### Article

Kyungpook Mathematical Journal 2017; 57(1): 1-84

**Published online** March 23, 2017

Copyright © Kyungpook Mathematical Journal.

### Some Properties of Fibonacci Numbers, Generalized Fibonacci Numbers and Generalized Fibonacci Polynomial Sequences

Alexandre Laugier^{1}

Manjil P. Saikia^{2}

Lycée professionnel Tristan Corbiére, 16 rue de Kerveguen - BP 17149, 29671 Morlaix cedex, France^{1}

Department of Mathematical Sciences, Tezpur University, Napaam - 784028, Assam, India^{2}Current Address: Fakultät für Mathematik, Universität Wien, Oskar-Morgenstern-Platz 1, Vienna 1090, Austria ^{2}

**Received**: March 26, 2014; **Accepted**: May 13, 2015

### Abstract

- Abstract
- 1. Preliminaries
- 2. Congruences of Fibonacci Numbers Modulo a Prime
- 3. Some Further Congruences of Fibonacci Numbers Modulo a Prime
- 4. Periods of the Fibonacci Sequence Modulo a Positive Integer
- 5. Some Results on Generalized Fibonacci Numbers
- 6. Some Results on Generalized Fibonacci Polynomial Sequences
- Acknowledgements
- References

In this paper we study the Fibonacci numbers and derive some interesting properties and recurrence relations. We prove some charecterizations for _{p}

**Keywords**: Fibonacci numbers, congruences, period of Fibonacci sequence

### 1. Preliminaries

- Abstract
- 1. Preliminaries
- 2. Congruences of Fibonacci Numbers Modulo a Prime
- 3. Some Further Congruences of Fibonacci Numbers Modulo a Prime
- 4. Periods of the Fibonacci Sequence Modulo a Positive Integer
- 5. Some Results on Generalized Fibonacci Numbers
- 6. Some Results on Generalized Fibonacci Polynomial Sequences
- Acknowledgements
- References

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 5

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

### Lemma 1.1.(Euclid)

### Remark 1.2

In particular, if gcd(

### Property 1.3

**Proof**

Let

If there exists an integer

Conversely, if

and

So, we have

and

Since |

and

It results 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

Moreover, denoting by

### Theorem 1.5.(Fermat’s Little Theorem)

^{p}^{−}^{1} ≡ 1 (mod

### Theorem 1.6

^{2} ≡ 1 (mod

**Proof**

If ^{2} ≡ 1 (mod

### Definition 1.7

Let ^{2} ≡

### Theorem 1.8.(Euler)

### Definition 1.9

Let

### Property 1.10

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

a /p ) ≡a ^{(}^{p}^{−}^{1)/2}(modp )(

ab /p ) = (a /p )(b /p )

### Remark 1.11

Taking

### Lemma 1.12.(Gauss)

### Corollary 1.13

### Theorem 1.14.(Gauss’ Quadratic Reciprocity Law)

### Corollary 1.15

Throughout this paper, we assume

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

### Theorem 1.16

(5/5

**Proof**

Clearly (5/5

For proofs of the above theorems the reader is suggested to see [2] or [6].

Let

From Theorem 1.6, we have either

or

Moreover, we can observe that

### Theorem 1.17

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

### Theorem 1.18

**Proof**

We have (5/5

Moreover, (5

Or, we have

If

r = 1, then using Theorem 1.14, (r /5) = 1.If

r = 2, then using Corollary 1.13, (r /5) = −1.If

r = 3, then using Theorem 1.14, (r /5) = −1.If

r = 4, then since (4/5) = (2^{2}/5) = 1 (see also Remark 1.11), (r /5) = 1.

### Theorem 1.19

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

We fix the notation [[1,

### Remark 1.20

Let 5

or equivalently

### Property 1.21

**Proof**

Notice that for

Or,

Multiplying these congruences we get

Therefore

Since (2

We have now the following generalization.

### Property 1.22

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

### Property 1.23

**Proof**

Notice that for

Or

Multiplying these congruences we get

Therfore

Since (2

We can generalize the above as follows.

### Property 1.24

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 (_{n}_{0} = 0, _{1} = 1, _{n}_{+2} = _{n}_{n}_{+1} for

From the definition of the Fibonacci sequences we can establish the formula for the

where

From binomial theorem, we have for

We set

So

Thus

We get, from (

Thus we have

### Theorem 1.25

### Property 1.26

### Theorem 1.27

_{k}_{+}_{l}_{l}_{k}_{+1} + _{l}_{−}_{1}_{k}

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

### Property 1.28

The above can be generalized to the following.

