EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 4, Article Number 6605
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

On Generalized Mersenne Numbers and Extended
Fermat Numbers

Shefa A. Bani Melhem1,∗, Ala’a Al-Kateeb1,∗, Afnan Dagher1

1 Department of Mathematics, Faculty of Science, Yarmouk University, Irbid, Jordan

Abstract. In this paper, we study a generalization of Mersenne numbers and we introduce an
extension of Fermat numbers, we find their generating functions binet formulas, related matrix
representation and many other properties. Also, we provide some applications in cryptography.

2020 Mathematics Subject Classifications: 11B39, 11B83

Key Words and Phrases: Mersenne and Fermat numbers, generating function, Binet formula,
key exchange and authentication protocols

1. Introduction

Fibonacci and Lucas integer sequences and their generalization/ extensions have many
interesting properties and have been heavily studied [1–5]. The Fibonacci/ Lucas se-
quences are given by the following recurrence relations:

Fn = Fn−1 + Fn−2, Ln = Ln−1 + Ln−2

where n ≥ 2, F0 = 0, F1 = 1 and L0 = 2, L1 = 1. There are other Fibonacci and Lucas
type sequences such as:

• Pell and Pell-Lucas numbers:

Pn = 2Pn−1 + Pn−2;Qn = 2Qn−1 + Qn−2,

where n ≥ 2, P0 = 0, P1 = 1, Q0 = Q1 = 1.

• Jacobsthal and Jacobsthal-Lucas numbers:

Jn = Jn−1 + 2Jn−2; jn = jn−1 + 2jn−2,

where n ≥ 2, J0 = 0, J1 = 1, j0 = j1 = 2.

∗Corresponding author.
∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v18i4.6605

Email addresses: shefa.bm@yu.edu.jo (Sh. A. Bani Melhem),
alaa.kateeb@yu.edu.jo (Al-Kateeb), afnand@yu.edu.jo (A. Dagher)

https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0)



Sh. A. Bani Melhem, Al-Kateeb, A. Dagher / Eur. J. Pure Appl. Math, 18 (4) (2025), 6605 2 of 15

All listed above sequences satisfy a set of common properties and identities, for example Bi-
net formulas, Catalan and Cassini’s identities. Recently, these sequences were generalized
or extended by many authors. Mersenne numbers are given by the formula Mn = 2n − 1
or using the recurrence relation Mn+2 = 3Mn+1−2Mn, where n ≥ 2,M0 = 0 and M1 = 1.
Also, Fermat numbers are given by Fn = 22

n
+1. Next we define the generalized Mersenne

and extended Fermat numbers, which are the main interests of this paper.

Definition 1. Let k ≥ 3, n ≥ 2 be two integers. We define the generalized Mersenne and
extended Fermat numbers respectively by

Mk,n = kMk,n−1 + (1 − k)Mk,n−2, Fk,n = kFk,n−1 + (1 − k)Fk,n−2

where Mk,0 = 0,Mk,1 = 1 and Fk,0 = 2, Fk,1 = 3.

Remark 1. The sequence Mk,n was introduced and studied before in [6].

The search for Mersenne primes is an active field in number theory, since each even
perfect number equals 2k−1Mk, where Mk is a prime Mersenne number. Fermat primes
are useful in generating pseudo-random sequences of numbers an important application
in computer science and cryptography, also they are important for some integer factor-
ization algorithms like in [7, 8], the new defined sequences maybe used to improve such
algorithms. In this paper we introduce and study a generalization to the Mersenne and
extended Fermat numbers. This paper is structured as follows in section 2 we introduce
the generalized Mersenne and extended Fermat numbers and derive their generating func-
tions and Binet formulas, in section 3 we find more other properties, in section 4 we find
the generating matrices of the generalized Mersenne and extended Fermat numbers, also
we present some results involving Mersenne numbers and some tridiagonal and Hessenberg
matrices, finally, in the last section we gave two applications of the generalized Mersenne
numbers matrices in cryptography, namely we present a key-exchange protocol and an
authentication scheme using matrices.

