case II: 1 ≤ |a| ≤ |b| and |b| is composite, and Case III: 1 ≤ |a| ≤ |b| and |b| is a prime.

Case I: |a| > |b|. To complete the proof of Theorem 1, in this case we take x_{0} = b^{4}−1, x_{1} = b^{4}. Clearly, x_{0} and x_{1} are both positive, relatively prime and composite. Moreover, x_{0} < x_{1}. Then x_{n} ≡ 0 (mod b) for all n ≥ 1, and by using Lemma 3 it follows that (|x_{n}|)_{n≥0} is strictly increasing. Hence, we have |x_{n}| ≥ x_{0} > |b| for all n ≥ 0, and therefore |x_{n}| is composite for all n ≥ 0.

Case II: 1 ≤ |a| ≤ |b| and |b| is composite. Choose x_{0} = 4b^{4} − 1, x_{1} = 2b^{2}. Clearly, gcd(x_{0}, x_{1}) = 1, and x_{0}, x_{1} are positive composite integers. It also follows that x_{n} ≡ 0 (mod b) for all n ≥ 1, and since |b| is composite, it follows that |x_{n}| is composite, unless x_{n} = 0 for some n. We will prove that this cannot happen.

To get a contradiction, suppose that x_{n+1} = 0 for some n ≥ 2. Then, by Lemma 2, we have x_{n}^{2} = (−b)^{n−1}(x_{1}^{2} − ax_{0}x_{1} − bx_{0}^{2}).

Since x_{0} = 4b^{4} − 1 and x_{1} = 2b^{2}, we have

x_{1}^{2} − ax_{0}x_{1} − bx_{0}^{2} = (−b)(16b^{8} + 8ab^{5} − 8b^{4} − 4b^{3} − 2ab + 1), which implies that

x_{n}^{2} = (−b)^{n}(16b^{8} + 8ab^{5} − 8b^{4} − 4b^{3} − 2ab + 1).  (6)

If n is even, this implies that 16b^{8} + 8ab^{5} − 8b^{4} − 4b^{3} − 2ab + 1 is a perfect square. If n is odd, this implies that (−b)(16b^{8} + 8ab^{5} − 8b^{4} − 4b^{3} − 2ab + 1) is a perfect square. We will prove that none of these are possible if a^{2} + 4b ≠ 0.

Note first that since |a| ≤ |b|, we have

(4b^{4} + ab − 2)^{2} < 16b^{8} + 8ab^{5} − 8b^{4} − 4b^{3} − 2ab + 1 < (4b^{4} + ab)^{2}.  (7)

Indeed, these inequalities are equivalent to the following ones:

8b^{4} − 2 − (4b^{3} + (ab − 1)^{2}) > 0, and 8b^{4} + 4b^{3} − 2 + (ab + 1)^{2} > 0.  (8)

We have

|a| ≤ |b| ⟹ |ab| ≤ b^{2} ⟹ |ab − 1| ≤ |ab| + 1 ≤ b^{2} + 1 ⟹ (ab − 1)^{2} ≤ b^{4} + 2b^{2} + 1. (9)

From this we obtain

|4b^{3} + (ab − 1)^{2}| ≤ 4|b|^{3} + (ab − 1)^{2} ≤ b^{4} + 4|b|^{3} + 2b^{2} + 1.  (10)

To prove the first inequality in (8), note that

8b^{4} − 2 − (4b^{3} + (ab − 1)^{2}) ≥ 8b^{4} − 2 − |4b^{3} + (ab − 1)^{2}| ≥
≥ 8b^{4} − 2 − (b^{4} + 4|b|^{3} + 2b^{2} + 1) = 7b^{4} − 4|b|^{3} − 2b^{2} − 3 =
= 7|b|^{4} − 4|b|^{3} − 2|b|^{2} − 3 = 4|b|^{3}(|b| − 1) + 2|b|^{2}(|b|^{2} − 1) + (|b|^{4} − 3) ≥
≥ 4|b|^{3} + 2|b|^{2} + |b|^{4} − 3 ≥ 4 · 2^{3} + 2 · 2^{2} + 2^{4} − 3 > 0.

The second inequality in (8) is much easier to prove:

8b^{4} + 4b^{3} − 2 + (ab + 1)^{2} ≥ 8b^{4} + 4b^{3} − 2 ≥ 8|b|^{4} − 4|b|^{3} − 2 ≥ 8 · 2^{4} − 4 · 2^{3} − 2 > 0.

Hence, the inequalities (7) hold.

If the middle term in (7) were to be a perfect square, the only option remaining is

16b^{8} + 8ab^{5} − 8b^{4} − 4b^{3} − 2ab + 1 = (4b^{4} + ab − 1)^{2}, which implies b^{2}(a^{2} + 4b) = 0.

