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116
IN A COURSE ON PROOFS THAT IS PRIMARILY FOR PRE-
service teachers, a number of problems deal with
identities for Fibonacci numbers. We consider some
of these identities in this article, and we discuss
general strategies used to help discover, prove, and
generalize these identities.
We start with a definition of Fibonacci numbers.
They satisfy
(1) Fn+1 = Fn + Fn–1, n = 1, 2, . . . , F0 = 0, F1 = 1.
The next few Fibonacci numbers are F2 = 1, F3 = 2,
F4 = 3, and F5 = 5. We encourage readers to make a
list of Fibonacci numbers up to F10 to use as a refer-
ence while reading this article.
We notice that
F2F0 – F1
2
= –1;
F3F1 – F2
2
= 2 – 1 = 1;
F4F2 – F3
2
= 3 – 4 = –1;
F5F3 – F4
2
= 10 – 9 = 1.
This result suggests that
(2) Fn+2Fn – Fn
2
+1 = (–1)n+1
.
One way to prove result (2) is by mathematical
induction. Since F2F0 – F1
2
= –1, the statement is
true when n = 0. To complete the argument, we
need to show that if statement (2) is true for n, it is
true for n + 1. What do we do with
Fn+3Fn+1 – Fn
2
+2?
When I give it as a problem, most students use
equation (1) in each term, getting
(Fn+2 + Fn+1)(Fn + Fn–1) – (Fn+1 + Fn)2
.
Although carrying through a proof with this
start is possible, doing so takes a great deal of cal-
culation. We also notice that this argument intro-
duces Fn–1, which we do not want. Making fewer
changes is usually better—we should be parsimo-
nious if possible. Fn+3 has to be removed, so we try
using equation (1) just to replace Fn+3:
Fn+3Fn+1 – Fn
2
+2 = (Fn+2 + Fn+1)Fn+1 – Fn
2
+2
= Fn+2Fn+1 + Fn
2
+1 – Fn
2
+2.
The term Fn
2
+1 is what we want. The other two
terms can be combined to get
Fn+2(Fn+1 – Fn+2) + Fn
2
+1 = Fn
2
+1 – Fn+2Fn.
Thus,
Fn+3Fn+1 – Fn
2
+2 = –[Fn+2Fn – Fn
2
+1]
= –[(–1)n+1
]
= (–1)n+2
,
which completes the induction, since we have
shown that an initial result is true for n = 0.
Some interesting curiosities are suggested by
equation (2). For example, n = 5 gives
13 • 5 – 82
= 1.
MATHEMATICS TEACHER
We
should be
parsimonious
if possible
DELVING DEEPER
Fibonacci and
Related Sequences
Richard A. Askey
Edited by Al Cuoco
alcuoco@edc.org
Center for Mathematics Education
Newton, MA 02458
E. Paul Goldenberg
pgoldenberg@edc.org
Education Development Center
Newton, MA 02458
Richard Askey is John Bascom Profes-
sor of Mathematics at the University of
Wisconsin. His book Special Functions,
written with George Andrews and
Ranjan Roy, was published by Cam-
bridge University Press. The courses
that he has taught range from arith-
metic for prospective elementary school
teachers to graduate-level courses.
This department focuses on mathematics content
that appeals to secondary school teachers. It provides
a forum that allows classroom teachers to share their
mathematics from their work with students, class-
room investigations and projects, and other experi-
ences. We encourage submissions that pose and solve
a novel or interesting mathematics problem, expand
on connections among different mathematical topics,
present a general method for describing a mathemat-
ical notion or solving a class of problems, elaborate
on new insights into familiar secondary school math-
ematics, or leave the reader with a mathematical
idea to expand. Please send submissions to “Delving
Deeper,” Mathematics Teacher, 1906 Association
Drive, Reston, VA 20191-1502.
Copyright © 2004 The National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved.
This material may not be copied or distributed electronically or in any other format without written permission from NCTM.
