Quaternion algebras and the generalized Fibonacci-Lucas quaternions

0.12 MB·English

by  Cristina Flaut

Quaternion algebras and the generalized
Fibonacci-Lucas quaternions
5
1
0
Cristina FLAUT and Diana SAVIN
2

r
a
M
Abstract. In this paper, we introduce the generalized Fibonacci-Lucas quater-
nions and we prove that the set of these elements is an order–in the sense of ring
6
1 theory– of a quaternion algebra. Moreover, we investigate some properties of these
elements.

]
A
Key Words: quaternion algebras; Fibonacci numbers; Lucas numbers.
R
. 2000 AMS Subject Classification: 15A24,15A06,16G30,1R52,11R37,
h
11B39.
t
a
m
[
1. Preliminaries

2
v
2 Let (fn)n 0 be the Fibonacci sequence:
7 ≥
7 f =0;f =1;f =f +f , n≥2
0 1 n n 1 n 2
1 − −
0 and (l ) be the Lucas sequence:
. n n 0
1 ≥
0 l =2;f =1;l =l +l , n≥2
0 1 n n 1 n 2
5 − −
1 In the paper [Ho; 61], A. F. Horadam generalized Fibonacci numbers in the
:
v following way:
i h =p,h =q,h =h +h , n≥2,
X 0 1 n n 1 n 2
− −
r where p and q are arbitrary integer numbers.
a
Inthepaper[Ho;63],A.F.HoradamintroducedtheFibonacciquaternionsand
generalized Fibonacci quaternions. In their work [Fl, Sh; 12], C. Flaut and V.
Shpakivskyi and later in the paper [Ak, Ko, To; 14], M. Akyigit, H. H Kosal,
M. Tosun found some properties of the generalized Fibonacci quaternions. In
the papers [Fl, Sa; 14] and [Fl, Sa, Io; 13], the authors gave the definitions of
theFibonaccisymbolelementsandLucassymbolelementsandtheydetermined
many properties of these elements. All these elements determine a Z-module
structure.
Inthispaper,wedefinethegeneralizedFibonacci-Lucasnumbersandthegener-
alized Fibonacci-Lucas quaternions and we provethat, in the second case, they
determine an order of a quaternion algebra.
1
For other results regardind above notions, the reader is referred to [Fl; 06],
[Sa, Fl, Ci; 09], [Fl, Sh; 13(1)], [Fl, Sa; 14], [Fl, Sa; 15],[Fl, St; 09].
2. Properties of the Fibonacci and Lucas numbers
From [Fib.], the following properties of Fibonacci numbers are known:
Proposition 2.1. Let (f ) be the Fibonacci sequence and let (l )
n n 0 n n 0
≥ ≥
be the Lucas sequence. Therefore the following properties hold:
i)
f2+f2 =f ,∀ n∈N;
n n+1 2n+1
ii)
f2 −f2 =f ,∀ n∈N ;
n+1 n 1 2n ∗
−
iii)
l2 −f2 =4f f ,∀ n∈N ;
n n n 1 n+1 ∗
−
iv)
l2 +l2 =5f ,∀ n∈N;
n n+1 2n+1
v)
l2 =l +2(−1)n,∀ n∈N ;
n 2n ∗
vi)
f +f =l ,∀ n∈N ;
n+1 n 1 n ∗
−
vii)
l +l =5f ,∀ n∈N;
n n+2 n+1
viii)
f +f =3f ,∀ n∈N;
n n+4 n+2
ix)
f l =f +(−1)m+1·f ,∀ m,p∈N;
m m+p 2m+p p
x)
f l =f +(−1)m·f ,∀ m,p∈N;
m+p m 2m+p p
xi)
1
f f = l +(−1)m+1·l ,∀ m,p∈N;
m m+p 2m+p p
5
(cid:16) (cid:17)
xii)
l l +5f f =2l ,∀ m,p∈N.
m p m p m+p
xiii)
f are even numbers if and only if n≡0 (mod 3).
n
2
In the next proposition, we will give other elementary properties of the Fi-
bonacciandLucasnumbers. Thesepropertieswillbenecessaryinthefollowing.
