This paper is divided into six sections. This section is devoted to stating few results that will be used in the remainder of the paper. We also set the notations to be used and derive few simple results that will come in handy in our treatment. In Section 2, we charecterize numbers 5k+2, which are primes with k being an odd natural number. In Section 3, we prove more general results than given in Section 2. In Section 4, we define the period of a Fibonacci sequence modulo some number and derive many properties of this concept. In Section 5, we devote to the study of a class of generalized Fibonacci numbers and derive some interesting results related to them. Finally, in Section 6, we define some generalized Fibonacci polynomial sequences and we obtain some results related to them.

We begin with the following famous results without proof except for some related properties.

Lemma 1.1.(Euclid)

If ab ≡ 0 (mod p) with a, b two integers and p a prime, then either p|a or p|b.

Remark 1.2

In particular, if gcd(a, b) = 1, p divides only one of the numbers a, b.

Property 1.3

Let a, b two positive integers, m, n two integers such that (|m|, |n|) = 1 and p a natural number. Then

if and only if there exists c ∈ ℤ such that

and

Remark 1.4

Using the notations given in the proof of Property 1.3, we can see that if there exists an integer c such that a ≡ nc (mod p) and b ≡ mc (mod p) with (|m|, |n|) = 1 and p a natural number, then we have

Moreover, denoting by g the gcd of a and b, if a = gn and b = gm, then ub + va = (um + vn)g = g, then g ≡ c (mod p).

Theorem 1.5.(Fermat’s Little Theorem)

If p is a prime and n ∈ ℕ relatively prime to p, then n^p−1 ≡ 1 (mod p).

Theorem 1.6

If x² ≡ 1 (mod p) with p a prime, then either x ≡ 1 (mod p) or x ≡ p − 1 (mod p).

Definition 1.7

Let p be an odd prime and gcd(a, p) = 1. If the congruence x² ≡ a (mod p) has a solution, then a is said to be a quadratic residue of p. Otherwise, a is called a quadratic nonresidue of p.

Theorem 1.8.(Euler)

Let p be an odd prime and gcd(a, p) = 1. Then a is a quadratic nonresidue of p if and only if ap-12≡-1   (mod p).

Definition 1.9

Let p be an odd prime and let gcd(a, p) = 1. The Legendre symbol (a/p) is defined to be equal to 1 if a is a quadratic residue of p and is equal to −1 is a is a quadratic non residue of p.

Property 1.10

Let p an odd prime and a and b be integers which are relatively prime to p. Then the Legendre symbol has the following properties:

• If a ≡ b (mod p), then (a/p) = (b/p).

• (a/p) ≡ a^(p−1)/2 (mod p)

• (ab/p) = (a/p)(b/p)

Remark 1.11

Taking a = b in (3) of Property 1.10, we have

Lemma 1.12.(Gauss)

Let p be an odd prime and let gcd(a, p) = 1. If n denotes the number of integers in the set S={a,2a,3a,…,(p-12)a}, whose remainders upon division by p exceed p/2, then

Corollary 1.13

If p is an odd prime, then

Theorem 1.14.(Gauss’ Quadratic Reciprocity Law)

If p and q are distinct odd primes, then

Corollary 1.15

If p and q are distinct odd primes, then

Throughout this paper, we assume k ∈ ℕ, unless otherwise stated.

From (1) of Property 1.10, Theorem 1.14, Corollary 1.13 and Corollary 1.15, we deduce the following result.

Theorem 1.16

(5/5k + 2) = −1.

Theorem 1.17

55k+12≡5k+1   (mod 5k+2) where 5k + 2 is a prime.

The proof of Theorem 1.17 follows very easily from Theorems 1.8, 1.16 and Property 1.10.

Theorem 1.18

Let r be an integer in the set {1, 2, 3, 4}. Then, we have

Theorem 1.19

Let r be an integer in the set {1, 2, 3, 4}. Then, if 5k + r is a prime, we have

The proof of Theorem 1.19 follows very easily from Theorems 1.8, 1.18 and Property 1.10.

We fix the notation [[1, n]] = {1, 2, …, n} throughout the rest of the paper. We now have the following properties.

Remark 1.20

Let 5k + r with r ∈ [[1,4]] be a prime number. Then

or equivalently

Property 1.21

(5k+12l+1)≡5k+1   (mod 5k+2), with l∈[[0,⌊5k2⌋]] and 5k + 2 is a prime.

Property 1.22

(5k+r-12l+1)≡-1   (mod 5k+r), with l∈[[0,⌊5k+r-22⌋]] and 5k + r is a prime such that r ∈ [[1, 4]] and k ≡ r + 1 (mod 2).