### Property 1.29

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

_{n}

### Corollary 1.31

_{p}_{5}_{k}_{+2}

In order to prove this assertion, it suffices to remark that

### Property 1.32

_{5}_{k}

### Property 1.33

_{n}

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.

### 2. Congruences of Fibonacci Numbers Modulo a Prime

- Abstract
- 1. Preliminaries
- 2. Congruences of Fibonacci Numbers Modulo a Prime
- 3. Some Further Congruences of Fibonacci Numbers Modulo a Prime
- 4. Periods of the Fibonacci Sequence Modulo a Positive Integer
- 5. Some Results on Generalized Fibonacci Numbers
- 6. Some Results on Generalized Fibonacci Polynomial Sequences
- Acknowledgements
- References

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

We now have the following properties.

### Property 2.1

_{5}_{k}_{+2} ≡ 5

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

**Proof**

From Theorems 1.17 and 1.25, we have

where we used the fact that

From Fermat’s Little Theorem, we have

We get _{5}_{k}_{+2} ≡ 5

### Property 2.2

_{5}_{k}_{+1} ≡ 1 (mod 5

**Proof**

From Theorem 1.25 and Property 1.21, we have

We have

We get from the above

Since

Notice that 5

From Theorem 1.17 we have

We can rewrite

Moreover, we have

Or,

We have

From Fermat’s Little Theorem, we have 5^{10}^{m}^{+6} ≡ 1 (mod 10

or equivalently

It follows that

Since 2 and 5

From Fermat’s Little Theorem, we have 2^{5}^{k}^{+1} ≡ 1 (mod 5

### Property 2.3

_{5}_{k}

**Proof**

From Theorem 1.25 and Property 1.23 we have

Also

where we have used the fact that for

So

Moreover since

Since 5^{5}^{m}^{+3} ≡ 10

Consequently

which implies

or equivalently for

Since 2 and 5

From Fermat’s Little Theorem, we have 2^{5}^{k}^{+1} ≡ 1 (mod 5

### Property 2.4

**Proof**

We prove the result by induction. We know that _{6} = 8. Or, 2(3+_{2}) = 2(3 + 1) = 2 × 4 = 8. So

Let us assume that

For

From the assumption above, we get

Thus the proof is complete by induction.

### Theorem 2.5

**Proof**

We prove the theorem by induction. We have, using Theorem 1.27

Using Properties 2.1, 2.2 and 2.3, we can see that

Also

So

Moreover, we have from Theorem 1.27, Property 2.2 and Property 2.3,

We have

So

Therefore

or equivalently

Let us assume that

and

Then, we have

Using Property 2.3 and _{5}_{k}_{−}_{1} ≡ 3 (mod 5

It gives

Moreover, we have

Using Properties 2.2 and 2.3 and the assumptions above, and since 25^{2} ≡ 4 (mod 5

Or,

Therefore

or equivalently

This completes the proof.

### Corollary 2.6

### Theorem 2.7

**Proof**

For

From Theorem 2.5, we have for

This completes the proof.

### Remark 2.8

In particular, if

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

### Lemma 2.10

**Proof**

We prove this lemma by induction. For _{5}_{k}_{+}_{r}_{5}_{k}_{+1} ≡ 1 (mod 5_{r}_{r}_{−}_{1} = _{1} − 2_{0} = _{1} ≡ 1 (mod 5

Let us assume that _{5}_{k}_{+}_{s}_{s}_{s}_{−}_{1} (mod 5

Thus the lemma is proved.

We can prove Lemma 2.10 as a consequence of Corollary 2.9 by taking

### Corollary 2.11

The above corollary can be deduced from Corollary 2.9.

### Lemma 2.12

**Proof**

For _{5}_{mk}_{3}_{m}_{0} = 0 ≡ 0 (mod 5_{5}_{mk}_{3}_{m}_{5}_{k}_{3} ≡ 5_{5}_{mk}_{3}_{m}

From Theorem 2.5, we have

From Property 2.4, we have _{5}_{mk}_{3}_{m}

We can prove Corollary 2.11 as a consequence of Lemma 2.12.

### Remark 2.13

We can observe that

and

By Properties 2.1, 2.2 and 2.3, we have

So, we have

or equivalently

Thus we have the following.