Proposition 2.2. Let (f ) be the Fibonacci sequence and (l ) be the
n n 0 n n 0
≥ ≥
Lucas sequence Then we have that
l l =l +(−1)m·l ,∀ m,p∈N.
m m+p 2m+p p
Proof. If we denote α = 1+√5 and β = 1 √5, using Binet’s formula, we have
2 −2
fn = ααn−ββn = √15(αn−βn), ∀ n∈N and ln =αn+βn, ∀ n∈N.
i) Let m−,p∈N, p≤m, therefore we have
l l −5f f =(αm+βm)·(αp+βp)−(αm−βm)·(αp−βp)=
m p m p
=2αpβp αm p+βm p =2(−1)p·l .
− − m p
−
p
It results lmlp−5fmfp =2(cid:0)(−1) ·lm p. A(cid:1)dding this equality with the equality
− p
from Proposition 2.1 (xii)) , we obtain l l = l +(−1) ·l ,∀ m,p ∈
m p m+p m p
N,p≤m. From here, it results that l l =l +(−1)m·l ,∀−m,p∈N.(cid:3)
m m+p 2m+p p
3. Generalized Fibonacci-Lucas numbers and generalized Fibonacci-
Lucas quaternions
Letnbeanarbitrarypositiveintegerandp,qbetwoarbitraryintegers. The
sequence g (n≥1), where
n
g =pf +ql , n≥0
n+1 n n+1
is called the generalized Fibonacci-Lucas numbers.
To emphasize the integer p and q, in the following we will use the notation gp,q
n
instead of g .
n
Let H(α,β) be the generalized real quaternion algebra, the algebra of the
elements of the form a=a ·1+a i+a j+a k, where a ∈R,i2 =α,j2 =β,
1 2 3 4 i
k = ij = −ji. We denote by t(a) and n(a) the trace and the norm of a real
quaterniona.Thenormofageneralizedquaternionhasthefollowingexpression
n(a)=a2−αa2−βa2+αβa2 andt(a)=2a .Itisknownthatfora∈H(α,β),
1 2 3 4 1
we have a2−t(a)a+n(a) = 0. The quaternion algebra H(α,β) is a division
algebra if for all a ∈ H(α,β), a 6= 0 we have n(a) 6= 0, otherwise H(α,β) is
called a split algebra.
Let H (α,β) be the generalized quaternion algebra over the rational field.
Q
We define the n-th generalized Fibonacci-Lucas quaternion to be the element of
the form
Gp,q =gp,q·1+gp,q ·i+gp,q ·j+gp,q ·k,
n n n+1 n+2 n+3
where i2 =α,j2 =β, k =ij =−ji.
3
In the following proposition, for α = −1 and β = p, we compute the norm
for the n-th generalized Fibonacci-Lucas quaternions.
Proposition 3.1. Let n,p be two arbitrary positive integers and q be an
arbitrary integer. Let Gp,q be the n-th generalized Fibonacci-Lucas quaternion.
n
Then the norm of the element Gp,q in the quaternion algebra H (−1,p)has the
n Q
form
i)
n(Gp,q)= p3+p2+8p2q+15pq2+2pq f +
n 2n 1
−
+ −3p3(cid:0)+5q2−22p2q−40pq2+2pq (cid:1)f ;
2n+1
or (cid:0) (cid:1)
ii)
n(Gp,q)=ga,b,
n 2n
where a=4p3+p2+30p2q+55pq2−5q2and b=−3p3+5q2−22p2q−40pq2+2pq.
Proof. i)
n(Gp,q)=Gp,qGp,q =(gp,q)2+ gp,q 2−p gp,q 2−p gp,q 2 =
n n n n n+1 n+2 n+3
=(pf +ql )2+(pf +ql )2−p(cid:0)(pf (cid:1) +ql(cid:0) )2−(cid:1) p(pf(cid:0) +(cid:1)ql )2 =
n 1 n n n+1 n+1 n+2 n+2 n+3
−
=p2 f2 +f2 −p3 f2 +f2 +q2 l2 +l2 −pq2 l2 +l2 +
n 1 n n+1 n+2 n n+1 n+2 n+3
−
(cid:0) +2pq(cid:1)(f l(cid:0) +f l )−(cid:1)2p2q((cid:0)f l (cid:1)+f l(cid:0) ) ((cid:1)3.1).
