Some new Fibonacci and Lucas identities by matrixmethods

Some new Fibonacci and Lucas identities by matrix methods
Refik Keskin and Bahar Demirtu rk *
Faculty of Science and Arts,Department of Mathematics,Sakarya University,54187
Sakarya,Turkey
(Received 23February 2009)
The aim of this article is to characterize the 2Â2matrices X satisfying X 2¼X þI and obtain some new identities concerning with Fibonacci and Lucas numbers.
Keywords:Fibonacci numbers;Lucas numbers;Fibonacci matrix AMS Subject Classifications:11B37;11B39;40C05
1.Introduction
The material of this article,Fibonacci and Lucas numbers,is a topic of some selective courses for undergraduate and graduate students.The contributions of these courses to all kinds of population of students can be evaluated separately.In undergraduate education,the students who want to specialize at number theory and algebra area prefer to select these courses.In this way,they achieve the ability to use the knowledge given in the linear algebra and number theory courses.Furthermore,the knowledge acquired in these courses is used in different disciplines such as computer sciences,economy,etc.By the abilities achieved from these courses,students can find opportunity to work in areas different from mathematics.In graduate education,in addition to the fundamental knowledge achieved in the undergraduate education,the students start to specialize in many areas of mathematics that use Fibonacci and Lucas numbers.
The Fibonacci sequence {F n }is defined by F 0¼0,F 1¼1and F n ¼F n À1þF n À2for n !2.F n is called the n th Fibonacci number.Fibonacci numbers for negative subscripts are defined as F Àn ¼(À1)n þ1F n for n !1and it is known that F n þ1¼F n þF n À1for every n 2Z .The Lucas sequence {L n }is defined as
L n ¼F n À1þF n þ1for every n 2Z .L n is called the n th Lucas number.It can be seen that L n ¼L n À1þL n À2for every n 2Z .It is well known that F n ¼ n À n
ðÞ=ffiffiffi5p and L n ¼ n þ n
for every n 2Z ,where ¼1þffiffiffi5p ÀÁ=2and ¼1Àffiffiffi5p ÀÁ=2:These are known as Binet’s formula.Moreover,it is well known that [1,2]for the matrices Q ¼Â1110Ãand R ¼Â0111Ã,we have Q n ¼ÂF n þ1F n F n F n À1Ãand R n
¼ÂF n À1F n F n F n þ1Ã:The matrices Q and R are called Fibonacci matrices.It is easily seen that Q 2¼Q þI and R 2¼R þI .Moreover,Q and R are similar matrices since
0110
"#1110
"#0110
"#À1
¼011
1"#:
*Corresponding author.Email:demirturk@.tr
International Journal of Mathematical Education in Science and Technology 379
DOI:10.1080/00207390903236426
Many identities for Fibonacci and Lucas numbers are proved by using Binet’s formula,induction or Fibonacci matrices[2–4].Since Q and R are similar matrices, they give the same identities.Some of the most known identities are as follows:
F nþk F nÀkÀF2
n ¼À1
ðÞnþkþ1F2
k
(Catalan’s identity),
F kþ1F nÀF k F nþ1¼À1
ðÞn F nÀk(d’Ocagne’s identity),
F kÀ1F nþF k F nþ1¼F nþk(Honsberger’s identity),
L2nþ2ðÀ1Þn¼L2
n
,
L2 n À5F2
n
¼4ðÀ1Þn:
In this study,we will characterize all the matrices X satisfying the relation X2¼XþI. Then we will obtain different identities by using this property.In fact,the matrices Q and R are special cases of the2Â2matrices X satisfying X2¼XþI.
We know that Binet’s formulas are obtained by using matrix methods such as diagonalization([5,pp.165–166]and[6,pp.252–254]).Therefore linear algebra course may include these materials.Moreover,any number theory course may include these materials[7,pp.259–274].
The recommendations regarding the teaching of the identities given in this article can be presented in two cases.The first is related to the pedagogical aspect. The identities require the use of mathematical knowledge learned in different courses.By this way we consider to affect students’motivation in a positive direction.The second recommendation is related to the qualification of these identities.Our identities are different from other known identities and our method exploits the properties of matrix algebra to find more general identities.
2.Main theorems
Lemma1:If X is a square matrix with X2¼XþI,then X n¼F n XþF nÀ1I for all n2Z.