However, this is impossible if a^{2}+4b ≠ 0. Hence, if one assumes that |a| ≤ |b| and a^{2}+4b ≠ 0, then 16b^{8} + 8ab^{5} − 8b^{4} − 4b^{3} − 2ab + 1 cannot be a perfect square.

Suppose next that (−b)(16b^{8} + 8ab^{5} − 8b^{4} − 4b^{3} − 2ab + 1) is a perfect square. Since the two factors are mutually prime, it follows that both have to be perfect squares. In particular, the second one has to be a perfect square, and we have already proved that it cannot be. It follows that the sequence (x_{n}) does not contain any terms equal to 0. This completes the proof of Theorem 1 in case II.

Case III: 1 ≤ |a| ≤ |b| and |b| is a prime.

We divide the proof of this case into three subcases: |a| = 1, |a| = |b| and 2 ≤ |a| < |b|.

Subcase IIIa: |a| = 1, |b| is a prime.

If a = 1 and b > 0, then take x_{0} = (2b^{2}−1)^{2}, x_{1} = b(b^{2}−1). Clearly, x_{0} and x_{1} are positive, relatively prime, composite integers. It follows immediately that x_{2} = b^{3}(4b^{2} − 3) is also composite. Finally, an easy induction shows that x_{n} ≡ 0 (mod b^{2}) and that \frac{x_{n}}{b^{2}} ≡ −1 (mod b) for all n ≥ 3. Hence, |x_{n}| is necessarily composite for all n ≥ 3.

The other situations can be dealt with similarly.

If a = −1 and b < 0, take x_{0} = (2b^{2} − 1)^{2}, x_{1} = −b(b^{2} − 1). Then x_{2} = b^{3}(4b^{2} − 3), and for n ≥ 3 one can show that x_{n} ≡ 0 (mod b^{2}) and \frac{x_{n}}{b^{2}} ≡ (−1)^{n+1} (mod b). Again, all |x_{n}| are composite.

If a = 1 and b < 0, we take x_{0} = (2b^{2} − 1)^{2}, x_{1} = −b(b^{2} + 1). Then x_{2} = b^{3}(4b^{2} − 5), and for all n ≥ 3 one can prove that x_{n} ≡ 0 (mod b^{2}) and \frac{x_{n}}{b^{2}} ≡ −1 (mod b). Hence, all |x_{n}| are composite.

Finally, if a = −1 and b > 0, select x_{0} = (2b^{2} − 1)^{2}, x_{1} = b(b^{2} + 1). Then x_{2} = b^{3}(4b^{2} − 5), and for all n ≥ 3 we have that x_{n} ≡ 0 (mod b^{2}) and \frac{x_{n}}{b^{2}} ≡ (−1)^{n+1} (mod b). All |x_{n}| are composite.

This completes the proof of Theorem 1 in subcase IIIa.

Subcase IIIb: |a| = |b|, |b| is a prime.

If a = b, take x_{0} = 4b^{4} − 1, x_{1} = 2b^{2}. Then x_{2} = b(4b^{4} + 2b^{2} − 1) is composite, and x_{n} ≡ 0 (mod b^{2}) for all n ≥ 3. It remains to show that x_{n} ≠ 0 for all n.

As in Case II, x_{n+1} = 0 implies that either 16b^{8} + 8b^{6} − 8b^{4} − 4b^{3} − 2b^{2} + 1 or (−b)(16b^{8} + 8b^{6} − 8b^{4} − 4b^{3} − 2b^{2} + 1) is a perfect square. One can use the same technique as in Case II to prove that this cannot happen if a^{2} + 4b = b^{2} + 4b ≠ 0.

If a = −b, take x_{0} = 4b^{4} − 1, x_{1} = 2b^{2} (same choice). Then x_{2} = b(4b^{4} − 2b^{2} − 1) is composite, and x_{n} ≡ 0 (mod b^{2}) for all n ≥ 3. It remains to show that x_{n} ≠ 0 for all n.

As in Case II, x_{n+1} = 0 implies that either 16b^{8} − 8b^{6} − 8b^{4} − 4b^{3} + 2b^{2} + 1 or (−b)(16b^{8} − 8b^{6} − 8b^{4} − 4b^{3} + 2b^{2} + 1) is a perfect square. The same approach as in Case II shows that this is impossible if a^{2} + 4b = b^{2} + 4b ≠ 0.

Subcase IIIc: 2 ≤ |a| < |b|, |b| is a prime.

It follows that gcd(a, b) = 1.