n 1 n n n+1 n+1 n+2 n+2 n+3
−
Using Proposition 2.1 (i; vi;viii; ix) and the relation (3.1), we obtain
n(Gp,q)=p2f +5q2f +2pq f +(−1)n+f +(−1)n+1 −
n 2n 1 2n+1 2n 1 2n+1
− −
h i
−p3f −5pq2f −2p2q f +(−1)n+2+f +(−1)n+3 =
2n+3 2n+5 2n+3 2n+5
h i
=p2f +5q2f −p3f −5pq2f +2pql −2p2ql .
2n 1 2n+1 2n+3 2n+5 2n 2n+4
−
From Fibonacci recurrence, we obtain
f =3f −f ,f =8f −3f .
2n+3 2n+1 2n 1 2n+5 2n+1 2n 1
− −
Usingtheseequalities,Proposition2.1(vi)andFibonaccirecurrence,weobtain
l =f +f and l =f +f =11f −4f .
2n 2n+1 2n 1 2n+4 2n+3 2n+5 2n+1 2n 1
− −
It results
n(Gp,q)= p3+p2+8p2q+15pq2+2pq f +
n 2n 1
−
+ −3p3(cid:0)+5q2−22p2q−40pq2+2pq (cid:1)f .
2n+1
ii) Using Propositio(cid:0)n 2.1 (vi), last equality becomes (cid:1)
n(Gp,q)= 4p3+p2+30p2q+55pq2−5q2 f +
n 2n 1
−
+ −3(cid:0)p3+5q2−22p2q−40pq2+2pq (cid:1)l .
2n
(cid:0) (cid:1)
4
If we denote a=4p3+p2+30p2q+55pq2−5q2 and b=−3p3+5q2−22p2q−
40pq2+2pq,thelastequalityisequivalentwithn(Gp,q)=af +bl =ga,b.(cid:3)
n 2n 1 2n 2n
−
Let Abe a Noetherianintegraldomainwith the field ofthe fractions K and
let H (α,β) be the generalized quaternion algebra. We recall that O is an
K
order in H (α,β) if O ⊆ H (α,β) is an A-lattice of H (α,β) (i.e a finitely
K K K
generated A submodule of H (α,β)) which is also a subring of H (α,β) (see
K K
[Vo; 10]).
In the following, we will built an order of a quaternion algebras using the gen-
eralized Fibonacci-Lucas quaternions. Also we will prove that Fibonacci-Lucas
quaternions can have an algebra structure over Q. For this, we make the fol-
lowing remarks.
Remark 3.1. Let nbean arbitrary positive integer and p,qbetwoarbitrary
integers. Let (gp,q) be the generalized Fibonacci-Lucas numbers. Then
n n 1
≥
pf +ql =gp,q+gp,0 ,∀ n∈N .
n+1 n n n+1 ∗
Proof.
pf +ql =pf +ql +pf =gp,q+pf =gp,q+gp,o .
n+1 n n 1 n n n n n n+1
−
(cid:3)
Remark 3.2. Let n be an arbitrary positive integer and p,q be two arbi-
trary integers. Let (gp,q) be the generalized Fibonacci-Lucas numbers and
n n 1
(Gp,q) be the generalized≥Fibonacci-Lucas quaternion elements. Then:
n n 1
≥
Gp,q =0 if and only if p=q =0.
n
Proof. ” ⇐” It is trivial.
” ⇒” If Gp,q =0, since {1,i,j,k} is a basis in H (α,β), we obtain that gp,q =
n Q n
gp,q = gp,q = gp,q = 0. It results gp,q = gp,q − gp,q = 0, ..., gp,q = 0,
n+1 n+2 n+3 n 1 n+1 n 2
gp,q = 0. But gp,q = pf +ql = 2q, there−fore q = 0. From gp,q = 0, we obtain
1 1 0 1 2
p=0.(cid:3)
Theorem 3.1. Let M be the set
n
M = 5Gpi,qi|n∈N ,p ,q ∈Z,(∀)i=1,n ∪{1}.
ni ∗ i i
( )
i=1
X
1) The set M has a ring structurewith quaternions addition and multiplication.