Proof:If n¼0,then the proof is obvious.It can be shown by induction that X n¼F n XþF nÀ1I for every n2N.We now show that XÀn¼FÀn XþFÀnÀ1I for every n2N.Let Y¼IÀX¼ÀXÀ1.Then
Y2¼IÀX
ðÞ2¼IÀ2XþX2¼IÀ2XþXþI¼IþIÀX¼IþY:
This shows that Y n¼F n YþF nÀ1I.That is,(ÀXÀ1)Àn¼F n(IÀX)þF nÀ1I.Therefore (À1)n XÀn¼ÀF n Xþ(F nþF nÀ1)I¼ÀF n XþF nþ1I.Thus
XÀn¼À1
ðÞnþ1F n XþÀ1
ðÞn F nþ1I¼FÀn XþFÀnÀ1I:h 380Classroom Notes
Theorem 1:Let X be an arbitrary 2Â2matrix .Then X 2¼X þI if and only if X is of the form
X ¼
a b c
1Àa
!
with det X ¼À1
or
X ¼ I ,where 2 , ÈÉ
:
Proof:Assume that X 2¼X þI .Then the minimum polynomial of X must be x À or x À or x 2Àx À1.In the first case X ¼ I ,in the second case X ¼ I and in the third,since X is a 2Â2matrix,its characteristic polynomial must also be x 2Àx À1.So its trace is 1and its determinant is À1.The argument reverses.œ
Corollary 1:If X ¼Âa b
c 1Àa is a matrix with det X ¼À1,then
X n
¼
aF n þF n À1bF n cF n F n þ1ÀaF n !for all n 2Z .
Proof:Since X 2¼X þI ,the result follows from Lemma 1.œ
Corollary 2: n ¼ F n þF n À1and n ¼ F n þF n À1for every n 2Z .
Corollary 3:Let Q ¼Â1110Ã:Then Q n
¼ÂF n þ1F n F n F n À1Ãfor every n 2Z .Corollary 4:Let S ¼1252
12
12
!:Then S n ¼L n 25F n
2F n 2
L n
2
"#for every n 2Z .Lemma 2:Assume that a ,b and a þb are nonzero integers .Then
X n j ¼0n j a j b n Àj F j þk ¼ðÀ1Þk þ1X n j ¼0n
j Àa ðÞj a þb ðÞn Àj F j Àk !,and
X n j ¼0n j a j b n Àj L j þk ¼ðÀ1Þk X n j ¼0
n
j Àa ðÞj a þb ðÞn Àj L j Àk !for all n 2N and k 2Z .
Proof:Let Z [ ]¼{a þb :a ,b 2Z }.It is well known from [8,9]that Z [ ]is the
algebraic integer ring of the real quadratic field Q ðffiffiffi5p Þ:Also,it is easy to show that :Z [ ]!Z [ ],given by (a þb )¼a þb ¼Àa þa þb ,is a ring isomorphism.On the other hand,we have
À F m þF m þ1¼ ð F m þF m À1Þ¼ ð m Þ¼ m ¼ðÀ1Þm Àm
for all m 2Z .Since
ðða þb Þn k Þ¼ ðða þb Þn Þ ð k Þ¼Àa þða þb ÞðÞn ðÀ1Þk Àk
International Journal of Mathematical Education in Science and Technology 381
¼ðÀ1Þk
X
n j ¼0
n j
!ðÀa Þj ða þb Þn Àj j Àk ¼ðÀ1Þk
X
n j ¼0
n j
!
ðÀa Þj ða þb Þn Àj F j Àk þF j Àk À1
ÀÁ¼ ðÀ1Þk X
n j ¼0n j !
ðÀa Þj ða þb Þn Àj F j Àk
!þðÀ1Þk
X
n j ¼0
n j
!ðÀa Þj ða þb Þn Àj F j Àk À1
!
and
ðða þb Þn k Þ¼ X n j ¼0n j a j b n Àj j ! k !¼ X n j ¼0
n j a j b n Àj
j þk
!
¼ X n j ¼0
n j a j b n Àj
ð F j þk þF j þk À1Þ
!¼ ÀX n j ¼0n j a j b n Àj F j þk !
þX n j ¼0
n j a j b n Àj
F j þk þ1,
we get the conclusions.