Vol. 97, No. 2 • February 2004 117
This result is the basis for an interesting geomet-
ric puzzle:
On a piece of graph paper, draw the following
figures. First, draw a square with vertices (0, 0),
(8, 0), (0, 8), and (8, 8). Then connect the points
(0, 5) and (8, 5), (3, 0) and (5, 5), and (0, 8) and
(8, 5). Second, draw a rectangle with vertices at
(0, 0), (13, 0), (0, 5), and (13, 5).
Now we rearrange the pieces of the square as
shown in the top part of figure 1. These pieces
seem to exactly cover both the square and the rec-
tangle. But the area of the rectangle is 65 and the
area of the square is 64. The seeming paradox is
resolved when we notice that the four points that
seem to be along the diagonal of the rectangle are
not collinear and that the two interior points are
not actually on the diagonal. Similar figures can be
drawn for each n. The larger n is, the smaller the
discrepancy is relative to the size of the square and
rectangle.
(4) Fn+3Fn – Fn+2Fn+1 = (–1)n+1
.
The following is a proof of statement (4):
Fn +3Fn – Fn+2Fn+1 = (Fn+2 + Fn+1)Fn – Fn+2Fn+1
= Fn+2Fn + Fn+1(Fn – Fn+2)
= Fn+2Fn – Fn
2
+1,
so statement (4) follows from statement (2).
In each of equations (2), (3), and (4), the subscripts
in each product add to the same number: 2n + 2
for (2), 2n + 4 for (3), and 2n + 3 for (4). We can use
that result to suggest other identities. However,
rather than trying to do so now, we start over with
a different way of looking at equation (1), which
leads to equations (2), (4), and more.
Matrices can be used to encode the information
in equation (1). One way to do so is as
1 1Fj+1 Fj+1 + Fj Fj+2
(5)    =   =  .
1 0 Fj   Fj+1  Fj+1
This equation can be iterated to give
1 1 n
Fj+1  Fn+j+1 
(6)     =  .
1 0   Fj   Fn+j 
Next, we use (6) for two columns to get
1 1 n
 Fj+1 F1   Fn+j+1 Fn+1 
(7)     =  .
1 0   +Fj1 F0   Fn+j Fn 
The determinant of the product of square matrices
of the same size is the product of the determinants,
so equation (7) implies that
(8) (–1)n
(0 – Fj) = (–1)n+1
Fj = Fn+j+1Fn – Fn+j Fn+1.
When j = 1, the rightmost equation is (2); and when
j = 2, it is (4).
A further generalization of equation (8) exists. I
was not able to prove it by using matrices, but I was
able to prove it by a completely different method. It is
(9) Fn+j+kFn – Fn+j Fn+k = (–1)n+1
Fj Fk.
I encourage readers to establish this result (or to
give it to students as a problem), perhaps by induc-
tion. A note giving the details would be a nice con-
tribution to “Delving Deeper.”
There are many other identities for Fibonacci
numbers. One that is easy to derive is
(10) Fn
2
+1 – F2
n = (Fn+1 – Fn)(Fn+1 + Fn)
= (Fn–1)(Fn+2).
If we compute the sum of the squares rather
than the difference, a surprise emerges:
n Fn
2
+1 + F2
n
0 11
1 12
2 15
3 13
The
subscripts
in each
product add
to the same
number
Fig. 1
A seeming paradox
A number of results are analogous to equation
(2). Instead of shifting by 1, we can shift by 2. A few
examples suggest that
(3) Fn+4Fn – Fn
2
+2 = (–1)n+1
.
We can use equation (1) on Fn+4 to get
Fn+4Fn – Fn
2
+2 = (Fn+3 + Fn+2)Fn – Fn
2
+2
= Fn+3Fn + Fn+2Fn – Fn
2
+2
= Fn+3Fn + Fn+2(–Fn+1)
= Fn+3Fn – Fn+2Fn+1.
