7/26/2019 elementary25-4

1/6

ELEMENTARY PROBLEMSANDSOLUTIONS

Edited by

A.

P.

HILLMAN

Please send

all

communications regarding

ELEMENTARY PROBLEMS

AN D

SOLUTIONS

to Dr. A. P. HILLMAN;709 SOLANO DR., S.E.;

ALBUQUERQUE,

NM 87108. Each solu-

tion or problem should be on a separate sheet (or sheets)

.

Preference will be

given to those typed with double spacing in the format used below. Solutions

should be received within four months of the publication date.

DEFINITIONS

The Fibonacci numbers F

n

and the

Lucas numbers L

n

satisfy

F

n +

2=

F

n

+

1 +

F

n>

F

o = >

F

l =

1

and

L

n

+ 2

=

L

n

+ 1

+

L

n>

L

Q

= 2

'

L

l

=

l

PROBLEMS PROPOSED

SN

THIS ISSUE

B-604

Proposed by Heinz-Jurgen Seiffert, Berlin, Germany

Let c

be a

fixed number

and

u

n

+ 2

=

cu

n+ 1

+

u

n

for

n

in

N ={0,

1

9

2,

...}.

Show that there exists

a

number hsuch that

U

n+ k

=

hU

n+ 3

hu

n+ l

+

U

n

f o r

n

i n

N

B-605

Proposed

by

Herta

T.

Freitag, Roanoke,

VA

Let

n

S(n)

=

L

2n

2i

_

1

.

i = l

Determine

the

positive integers

n, if

any,

for

which S(n)

is

prime.

B~606

Proposed by L. Kuipers, Sierre, Switzerland

Simplify

the

expression

2

+i

+ 2

,L

M1

-

25F

2

+

L

2

1

.

n + l n -1

n

+ 1 n n- 1

B-607

Proposed

by

Charles

R.

Wall, Trident Technical College, Charleston,

SC

Let

=

0

\k)

F

k

L

n-k-

Show that C

n

/2

n

is anintegerfornin{0,1, 2,...} .

370

[Nov.

7/26/2019 elementary25-4

2/6

ELEMENTARY PROBLEMS AND SOLUTIONS

B-608 Proposed by Plero Filipponi,

Fond.

U. Bordoni, Rome,, Italy

For k = {2, 3

9

...} and n in N ={0

9

1, 2

9

. . . } ,

let

,c+ c -1

denote the quadratic mean taken over kconsecutive Fibonacci numbers of which

the first is F

n

. Find the smallest such k > 2for which

S

nak

is an integer for

all n in 71/.

B-609 Proposed by Adina DiPorto & Piero Filipponi,

Fond, U. Bordoni, Rome, Italy

Find a closed form expression for

k = i

and show that S

n

En(-l)

n

(mod F

n

).

SOLUTIONS

Nondivisors of the Ly.

B 580

Proposed by Valentina Bakinova, Rondou t Valley, NY

What are the three smallest positive integers d such that no Lucas number

L

n

is an integral multiple of dP.

Solution by J. Suck, Essen, German y

They are 5

9

8

9

10. Since

l\L

ns

2\L

QS

3\L

2S

4| L

3

,

6|L

6S

l\L

h

,

9|L

69

it re-

mains to show that 5\L

n

and 8J(L

n

for all n - 0

9

1, 2, ... . This follows from

the fact that the Lucas sequence modulo 5 or 8 is periodic with period 2 , 1

9

3,

4 or 2, 1, 3

9

4

9

7

9

3

9

2

9

5

9

7

9

4

s

3

9

7

9

respectively.

Also solved by Paul S. Bruckman , L. A. G. Dresel, Piero Filipponi, Herta T.

Freitag, L. Kuipers, Bob Prielipp, H.-J. Seiffert, Lawrence Somer, and the

proposer.

Third Degree Representations for F

B-581 Proposed by Antal Bege, University of Cluj, Roman ia

Prove that, for every positive integer n

9

there are at least

[n/2]

ordered

6-tuples

(a, b

5

o

5

x

9

y

3

z) such that

F

n

= ax

2

-\-by

2

- cz

2

and each of a, b

9

c

3

x

s

y

3

z is a Fibonacci number. Here [t] is the greatest

integer in t .