If a > 0, b > 0, take x_{0} = a^{3}, x_{1} = b(b^{2} − a^{2}). Then x_{2} = ab^{3}. For n ≥ 3, x_{n} ≡ 0 (mod b^{2}) and \frac{x_{n}}{b^{2}} ≡ −a^{n−1} (mod b). Since gcd(a, b) = 1, x_{n} ≠ 0.

If a < 0, b < 0, take x_{0} = −a^{3}, x_{1} = −b(b^{2} − a^{2}). Then x_{2} = −ab^{3}. For n ≥ 3, x_{n} ≡ 0 (mod b^{2}) and \frac{x_{n}}{b^{2}} ≡ a^{n−1} (mod b). Hence, x_{n} ≠ 0 for all n.

If a < 0, b > 0, take x_{0} = −a^{3}, x_{1} = b(b^{2} + a^{2}). Then x_{2} = ab^{3}, and for n ≥ 3, x_{n} ≡ 0 (mod b^{2}) and \frac{x_{n}}{b^{2}} ≡ a^{n−1} (mod b). So, x_{n} ≠ 0 for all n.

Lastly, if a > 0, b < 0, take x_{0} = a^{3}, x_{1} = −b(b^{2} + a^{2}). Then x_{2} = −ab^{3}. For n ≥ 3, x_{n} ≡ 0 (mod b^{2}) and \frac{x_{n}}{b^{2}} ≡ −a^{n−1} (mod b). Since gcd(a, b) = 1, we have x_{n} ≠ 0 for all n.

Hence, in each case one can choose x_{0} and x_{1} to be positive, composite and relatively prime so that x_{2} is composite, and for all n ≥ 3 we have that x_{n} ≡ 0 (mod b^{2}) and x_{n} ≠ 0. It follows that |x_{n}| is composite for all n ≥ 0, and this completes the proof of Theorem 1 in the case |b| ≥ 2.

5 The case |b| = 1

The main idea behind the proof of Theorem 1 in this particular case can be summarized as follows: we want to find a finite set of primes p_{1}, p_{2}, …, p_{t} such that for every n ≥ 0 the number |x_{n}| is divisible by at least one of these primes.

We start the analysis with the simple case in which |a| has at least two distinct prime factors.

Lemma 5. Let |b| = 1, and suppose that |a| has at least two distinct prime factors: p_{1} and p_{2}, with p_{1} < p_{2}. Then the sequence given by (1) and the initial terms x_{0} = p_{1}^{2}, x_{1} = p_{2}^{2} satisfies the conditions in Theorem 1.

Proof. Clearly, x_{0} and x_{1} are positive relatively prime composite integers. An easy induction shows that x_{n} ≡ 0 (mod p_{1}) if n is even and x_{n} ≡ 0 (mod p_{2}) if n is odd. Since |x_{0}| < |x_{1}| and 1 = |b| < |a|, the hypotheses of Lemma 3 are satisfied; hence (|x_{n}|)_{n≥0} is strictly increasing. It follows that x_{n} is composite for all n ≥ 0, and therefore Theorem 1 is verified in this case.

Remark 6. In all what follows we will assume that |a| = p_{1}^{s} for some prime p_{1} and some nonnegative integer s ≥ 0. Note that this allows the possibility of |a| = 1.

Next, we introduce a special sequence (u_{n})_{n≥0} given by

u_{0} = 0, u_{1} = 1, u_{n+1} = au_{n} + bu_{n−1} for all n ≥ 1.  (11)

This sequence is called the Lucas sequence of the first kind.

One can prove that (u_{n})_{n≥0} is a divisibility sequence; that is, u_{m} divides u_{n} whenever m divides n. Indeed, suppose that a^{2} + 4b ≠ 0, which implies that the roots α and β of the characteristic equation are distinct, α ≠ β. Assume that n = mk. Then

u_{n} = \frac{α^{n} − β^{n}}{α − β} = \frac{α^{mk} − β^{mk}}{α − β} = \frac{α^{m} − β^{m}}{α − β} · (α^{m(k−1)} + α^{m(k−2)}β^{m} + ⋯ + β^{m(k−1)})
= u_{m} (α^{m(k−1)} + α^{m(k−2)}β^{m} + ⋯ + β^{m(k−1)}) .

The second factor of the last term in the equality above is a symmetric function in α and β, and therefore it can be written as a polynomial function of α + β = a and αβ = −b. It follows that u_{n}/u_{m} is an integer, as claimed.

Of particular interest in the sequel are the values of u_{4} and u_{6}:

u_{4} = a(a^{2} + 2b) and u_{6} = a(a^{2} + b)(a^{2} + 3b).  (12)

The next two results give information regarding the prime factors of u_{4} and u_{6} when |b| = 1.