Thus, equation (3) is true if and only if
118 MATHEMATICS TEACHER
The second column contains Fibonacci numbers,
but only half of them. These data suggest that
(11) Fn
2
+1 + F2
n = F2n+1.
This outcome can be proved using the still-
unproved result (9) in two different ways. The easi-
er way is to consider
F2n+1 – Fn
2
+1 = F2n+1F1 – Fn
2
+1.
We next make some substitutions in result (9).
We first replace n by 1 and then replace j and k by n.
The result is
F2n+1F1 – Fn
2
+1 = F2
n.
Hence,
F2n+1 – Fn
2
+1 = F2
n
and
F2n+1 = F2
n + Fn
2
+1.
A more interesting way to prove equation (11)
starts with the left-hand side. We first need to
extend the definition of Fibonacci numbers to nega-
tive values of n.
One way to do so is to force definition (1) to work
for negative values of n. For example, if we require
equation (1), that is,
Fn+1 = Fn + Fn–1, n = 1, 2, . . . , F0 = 0, F1 = 1,
for n = 0, we get F1 = F0 + F–1, or 1 = 0 + F–1, mak-
ing F–1 = 1. Then applying equation (1) to n = –1, we
have F0 = F–1 + F–2, or 0 = 1 + F–2, so that F–2 = –1.
Next, we let n = –2, and we get F–1 = F–2 + F–3, or
1 = –1 + F–3, so that F–3 = 2. If we keep working
down in this way, defining F–n by
(12) F–n = (–1)n+1
Fn, n = 0, 1, . . .
seems to extend equation (1) to all negative inte-
gers. This result is valid, and it can be proved by
induction.
Using (12) as a definition, we have
Fn
2
+1 + F2
n = (–1)n
Fn+1F–n–1 + (–1)n+1
FnF–n
= (–1)n+1
(F–nFn – Fn+1F–n–1).
Then we use result (9) when j = 1 and k = –2n – 1
to get
Fn
2
+1 + F2
n = (–1)n+1
(–1)n+1
F1 F–2n–1
= F1 F2n+1
= F2n+1.
We used result (9) when k < 0. Readers should veri-
fy that this use is valid.
Lucas numbers were mentioned in the title of
this article, but we have not yet introduced them.
Lucas numbers are related to Fibonacci numbers.
They are denoted by Ln, and they satisfy
(13) Ln+1 = Ln + Ln–1, n = 1, 2, . . . ,
L0 = 2,
L1 = 1.
Everything that we have done has analogues for
Lucas numbers, and five results analogous to (9)
involve Lucas numbers and Fibonacci numbers.
They will be discussed in a future article, but
readers might try to figure out what they are and
how to prove them. As a hint, one extension of
expression (2) involves
Ln+2Ln – 5Fn
2
+1.
EDITORS’ NOTE
The whole topic of Fibonacci numbers is full of
interesting facts and identities that have fascinated
people for centuries. Richard Askey has provided
us with a glimpse of some of these beautiful facts
and, more important, with a collection of strategies
for discovering and proving some of them. He
promises further articles on this topic that will
make connections with trigonometry, complex
numbers, and more. And we invite other readers to
write up their own Fibonacci investigations for
publication in “Delving Deeper.” In addition to the
topics suggested in Askey’s article, here are some
others to think about:
• The Fibonacci numbers are integers, so we can
investigate arithmetic properties. For example,
what can we say about the greatest common divi-
sor of two Fibonacci numbers?
• Askey defines the Fibonacci numbers recursively.
Is there a closed form that produces them?
• Do any connections exist between Fibonacci
numbers and other topics that have been dis-
cussed in “Delving Deeper”? Pythagorean
triples? Pascal’s triangle?
Another question is raised by a statement in
Askey’s article: “The determinant of the product of
square matrices of the same size is the product of the
determinants.” An article that delves into this fact
could be interesting.