### Property 2.14

**Proof**

We have

and

So, _{5}_{k}_{0} − 2_{1} (mod 5

Moreover, we know that

Let us assume that

for

Thus the proof is complete by induction.

### Property 2.15

**Proof**

We prove the result by induction. We have for _{3}_{m}_{+1} = _{3}_{×}_{2+1} = _{7} = 13 and _{7} ≡ 3^{2} + 2_{3} ≡ 13 (mod 5

Let us assume that for

Thus the induction hypothesis holds.

### Corollary 2.16

**Proof**

It stems from the recurrence relation of the Fibonacci sequence which implies that _{3}_{m}_{+2} = _{3}_{m}_{3}_{m}_{+1} and _{3}_{k}_{+1} = _{3}_{k}_{3}_{k}_{−}_{1} and Property 2.4 and Property 2.15.

### 3. Some Further Congruences of Fibonacci Numbers Modulo a Prime

- Abstract
- 1. Preliminaries
- 2. Congruences of Fibonacci Numbers Modulo a Prime
- 3. Some Further Congruences of Fibonacci Numbers Modulo a Prime
- 4. Periods of the Fibonacci Sequence Modulo a Positive Integer
- 5. Some Results on Generalized Fibonacci Numbers
- 6. Some Results on Generalized Fibonacci Polynomial Sequences
- Acknowledgements
- References

In this section we state and prove some more results of the type that were proved in the previous section. These results generalizes some of the results in the previous section and in [1] and [5].

Let

We have the following properties.

### Property 3.1

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

**Proof**

From Theorems 1.17 and 1.25 we have

where we used the fact that

From Theorem 1.5, we have

We get

### Corollary 3.2

**Proof**

We can notice that _{2} = 1 ≡ 1 (mod 2) and 2 ≡ 2 (mod 5). Moreover, we can notice that _{3} = 2 ≡ 2 (mod 3) and 3 ≡ 3 (mod 5). So, Corollary 3.2 is true for

We can observe that the result of Corollary 3.2 doesn’t work for _{5} = 5 ≡ 0 (mod 5).

The Euclid division of a prime number

It completes the proof of this corollary.

### Property 3.3

Some parts of this result is stated in [1] in a different form. We give an alternate proof of the result below.

**Proof**

From Theorem 1.25 and Property 1.22 we have

It comes that

From Theorem 1.5, we have

So, since 5

Since 5

where we used the property that ⌊

It follows that

The case

From Theorem 1.19, if

From Theorem 1.19, if

From Theorem 1.19, if

The following two results are easy consequences of Properties 3.1 and 3.3.

### Property 3.4

### Property 3.5

The following is a consequence of Properties 3.1 and 3.5.

### Property 3.6

Some of the stated properties above are given in [1] and [5] also, but the methods used here are different.

### 4. Periods of the Fibonacci Sequence Modulo a Positive Integer

- Abstract
- 1. Preliminaries
- 2. Congruences of Fibonacci Numbers Modulo a Prime
- 3. Some Further Congruences of Fibonacci Numbers Modulo a Prime
- 4. Periods of the Fibonacci Sequence Modulo a Positive Integer
- 5. Some Results on Generalized Fibonacci Numbers
- 6. Some Results on Generalized Fibonacci Polynomial Sequences
- Acknowledgements
- References

Notice that _{1} = _{2} ≡ 1 (mod

### Definition 4.1

The Fibonacci sequence (_{n}_{m }

The number _{m}_{n}

### Remark 4.2

For _{m}_{m}_{3} = 2.

From Theorem 1.27 we have

Since _{ℓm} + _{ℓm−1} = _{1+ℓm}, we get

Therefore we have the following.

### Property 4.3

_{ℓm} ≡ 0 (mod

Moreover, from Theorem 1.27 we have

Since _{1+}_{ℓm} ≡ 1 (mod

### Property 4.4

_{ℓm−1} ≡ 1 (mod

Besides, using the recurrence relation of the Fibonacci sequence, from Property 4.3 we get

Using Property 4.4, we obtain

### Property 4.5

_{ℓm−2} ≡

### Remark 4.6

From Theorem 1.27 we have for

and

From this we get

and

### Theorem 4.7

**Proof**

Using the recurrence relation of the Fibonacci sequence, and from Properties 2.1, 2.2 and 2.3, we have

Taking _{2}_{m}_{+3} and _{2}_{m}_{+4}, we have

and

Thus

or equivalently

We deduce that a period of the Fibonacci sequence modulo 5_{5}_{k}_{+2} = 2(5

We can generalize the above result as follows.