2. Basic Properties

Theorem 1 (Generating functions). The generating functions of the sequences Mk,n and
Fk, n respectively are

(i) M(x) = x
1−kx+(k−1)x2

(ii) F (x) = 2+(3−2k)x
1−kx+(k−1)x2

Proof. Let M(x) represents the generating functions of Mk,n. Note,

M(x) =
∞∑
n=0

Mk,nx
n



Sh. A. Bani Melhem, Al-Kateeb, A. Dagher / Eur. J. Pure Appl. Math, 18 (4) (2025), 6605 3 of 15

= Mk,0 + Mk,1x +

∞∑
n=2

Mk,nx
n

= x +
∞∑
n=2

(kMk,n−1 + (1 − k)Mk,n−2)x
n

= x + kx
∞∑
n=0

Mk,nx
n + (1 − k)x2

∞∑
n=0

Mk,nx
n

Thus, x = (1− kx− (1− k)x2)M(x) ⇒ M(x) = x
1−kx+(k−1)x2 . Similarly we can prove the

theorem for F (x).

F (x) =
∞∑
n=0

Fk,nx
n

= Fk,0 + Fk,1x +

∞∑
n=2

Fk,nx
n

= 2 + 3x +
∞∑
n=2

(kFk,n−1 + (1 − k)Fk,n−2)x
n

= 2 + 3x− 2kx + kx

∞∑
n=0

Fk,nx
n + (1 − k)x2

∞∑
n=0

Fk,nx
n

Thus, 2 + (3 − 2k)x = (1 − kx− (1 − k)x2)F (x) ⇒ F (x) = 2+(3−2k)x
1−kx+(k−1)x2 .

Theorem 2 (Binet formula). The n-th terms of the generalized Mersenne and extended
Fermat sequences are given by

Mk,n =
(k − 1)n − 1

k − 2

and

Fk,n =
(k − 1)n + 2k − 5

k − 2

Proof. A proof for Mk,n can be found in [6], the formula for Fk,n can be proved easily
by induction.

Proposition 1. For n > 1 we have Fk,n = Mk,n + 2

Proof. Immediate from the Binet formulas.

Theorem 3 (Catalan’s identity). We have



Sh. A. Bani Melhem, Al-Kateeb, A. Dagher / Eur. J. Pure Appl. Math, 18 (4) (2025), 6605 4 of 15

• Mk,n−rMk,n+r −M2
k,n = −(k − 1)n−rM2

k,r

• Fk,n−rFk,n+r − F 2
k,n = (2k − 5)(k − 1)n−rM2

k,r

Proof.

•

Mk,n−rMk,n+r −M2
k,n =

(k − 1)n−r − 1

(k − 2)

(k − 1)n+r − 1

(k − 2)
− (

(k − 1)n − 1

(k − 2)
)2

=
(k − 1)2n − (k − 1)n−r − (k − 1)n+r + 1

(k − 2)2
− (k − 1)2n − 2(k − 1)n + 1

(k − 2)2

=
−(k − 1)n−r − (k − 1)n+r + 2(k − 1)n

(k − 2)2

= −(k − 1)n−r 1 + (k − 1)2r − 2(k − 1)r

(k − 2)2

= −(k − 1)n−rM2
k,r

•

Fk,n−rFk,n+r − F 2
k,n =

(k − 1)n−r + 2k − 5

k − 2

(k − 1)n+r + 2k − 5

k − 2
− (

(k − 1)n + 2k − 5

k − 2
)2

=
(k − 1)2n + (2k − 5)(k − 1)n−r + (2k + 5)(k − 1)n+r + (2k + 5)2

(k − 2)2

− (k − 1)2n + 2(2k − 5)(k − 1)n + (2k − 5)2

(k − 2)2

=
(2k − 5)(k − 1)n−r + (2k + 5)(k − 1)n+r − 2(2k − 5)(k − 1)n

(k − 2)2

= (2k − 5)(k − 1)n−r 1 + (k − 1)2r − 2(k − 1)r

(k − 2)2

= (2k − 5)(k − 1)n−rM2
k,r.

Theorem 4 (d’Ocagne’s identity). If ℓ ≥ n, then

• Mk,ℓMk,n+1 −Mk,ℓ+1Mk,n = (k − 1)nMk,ℓ−n.

• Fk,ℓFk,n+1 − Fk,ℓ+1Fk,n = −(5 − 2k)(k − 1)nMk,ℓ−n.

Proof.