2) The set M is an order of the quaternion algebra H (α,β).
Q
3)Theset M = n 5Gp′i,qi′|n∈N ,p ,q ∈Q,(∀)i=1,n ∪{1}isaQ−algebra.
′ ni ∗ ′i i′
(cid:26)i=1 (cid:27)
P
Proof. 2) First, we remark that 0∈M (using Remark 3.2).
Now we prove that M is a Z− submodule of H (α,β).
Q
′ ′
Let n,m∈N , a,b,p,q,p ,q ∈Z. It is easy to prove that
∗
′ ′ ′ ′
agp,q+bgp ,q =gap,aq+gbp ,bq .
n m n m
5
This implies that
′ ′ ′ ′
aGp,q+bGp ,q =Gap,aq +Gbp ,bq .
n m n m
From here, we get immediately that M is a Z− submodule of the quaternion
algebra H (α,β). Since {1,i,j,k} is a basis for this submodule, it results that
Q
M is a free Z− module of rank 4.
Now, we prove that M is a subring of H (α,β). It is enough to show that
Q
′ ′
5Gp,q·5Gp ,q ∈M. For this, if m<n, we calculate
n m
5gp,q·5gp′,q′ =5(pf +ql )·5 p′f +q′l =
n m n 1 n m 1 m
− −
(cid:16) (cid:17)
′ ′ ′ ′
=25pp f f +25pq f l +25p qf l +25qq l l (3.2).
n 1 m 1 n 1 m m 1 n n m
− − − −
Using Proposition2.1 (ix, x, xi), Proposition2.2 , Remark 3.1 and the equality
(3.2) we obtain:
5gp,q·5gp′,q′ =5pp′[l +(−1)m·l ]+25pq′[f +(−1)m·f ]+
n m m+n 2 n m m+n 1 n m 1
− − − − −
′ m ′ m
+25p q[f +(−1) ·f ]+25qq [l +(−1) ·l ]=
m+n 1 n m+1 m+n n m
− − −
′ ′ ′ m ′ m
=5 pp l +5p qf +5 5p q(−1) ·f +pp (−1) ·l +
m+n 2 m+n 1 n m+1 n m
− − − −
(cid:16) (cid:17) h i
′ ′ ′ m ′ m
+25 pq f +qq l +25 pq ·(−1) ·f +qq ·(−1) ·l =
m+n 1 m+n n m 1 n m
− − − −
(cid:16) =5gm5p+′qn,pp2′ +5gm5p+′qn(cid:17),01+5hgn5p′mq·(−1)m,pp′·(−1)m +5gn5p′mq·(+−11)m,0+ i
− − − −
+5gm5p+q′n,5qq′ +5gn5pqm′·(−1)m,5qq′·(−1)m.
−
′ ′
Therefore, we obtain that 5Gp,q·5Gp ,q ∈M.
n m
From here, it results that M is an order of the quaternion algebra H (α,β).
Q
1) and 3) are obviously.(cid:3)
Remark 3.3. For α = β = −1, we have that M is included in the set of
Hurwitz quaternions,
1
H={q =a +a i+a j+a k ∈H (−1,−1),a ,a ,a ,a ∈Z or Z+ },
1 2 3 4 Q 1 2 3 4
2
which is a maximal order in H (−1,−1).
Q
From [Lam; 04], [Sa; 14], it is know that if p is an odd prime positive in-
teger, the algebra H (−1,p) is a split algebra if and only if p ≡ 1 (mod 4).
Q
In the following, we will show that there are an infinite numbers of generalized
Fibonacci-Lucas quaternion elements which are invertible in this algebra.
Proposition 3.2. Let n be an arbitrary positive integer. Let (f ) be
n n 0
≥
the Fibonacci sequence and (l ) be the Lucas sequence. Let p be an odd
n n 0
≥
6
prime positive integer, p≡1 (mod 4), q be an arbitrary integer. Let Gp,q be the
n
n-th generalized Fibonacci-Lucas quaternion and H (−1,p) be the quaternion
Q
algebra. The following statements are true:
i) n(Gp,q)6=0, (∀) (n,q)∈N ×N;
n ∗
ii) If p=5, then n G5,q 6=0, (∀) (n,q)∈N ×Z , (n,q)6=(1,−2);
n ∗
iii) if p>5 and n≡0 (mod 3), then n(Gp,q)6=0−, (∀) q∈Z.