### Theorem 4.8

**Proof**

Using the formula for _{2}_{m}

From Properties 3.1, 3.5 and 3.6, we obtain

Using the formula for _{2}_{m}_{+1} given in Remark 4.6, taking

From Properties 3.1 and 3.5, we obtain

Using the recurrence relation of the Fibonacci sequence, we have _{2(5}_{k}_{+}_{r}_{+1)+2} = _{2(5}_{k}_{+}_{r}_{+1)} + _{2(5}_{k}_{+}_{r}_{+1)+1}. So

Therefore, when 5_{2(5}_{k}_{+}_{r}_{+1)} ≡ 0 (mod 5_{2(5}_{k}_{+}_{r}_{+1)+1} ≡ _{2(5}_{k}_{+}_{r}_{+1)+2} ≡ 1 (mod 5

### Theorem 4.9

**Proof**

Using the formula for _{2}_{m}

From Properties 3.1, 3.3 and 3.4, we obtain

Using the formula for _{2}_{m}_{+1} given in Remark 4.6, taking

From Properties 3.1 and 3.3, we obtain

Using the recurrence relation of the Fibonacci sequence, we have _{2(5}_{k}_{+}_{r}_{−}_{1)+2} = _{2(5}_{k}_{+}_{r}_{−}_{1)} + _{2(5}_{k}_{+}_{r}_{−}_{1)+1}. So

Therefore, when 5_{2(5}_{k}_{+}_{r}_{−}_{1)} ≡ 0 (mod 5_{2(5}_{k}_{+}_{r}_{−}_{1)+1} ≡ _{2(5}_{k}_{+}_{r}_{−}_{1)+2} ≡ 1 (mod 5

### Corollary 4.10

Corollary 4.10 follows from Theorems 4.8 and 4.9.

### Corollary 4.11

**Proof**

The Euclid division of

### Property 4.12

**Proof**

From Property 1.32, we know that _{5}_{k}_{5}_{k}_{+1} ≡ _{5}_{k}_{+2} (mod 5). So, it is relevant to search a period as an integer multiple of 5. Trying the first non-zero values of

### Property 4.13

**Proof**

Since _{5}_{k}_{5}_{k}_{+2} = _{5}_{k}_{+1} + _{5}_{k}_{5}_{k}_{+1} (mod 5).

If _{5}_{k}_{+1} = _{20}_{m}_{+1} ≡ _{1} ≡ 1 (mod 5).

If

So, if _{6} = 8 ≡ 3 (mod 5), _{5} = 5 ≡ 0 (mod 5) and since 20 is a period of the Fibonacci sequence modulo 5 (see Property 4.12), then we have _{5}_{k}_{+1} ≡ _{6}_{1} ≡ 3 (mod 5).

If

So, if _{11} = 89 ≡ 4 (mod 5), _{10} = 55 ≡ 0 (mod 5) and since 20 is a period of the Fibonacci sequence modulo 5 (see Property 4.12), then we have _{5}_{k}_{+1} ≡ _{11}_{1} ≡ 4 (mod 5).

If

So, if _{16} = 987 ≡ 2 (mod 5), _{15} = 610 ≡ 0 (mod 5) and since 20 is a period of the Fibonacci sequence modulo 5 (see Property 4.12), we have

### Property 4.14

Property 4.14 stems from the recurrence relation of the Fibonacci sequence and Property 4.13.

### Corollary 4.15

Corollary 4.15 stems from Euclid division, Properties 4.12, 4.13 and 4.14.

### Property 4.16

**Proof**

Let prove the property by induction on the integer

We have _{0} = 0 ≡ 0 (mod 5_{1} = 1 ≡ 1 (mod 5

Moreover, from Properties 3.1 and 3.3, we can notice that

and

Let assume that for a positive integer

and

Then, using the assumption, Theorem 1.27 and Properties 3.1 and 3.3, we have

and

This completes the proof by induction on the integer

### Property 4.17

This is a direct consequence of Property 4.16.

### Property 4.18

**Proof**

From Properties 3.1, 3.3 and 3.5, we have

and

So

It results that 5

### Corollary 4.19

Corollary 4.19 stems from Corollary 4.11 and Properties 4.17 and 4.18.