•

Mk,ℓMk,n+1 −Mk,ℓ+1Mk,n =
(k − 1)ℓ − 1

(k − 2)

(k − 1)n+1 − 1

(k − 2)
− (k − 1)ℓ+1 − 1

(k − 2)

(k − 1)n − 1

(k − 2)



Sh. A. Bani Melhem, Al-Kateeb, A. Dagher / Eur. J. Pure Appl. Math, 18 (4) (2025), 6605 5 of 15

=
(k − 1)ℓ+1 − (k − 1)ℓ − (k − 1)n+1 + (k − 1)n

(k − 2)2

=
(k − 1)ℓ(k − 2) − (k − 1)n(k − 2)

(k − 2)2

= (k − 1)n
(k − 1)ℓ−n − 1

k − 2

= (k − 1)nMk,ℓ−n

•

Fk,ℓFk,n+1 − Fk,ℓ+1Fk,n =
(k − 1)ℓ + 2k − 5

(k − 2)

(k − 1)n+1 + 2k − 5

(k − 2)

− (k − 1)ℓ+1 + 2k − 5

(k − 2)

(k − 1)n + 2k − 5

(k − 2)

= (2k − 5)
−(k − 1)ℓ+1 + (k − 1)ℓ + (k − 1)n+1 − (k − 1)n

(k − 2)2

= (2k − 5)
(k − 1)ℓ(2 − k) + (k − 1)n(k − 2)

(k − 2)2

= (2k − 5)
(k − 1)n − (k − 1)ℓ

(k − 2)
= (2k − 5)(k − 1)n

1 − (k − 1)ℓ−n

(k − 2)

= −(5 − 2k)(k − 1)nMk,ℓ−n.

Theorem 5 (Vajda’s identity).

(i) (Formulation 1) Mk,n+iMk,n+j −Mk,nMk,n+i+j = (k − 1)nMk,iMk,j

(ii) (Formulation 2) Mk,n+jMk,m−j −Mk,nMk,m = (k − 1)nMk,m−n−jMk,j

Proof.

(i) (Formulation 1)

Mk,n+iMk,n+j −Mk,nMk,n+i+j =
(k − 1)n+i − 1

(k − 2)

(k − 1)n+j − 1

(k − 2)
− (k − 1)n − 1

(k − 2)

(k − 1)n+i+j − 1

(k − 2)

=
(k − 1)n − (k − 1)n+i − (k − 1)n+j + (k − 1)n+i+j

(k − 2)2

= (k − 1)n
1 − (k − 1)i − (k − 1)j + (k − 1)i+j

(k − 2)2

= (k − 1)n
1 − (k − 1)i − (k − 1)j(1 − (k − 1)i)

(k − 2)2



Sh. A. Bani Melhem, Al-Kateeb, A. Dagher / Eur. J. Pure Appl. Math, 18 (4) (2025), 6605 6 of 15

= (k − 1)n
(1 − (k − 1)i)(1 − (k − 1)j)

(k − 2)2

= (k − 1)nMk,iMk,j

(ii) (Formulation 2)

Mk,n+iMk,m−j −Mk,nMk,m =
(k − 1)n+j − 1

(k − 2)

(k − 1)m−j − 1

(k − 2)
− (k − 1)n − 1

(k − 2)

(k − 1)m − 1

(k − 2)

=
(k − 1)m − (k − 1)m−j − (k − 1)n+j + (k − 1)n

(k − 2)2

= (k − 1)n
1 − (k − 1)m−n−j + (k − 1)j((k − 1)m−n−j − 1)

(k − 2)2

= (k − 1)nMk,m−n−jMk,j

Theorem 6 (Sum of terms). If n ≥ 2, then

(i)
∑n

i=0Mk,i = 1
k−2(Mk,n+1 − n− 1)

(ii)
∑n

i=0 Fk,i = 1
k−2(Fk,n+1 − 2 + (n + 1)(2k − 5)) = 1

k−2(Mk,n+1 + (n + 1)(2k − 5))

(iii)
∑n

i=0Mk,2i = 1
k−2( 1kMk,2(n+1) − n− 1)

(iv)
∑n

i=0 Fk,2i = 1
k−2( 1kMk,2(n+1)+(n+1)(2k−5)) = 1

k−2( 1k (Fk,2(n+1)−2)+(n+1)(2k−5))

Proof.