(cid:0) (cid:1) n
Proof. We know that an element x 6= 0 from a quaternion algebra is
invertible in this algebra if the norm n(x)6=0.
i) From Proposition 3.1, we know that
n(Gp,q)= p3+p2+8p2q+15pq2+2pq f +
n 2n 1
−
+ −3p3(cid:0)+5q2−22p2q−40pq2+2pq (cid:1)f (3.3).
2n+1
If q ∈N, since p∈N∗(cid:0), it results (cid:1)
p3+p2+8p2q+15pq2+2pq <3p3−5q2+22p2q+40pq2−2pq.
Using the inequality f <f , we obtain that n(Gp,q)<0. So n(Gp,q)6=
2n 1 2n+1 n n
−
0.
In the case when q ∈Z , if there exist generalized Fibonacci-Lucas quaternion
elements with n(Gp,q)−=0, using relation (3.3), we obtain that
n
p p2+p+8pq+15q2+2q f + −3p2−22pq−40q2+2q f =
2n 1 2n+1
−
(cid:2)(cid:0) (cid:1) =−5q2f(cid:0) , (cid:1) (cid:3)(3.4).
2n+1
therefore,p|5q2f .Butweknowfromthehypothesisthatpisaprimepositive
2n+1
integer, p≡1 (mod 4), so we consider two cases:
ii) The case p = 5. Using relation (3.3) and making some calculations we get
that
n G5,q =0⇔ 5q2+10q+30 f − 39q2+108q+75 f =0 (3.5).
n 2n 1 2n+1
−
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
Therefore
0<5q2+10q+30<39q2+108q+75, (∀)q ∈Z,q 6=−2;−1.
Using the fact that f <f , (∀)n∈N , it results
2n 1 2n+1 ∗
−
n G5,q <0 (∀)n∈N , (∀)q ∈Z ,q 6=−2;−1.
n ∗
−
Inthefollowing,w(cid:0)ewil(cid:1)lfindiftheequationsn G5, 1 =0respectivelyn G5, 2 =
n− n−
0 have solutions for n∈N .
∗
(cid:0) (cid:1) (cid:0) (cid:1)
Using the relation (3.5) and the Fibonacci recurrence, we have
n G5, 1 =0⇔25f −6f =0⇔7f +6f =0.
n− 2n 1 2n+1 2n 1 2n 3
− − −
(cid:0) (cid:1)
7
Since f ,f > 0 it results that does not exist n∈N such that 7f +
2n 1 2n 3 ∗ 2n 1
− − −
6f =0.
2n 3
−
Using relation (3.5) and the Fibonacci recurrence, it results
n G5n,−2 =0⇔30f2n 1−15f2n+1 =0⇔f2n 2 =0⇔n=1.
− −
iii) when (cid:0)p > 5,(cid:1)since n ≡ 0 ( mod 3) it results that 2n−1 ≡ 2 ( mod 3) and
2n+1≡1 ( mod 3). Applying Applying Proposition 2.1 (xiii), we obtain that
f and f are odd numbers. Since p is an odd number, we obtain that
2n 1 2n+1
−
the equality (3.4) cannot be held (is easy to prove that the equality (3.4) holds
only when f is an even number or f is an even number).(cid:3)
2n 1 2n+1
−
For a∈H (α,β). The centralizer of the element a is
Q
C(a)={x∈H (α,β) / ax=xa}.
Q
From [Fl; 01], Proposition 2.13, we know that the equation
ax=bx,a,b∈H (α,β), (3.6.)
K
where K is anarbitraryfield of charK 6=0,a,b∈/ K,a6=b, has the solutions of
the form
x=λ (a-t(a)+b-t(b))+λ (n(a-t(a))-(a-t(a))(b-t(b))),λ ,λ ∈K (3.7.)