### Property 4.20

**Proof**

We prove this result by induction on the integer

From Properties 3.3 and 3.4, we have

and

So, we verify that (_{0} = 0 ≡ _{5}_{k}

Let us assume for an integer _{5}_{k}_{−}_{i}^{i}^{+1}_{i}

since (−1)^{2} = 1. It achieves the proof of Property 4.20 by induction on the integer

### Remark 4.21

Property 4.20 implies that we can limit ourself to the integer interval [1,

Indeed, for instance, if 5_{m}

### Theorem 4.22

**Proof**

If

and

Since 2 and 5

### Remark 4.23

We can observe that

and

Using induction we can show the following two properties.

### Property 4.24

### Property 4.25

### Remark 4.26

We can notice that

and

And for

Furthermore, we have for

Besides, we have for

We can state the following property, the proof of which follows from the above remark and by using induction.

### Property 4.27

_{5}_{k}_{+}_{n}_{n}_{−}_{3} (mod 5

### Theorem 4.28

**Proof**

The proof of the theorem will be done by induction. We have _{0} ≡ 0 (mod 5

Let us assume that

We have

Since _{5}_{k}_{+3} ≡ 0 (mod 5

The following follows very easily from the above theorem.

### Corollary 4.28

_{m}

### Property 4.30

**Proof**

Let us prove Property 4.30 by induction on the integer

and

Using the recurrence relation of the Fibonacci sequence, it comes that

and

So, we verify (

Notice that (_{0} = 0 ≡ 0 (mod 5_{5}_{k}_{+3} ≡ 0 (mod 5

Let assume for an integer _{5}_{k}_{−}_{i}^{i}^{+1}_{i}_{+3} (mod 5

since (−1)^{2} = 1. It achieves the proof of Property 4.30 by induction on the integer

Notice that Property 4.30 is also true for

### Remark 4.31

In general, the number 2(5

### Theorem 4.32

**Proof**

Since 5

Using Property 4.30 and taking

or,

Finally,

since 2 and 5

### Theorem 4.33

**Proof**

If _{5}_{k}_{+3} ≡ 0 (mod 5_{15}_{m}_{+3} ≡ 0 (mod 15

Or, from Remark 4.6, we have

and

We have also

So

So, the congruence _{15}_{m}_{+3} ≡ 0 (mod 15

or

If _{5}_{m}_{+1} ≡ 0 (mod 15

Using the recurrence relation of the Fibonacci sequence, it implies also that _{5}_{m}_{5}_{m}_{+2} (mod 15