Lemma 7. Suppose that b = −1, |a| ≥ 4 and |a| = p_{1}^{s} for some prime p_{1} and some s ≥ 1. Then u_{6} = a(a^{2} − 1)(a^{2} − 3) has at least three additional distinct prime factors, p_{2}, p_{3} and p_{4}, all three different from p_{1}.

Proof. Note that a^{2} − 3 is not divisible by 4 or 9. Since a^{2} − 3 ≥ 13, it follows that a^{2} − 3 has an odd prime factor p_{4} different from 3, and since gcd(a, a^{2} − 3) ∈ {1, 3} we have that p_{4} ≠ p_{1}. Clearly, neither p_{1} nor p_{4} are factors of a^{2} − 1. Note that a^{2} − 1 has at least two distinct prime factors. Indeed, if one assumes the opposite, then both |a − 1| and |a + 1| are powers of some prime. But this cannot happen, since |a| ≥ 4. It follows that a^{2} − 1 has at least two distinct prime factors, p_{2} and p_{3}, both different from p_{1} and p_{4}.

To illustrate Lemma 7, let us take (a, b) = (−9, −1). Then a^{2} − 3 = 78 = 2 · 3 · 13 while a^{2} − 1 = 80 = 2^{4} · 5. Hence, one has p_{1} = 3, p_{4} = 13, p_{2} = 2, and p_{3} = 5.

Lemma 8. Suppose that b = 1, |a| ≥ 6 and |a| = p_{1}^{s} for some prime p_{1} and some s ≥ 1.

If p_{1} ≠ 3, then u_{4} = a(a^{2} + 2) has at least two additional distinct prime factors, p_{2} and p_{3}, both different from p_{1}.

If |a| = 3^{s} for some s ≥ 2, then u_{6} = a(a^{2} + 1)(a^{2} + 3) has at least three additional distinct prime factors, p_{2}, p_{3} and p_{4}, all three different from p_{1}.

Proof. Suppose first that p_{1} = 2; that is, |a| = 2^{s} for some s ≥ 3. Note that a^{2} + 2 ≡ 0 (mod 3), hence one can choose p_{2} = 3. Note that a^{2} = 2(2^{2s−1} + 1) must have a prime factor different from 2 and 3. Indeed, if one assumes the opposite, then 2^{2s−1}+1 = 3^{t} for some t ≥ 2. However, under the assumption that 2s − 1 ≥ 2 and t ≥ 2, the Catalan-Mihăilescu theorem implies that the only solution of this equation is s = 2, t = 2. But this implies that |a| = 2^{2} = 4, a contradiction.

Consider next the case |a| = p_{1}^{s} with p_{1} > 3. In particular, a^{2} + 2 is odd and therefore gcd(a, a^{2} + 2) = 1. Moreover, since a^{2} + 2 ≡ 0 (mod 3) one can safely take p_{2} = 3. We claim that a^{2} + 2 has at least one other prime factor, p_{3}, different from p_{1} and p_{2}. Indeed, if one assumes otherwise, then a^{2} + 2 = p_{1}^{2s} + 2 = 3^{t} for some t ≥ 2. However, it was shown by Ljunggren [7] that the more general equation x^{2} + 2 = y^{n}, n ≥ 2 has the unique solution x = 5, y = 3, n = 3. This would give |a| = 5, which is excluded from our analysis.

Finally, suppose that p_{1} = 3 and therefore |a| = 3^{s} for some s ≥ 2. Then both a^{2} + 1 and a^{2} +3 are even, so one can take p_{2} = 2. Note that a^{2} +1 = 9^{s} +1; hence a^{2} +1 ≡ 10 (mod 72) and a^{2}+3 ≡ 12 (mod 72). Hence, there exists a positive integer c such that a^{2}+1 = 2(36c+5) and a^{2} + 3 = 12(6c + 1). Let p_{3} be a prime factor of 36c + 5 and let p_{4} be a prime factor of 6c + 1. It follows immediately that p_{1} = 3, p_{2} = 2, p_{3}, and p_{4} are all distinct.

We present a couple of particular instances covered by the above lemma.

Suppose first that a = 8 and b = 1. Then u_{2} = a = 2^{3} while u_{4} = 2^{4} · 3 · 11. Thus, in this case one can take p_{1} = 2, p_{2} = 3, p_{3} = 11.

Suppose next that a = −49 and b = 1. Then u_{2} = a = −7^{2}, while u_{4} = −7^{2} · 3^{3} · 89. It follows that we can choose p_{1} = 7, p_{2} = 3, p_{3} = 89.

Finally, let us assume that a = 9 and b = 1. Then u_{2} = a = 3^{2}, while a^{2} + 1 = 2 · 41, a^{2} + 3 = 2^{2} · 3 · 7. Hence, in this case we have p_{1} = 3, p_{2} = 2, p_{3} = 41, p_{4} = 7.