References about Fibonacci numbers abound.
Some good ones are the following:
Koshy, Thomas. Fibonacci and Lucas Numbers with
Applications. New York: Wiley, 2001.
Vajda, Steven. Fibonacci and Lucas Numbers and the
Golden Section, New York: Wiley, 1989.
Vorobev, N. N. Fibonacci Numbers. New York:
Blaisdell, 1961.
The June 2003 issue of Mathematics Magazine
has three interesting articles about Fibonacci num-
bers, and the book Concrete Mathematics by Donald
Knuth (Addison Wesley, 1989) has an extensive
chapter on Fibonacci numbers. Once again, we
invite readers to send us their favorite Fibonacci
references. The following is an abbreviated bibliog-
raphy of materials that have appeared in the
Mathematics Teacher.
Vol. 97, No. 2 • February 2004 119
FURTHER READINGS
Andreasen, Corey. “Fibonacci and Pascal Together
Again: Pattern Exploration in the Fibonacci
Sequence.” Mathematics Teacher 91 (March 1998):
250–53.
Bradley, Sean. “Generalized Fibonacci Sequences.”
Mathematics Teacher 93 (October 2000): 604–6.
Brown, Stephen I. “From the Golden Ratio and
Fibonacci to Pedagogy and Problem Solving.” Mathe-
matics Teacher 69 (February 1976): 180–88.
Crawford, John, and Calvin Long. “Guessing, Mathe-
matical Induction, and a Remarkable Fibonacci
Result.” Mathematics Teacher 72 (November 1979):
613–19.
Gannon, Gerald E., and Cherlyn Converse. “Extending
a Fibonacci Number Trick.” Mathematics Teacher 80
(December 1987): 744–47.
Greenbury, Garnet. “Reader Reflections: Tribonacci
triangle.” Mathematics Teacher 78 (January 1985):
14, 16.
Kelly, Lucille A. “A Generalization of the Fibonacci
Formulae.” Mathematics Teacher 75 (November
1982): 664–65.
Martin, J. Susan. “Activities: The Fibonacci
Sequence.” Mathematics Teacher 74 (January 1981):
39–42. Correction. Mathematics Teacher 74 (April
1981): 303.
Nygaard, P. H. “Fibonacci-type Sequences.” Mathe-
matics Teacher 63 (December 1970): 671–72.
Patterson, Walter M. “Reader Reflections: The nth
Fibonacci number.” Mathematics Teacher 80
(October 1987): 512.
Pruitt, Kenny. “Reader Reflections: Tribonacci formula.”
Mathematics Teacher 76 (December 1983): 715.
Rulf, Benjamin. “A Geometric Puzzle That Leads to
Fibonacci Sequences.” Mathematics Teacher 91
(January 1998): 21–23.
Schielack, Vincent P. “The Fibonacci Sequence and the
Golden Ratio.” Mathematics Teacher 80 (May 1987):
357–58.
Schwartzman, Steven. “Factoring Polynomials and
Fibonacci.” Mathematics Teacher 79 (January 1986):
54–56, 65.
———. “Reader Reflections: Fibonacci, tribonacci, . . .”
Mathematics Teacher 73 (September 1980): 408, 410.
Shaw, Kenneth L., and Leslie Aspenwall. “The Recur-
ring Fibonacci Sequence: Using a Pose-and-Probe
Rubric.” Mathematics Teacher 92 (March 1999):
192–96.
‰
Share Your Activities
Do you have an idea for a mathematics activity that could interest young mathe-
matics students? Is it within the reach of bright fifth graders? Can it be extended
to challenge sophomores or juniors in high school?
The Editorial Panel of Student Math Notes is seeking manuscripts to include in
future issues. Please share your ideas with the audience of Student Math Notes.
We know that mathematics students respond to these interesting and challenging
activities.