Or, we have

Since _{5}_{k}_{+2} ≡ 5

with 15

and

it implies that

We get

and

So, either

or

If _{5}_{m}_{+1} ≡ 0 (mod 15_{5}_{m}

then

It results that the number 10

If _{5}_{m}_{+1} ≡ 0 (mod 15

then since

Notice that in this case, we cannot have

since 3 ≢ 0 (mod 15

then the number 10

which implies that

Since

But, since 15_{5}_{m}

Moreover, if _{5}_{m}_{+1} ≡ 0 (mod 15

which implies that

So, either

or

Since we have (

If

So

or equivalently (_{5}_{m}_{+2} = _{5}_{m}_{+1} + _{5}_{m}

So, if the number 10

and

Since _{10}_{m}_{+1} + _{5}_{m}_{+1}_{5}_{m}

and

So, either

or

If _{5}_{m}_{5}_{m}_{+2} (mod 15

and

where we used the fact that

and (3, 15

then using the recurrence relation of the Fibonacci sequence, we must have

and so

where we used the fact that

It implies that

and using Theorem 1.5, it gives

since 5^{15}^{m}^{+1} ≡ 1 ≡ 30^{15}^{m}

and

it results that

and so 4(6

Therefore, when 5

if and only if

Since _{15}_{m}_{+3} = _{5}_{k}_{+3} ≡ 0 (mod 5

is also true when

Thus, if 10

if and only if

if and only if

Besides,

implies that

Reciprocally, if _{10}_{m}_{+2} ≡ 0 (mod 15

So, either

or

If _{5}_{m}_{5}_{m}_{+2} (mod 15

and since _{5}_{m}_{+1} ≡ 0 (mod 15

But, then, if _{10}_{m}_{+2} ≡ 0 (mod 15

Or,

It leads to a contradiction meaning that

is not possible. So, if _{10}_{m}_{+2} ≡ 0 (mod 15

which implies the congruence

and so which translates the congruence

into the congruence

which has at least one solution. So, if 10

if and only if

if and only if

if and only if

Since

### Property 4.34

**Proof**

Let us prove Property 4.34 by induction on the integer

From Properties 3.1 and 3.3, we have

and

Using the recurrence relation of the Fibonacci sequence, it comes that

and

So, we verify (

Notice that (_{0} = 0 ≡ 0 (mod 5_{5}_{k}_{+4} ≡ 0 (mod 5

Let us assume for an integer _{5}_{k}_{−}_{i}^{i}_{i}_{+4} (mod 5

since (−1)^{2} = 1. It achieves the proof of Property 4.34 by induction on the integer

Notice that Property 4.34 is also true for

### Remark 4.35

It can be noticed that for

### Theorem 4.36

**Proof**

Since 5

Using Property 4.34 and taking

or,

and finally,

since 2 and 5

### Theorem 4.37

**Proof**

If

with 5

with 15

Or, from Remark 4.6, we have

and

We have also

So

So, the congruence

with

or

If _{5}_{m}_{+3} ≡ 0 (mod 15

Using the recurrence relation of the Fibonacci sequence, it implies also that

Moreover, we have

Or we have,

Since _{5}_{k}_{+3} ≡ 5

with 15_{10}_{m}_{+6} ≡ 0 (mod 15

It comes that

or,

So, either

or

If _{5}_{m}_{+3} ≡ 0 (mod 15_{5}_{m}_{+2} + 1 ≡ 0 (mod 15

then

It results that the number 10

If _{5}_{m}_{+3} ≡ 0 (mod 15

then since

Notice that in this case, we cannot have

since 3 ≢ 0 (mod 15

which implies that

Since _{5}_{m}_{+2} ≢ −1 (mod 15

Moreover, if _{5}_{m}_{+3} ≡ 0 (mod 15

which implies that

So, either

or

Since we have also

(see above), it remains only one possibility, that is to say

in addition to the condition

If

So

or equivalently (_{5}_{m}_{+4} = _{5}_{m}_{+3} + _{5}_{m}_{+2} and

So, if the number 10

and

Since _{10}_{m}_{+6} ≡ 0 (mod 15

and

So, either

or

If _{5}_{m}_{+2} ≡ _{5}_{m}_{+4} (mod 15

and

where we used the fact that

and (3, 15

then using the recurrence relation of the Fibonacci sequence, we must have

and so

where we used (

and using Theorem 1.5, it gives

since 5^{15}^{m}^{+7} ≡ 1 ≡ 45^{15}^{m}^{+6} ≡ 9_{5}_{m}_{+3} ≡ −2_{5}_{m}_{+2} (mod 15

and so 4(9

Therefore, when 5

if and only if

Since _{15}_{m}_{+9} = _{5}_{k}_{+4} ≡ 0 (mod 5

is also true when

Thus, if 10

if and only if

if and only if

Besides,

implies that

Reciprocally, if _{10}_{m}_{+6} ≡ 0 (mod 15

So, either

or

If _{5}_{m}_{+2} ≡ −_{5}_{m}_{+4} (mod 15

and since _{5}_{m}_{+3} ≡ 0 (mod 15

But, then, if _{10}_{m}_{+6} ≡ 0 (mod 15

Or,

It leads to a contradiction meaning that

is not possible. So, if _{10}_{m}_{+6} ≡ 0 (mod 15

which implies the congruence

and so which translates the congruence

into the congruence

which has at least one solution. So, if 10

if and only if

if and only if

if and only if

Since

### Property 4.38

**Proof**

From Properties 3.3 and 3.4, we have

and

Then, using the recurrence relation of the Fibonacci sequence, it comes that

and

So, we verify (

Notice that (_{0} = 0 ≡ 0 (mod 5_{5}_{k}_{+3} ≡ 0 (mod 5

Let assume for an integer _{5}_{k}_{−}_{i}^{i}_{i}_{+3} (mod 5

since (−1)^{2} = 1. It achieves the proof of Property 4.38 by induction on the integer