œ
Lemma 3:Let m ,k 2Z with m ¼1and m ¼0.Then F mn þk ¼X n j ¼0n j F j m F n Àj m À1F j þk ,
and
L mn þk
¼X n j ¼0
n j F j m F n Àj m À1L j þk
for all n 2N .Proof:Since
S mn þk
¼S m ðÞn S k ¼F m S þF m À1I ðÞn S k ¼X n j ¼0
n j F j m
F n Àj m À1S j
!
S k ¼X n j ¼0
n j F j m
F n Àj m À1S j þk ,the results follow from Corollary 4.
œ
In view of Lemma 2and Corollary 4,the following corollary can be given.Corollary 5:Let m ,k 2Z with m ¼1and m ¼0.Then
F mn þk ¼ðÀ1Þk þ1
X
n j ¼0
n j
!ðÀF m Þj F n Àj
m þ1F j Àk
!
,
L mn þk ¼ðÀ1Þk
X
n j ¼0
n j
!ðÀF m Þj F n Àj
m þ1L j Àk
!
for all n 2N .
382Classroom Notes
Theorem2:For all m,k2Z and n2N,we have
5n F2nmþk¼
X2n
j¼02n
j
!
L j
m
L2nÀj
mÀ1
F jþk,
5n L2nmþk¼
X2n
j¼02n
j
!
L j
m
L2nÀj
mÀ1
L jþk,
5nþ1Fð2nþ1Þmþk¼X
2nþ1
j¼0
2nþ1
j
!
L j
m
L2nþ1Àj
mÀ1
L jþk,
5n Lð2nþ1Þmþk¼X
2nþ1
j¼0
2nþ1
j
!
L j
m
L2nþ1Àj
mÀ1
F jþk:
Proof:It is clear that
12 2À1!F
mþ1
F m
F m F mÀ1
!
¼
F mþ1F m
F m F mÀ1
!12
2À1
!
¼
L mþ1L m
L m L mÀ1
!
and
12 2À1!2
¼5I,
we get
5n Q2nm¼5n F2nmþ1F2nm
F2nm F2nmÀ1
!
¼
L mþ1L m
L m L mÀ1
!2n
:
Hence it follows that
5n Q2nmþk¼
L mþ1L m
L m L mÀ1
"#2n
Q k¼ðL m QþL mÀ1IÞ2n
Q k¼
X2n
j¼02n
j
!
L j
m
L2nÀj
mÀ1
Q jþk:
Then the first two identities follow and the others are proved in a similar way.œTheorem3:Let n be a positive integer and m be a nonzero integer.Then
2n L mnþk¼X n=2
b c
j¼0
n
2j
F2j
m
L nÀ2j
m
5j L kþ5
X
nÀ1
ðÞ=2
b c
j¼0
n
2jþ1
F2jþ1
m
L nÀ2jÀ1
m
5j F k
and
2n F mnþk¼X n=2
b c
j¼0
n
2j
F2j
m
L nÀ2j
m
5j F kþ
X
nÀ1
ðÞ=2
b c
j¼0
n
2jþ1
F2jþ1
m
L nÀ2jÀ1
m
5j L k:
International Journal of Mathematical Education in Science and Technology383
Proof:Let K¼Â05
10
Ã
:Then K2¼5I and it follows that K2j¼5j I and K2jþ1¼5j K
for all j2N.Thus
S m¼L m=25F m=2
F m=2L m=2
!
¼
1
2
ðL m IþF m KÞ
and therefore
S mn¼1
2n
ðL m IþF m KÞn¼
1
2n
X n
j¼0
n
j
F j
m
L nÀj
m
K j
¼1
2
X n=2
b c
j¼0
n
2j
!
F2j
m
L nÀ2j
m
K2jþ
X
nÀ1
ðÞ=2
b c
j¼0
n
2jþ1
!
F2jþ1
m
L nÀ2jÀ1
m
K2jþ1 0
@
1
A
¼1
2n
X n=2
b c
j¼0
n
2j
!
F2j
m
L nÀ2j
m
5j Iþ
X
nÀ1
ðÞ=2
b c
j¼0
n
2jþ1
!