What should you do? Put your ideas on paper, and send them to NCTM, where
they will be directed to the Editorial Panel of Student Math Notes. The panel consid-
ers all manuscripts received. If a manuscript has good potential, the panel puts it
in final form. You will not have to do any rewriting; the panel takes your idea and
develops it into a publishable article, but you get the credit.
Send your ideas or articles to NCTM, Student Math Notes, 1906 Association
Drive, Reston, VA 20191-1502.

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Fibonacci and Related Sequences

  • 1. 116 IN A COURSE ON PROOFS THAT IS PRIMARILY FOR PRE- service teachers, a number of problems deal with identities for Fibonacci numbers. We consider some of these identities in this article, and we discuss general strategies used to help discover, prove, and generalize these identities. We start with a definition of Fibonacci numbers. They satisfy (1) Fn+1 = Fn + Fn–1, n = 1, 2, . . . , F0 = 0, F1 = 1. The next few Fibonacci numbers are F2 = 1, F3 = 2, F4 = 3, and F5 = 5. We encourage readers to make a list of Fibonacci numbers up to F10 to use as a refer- ence while reading this article. We notice that F2F0 – F1 2 = –1; F3F1 – F2 2 = 2 – 1 = 1; F4F2 – F3 2 = 3 – 4 = –1; F5F3 – F4 2 = 10 – 9 = 1. This result suggests that (2) Fn+2Fn – Fn 2 +1 = (–1)n+1 . One way to prove result (2) is by mathematical induction. Since F2F0 – F1 2 = –1, the statement is true when n = 0. To complete the argument, we need to show that if statement (2) is true for n, it is true for n + 1. What do we do with Fn+3Fn+1 – Fn 2 +2? When I give it as a problem, most students use equation (1) in each term, getting (Fn+2 + Fn+1)(Fn + Fn–1) – (Fn+1 + Fn)2 . Although carrying through a proof with this start is possible, doing so takes a great deal of cal- culation. We also notice that this argument intro- duces Fn–1, which we do not want. Making fewer changes is usually better—we should be parsimo- nious if possible. Fn+3 has to be removed, so we try using equation (1) just to replace Fn+3: Fn+3Fn+1 – Fn 2 +2 = (Fn+2 + Fn+1)Fn+1 – Fn 2 +2 = Fn+2Fn+1 + Fn 2 +1 – Fn 2 +2. The term Fn 2 +1 is what we want. The other two terms can be combined to get Fn+2(Fn+1 – Fn+2) + Fn 2 +1 = Fn 2 +1 – Fn+2Fn. Thus, Fn+3Fn+1 – Fn 2 +2 = –[Fn+2Fn – Fn 2 +1] = –[(–1)n+1 ] = (–1)n+2 , which completes the induction, since we have shown that an initial result is true for n = 0. Some interesting curiosities are suggested by equation (2). For example, n = 5 gives 13 • 5 – 82 = 1. MATHEMATICS TEACHER We should be parsimonious if possible DELVING DEEPER Fibonacci and Related Sequences Richard A. Askey Edited by Al Cuoco alcuoco@edc.org Center for Mathematics Education Newton, MA 02458 E. Paul Goldenberg pgoldenberg@edc.org Education Development Center Newton, MA 02458 Richard Askey is John Bascom Profes- sor of Mathematics at the University of Wisconsin. His book Special Functions, written with George Andrews and Ranjan Roy, was published by Cam- bridge University Press. The courses that he has taught range from arith- metic for prospective elementary school teachers to graduate-level courses. This department focuses on mathematics content that appeals to secondary school teachers. It provides a forum that allows classroom teachers to share their mathematics from their work with students, class- room investigations and projects, and other experi- ences. We encourage submissions that pose and solve a novel or interesting mathematics problem, expand on connections among different mathematical topics, present a general method for describing a mathemat- ical notion or solving a class of problems, elaborate on new insights into familiar secondary school math- ematics, or leave the reader with a mathematical idea to expand. Please send submissions to “Delving Deeper,” Mathematics Teacher, 1906 Association Drive, Reston, VA 20191-1502. Copyright © 2004 The National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in any other format without written permission from NCTM.