The next lemma uses the concept of covering system introduced by Erdős in [4].

Definition 9. A collection of residue classes r_{i} (mod m_{i}), 0 ≤ r_{i} < m_{i}, where 1 ≤ i ≤ t is said to be a covering system if every integer n satisfies at least one equality n ≡ r_{i} (mod m_{i}).

In the proof of the theorem when |b| = 1, except for finitely many values of a we will use one of the following two covering systems

{0 (mod 2), 1 (mod 6), 3 (mod 6), 5 (mod 6)} or {0 (mod 2), 1 (mod 4), 3 (mod 4)}.

Lemma 10. Let a, b be two integers such that |a| ≥ 2 and |b| = 1. Let (u_{n})_{n≥0} be the Lucas sequence defined in (11), and suppose that there exists a finite collection of triples (p_{i}, m_{i}, r_{i}), 1 ≤ i ≤ t with the following properties:

(i) All primes p_{i} are distinct.

(ii) The residue classes r_{i} (mod m_{i}) form a covering system.

(iii) p_{i} | u_{m_{i}} for all 1 ≤ i ≤ t.

Then there exist two relatively prime positive integers x_{0} and x_{1} such that each term of the sequence (x_{n})_{n≥0} defined in (1) is composite.

Proof. Let P = p_{1}p_{2} … p_{t}.

By the Chinese remainder theorem, there exist y, z ∈ {0, 1, … P − 1} satisfying

y ≡ u_{m_{i}−r_{i}} (mod p_{i}),
z ≡ u_{m_{i}−r_{i}+1} (mod p_{i}),  (13)

for i = 1, 2, …, t. Note that there is no prime which divides y, z, and P simultaneously. Indeed, if such a prime p_{j} were to exist, then there would be two consecutive terms u_{n} and u_{n+1} both divisible by p_{j}. Since u_{n+1} = au_{n}+bu_{n−1} and |b| = 1, then p_{j} | u_{n−1}. By induction, it follows that p_{j} | u_{1} which is impossible since u_{1} = 1.

Let x_{0} ≡ y (mod P) and x_{1} ≡ z (mod P).

Then we have x_{0} ≡ u_{m_{i}−r_{i}} (mod p_{i}) and x_{1} ≡ u_{m_{i}−r_{i}+1} (mod p_{i}) for all i = 1, 2, …, t. By induction on n, we obtain x_{n+1} ≡ u_{m_{i}−r_{i}+n} (mod p_{i}) for every n ≥ 0 and every 1 ≤ i ≤ t.

Since the residue classes r_{i} (mod m_{i}) form a covering system, each nonnegative integer n belongs to one of these classes, say n = r_{i} + k m_{i} for some k ≥ 0 and some i ∈ {1, 2, …, t}. This implies that

x_{n+1} ≡ u_{m_{i}−r_{i}+n} (mod p_{i}) ≡ u_{m_{i}(k+1)} (mod p_{i}) ≡ 0 (mod p_{i}),  (14)

since p_{i} | u_{m_{i}} and u_{m_{i}} | u_{m_{i}(k+1)}. Hence, every term of the sequence (x_{n})_{n} is divisible by some prime p_{i}. It remains to choose x_{0} and x_{1}, relatively prime positive integers x_{0} ≡ y (mod P) and x_{1} ≡ z (mod P) such that |x_{n}| ≥ P for every n ∈ ℕ.

In order to achieve this, take x_{0} = y+P and x_{1} = z+kP where k ≥ 2 and gcd(x_{0}, x_{1}) = 1. Using Lemma 4 with n_{1} = y + P, n_{2} = z and n_{3} = P shows that such a choice is always possible. Recall that we proved earlier that gcd(y, z, P) = 1. Such a choice implies that 0 < P ≤ x_{0} < x_{1}, and since |a| > |b| = 1, Lemma 3 implies that (|x_{n}|)_{n} is a strictly increasing sequence. It follows that |x_{n}| ≥ P for all n ≥ 0, and therefore each such x_{n} is composite.

We can now prove the theorem if |b| = 1.

Suppose first that b = −1 and |a| = p_{1}^{s} ≥ 4. Then, by Lemma 7, there are four distinct primes p_{1}, p_{2}, p_{3}, p_{4} dividing u_{2}, u_{6}, u_{6}, u_{6}, respectively. The theorem follows after using Lemma 10 for the triples (p_{1}, 2, 0), (p_{2}, 6, 1), (p_{3}, 6, 3), (p_{4}, 6, 5).