F2jþ1
m
L nÀ2jÀ1
m
5j K
@
1
A:ð2:1Þ
Multiplying each side of(2.1)by S k,we get
S mnþk¼1X n=2
b c
j¼0
n
2j
F2j
m
L nÀ2j
m
5j S kþ
X
nÀ1
ðÞ=2
b c
j¼0
n
2jþ1
F2jþ1
m
L nÀ2jÀ1
m
5j KS k
@
1
A:
By using the identities
KS k¼5F k=25L k=2 L k=25F k=2
!
and
S mnþk¼L mnþk=25F mnþk=2 F mnþk=2L mnþk=2
!
,
we obtain the results.œWe now give some identities from[2,3].
L2 n À5F2
n
¼4ðÀ1Þnð2:2Þ
L n L mÀ5F n F m¼2ðÀ1Þn L nÀmð2:3ÞL m F nþL n F m¼2F mþnð2:4ÞF n L mÀL n F m¼2À1
ðÞm F nÀmð2:5Þ
L2 nþ1ÀL nþ1L nÀL2
n
¼5À1
ðÞnþ1ð2:6Þ
F mÀ1F nþF m F nþ1¼F nþmð2:7Þ384Classroom Notes
F m þ1F n ÀF m F n þ1¼À1ðÞm F n Àm
ð2:8ÞF 2n þ1ÀF n þ1F n ÀF 2
n ¼À1ðÞn :
ð2:9Þ
All the above identities can be shown by using the previously obtained formulae for
Q n and S n .Using these identities,we can give the following theorems and corollaries.Theorem 4:For all n 2N and r ,k 2Z with r ¼k and k ¼0,we get
X n j ¼0L rn þj ðk Àr Þ¼
L rn ÀL k ðn þ1ÞÀr ÀÀ1ðÞk þr L r ðn þ1ÞÀk ÀL kn
ÀÁ1þÀ1ðÞÀL k Àr and
X n j ¼0
F rn þj ðk Àr Þ¼
F rn ÀF k ðn þ1ÞÀr ÀÀ1ðÞk þr F r ðn þ1ÞÀk ÀF kn
ÀÁ
1þÀ1ðÞÀL k Àr
:
Proof:We see that det(S r ÀS k )¼(À1)r À(À1)k L r Àk þ(À1)k ,by using the identities (2.2)and (2.3).Then from the hypothesis it follows that det (S r ÀS k )¼0.Moreover,
S r ðÞ
n þ1
ÀS k ÀÁn þ1¼S r ÀS k ÀÁX n j ¼0
S r ðn Àj ÞS kj ¼S r ÀS k ÀÁX n j ¼0
S rn þj ðk Àr Þ
and therefore
S r
ÀS
k ÀÁÀ1S
r ðn þ1Þ
ÀS
k ðn þ1Þ
ÀÁ
¼
X n j ¼0
S rn þj ðk Àr Þ
¼12X n
j ¼0L rn þj ðk Àr Þ
5
X
n j ¼0
F rn þj ðk Àr ÞX
n j ¼0
F rn þj ðk Àr Þ
X n j ¼0L rn þj ðk Àr Þ
2666643
7
7775:
Using the identity (2.5)and Corollary 4for the rest of the proof,we get the
results.œTheorem 5:Let n ,m ,k 2Z .Then
L k L n þm þk ¼L n þk L m þk þ5À1ðÞk F n F m ,L k F n þm ¼L n þk F m þÀ1ðÞk L m Àk F n ,
and
L k L n þm Àk ¼5F n F m þÀ1ðÞk L n Àk L m Àk :
Proof:Let a ¼L k þ1
k
and consider A ¼Âa b c 1Àa Ãwith det A ¼À1.Then by Corollary 1,we get
A n ¼
aF n þF n À1
bF n cF n
F n þ1ÀaF n
!
¼
L k þ1L k
F n þF n À1bF n
cF n
F n þ1ÀL k þ1
L k
F n
"#:
International Journal of Mathematical Education in Science and Technology 385
By using (2.4)and (2.5)we see that
A n
¼L n þk L k
bF n
cF n
À1ðÞk L n Àk
k
24
35:
Since det A ¼À1and a ¼L k þ1
k
,it follows that bc ¼ÀL 2k þ1ÀL k þ1L k ÀL 2k
ÀÁL k
¼5À1ðÞk L k by (2.6).If we consider the matrix multiplication A n A m ¼A n þm ,we get the
results.œWe now give the following identity similar to Catalan’s identity.