(i)

n∑
i=0

Mk,i =

n∑
i=0

(k − 1)i − 1

k − 2

=
1

(k − 2)

(
1 − (k − 1)n+1

2 − k
− (n + 1)

)
, geometric series

=
1

k − 2
(Mk,n+1 − n− 1)

(ii)

n∑
i=0

Fk,i =

n∑
i=0

(k − 1)i + (2k − 5)

k − 2

=
1

(k − 2)

(
1 − (k − 1)n+1

2 − k
+ (n + 1)(2k − 5)

)
, geometric series

=
1

k − 2
(Fk,n+1 − 2 + (n + 1)(2k − 5)) =

1

k − 2
(Mk,n+1 + (n + 1)(2k − 5))



Sh. A. Bani Melhem, Al-Kateeb, A. Dagher / Eur. J. Pure Appl. Math, 18 (4) (2025), 6605 7 of 15

(iii)

n∑
i=0

Mk,2i =
n∑

i=0

((k − 1)2)i − 1

k − 2

=
1

(k − 2)

(
1 − ((k − 1)2)n+1

1 − (k − 1)2
− (n + 1)

)
, geometric series

=
1

(k − 2)

(
1 − ((k − 1)2n+2

k(2 − k)
− (n + 1)

)
=

1

k − 2
(
1

k
Mk,n+1 − n− 1)

(iv)

n∑
i=0

Fk,2i =
n∑

i=0

((k − 1)2)i + (2k − 5)

k − 2

=
1

(k − 2)

(
1 − ((k − 1)2)n+1

1 − (k − 1)2
+ (n + 1)(2k − 5)

)
, geometric series

=
1

(k − 2)

(
1 − ((k − 1)2n+2

k(2 − k)
+ (n + 1)(2k − 5)

)
=

1

k − 2
(
1

k
Mk,2(n+1) + (n + 1)(2k − 5))

=
1

k − 2
(
1

k
(Fk,2(n+1) − 2) + (n + 1)(2k − 5))

Lemma 1. We have limn→∞
Mk,n+1

Mk,n
= k − 1.

Proof. limn→∞
Mk,n+1

Mk,n
= limn→∞

(k−1)n+1−1
k−2 · k−2

(k−1)n−1 = limn→∞
(k−1)n+1(1− 1

(k−1)n+1 )

(k−1)n(1− 1
(k−1)n

)
=

k − 1

Lemma 2. The series
∑∞

n=0
1

Mk,n
is a convergent series.

Proof. By ratio test limn→∞
Mk,n

Mk,n+1
= 1

k−1 < 1, so the series is convergent,

Lemma 3. The series
∑∞

n=1
Mk,n

kn = k
k−1 .

Proof. Let S =
∑∞

n=1
Mk,n

kn . Then

∞∑
n=1

Mk,n

kn
=

1

k
+

∞∑
n=2

Mk,n

kn



Sh. A. Bani Melhem, Al-Kateeb, A. Dagher / Eur. J. Pure Appl. Math, 18 (4) (2025), 6605 8 of 15

=
1

k
+

∞∑
n=1

Mk,n+1

kn+1

=
1

k
+

∞∑
n=1

kMk,n + (1 − k)Mk,n−1

kn+1

=
1

k
+

∞∑
n=1

Mk,n

kn
+

(1 − k)

k2

∞∑
n=1

Mk,n−1

kn−1

=
1

k
+ S +

(1 − k)

k2

∞∑
n=0

Mk,n

kn

=
1

k
+ S +

(1 − k)

k2
S

Thus, S = k
k−1 .

We need the next remark in the proof of the next theorem. In fact it is exercise 16 in
section 3.3 in [9].

Remark 2. For any three integers a, b and c such that gcd(a, b) = gcd(a, c) = 1, we have
gcd(a, bc) = 1

Theorem 7. For n ≥ 1, we have

(i) gcd(Mk,n, k − 1) = 1

(ii) gcd(Mk,n, k) =

{
1, if n is odd

k, if n is even

(iii) gcd(Mk,n,Mk,n+1) = 1

(iv) gcd(Fk,n, k − 1) = 1

(v) gcd(Fk,n, k) =

{
gcd(3, k), if n is odd

gcd(2, k), if n is even

(vi) gcd(Fk,n, Fk,n+1) = 1

Proof.