As a numerical illustration, suppose that a = −9, b = −1. Then, as described in the paragraph following the proof of Lemma 7, we have p_{1} = 3, p_{2} = 2, p_{3} = 5 and p_{4} = 13.

The system (13) becomes

y ≡ u_{2} (mod 3) z ≡ u_{3} (mod 3)
y ≡ u_{5} (mod 2) z ≡ u_{6} (mod 2)
y ≡ u_{3} (mod 5) z ≡ u_{4} (mod 5)
y ≡ u_{1} (mod 13) z ≡ u_{2} (mod 13)

and its solution is P = p_{1}p_{2}p_{3}p_{4} = 390, y = 105, z = 134. Since y < z and gcd(y, z) = 1, one can safely take x_{0} = y = 105 and x_{1} = z = 134. Then the sequence (|x_{n}|)_{n≥0} is strictly increasing, and for every n ≥ 0 we have that x_{2n} ≡ 0 (mod 3), x_{6n+1} ≡ 0 (mod 2), x_{6n+3} ≡ 0 (mod 5) and x_{6n+5} ≡ 0 (mod 13). It follows that all terms of the sequence are composite.

Next suppose that b = 1 and |a| = p_{1}^{s} ≥ 6 for some prime p_{1}. If p_{1} ≠ 3, then by Lemma 8 there exist three distinct primes p_{1}, p_{2}, p_{3}, dividing u_{2}, u_{4}, u_{4}, respectively.

The theorem follows after using Lemma 10 for the triples (p_{1}, 2, 0), (p_{2}, 4, 1), (p_{3}, 4, 3). As in the case b = −1, we present the details in a couple of particular cases.

Suppose first that a = 8, b = 1. Then, as described in the paragraph following the proof of Lemma 8, we have p_{1} = 2, p_{2} = 3, p_{3} = 11. The system (13) becomes

y ≡ u_{0} (mod 2) z ≡ u_{1} (mod 2)
y ≡ u_{3} (mod 3) z ≡ u_{4} (mod 3)
y ≡ u_{1} (mod 11) z ≡ u_{2} (mod 11)

and its solution is P = p_{1}p_{2}p_{3} = 66, y = 56, z = 63. Note that in this case gcd(y, z) = 7 > 1. Still, one can safely take x_{0} = y = 56 and x_{1} = z + P = 129, and now gcd(x_{0}, x_{1}) = 1 as desired. Then, since 0 < x_{0} < x_{1}, the sequence (|x_{n}|)_{n≥0} is strictly increasing and for every n ≥ 0 we have that x_{2n} ≡ 0 (mod 2), x_{4n+1} ≡ 0 (mod 3), and x_{4n+3} ≡ 0 (mod 11). It follows that all terms of the sequence are composite.

Next assume that a = −49, b = 1. Then p_{1} = 7, p_{2} = 3, p_{3} = 89, and the system (13) becomes

y ≡ u_{0} (mod 7) z ≡ u_{1} (mod 7)
y ≡ u_{3} (mod 3) z ≡ u_{4} (mod 3)
y ≡ u_{1} (mod 89) z ≡ u_{2} (mod 89).

The solution is P = p_{1}p_{2}p_{3} = 1869, y = 980, z = 1464. Notice that in this case gcd(y, z) = 4 > 1. Still, one can safely take x_{0} = y = 980 and x_{1} = z+P = 3333 and now gcd(x_{0}, x_{1}) = 1 as desired. Moreover, since 0 < x_{0} < x_{1}, the sequence (|x_{n}|)_{n≥0} is strictly increasing, and for every n ≥ 0 we have that x_{2n} ≡ 0 (mod 7), x_{4n+1} ≡ 0 (mod 3), and x_{4n+3} ≡ 0 (mod 89). It follows that all terms of the sequence are composite.

For the case b = 1 and |a| = 3^{s} with s ≥ 2, we use the second part of Lemma 8 to conclude that there are four distinct primes p_{1}, p_{2}, p_{3}, p_{4}, dividing u_{2}, u_{6}, u_{6}, u_{6}, respectively. The theorem follows after using Lemma 10 for the triples (p_{1}, 2, 0), (p_{2}, 6, 1), (p_{3}, 6, 3), (p_{4}, 6, 5).

We show the full details if a = 9, b = 1. Then, as mentioned in the paragraph following the proof of Lemma 8, we have p_{1} = 3, p_{2} = 2, p_{3} = 41, p_{4} = 7. The system (13) becomes

y ≡ u_{2} (mod 3) z ≡ u_{3} (mod 3)
y ≡ u_{5} (mod 2) z ≡ u_{6} (mod 2)
y ≡ u_{3} (mod 41) z ≡ u_{4} (mod 41)
y ≡ u_{1} (mod 7) z ≡ u_{2} (mod 7).