Notice that Property 4.38 is also true for

### Remark 4.39

Property 4.38 implies that we can limit ourself to the integer interval [1, _{m}

### Theorem 4.40

**Proof**

Since 5

Using Property 4.38 and taking

or,

finally,

since 2 and 5

### Theorem 4.41

**Proof**

If _{5}_{k}_{15}_{m}

From Remark 4.6, we have

So

So, the congruence _{15}_{m}

or

If _{5}_{m}

Using the recurrence relation of the Fibonacci sequence, it implies also that _{5}_{m}_{+1} ≡ _{5}_{m}_{−}_{1} (mod 15

Or, using Theorem 1.27, we have

Since _{5}_{k}_{+1} ≡ 1 (mod 5

with 15

and

it implies that

We get

and

So, either

or

If _{5}_{m}_{5}_{m}_{+1} − 1 ≡ 0 (mod 15

then

It results that the number 10_{5}_{m}

then since

Notice that in this case, we cannot have

since 3 ≢ 0 (mod 15

which implies that

Since

But, since 15_{5}_{m}_{+1} ≢ 1 (mod 15_{5}_{m}

which implies that

So, either

or

Since we have (

If

or equivalently (_{5}_{m}_{+1} = _{5}_{m}_{5}_{m}_{−}_{1} and

So, if the number 10

and

Since _{10}_{m}_{−}_{1}+_{5}_{m}_{−}_{1}_{5}_{m}

and

So, either

or

If _{5}_{m}_{+1} ≡ _{5}_{m}_{−}_{1} (mod 15

and

where we used the fact that

and (3, 15

then using the recurrence relation of the Fibonacci sequence, we must have

and so

where we used the fact that

It implies that

and using Theorem 1.5, it gives

since 5^{15}^{m}^{15}^{m}^{−}^{1} ≡ 12_{5}_{m}_{5}_{m}_{−}_{1} (mod 15

and so 4(12

Therefore, when 5

if and only if

Since _{15}_{m}_{5}_{k}

is also true when

Thus, if 10

if and only if

if and only if

Besides,

implies that

Reciprocally, if _{10}_{m}

So, either

or

If _{5}_{m}_{+1} ≡ −_{5}_{m}_{−}_{1} (mod 15

and since _{5}_{m}

But, then, if _{10}_{m}

Or,

It leads to a contradiction meaning that

is not possible. So, if _{10}_{m}

which implies the congruence

and so which translates the congruence

into the congruence

which has at least one solution. So, if 10

if and only if

if and only if

if and only if

Since

### Theorem 4.42

**Proof**

The proof is very similar to the proof of Theorem 4.33.

The next theorem below is a generalization of Theorems 4.33, 4.37 and Theorem 4.41 and 4.42 given above. The number _{5}_{k}_{+}_{r}

### Theorem 4.43

**Proof**

The results stated in Theorem 4.43 can be deduced from Theorems 4.33, 4.37 and Theorems 4.41 and 4.42 given above.

### 5. Some Results on Generalized Fibonacci Numbers

- Abstract
- 1. Preliminaries
- 2. Congruences of Fibonacci Numbers Modulo a Prime
- 3. Some Further Congruences of Fibonacci Numbers Modulo a Prime
- 4. Periods of the Fibonacci Sequence Modulo a Positive Integer
- 5. Some Results on Generalized Fibonacci Numbers
- 6. Some Results on Generalized Fibonacci Polynomial Sequences
- Acknowledgements
- References

In this section, we deduce some small results related to the generalized fibonacci numbers as defined below.

### Definition 5.1

Let _{n}_{,2}(

with

In particular, we have

### Remark 5.2

This sequence can be defined from _{2,2}(

### Proposition 5.3

_{n}_{,2}(_{n}_{,2}(1_{n}

**Proof**

Let

Let us assume that this proposition is true up to _{n}_{,2}(_{n}_{,2}(1, 0,−1)) and (_{n}

Thus by induction, the proof is complete.

### Proposition 5.4

_{n}_{,2}(1, 0,−1)) _{n}

This result can be easily verifies using mathematical induction and Theorem 1.26 and Proposition 5.4. We shall omit the details here.

The theorem below appears in any standard linear algebra textbook.

### Proposition 5.5. (i)

_{n}_{n}_{≥}_{0}

_{1}, _{2}_{0}_{1}.

(ii) _{n}_{n}_{≥}_{0}, (_{n}_{n}_{≥}_{0}

_{n}_{n}_{≥}_{0}

**Proof**

The statement (i) is proved by induction.