Solution by Paul S. Bruckman , Fair Oaks, CA

1387]

371

7/26/2019 elementary25-4

3/6

ELEMENTARY PROBLEMSANDSOLUTIONS

We first prove

the

following relations:

TP

=

TP TP

2

. L . TP TP ^ TP jp 2

. / 1\

2 n

r

2s+l

r

n-s+l

r

2s n-s

r

2s -

1

n

-

s -

1 '

v J

F = F F

2

+ F F

2

- F F

2

.

(2)

L

2n+1

L

2s + 2

n~s+l

L

2s + l

L

n-s

2s

L

n-s-l

9

v^ '

v a l i d f o r a l l i n t e g e r s

s

a n d

n.

P r o o f o f (1 ) a nd ( 2 ) : We u s e t h e f o l l o w i n g r e l a t i o n s r e p e a t e d l y :

F

u

F

v = ^

F

2 v

+

u -

(-V

F

2v-u

-

H-lfF

u

), (3)

w h i c h i s r e a d i l y p r o v e n f r om t h e B i n e t f o r m u l a s a n d i s g i v e n w i t h o u t

p r o o f

M u l t i p l y i n g t h e r i g h t m e m be r o f ( 1 ) b y 5 , w e a p p l y ( 3 ) t o t r a n s f o r m t h e

r e s u l t a s f o l l o w s :

(

F

2n

+3

+

F

2n

-*s

+

i

+ 2 ( - D -

8

F

2 8 + 1

) +

(F

2n

-

F

2n

_

h8

- 2(-l)

n

-

s

F

Zs

)

~ (

F

2n -

3

+

F

2n^s-1

+

2(-l) -

S

F

2s

_

1

)

= (F - F + F ) + (F - F - F )

K

2n+3 2n-3 2n

J v

2n-ks+l 2n-hs 2n-hs-l

J

+ 2(-l) -

s

(F

2s

-F

2B

- F

2 s

_

x

=

^

3

F

2n

+ F

z

)

+ 0 + 0 = 5F

2n

.

This proves(1).

Likewise, multiplying theright memberof (2) by 5yields:

(F - F

+

2(-l)

n

~

s

F )

+

(F + F - 2(-l)

n

~

8

F )

KL

2n + k

L

2n-hs ^ ^

K L)

2s + 2

J

^

y

2n + l

L

2n -ks -1

^

V

x

/

L

2s + l

J

- (F

2

n-2 ~

F

2i^-hs-2

+ 2 ( - l )

F

2s

)

=

^2n + h ~ &\n -2

+

F

2n+l)

~

(^2n -hs ~ ^'2n -hs

- 1 ~

^2n -ks

- 2 /

+ 2 ( - l ) -

e

( F

2 s + 2

-

F

2s+l

- F

ls

)

=

(L,F

2n+1

+ F

2n + 1

)

- 0 + 0 = 5F

2n+1

.

T h i s p r o v e s ( 2 ) .

We m ay c o m b i n e ( 1 ) a n d ( 2 ) i n t o t h e s i n g l e f o r m u l a :

F = F F

2

+ F F

2

- F F

2

, (4 )

n

x

2s + l +o

n

m -s + 1 2s +o

n

m-s 2s -1 + o

n

m -s -1 '

v

^ '

w h e r e

m-= mm,

0 B

E

i . - i n / 2 . { J ; ^

Weseethatthe6-tuples

(a, b

9

e, x

9

y

9

z)

~ ^ 2S + 1+ o

n

> 2s + o

n

* 2s

-1 + o

n

' ^m-s+1*

^m-s

9

^m-s-1'

'

'

are solutionsof theproblem

3

,as s isallowedtovary. For atleastthevalues

s

= 0, 1,

...

9

m

- 1,

different 6-tuples

are

produced

in

(5). Hence, there

are

at least

m= [n/2]

distinct 6-tuples solving

the

problem,

Also solved by the proposer

.

372

[Nov.

7/26/2019 elementary25-4

4/6

ELEMENTARY PROBLEMS AND SOLUTIONS

Zeckendorf Representations

B-582

Proposed by Piero Filipponi, Fond. U. Bordoni, Rome, Italy

It is known that every positive integer

N

can be represented uniquely as a

sum of distinct nonconsecutive positive Fibonacci numbers. Let f(N) be the

number of Fibonacci addends in this representation, a= (1 +v5)/2,and [x] be

the greatest integer in

x

Prove that

f([aF

2

]) = [(n + l)/2] for n =1, 2, ... .