Solving, we obtain P = p_{1}p_{2}p_{3}p_{4} = 1722, y = 1107, z = 1444. In this case we can simply choose x_{0} = y = 1107 and x_{1} = z = 1144.

Then gcd(x_{0}, x_{1}) = 1, and since 0 < x_{0} < x_{1}, the sequence (|x_{n}|)_{n≥0} is strictly increasing. Moreover, for every n ≥ 0 we have that x_{2n} ≡ 0 (mod 3), x_{6n+1} ≡ 0 (mod 2), x_{6n+3} ≡ 0 (mod 41), and x_{6n+5} ≡ 0 (mod 7). It follows that all terms of the sequence are composite as desired.

At this point we have proved the main theorem when |b| = 1 for all but finitely many values of a. We still have to study what happens when b = −1 and |a| ≤ 3 as well as the cases when b = 1 and |a| ≤ 5. Recall that the cases a = 0 and (a, b) = (±2, −1) were already handled in section 2.

For most of these cases, we will still use Lemma 10; the only difference is that the set of triples {p_{i}, m_{i}, r_{i}}_{i=1}^{i=t} is occasionally going to be slightly more numerous.

We summarize our findings in the table below. We invite the reader to verify that the collections {p_{i}, m_{i}, r_{i}}_{i=1}^{i=t} do indeed satisfy the three conditions in Lemma 10. Note that in each case we have 0 < x_{0} < x_{1}, and since |a| > |b| = 1, Lemma 3 implies that (|x_{n}|)_{n≥0} is strictly increasing.

It remains to see what happens when |a| = |b| = 1, as in these cases Lemma 10 does not apply.

\begin{matrix}a & b & \{(p_{i}, m_{i}, r_{i})\} & x_{0} & x_{1} \\ 5 & 1 & (5, 2, 0), (2, 6, 1), (7, 6, 3), (13, 6, 5) & 495 & 1136 \\ −5 & 1 & (5, 2, 0), (2, 6, 1), (7, 6, 3), (13, 6, 5) & 495 & 866 \\ 4 & 1 & (2, 2, 0), (3, 4, 1), (7, 8, 3), (23, 8, 7) & 116 & 165 \\ −4 & 1 & (2, 2, 0), (3, 4, 1), (7, 8, 3), (23, 8, 7) & 116 & 801 \\ 3 & 1 & (3, 2, 0), (11, 4, 1), (7, 8, 3), (17, 8, 7) & 1803 & 3454 \\ −3 & 1 & (3, 2, 0), (11, 4, 1), (7, 8, 3), (17, 8, 7) & 1803 & 3091 \\ 2 & 1 & (2, 2, 0), (5, 3, 0), (3, 4, 1), (7, 6, 5), (11, 12, 7) & 260 & 807 \\ −2 & 1 & (2, 2, 0), (5, 3, 0), (3, 4, 1), (7, 6, 5), (11, 12, 7) & 260 & 1503 \\ 3 & −1 & (3, 2, 0), (2, 3, 0), (7, 4, 3), (47, 8, 5), (23, 12, 5), (1103, 24, 1) & 7373556 & 2006357 \\ −3 & −1 & (3, 2, 0), (2, 3, 0), (7, 4, 3), (47, 8, 5), (23, 12, 5), (1103, 24, 1) & 7373556 & 14686445 \end{matrix}

Table 1: Covering triples for the cases b = 1, a = ±2, ±3, ±4, ±5 and b = −1, a = ±3

If a = −1, b = −1, then it can be easily verified that the sequence given by the recurrence x_{n+1} = −x_{n} − x_{n−1} has period 3. Hence, if one chooses x_{0} = 8 and x_{1} = 27, then x_{2} = −35, and due to the periodic behavior all terms of the sequence are composite.

Similarly, if a = 1, b = −1, then the sequence given by the recurrence x_{n+1} = x_{n} − x_{n−1} has period 6. Again, if one chooses x_{0} = 8 and x_{1} = 35, then the first few terms of the sequence are 8, 35, 27, −8, −35, −37, 8, 35, 27, …; that is, x_{n} is always composite.