(i) We prove this property using mathematical induction. At first gcd(Mk,1, k − 1) =
gcd(1, k − 1) = 1. Assume that gcd(Mk,n, k − 1) = 1, consider gcd(Mk,n+1, k − 1) =
gcd(kMk,n + (1−k)Mk,n−1, k−1) = gcd(kMk,n, k−1) = gcd(Mk,n, k−1) = 1, using
the well-known property gcd(a + bc, b) = gcd(a, b)



Sh. A. Bani Melhem, Al-Kateeb, A. Dagher / Eur. J. Pure Appl. Math, 18 (4) (2025), 6605 9 of 15

(ii) We prove this property using mathematical induction. At first gcd(Mk,1, k) = 1 and

gcd(Mk,2, k) = 2. Assume that gcd(Mk,n, k) =

{
1, if n is odd

k, if n is even
, now consider

gcd(Mk,n+1, k) = gcd(kMk,n + (1 − k)Mk,n−1, k)

= gcd((1 − k)Mk,n−1, k)

= gcd(Mk,n−1, k)

=

{
1, if n− 1 is odd

k, if n− 1 is even

=

{
1, if n + 1 is odd

k, if n + 1 is even

as desired.

(iii) The result is clear for n = 1, so we proceed by mathematical induction. Let
d = gcd(Mk,n+1,Mk,n+2). We have d|Mk,n+1 and d|Mk,n+2 so it divides any lin-
ear combination of them that is

d|(Mk,n+2 − kMk,n+1) = (1 − k)Mk,n

Thus, d|(gcd(Mk,n+1, (1 − k)Mk,n)). Now, from Remark 2 and part 1 we have

gcd(Mk,n+1, (1 − k)Mk,n) = 1

which leads to the fact that d = 1.

(iv) Similar to number 1 above.

(v) Similar to number 2 above.

(vi) Similar to number 3 above.

Lemma 4. If k is even, then gcd(Mk,n, Fk,n) =

{
1, if n is odd

2, if n is even

Proof. Recall that Fk,n = Mk,n + 2 (Proposition 1), also if k is even we have Mk,n ≡ n

mod 2, thus gcd(Mk,n, Fk,n) =

{
1, if n is odd

2, if n is even
, as desired.



Sh. A. Bani Melhem, Al-Kateeb, A. Dagher / Eur. J. Pure Appl. Math, 18 (4) (2025), 6605 10 of 15

3. Matrix representations

Theorem 8 (Matrix of generalized Mersenne numbers). Let Qk =

[
0 1

1 − k k

]
. Then for

n ≥ 2 we have

Qn
k =

[
(1 − k)Mk,n−1 Mk,n

(1 − k)Mk,n Mk,n+1

]
Proof. We prove the result by mathematical induction:

• If n = 2, we have

Q2
k =

[
1 − k k

k(1 − k) k2 + (1 − k)2

]
=

[
(1 − k)Mk,1 Mk,2

(1 − k)Mk,2 Mk,3

]
• Assuming that the result is true for all 2 ≤ m ≤ n.

• Consider Qn+1
k = QkQ

n
k =

[
0 1

1 − k k

] [
(1 − k)Mk,n−1 Mk,n

(1 − k)Mk,n Mk,n+1

]
=

[
(1 − k)Mk,n Mk,n+1

(1 − k)Mk,n+1 Mk,n+2

]
as desired.

Lemma 5. For any integers m and n and k ≥ 2 we have

(i) Mk,n+m = (1 − k)Mk,n−1Mk,m + Mk,nMk,m+1.

(ii) Mk,n+m+1 = (1 − k)Mk,nMk,m + Mk,n+1Mk,m+1.

(iii) Mk,2n = (1 − k)Mk,n−1Mk,n + Mk,nMk,n+1.

(iv) Mk,2n+1 = (1 − k)M2
k,n + M2

k,n+1.