  • 2. Vol. 97, No. 2 • February 2004 117 This result is the basis for an interesting geomet- ric puzzle: On a piece of graph paper, draw the following figures. First, draw a square with vertices (0, 0), (8, 0), (0, 8), and (8, 8). Then connect the points (0, 5) and (8, 5), (3, 0) and (5, 5), and (0, 8) and (8, 5). Second, draw a rectangle with vertices at (0, 0), (13, 0), (0, 5), and (13, 5). Now we rearrange the pieces of the square as shown in the top part of figure 1. These pieces seem to exactly cover both the square and the rec- tangle. But the area of the rectangle is 65 and the area of the square is 64. The seeming paradox is resolved when we notice that the four points that seem to be along the diagonal of the rectangle are not collinear and that the two interior points are not actually on the diagonal. Similar figures can be drawn for each n. The larger n is, the smaller the discrepancy is relative to the size of the square and rectangle. (4) Fn+3Fn – Fn+2Fn+1 = (–1)n+1 . The following is a proof of statement (4): Fn +3Fn – Fn+2Fn+1 = (Fn+2 + Fn+1)Fn – Fn+2Fn+1 = Fn+2Fn + Fn+1(Fn – Fn+2) = Fn+2Fn – Fn 2 +1, so statement (4) follows from statement (2). In each of equations (2), (3), and (4), the subscripts in each product add to the same number: 2n + 2 for (2), 2n + 4 for (3), and 2n + 3 for (4). We can use that result to suggest other identities. However, rather than trying to do so now, we start over with a different way of looking at equation (1), which leads to equations (2), (4), and more. Matrices can be used to encode the information in equation (1). One way to do so is as 1 1Fj+1 Fj+1 + Fj Fj+2 (5)    =   =  . 1 0 Fj   Fj+1  Fj+1 This equation can be iterated to give 1 1 n Fj+1  Fn+j+1  (6)     =  . 1 0   Fj   Fn+j  Next, we use (6) for two columns to get 1 1 n  Fj+1 F1   Fn+j+1 Fn+1  (7)     =  . 1 0   +Fj1 F0   Fn+j Fn  The determinant of the product of square matrices of the same size is the product of the determinants, so equation (7) implies that (8) (–1)n (0 – Fj) = (–1)n+1 Fj = Fn+j+1Fn – Fn+j Fn+1. When j = 1, the rightmost equation is (2); and when j = 2, it is (4). A further generalization of equation (8) exists. I was not able to prove it by using matrices, but I was able to prove it by a completely different method. It is (9) Fn+j+kFn – Fn+j Fn+k = (–1)n+1 Fj Fk. I encourage readers to establish this result (or to give it to students as a problem), perhaps by induc- tion. A note giving the details would be a nice con- tribution to “Delving Deeper.” There are many other identities for Fibonacci numbers. One that is easy to derive is (10) Fn 2 +1 – F2 n = (Fn+1 – Fn)(Fn+1 + Fn) = (Fn–1)(Fn+2). If we compute the sum of the squares rather than the difference, a surprise emerges: n Fn 2 +1 + F2 n 0 11 1 12 2 15 3 13 The subscripts in each product add to the same number Fig. 1 A seeming paradox A number of results are analogous to equation (2). Instead of shifting by 1, we can shift by 2. A few examples suggest that (3) Fn+4Fn – Fn 2 +2 = (–1)n+1 . We can use equation (1) on Fn+4 to get Fn+4Fn – Fn 2 +2 = (Fn+3 + Fn+2)Fn – Fn 2 +2 = Fn+3Fn + Fn+2Fn – Fn 2 +2 = Fn+3Fn + Fn+2(–Fn+1) = Fn+3Fn – Fn+2Fn+1. Thus, equation (3) is true if and only if