If a = b = 1, then Vsemirnov’s pair v_{0} = 106276436867, v_{1} = 35256392432 shows that all the numbers

v_{n} = v_{n−1} + v_{n−2} = v_{1}F_{n} + v_{0}F_{n−1}  (15)

are composite. Here, F_{n} is the nth Fibonacci number, where F_{−1} = 1, F_{0} = 0, F_{1} = 1. For the case a = −1, b = 1, we follow the solution in [3]. It can be easily checked that the general term of the sequence x_{n+1} = −x_{n} + x_{n−1} can be

written as

x_{n} = (−1)^{n+1}x_{1}F_{n} + (−1)^{n}x_{0}F_{n−1}, n ≥ 0.  (16)

We choose x_{0} = v_{0} −v_{1} = 71020044435 and x_{1} = v_{0} = 106276436867. It is easy to check that x_{0} and x_{1} are relatively prime composite integers. Moreover, from (15) and (16) we obtain that

x_{n} = (−1)^{n+1}v_{0}F_{n} + (−1)^{n}(v_{0} − v_{1})F_{n−1}+ = (−1)^{n+1}v_{1}F_{n−1} + (−1)^{n+1}v_{0}(F_{n} − F_{n−1}) =
= (−1)^{n+1}v_{1}F_{n−1} + (−1)^{n+1}v_{0}F_{n−2} = (−1)^{n+1} (v_{1}F_{n−1} + v_{0}F_{n−2}) = (−1)^{n−1}v_{n−1}.

Hence, |x_{n}| = v_{n−1} is composite for all n ≥ 0. The proof of the theorem is now complete.

6 A surprising result

In this section we prove the following:

Theorem 11. Consider the integers a, b such that |a| ≥ 3 and b = −1. Let u_{0} = 0, u_{1} = 1 and u_{n+1} = au_{n} + bu_{n−1} = au_{n} − u_{n−1} be the Lucas sequence of the first kind associated with a and −1. Then |u_{n}| is composite for all n ≥ 3.

Proof. One has u_{3} = a^{2} − 1, and u_{4} = a^{3} − 2a, which are obviously composite. Since |a| > |b| = 1, Lemma 3 implies that the sequence (|u_{n}|)_{n≥0} is strictly increasing. Suppose for the sake of contradiction that there exists an n ≥ 2 such that |u_{n+1}| = p, where p is some prime number. Since |u_{n+1}| ≥ |u_{3}| = a^{2} − 1 it follows that necessarily p > |a|.

Now using Lemma 2 for the sequence (u_{n})_{n≥0}, equality (4) becomes

u_{n+1}^{2} − au_{n}u_{n+1} + u_{n}^{2} = u_{1}^{2} − au_{1}u_{0} + u_{0}^{2},

and since u_{0} = 0, u_{1} = 1, and |u_{n+1}| = p, we obtain that

u_{n}^{2} ± apu_{n} + p^{2} − 1 = 0.  (17)

Regard the above equation as a quadratic in u_{n}. Since u_{n} ∈ ℤ, it is necessary that the discriminant is a perfect square, that is, there exist a nonnegative integer c such that

a^{2}p^{2} − 4(p^{2} − 1) = c^{2} from which (a^{2} − 4)p^{2} = c^{2} − 4 = (c − 2)(c + 2).  (18)

Since |a| ≥ 3, one can assume that c ≥ 3. Since p is a prime and p^{2} divides (c − 2)(c + 2), we have two possibilities. If p divides both c − 2 and c + 2 then p divides 4, which means that p = 2. However, this is impossible since p > |a| ≥ 3. Otherwise, p^{2} divides either c − 2 or c + 2. In either case we obtain that c + 2 ≥ p^{2}. Using this inequality in (18), it follows that

(a^{2} − 4)p^{2} = (c − 2)(c + 2) ≥ p^{2}(p^{2} − 4) ⟹ |a| ≥ p, a contradiction.

This completes the proof.

In particular, the above theorem holds if a = 4 and b = −1, thus answering a question of Vos Post (see \underline{A001353} in [14]).

What is remarkable about this situation is that while u_{n} is composite for every n ≥ 3, it seems likely that there is no finite set of primes p_{1}, p_{2}, … p_{t} such that every u_{n} is divisible by some p_{i}, 1 ≤ i ≤ t. In fact, we suspect the following is true.

Conjecture 12. Let (u_{n})_{n≥0} be the Lucas sequence of the first kind associated with some |a| ≥ 3 and b = −1. Then for any two different primes p and q, u_{p} and u_{q} are relatively prime.

If true, this conjecture would immediately imply that there is no finite set of primes p_{1}, p_{2}, … p_{t} such that every u_{n} is divisible by some p_{i}, 1 ≤ i ≤ t.

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2010 Mathematics Subject Classification: Primary 11B39; Secondary 11B37, 11A07, 11Y55. Keywords: second-order linear recurrence, Lucas sequence of the first kind, composite number, covering system of congruences.

(Concerned with sequences \underline{A001906}, \underline{A001353}, \underline{A004254}, and \underline{A001109}.)

Received January 7 2019; revised versions received February 4 2019; September 8 2019; September 24 2019. Published in Journal of Integer Sequences, September 24 2019.

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