The proof of Property 1.22 is very similar to the proof of Property 1.21.

Property 1.23

(5k2l+1)≡5k-2l≡-2(l+1)   (mod 5k+2)with l∈[[0,⌊5k2⌋]] and 5k + 2 is a prime.

Property 1.24

(5k+r-22l+1)≡-2(l+1)   (mod 5k+r), with l∈[[0,⌊5k+r-32⌋]] and 5k + r is a prime such that r ∈ [[1, 4]] and k ≡ r + 1 (mod 2)

The proof of Property 1.24 is very similar to the proof of Property 1.23.

In the memainder of this section we derive or state a few results involving the Fibonacci numbers. The Fibonacci sequence (F_n) is defined by F₀ = 0, F₁ = 1, F_n+2 = F_n + F_n+1 for n ≥ 0.

From the definition of the Fibonacci sequences we can establish the formula for the nth Fibonacci number,

where ϕ=1+52 is the golden ratio.

From binomial theorem, we have for a ≠ 0 and n ∈ ℕ,

We set

So

Thus

We get, from (1.1),

Thus we have

Theorem 1.25

Fn=12n-1∑l=0⌊n-12⌋(n2l+1)5l.

Property 1.26

Fk+2=1+∑i=1kFi.

Theorem 1.27

F_k+l = F_lF_k+1 + F_l−1F_k with k ∈ ℕ and l ≥ 2.

The proofs of the above two results can be found in [6].

Property 1.28

Let m be a positive integer which is greater than 2. Then, we have

The above can be generalized to the following.

Property 1.29

Let m be a positive integer which is greater than 2. Then, we have

The above three results can be proved in a straighforward way using the recurrence relation of Fibonacci numbers.

We now state below a few congruence satisfied by the Fibonacci numbers.

Property 1.30

F_n ≡ 0 (mod 2) if and only if n ≡ 0 (mod 3).

Corollary 1.31

If p = 5k + 2 is a prime which is strictly greater than 5 (k ∈ ℕ and k odd), then F_p = F_5k+2is an odd number.

In order to prove this assertion, it suffices to remark that p is not divisible by 3.

Property 1.32

F_5k ≡ 0 (mod 5) with k ∈ ℕ.

Property 1.33

F_n ≥ n with n ∈ ℕ and n ≥ 5.

The proofs of the above results follows from the principle of mathematical induction and Theorem 1.27 and Proposition 1.26. For brevity, we omit them here.

In this section, we give some new congruence relations involving Fibonacci numbers modulo a prime. The study in this section and some parts of the subsequent sections are motivated by some similar results obtained by Bicknell-Johnson in [1] and by Hoggatt and Bicknell-Johnson in [5].

Let p = 5k + 2 be a prime number with k a non-zero positive integer which is odd. Notice that in this case, 5k ± 1 is an even number and so

We now have the following properties.

Property 2.1

F_5k+2 ≡ 5k + 1 (mod 5k + 2) with k ∈ ℕ and k odd such that 5k+2 is prime.

This result is also stated in [5], but we give a different proof of the result below.

Property 2.2

F_5k+1 ≡ 1 (mod 5k + 2) with k ∈ ℕ and k odd such that 5k + 2 is prime.

Property 2.3

F_5k ≡ 5k (mod 5k + 2) with k ∈ ℕ and k odd such that 5k + 2 is prime.

Property 2.4

Let 5k + 2 be a prime with k odd and let m be a positive integer which is greater than 2. Then, we have

Theorem 2.5

Let 5k +2 be a prime with k an odd integer and let m be a positive integer which is greater than 2. Then

and

Corollary 2.6

Let 5k + 2 be a prime with k an odd integer and m be a positive integer which is greater than 2. Then

and

Theorem 2.7

Let 5k+2 be a prime with k an odd positive integer, m be a positive integer which is greater than 2 and r ∈ ℕ. Then

Remark 2.8

In particular, if r = 3, we know that

This congruence can be deduced from Property 2.4 and Theorem 2.7. Indeed, using Theorem 2.7, we have

Using Property 2.4, we get

Corollary 2.9

Let 5k +2 be a prime with k an odd positive integer and m, r ∈ ℕ. Then

Lemma 2.10

Let 5k + 2 be a prime with k an odd positive integer and r ∈ ℕ. Then

Corollary 2.11

Let 5k + 2 be a prime with k an odd positive integer, let m be a positive integer and r ∈ ℕ. Then

The above corollary can be deduced from Corollary 2.9.

Lemma 2.12

Let 5k +2 be a prime with k an odd positive integer and let m be a positive integer. Then

Remark 2.13

We can observe that

and

By Properties 2.1, 2.2 and 2.3, we have