  • 3. 118 MATHEMATICS TEACHER The second column contains Fibonacci numbers, but only half of them. These data suggest that (11) Fn 2 +1 + F2 n = F2n+1. This outcome can be proved using the still- unproved result (9) in two different ways. The easi- er way is to consider F2n+1 – Fn 2 +1 = F2n+1F1 – Fn 2 +1. We next make some substitutions in result (9). We first replace n by 1 and then replace j and k by n. The result is F2n+1F1 – Fn 2 +1 = F2 n. Hence, F2n+1 – Fn 2 +1 = F2 n and F2n+1 = F2 n + Fn 2 +1. A more interesting way to prove equation (11) starts with the left-hand side. We first need to extend the definition of Fibonacci numbers to nega- tive values of n. One way to do so is to force definition (1) to work for negative values of n. For example, if we require equation (1), that is, Fn+1 = Fn + Fn–1, n = 1, 2, . . . , F0 = 0, F1 = 1, for n = 0, we get F1 = F0 + F–1, or 1 = 0 + F–1, mak- ing F–1 = 1. Then applying equation (1) to n = –1, we have F0 = F–1 + F–2, or 0 = 1 + F–2, so that F–2 = –1. Next, we let n = –2, and we get F–1 = F–2 + F–3, or 1 = –1 + F–3, so that F–3 = 2. If we keep working down in this way, defining F–n by (12) F–n = (–1)n+1 Fn, n = 0, 1, . . . seems to extend equation (1) to all negative inte- gers. This result is valid, and it can be proved by induction. Using (12) as a definition, we have Fn 2 +1 + F2 n = (–1)n Fn+1F–n–1 + (–1)n+1 FnF–n = (–1)n+1 (F–nFn – Fn+1F–n–1). Then we use result (9) when j = 1 and k = –2n – 1 to get Fn 2 +1 + F2 n = (–1)n+1 (–1)n+1 F1 F–2n–1 = F1 F2n+1 = F2n+1. We used result (9) when k < 0. Readers should veri- fy that this use is valid. Lucas numbers were mentioned in the title of this article, but we have not yet introduced them. Lucas numbers are related to Fibonacci numbers. They are denoted by Ln, and they satisfy (13) Ln+1 = Ln + Ln–1, n = 1, 2, . . . , L0 = 2, L1 = 1. Everything that we have done has analogues for Lucas numbers, and five results analogous to (9) involve Lucas numbers and Fibonacci numbers. They will be discussed in a future article, but readers might try to figure out what they are and how to prove them. As a hint, one extension of expression (2) involves Ln+2Ln – 5Fn 2 +1. EDITORS’ NOTE The whole topic of Fibonacci numbers is full of interesting facts and identities that have fascinated people for centuries. Richard Askey has provided us with a glimpse of some of these beautiful facts and, more important, with a collection of strategies for discovering and proving some of them. He promises further articles on this topic that will make connections with trigonometry, complex numbers, and more. And we invite other readers to write up their own Fibonacci investigations for publication in “Delving Deeper.” In addition to the topics suggested in Askey’s article, here are some others to think about: • The Fibonacci numbers are integers, so we can investigate arithmetic properties. For example, what can we say about the greatest common divi- sor of two Fibonacci numbers? • Askey defines the Fibonacci numbers recursively. Is there a closed form that produces them? • Do any connections exist between Fibonacci numbers and other topics that have been dis- cussed in “Delving Deeper”? Pythagorean triples? Pascal’s triangle? Another question is raised by a statement in Askey’s article: “The determinant of the product of square matrices of the same size is the product of the determinants.” An article that delves into this fact could be interesting. References about Fibonacci numbers abound. Some good ones are the following: Koshy, Thomas. Fibonacci and Lucas Numbers with Applications. New York: Wiley, 2001. Vajda, Steven. Fibonacci and Lucas Numbers and the Golden Section, New York: Wiley, 1989. Vorobev, N. N. Fibonacci Numbers. New York: Blaisdell, 1961. The June 2003 issue of Mathematics Magazine has three interesting articles about Fibonacci num- bers, and the book Concrete Mathematics by Donald Knuth (Addison Wesley, 1989) has an extensive chapter on Fibonacci numbers. Once again, we invite readers to send us their favorite Fibonacci references. The following is an abbreviated bibliog- raphy of materials that have appeared in the Mathematics Teacher.