1 2 1 2
ifH (α,β)is adivisionquaternionalgebraorifH (α,β)is asplitquaternion
K K
algebra and n(a)6=0,n(b)6=0.
Proposition 3.3. Let n be a positive integer,n ≡ 0 (mod 3). Let (f )
n n 0
≥
be the Fibonacci sequence and (l ) be the Lucas sequence. Let p be an odd
n n 0
≥
prime positive integer, p≡1 (mod 4), q be an arbitrary integer. Therefore, the
centralizer of the element Gp,q ∈ H (−1,p) with p > 5, q arbitrary integer is
n Q
the set
C(Gp,q)={Gγ,δ+Λ,Λ∈Q},
n n
w2λheAre,γwi=th2λλ1,pλ,δ =∈Q2λa1nq,dΛA==gn−2p22λ1(p−,−12)λn1+q+2gq222nλ(2−a−14)λn2p+q,22λp2qb−(−2λ12)qn2-.2λ2p52l2n−2-
2 1 2 5
Proof. Since C(Gp,q)={x∈H (−1,p) / Gp,qx=xGp,q}, using relations
n Q n n
(3.6) and (3.7) for a = b, we obtain that the equation Gp,qx = xGp,q has the
n n
solutions of the form
x=2λ (Gp,q-t(Gp,q))+2λ n(Gp,q-t(Gp,q)),λ ,λ ∈Q. (3.8.)
1 n n 2 n n 1 2
For Gp,q =gp,q·1+gp,q ·i+gp,q ·j+gp,q ·k, we have t(Gp,q)=gp,q and
n n n+1 n+2 n+3 n n
n(Gp,q) = ga,b, with a and b as in Proposition 3.1. From here, we have that
n 2n
n(Gp,q-t(Gp,q)) = ga,b−gp,q = ga,b−(pf +ql )2. Using Proposition 2.1,
n n 2n n 2n n 1 n
−
relations v), ix) and xii), it results
n(Gp,q-t(Gp,q))=ga,b-p2(l -(−1)nl )-q2(l +2(−1)n)-2pq(f +(−1)n)=
n n 2n 5 2n−2 0 2n 2n−1
8
=ga,b− p2l − 2p2(−1)n−q2l −2q2(−1)n−2pqf −2pq(−1)n =
===relggga2222aaatnnnn−−,ibo22n+ppqq(,,g3bb52−−−.8n2qq2)p22n,q−−−,w−2peq5p2522ol2−lb2ntn5−ap−5i22n2−l2−xnA−=,225pw22−hλ(−e12r5p1(e2G)2A(nnpn−,−=q1-)g2n2npq5p,2−q2)((+−2−q2112λ))(nn2−+−1g2)22aqnnp−2q−2(p(−q2−2,1pnb1−)q−)nq1(n+2−=−21p)npq52(=l−2n1−)n2−.UAsing=
=Q.G(cid:3)n2λ1p,2λ1q+gn−2λ1p,−2λ1q+g22nλ2a−4λ2pq,2λ2b−2λ2q(cid:16)2−2λ2p52l2n−2−2λ2A,λ1,λ2(cid:17)∈
Conclusions. Inthis paper,startingfromaspecialsetofelements,namely
Fibonacci-Lucas quaternions, we proved that this set is an order –in the sense
of ring theory– of the quaternion algebra H (α,β) and it can have an algebra
Q
structure overQ. It willbe very interestingto find allproperties ofthis algebra
and to find conditions under which it is a division algebra or a split algebra.
Acknowledgments
The authors thank ProfessorRafal Abl amowicz and anonymous referees for
their comments, suggestions and ideas which helped us to improve this paper.
References
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Cristina FLAUT
Faculty of Mathematics and Computer Science, OvidiusUniversity,
Bd. Mamaia 124, 900527, CONSTANTA,ROMANIA
http://cristinaflaut.wikispaces.com/; http://www.univ-ovidius.ro/math/
e-mail: cfl[email protected];cristina fl[email protected]
Diana SAVIN
Faculty of Mathematics and Computer Science,
OvidiusUniversity,
Bd. Mamaia 124, 900527, CONSTANTA,ROMANIA
http://www.univ-ovidius.ro/math/
e-mail: [email protected], [email protected]
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