Solution by L. A. G. Dresel, University of Reading, England

S i n c e

F

2

- F

2

0

= (F - F J (F + F ) = F L = F

T

r-2.

v

r

v-z

JK

r v-2

J

v - \ v - \

2 ( r - 1 )

w e h a v e , s um m i ng f o r

even

v a l u e s

v

= 2 ,

t =

1

9

2

5

. . . , 777,

F

2m

0 = F ^

_

2

+

F

hm

_

6

+

e

+

F

2 s

a n d s u m m i n g f o r

od d

v a l u e s r = 2 t + l

9

t = 1 , 2 , . . . 5 / 7 ? ,

n

2m+l

L r

km ^

r

hm-h ^ ^

r

k

9

L e t

a = h(l +

A ) a n d & = % ( 1 - > / 5 ) , t h e n

a F

2 g

= ( a

2 s + 1

- a Z ?

2 s

) / / 5 =

F

2g + 1

+ (6 -

a)b

2s

//5 = F

2s + 1

- b

2s

Applying the formula for F

2

, we obtain

aF? = F

u

, + F

u

, + + F - O ^

- 2

+ Z^

6

+ '+

2

)

2m

km -

1 4m

- 5

3

and since 0 1)

withtheusual convention0

:

= 1. Considerthefunction

Fix, n) := S

n

,

s

~y

.

(2)

s = 0

S

-

Since0 ^ S

n

,s ^ 2

n

n

s

, theabove series convergesfor allreal x. Using(1),

one obtains

H*.

n)

=

t E O ^

E (I) ^

E )**

= 0 k = 0

XK/ S

k=

o

V K /

s

= 0

s

- k = 0^

K/

or

F(x,n) = (e* + l

n

, (3)

which yields

F(x

9

m+ n) = FGr, m)F(x, n). (4)

374 [Nov.

7/26/2019 elementary25-4

6/6

ELEMENTARY PROBLEMS

AND

SOLUTIONS

Cauchy

T

s product leads to

F(x, m)F(x, n) = Z L ~JT (

s

- k) I

(5)

From (2)

9

(4),

and (5), and by comparing coefficients, one obtains the equation

as stated in the proposal for the S

n s

defined in (1).

Also solved by Paul S. Bruckman, Odoardo Brugla Plero Filipponi, L. A. G.

Dresel, L. Kuipers, Fuchln He, J. Suck, Nicola Treitzenberg, Paul Tzermias,

TadP. White, and the proposer.

Combinatorial Interpretation of the

B-585 Proposed by Constant in Gonciulea Nicolae Gonciulea, Trian College,

Drobeta Turnu-Severin, Romania

For each subset

A of X = {I

s

2

9 oe

.

9

n}

9

let

r(A)

be the number of j such

that {j,

j

+ 1} c

A.

Show that

A CX

Solution by

J.

Suck, Essen, Germany

Let us supplement the proposal by

and 2

rW

= F .

neJcX

2n

We now have a beautiful combinatorial interpretation of the Fibonacci sequence,

The two identities help each other in the following induction proof*

For

n =

1,

A

= 0 or

X, r(A) =

0.

Thus

9

both identities hold here. Suppose

they hold for k =1, . . . , n. Consider Y

1

={1, . . .

9

n

9

n +1}. If {n

9

n +1} c

Bc J

9

r

(B) = r(B\{n + 1}) + 1. If n t Bc J

9

r(5) = r(S\{n + 1}). Thus

9

E 2

r (B)

= E

2

r ( i 4 )

+ 1

+ E

2

r C 4 )

(

t h e l a s t s u m i s

l

f o r

n+ UB c y n e ^ c i ^cz\{n} th e st ep 1 -> 1 + 1)

= 2

^ 2 n

+ F

2(n-l) + l

= F

2n

+ F

2n+1

=

^2(n+ l)

and

2

(B )

= 2^

(B)

+ 2 ^ ) =f

20i+i

+

F2n+1

=

P Z ( B + 1 )

x

.

j ?cy n+ ie s c y ^ c j

Also solved by Paul S. Bruckman, L. A. G. Dresel, N* J. Kuenzi Bob Prielipp,

Paul Tzermias, Tad P. White, and the proposer.

1987]

375