The statement (ii) can be proved from (i) and from the Cramer’s rule for system of linear equations.

### Definition 5.6

Let _{0}, …, _{k}_{−}_{1 }be _{n,k}_{0}, …, _{k}_{−}_{1})) for

with

The sequence (_{n,k}_{0}, …, _{k}_{−}_{1})) is called the _{0}, …, _{k}_{−}_{1}.

### Proposition 5.7

_{0}, _{1}_{n,}_{2}(_{0}_{1}))

**Proof**

Let _{0}, _{1} be two numbers. Using the relation of recurrence of the sequence (_{n}_{,2}(_{0}, _{1})) and taking the Ansatz _{n}_{,2}(_{0}, _{1}) = ^{n}

For ^{2} −

We can notice that any linear combination of ^{n}^{n}^{n}^{n}^{−}^{1} + ^{n}^{−}^{2} for ^{n}^{n}_{n}_{,2}(_{0}, _{1})) form a vector subspace of the set of complex sequences. Given _{0}, _{1}, from Theorem 5.5 above, since

we deduce that there exist two numbers

Since

the coefficients

So:

where

So:

### Proposition 5.8

_{0}, _{1}

**Proof**

From Proposition 5.7, we have

### Proposition 5.9

This is a standard result and we omit the proof here.

### Example 5.10

Applying Proposition 5.9 when

Thus

### Proposition 5.11

_{0}_{1}

**Proof**

Let

and

So, the formula of Proposition 5.11 is true for

So, using Proposition 5.9, we have (

Or (

where we used the fact that _{0} = 0.

It follows that (

From Proposition 5.9, it results that (

Since this relation is also true for

### Example 5.12

Applying Proposition 5.11 when _{0} = 2, _{1} = 1 and _{n}_{,2}(2, 1) = _{n}

Therefore

### Proposition 5.13

This result can be derived routinely using the results we have derived so far. Although the proof is a little involved, but it follows essentially the same pattern as the previous result. So for the sake of brevity we shall omit it here.

### Example 5.14

Applying Proposition 5.13 when

So,

Applying Proposition 5.13 when

So,

### Proposition 5.15

_{0}, _{1}

**Proof**

Let _{0}, _{1} be two numbers. From Proposition 5.8, we know that for

Using Theorem 1.27, we have

Using Proposition 5.8, we get

In a similar way we can obtain the following result by using the corresponding results dervide so far.

### Proposition 5.16

We now have the following more general results.

### Theorem 5.17

_{0}, _{1}

The proof is an easy application of Proposition 5.9 and we shall omit it here.

### Theorem 5.18

Using Proposition 5.5 and the principle of mathematical induction the above result can be verified. We omit the details here.

### Remark 5.19

Using Proposition 5.4 and using Proposition 5.8, we can notice that

Indeed, we have (

Using the definition of the Fibonacci sequence, we have for

Since _{0,2}(_{0,2}(_{1,2}(_{1,2}(

Taking

and so (

using the relation (

So (

Moreover, from Theorem 5.18, we have (

Therefore (

which is equivalent to Theorem 5.17 when _{0} is replaced by _{1} is replaced by _{n}_{,2}(0, 1) = _{n}

### Definition 5.20

Let _{n,l}_{n,l}_{n,l}

and for

In the following, when there is no ambiguity and when it is possible, we will abbreviate the notations used for terms of sequences (_{n,l}_{n,l}_{n,l}_{n,l}_{n,l}_{n,l}_{n,l}_{n,l}_{n,l}_{n,l}_{n,l}_{n,l}

### Proposition 5.21

**Proof**

This proposition is a direct consequence of Definition 5.1, Definition 5.20 and Theorem 5.18.

### Proposition 5.22

**Proof**

In the following,

So

Moreover, we have (

So

Taking (_{n,l}_{n,l}_{n,l}_{n,l}

and so

It proves the first part of Proposition 5.22. The second part of Proposition 5.22 follows from its first part. Indeed, from the first part of Proposition 5.22, we have (

It proves the second part of Proposition 5.22.

### Definition 5.23

Let be a field. Let _{l}

### Remark 5.24

From Definition 5.20 and from Definition 5.23, we have (

So, from Proposition 5.21, we have

### Proposition 5.25

_{l}

**Proof**

Let

Moreover, from Proposition 5.4, using the definition of the Fibonacci sequence, we have (

So

Using Definition 5.20 and using