Corollary 6:L n þk L n Àk À5F 2n
¼À1ðÞn þk L 2
k for all k ,n 2Z .Proof:Since det A ¼À1,det A n ¼(det A )n ¼(À1)n .Moreover,since
det A n
¼L n þk L k À1ðÞk L n Àk L k ÀbcF 2
n ¼À1ðÞk L n þk L n Àk L 2k À5À1ðÞk F 2
n L 2k
,it follows that
À1ðÞk L n þk L n Àk À5À1ðÞk F 2n
L k
¼À1ðÞn :This shows that L n þk L n Àk À5F 2n
¼À1ðÞn þk L 2
k :œ
Theorem 6:Let m ,n ,k 2Z with k ¼0.Then
F k F n þm þk ¼F n þk F m þk þÀ1ðÞk À1F n F m ,F k F n þm ¼F m F n þk þÀ1ðÞk À1F m Àk F n
and
F k F n þm Àk ¼F n F m þÀ1ðÞk À1F n Àk F m Àk :
Proof:Take a ¼F k þ1
F k
and consider B ¼Âa b c 1Àa Ãwith det B ¼À1.Then by using (2.7)and (2.8)we see that
B n
¼F n þk F k
bF n
cF n
À1ðÞk À1F n Àk
F k
24
35:
Since det B ¼À1,it follows that bc ¼ÀF 2k þ1
ÀF k þ1F k ÀF 2k ðÞF 2k
¼À1
ðÞ
k À1
F 2
k by (2.9).Since
B n B m ¼B n þm ,the result follows.
œ
We now give the Catalan’s identity easily in the following corollary.
Corollary 7:F n þk F n Àk ÀF 2n ¼À1ðÞn þk þ1F 2
k for all k ,n 2Z .
Proof:If k ¼0,then the result is obvious and in other case it is immediate from the
above formulae for B n together with the fact that det B n ¼(À1)n .
œ386Classroom Notes
International Journal of Mathematical Education in Science and Technology387
3.Concluding remark
We think that finding new Fibonacci and Lucas identities is a sphere of interest which keeps up to date.Also,a book that proves an enormous amount of Fibonacci and Lucas identities by dominos or,in other words,phase tilling has been written [10].
In this note,it has been shown that all X matrices satisfying X2¼XþI are useful to obtain many identities related to Fibonacci and Lucas numbers.Fibonacci and Lucas identities can be used in different areas concerning Fibonacci and Lucas numbers,such as divisibility properties of these numbers(e.g.if m|n,then F m|F n and L m|L m(2nþ1)for m!1).
References
[1]J.R.Silvester,Fibonacci properties by matrix methods,Math.Gazette63(1979),
pp.188–191.
[2]T.Koshy,Fibonacci and Lucas Numbers with Applications,John Wiley and Sons,
New York,Toronto,2001.
[3]S.Vajda,Fibonacci and Lucas Numbers and the Golden Section,Ellis Horwood Limited
Publ.,England,1989.
[4]R.C.Johnson,Matrix methods for Fibonacci and related sequences.Available at
/bob.johnson/fibonacci/.
[5]T.S.Blyth and E.F.Robertson,Basic Linear Algebra,2nd ed.,Springer,London,2000.
[6]G.Strang,Introduction to Linear Algebra,MIT,Wellesley,MA,1993.
[7]D.M.Burton,Elementary Number Theory,McGraw-Hill Comp.Inc.,New York,1998.
[8]G.H.Hardy and E.M.Wright,An Introduction to the Theory of Numbers,Oxford
University Press,USA,1980.
[9]K.W.Yang,Fibonacci with golden ring,Math.Mag.70(1997),pp.131–135.
[10]A.T.Benjamin and J.J.Quinn,Proofs That Really Count:The Art of Combinatorial Proof,
The Mathematical Association of America,Washington,DC,2003.
Promoting number theory in high schools or birthday
problem and number theory
V.K.Srinivasan*
Department of Mathematical Sciences,University of Texas at El Paso,
1830Rue de Montreal,Tucker,GA30084,USA
(Received14April2009)
The author introduces the birthday problem in this article.This can amuse
willing members of any birthday party.This problem can also be used as
*Email:vilappakkam@
DOI:10.1080/00207390903236426
Copyright of International Journal of Mathematical Education in Science & Technology is the property of Taylor & Francis Ltd and its content may not be copied or emailed to multiple sites or posted to a listserv without the copyright holder's express written permission. However, users may print, download, or email articles for individual use.。