Proof. Consider

Qn+m
k = Qn

kQ
m
k

=

[
(1 − k)Mk,n+m+1 Mk,n+m

(1 − k)Mk,n+m Mk,n+m+1

]
=

[
(1 − k)Mk,n−1 Mk,n

(1 − k)Mk,n Mk,n+1

] [
(1 − k)Mk,m−1 Mk,m

(1 − k)Mk,m Mk,m+1

]
we get 1 and 2 from equating the corresponding entries of the equal matrices. Also, we get
3 and 4 by letting m = n in 1 and 2.

Lemma 6 (Matrix of Extended Fermat Numbers ). Let Rk =

[
2 3
3 2 + k

]
. Then

Qn
kRk =

[
Fk,n Fk,n+1

Fk,n+1 Fk,n+2

]
Proof. Can be proved easily by mathematical induction.



Sh. A. Bani Melhem, Al-Kateeb, A. Dagher / Eur. J. Pure Appl. Math, 18 (4) (2025), 6605 11 of 15

4. Special Kind of Tridiagonal and Hessenberg matrices

A tridiagonal matrix is a matrix that has nonzero elements only on the main diagonal
and on the first diagonal below and above the main diagonal. In this subsection we
represent some tridiagonal matrices whose determinant or permanent is a generalized
Mersenne number.

Lemma 7. Suppose that n ≥ 1 is an integer.

(i) Let Nn(k) =



k k − 1 0 · · · · · · · · · · · · 0
−1 k 1 − k 0 · · · · · · · · · 0
0 −1 k 1 − k · · · · · · · · · 0
0 0 −1 k 1 − k · · · · · · 0
...

...
...

...
...

...
...

...
0 ... −1 k


Then det(Nn) = Mk,n+1.

(ii) Let Hn(k) =



0 1 0 · · · · · · · · · · · · 0
−1 0 1 − k 0 · · · · · · · · · 0
0 −1 k 1 − k · · · · · · · · · 0
0 0 −1 k 1 − k · · · · · · 0
...

...
...

...
...

...
... 0

0 · · · · · · · · · · · · · · · −1 k


.

Then det(Hn) = Mk,n.

Proof. Clear from Section 2.1 and 2.2 in [3]

Lemma 8. Suppose that n ≥ 1 is an integer. Let Tn(k) =


k 1 0 · · · · · · · · · · · · 0

k − 1 k 1 0 · · · 0
0 k − 1 k 1 0 ... 0
...

...
...

...
...

0 ... k − 1 k

.

Then det(Tn) = Mk,n+1.

Proof. Let α = k − 1, β = 1. Then Mk,n = αn−βn

α−β , following the work of Kilic and
Tasci in [10, 11] we have det(Tn) = Mk,n+1.

Finally, we use Theorems 1-3 from [10] to get the following result.

Lemma 9. We have

(i) Let An(k) =


k 1 − k 0 ... ... ... 0
1 k 1 − k 0 ... 0
0 1 k 1 − k 0 ... 0
...

...
...

...
...

0 ... 1 k

.

Then per(An) = Mk,n+1.



Sh. A. Bani Melhem, Al-Kateeb, A. Dagher / Eur. J. Pure Appl. Math, 18 (4) (2025), 6605 12 of 15

(ii) Let Hn(k) =


k 1 − k 0 ... ... 0
1 0 0 0 0
0 0 An(k) 0 · · · 0 0
0 · · · 0

.

Then per(Hn) =
∑i=0

n Mk,i where n ≥ 2 and An(k) defined as the last part.

(iii) Let Gn(k) =


1 1 1 ... ... ... 1
−1 k 1 − k 0 ... 0
0 −1 k 1 − k 0 ... 0
...

...
...

...
...

0 ... −1 k

.

Then det(Gn) =
∑i=0

n Mk,i where n ≥ 2.

Proof. Using α = k − 1, β = 1 and Mk,n = αn−βn

α−β

(i) Statement 1 is true by Theorem 1 from [10].

(ii) Statement 2 is true by Theorem 2 from [10].

(iii) Statement 3 is true by Theorem 3 from [10].

5. Some Applications

In this section we give two applications in cryptography for the matrix representation
of Mk,m, which was found in Section 3.

5.1. An authentication protocol

In this section we will introduce an authentication protocol based on matrices. At first,
Alice chooses a large integer n = pq a product of two primes and an integer k > 2. Then
Alice chooses two matrices A = Qn1

k and B = Qn2
k and computes their squares modulo n.