  • 4. Vol. 97, No. 2 • February 2004 119 FURTHER READINGS Andreasen, Corey. “Fibonacci and Pascal Together Again: Pattern Exploration in the Fibonacci Sequence.” Mathematics Teacher 91 (March 1998): 250–53. Bradley, Sean. “Generalized Fibonacci Sequences.” Mathematics Teacher 93 (October 2000): 604–6. Brown, Stephen I. “From the Golden Ratio and Fibonacci to Pedagogy and Problem Solving.” Mathe- matics Teacher 69 (February 1976): 180–88. Crawford, John, and Calvin Long. “Guessing, Mathe- matical Induction, and a Remarkable Fibonacci Result.” Mathematics Teacher 72 (November 1979): 613–19. Gannon, Gerald E., and Cherlyn Converse. “Extending a Fibonacci Number Trick.” Mathematics Teacher 80 (December 1987): 744–47. Greenbury, Garnet. “Reader Reflections: Tribonacci triangle.” Mathematics Teacher 78 (January 1985): 14, 16. Kelly, Lucille A. “A Generalization of the Fibonacci Formulae.” Mathematics Teacher 75 (November 1982): 664–65. Martin, J. Susan. “Activities: The Fibonacci Sequence.” Mathematics Teacher 74 (January 1981): 39–42. Correction. Mathematics Teacher 74 (April 1981): 303. Nygaard, P. H. “Fibonacci-type Sequences.” Mathe- matics Teacher 63 (December 1970): 671–72. Patterson, Walter M. “Reader Reflections: The nth Fibonacci number.” Mathematics Teacher 80 (October 1987): 512. Pruitt, Kenny. “Reader Reflections: Tribonacci formula.” Mathematics Teacher 76 (December 1983): 715. Rulf, Benjamin. “A Geometric Puzzle That Leads to Fibonacci Sequences.” Mathematics Teacher 91 (January 1998): 21–23. Schielack, Vincent P. “The Fibonacci Sequence and the Golden Ratio.” Mathematics Teacher 80 (May 1987): 357–58. Schwartzman, Steven. “Factoring Polynomials and Fibonacci.” Mathematics Teacher 79 (January 1986): 54–56, 65. ———. “Reader Reflections: Fibonacci, tribonacci, . . .” Mathematics Teacher 73 (September 1980): 408, 410. Shaw, Kenneth L., and Leslie Aspenwall. “The Recur- ring Fibonacci Sequence: Using a Pose-and-Probe Rubric.” Mathematics Teacher 92 (March 1999): 192–96. ‰ Share Your Activities Do you have an idea for a mathematics activity that could interest young mathe- matics students? Is it within the reach of bright fifth graders? Can it be extended to challenge sophomores or juniors in high school? The Editorial Panel of Student Math Notes is seeking manuscripts to include in future issues. Please share your ideas with the audience of Student Math Notes. We know that mathematics students respond to these interesting and challenging activities. What should you do? Put your ideas on paper, and send them to NCTM, where they will be directed to the Editorial Panel of Student Math Notes. The panel consid- ers all manuscripts received. If a manuscript has good potential, the panel puts it in final form. You will not have to do any rewriting; the panel takes your idea and develops it into a publishable article, but you get the credit. Send your ideas or articles to NCTM, Student Math Notes, 1906 Association Drive, Reston, VA 20191-1502.