Namely, C = A2 = Q2n1
k mod n and D = B2 = Q2n2

k mod n. Alice publishes C and D.
The method works as follows:

Algorithm 1. (i) Alice chooses two random integers m1,m2 and finds the matrices
X1 = Qm1

k , X2 = Qm2
k . Then she computes Y1 = X2

1 mod n and Y2 = X2
2 mod n

and sends Y1, Y2 to Bob.

(ii) Bob chooses two random numbers v1, v2 ∈ {0, 1} and sends them to Alice.

(iii) Alice computes Z = X1X2A
v1Bv2 mod n and send it to Bob.

(iv) Bob verifies the identity of Alice by checking that Z2 = Y1Y2C
v1Dv2

(v) Bob asks Alice to send him one of X1 or X2 and he checks that Y1 = X2
1 or Y2 = X2

2 .



Sh. A. Bani Melhem, Al-Kateeb, A. Dagher / Eur. J. Pure Appl. Math, 18 (4) (2025), 6605 13 of 15

Example 1. Let n = 11 · 13 = 143, and k = 5. If Alice takes on n1 = 2 and n2 = 3. Then

A =

[
139 5
123 21

]
and B =

[
123 21
59 85

]
. Also, we have

C = A2 =

[
59 85
89 55

]
and

D = B2 =

[
66 78
117 27

]
(i) Alice chooses two random integers m1 = 3,m2 = 4 and finds the matrices

X1 = Qm1
5 =

[
123 21
59 85

]

X2 = Qm2
5 =

[
59 85
89 55

]
Then she computes Y1 = X2

1 =

[
66 78
117 27

]
and Y2 = X2

2 =

[
35 109
136 8

]
and sends

Y1, Y2 to Bob.

(ii) Bob chooses two random numbers v1 = 0, v2 = 1 and sends them to Alice.

(iii) Alice computes Z = X1X2A
v1Bv2 mod n =

[
111 33
11 133

]
and send it to Bob.

(iv) Bob verifies the identity of Alice by checking that Z2 = Y1Y2C
v1Dv2 =

[
100 44
110 34

]
(v) Bob asks Alice to send him one of X1 or X2 and he checks that Y1 = X2

1 or Y2 = X2
2 .

We should mention that it is very hard to know the matrices A and B from C and D.
Also, be choosing a very large number the problem will be more and more harder.

5.2. A Key Exchange Protocol

In this section we propose a key exchange protocol based on matrices At first Alice
and Bob choose and integer n = pq and k > 2 . Then they choose a 2 × 2 matrix A The
protocol works as follows:

Algorithm 2. (i) Alice selects an integer t > 0 and a secret matrix B = Qn1
k mod n

that doesn’t commute with A. She computes X1 = AtB mod n and send it to Bob.

(ii) Bob chooses an integer s > 0 and a secret matrix C = Qn2
k mod n that doesn’t

commute with A and computes X2 = AsC mod n and send it to Alice.

(iii) Alice computes KA = AtX2B mod n.



Sh. A. Bani Melhem, Al-Kateeb, A. Dagher / Eur. J. Pure Appl. Math, 18 (4) (2025), 6605 14 of 15

(iv) Bob computes KB = AsX1C mod n

we have KA = KB = K.

Example 2. Let k = 4 and A =

[
1 2
3 4

]
.

(i) Alice selects an integer t = 3 and computes B = Q3
k =

[
131 13
104 40

]
which doesn’t

commute with A. She computes X1 = AtB =

[
24 67
3 53

]
and send it to Bob.

(ii) Bob chooses s = 7 and find C = Q5
k mod n =

[
23 121
66 78

]
which doesn’t commute

with A and computes X2 = AsC =

[
130 42
85 72

]
and send it to Alice.

(iii) Alice computes KA = AtX2B =

[
1 112

111 29

]
.

(iv) Bob computes KB = AsX1C =

[
1 112

111 29

]

6. Conclusion

In this paper we consider a generalization of Merssen and Fermat numbers into complex
numbers, we derive some properties of the new sequence. For future work we may consider
more generalizations and extensions of such sequence and try to relate them with real life
applications.

Acknowledgements

The publication of this paper was supported by the Yarmouk University Research
Council.

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