advanced problems

7/24/2019 Advanced problems

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ADVANCED PROBLEMS AND SOLUTIONS

dited by

RAYMOND E. WHITNEY

Lock Haven State College, Lock Haven, PA 17745

Send all comm unications concerning ADVANCED PROBLEMS AND SOLUTIONS to

RAYMOND E. WHITNEY, MATHEMATICS DEPARTMENT, LOCK HAVEN STATE C OLLEGE, LOCK

HAVEN, P-A 17745, This department especially welcomes problems believed to

be new or extending old results. Proposers should submit solutions or other

information that will assist the editor. To facilitate their consideration,

solutions should be submitted on separate signed sheets within two months of

publication of the problems,

H 307

Proposed by Larry Taylor, Briarwood, NY

(A) If p = 1 (mod 10) is prime, x =

/B ,

and a = . 7 (modp ,prove

that

a,

a + 1, a + 2, a + 3, and

a +

4 have the same quadratic character mod-

ulo p if and only if 11 < p = 1 or 11 (mod 60) and

(-2x/p)

= 1.

(B) If p = 1 (mod

60), (2x/p)

= 1, and

b

= ~

2

^ J ^

5 )

(mod

p ) ,

then

b,

b

+ 2,

b

+ 3, and

b +

4 have the same quadratic character modulo p. Prove

that (llab/p) = 1 .

H 3O8

Proposed by Paul Bruckman , Concord, CA

K ?n

( a

i'

a

2>

U

n)

L e t

[a

, a , . . . ,

a

n

]

= 7 - r d e n o t e t h e n t h c o n v e r -

g e n t o f t h e i n f i n i t e s i m p l e c o n t i n u e d f r a c t i o n

[a

l9

a

2

,

. . . ] ,

n

= 1 , 2 , . . . .

A l s o , d e f i n e p

0

= 1, ^

0

= 0 . F u r t h e r , d e f i n e

(1 )

W

n>k

= p

n

( a

x

, a

2

, . . . , a

n

) q

f e

( a

1 5

a

2

, . . . , a

fc

)

-

p

k

( a

x

, a

2

, . . . , ^

f e

) ^

n

t o

1 5

a

z

*

a

^ )

P

n

^ P

k

0 < f c < n .

Find a general formula for W

n>

^.

H 309

Proposed by David Singmaster, Polytechnic of the South Bank, London,

England

Let f be apermutation of

{l,

2, ..., m- 1} such that the terms i+f(i)

are all distinct (mod m). Characterize and/or enumerate such /. [Each such

/ gives a decomposition of the

m( m

+ 1) m-nomial coefficients, which are the

nearest neighbors of a given m-nomial coefficient, into msets of m+ 1 coef-

ficients which have equal products and are congruent by rotationsee Hoggatt

& Alexanderson, A Property of Multinomial Coeffieients,

The Fibonacci Quar-

terly 9, No. 4 (1971):351-356, 420-421.]

I have run a simple program to generate and enumerate such /, but can

see no pattern. The number Nof such permutations is given below for m


2/4

Dec.1979

ADVANCED PROBLEMSANDSOLUTIONS

375

m 2 3 4 5 6 7 8 9 10

N

1 1 2 3 8 19 64 225 928

H-310 Proposed by V. E. Hoggatt, Jr., San Jose State University, San Jose, CA

Leta = (1 +

/5)/2

5

[na] =

a

n9

and

[na

2

] = b

n

.

Clearly,

a

n

+ n

=

b

n

.

a) Show that

if n=F

2 m + 1

,

then

a

n

= F

2m +2

and

n

=

F

2m+S

.

b) Show thatif n=F

2 m

, thena

n

= F

2m +1

*

anc

* ^

=

^im

i ~*

e) Show thatif n =-^2m

s

thena

n

= L

2m 1

and2?

n

=

^2m +2'

d) Show that if n =

2

m + l

5 t n e n

^ n

=

^ 2 ^+2 ~ *

a n d

^

=

^2m +

3

~

1 B

SOLUTIONS

EdAJtohslcit

Uotzt Starting with this issue, we shall indicate the issue and

date when each problem was proposed.

Continue

H-278 Proposed by V.E. Hoggatt, Jr., San Jose State University, San Jose, CA

(Vol. 16, No. 1, Feb. 1978)

Show J =


3/4

376

ADVANCED PROBLEMSANDSOLUTIONS

[Dec

A Rare Mixture

H-279

Proposed by G. Wulczyn, Bucknell University, Lewisburg, PA

(Vol. 16, No. 1, Feb. 1978)

Establish the

F-L

identities:

(a) ^

n + 6 r

~ ^hr

+

* ^n

hr ~

n.

2r *

=

^

2vi 2> 6r^hn

12r

( b ) ^n+er 3

+

(k'tr

+ 2

*~

*'

^n+hr+2 ~ ^n 2r l^ ""

n

jp p ~p jp

2r

l

r

hr

2 Sr+ 3 hn

12r+B

Solution

by

Paul Bruckman, Concord,

CA

LmmcL 1: L

3m

- (-l)

m

L

m

=

5F

m

F

2m

.

Vtiooji L

3m

- (-l)

m

L

m

= a

3m

+ b

3m

- (ab)

m

(a

m

+ b

m

)

=

(a

m

-

b

m

)(a

2m

-b

2m

)

= 5F

m

F

2m

.

Imma 2: 5(F* - * * ) = F

U

_

V

F

U+ V

(L

U

_

V

L

U + V

-

4 ( - l )

w

) .

VKooji 25F

h

u

= (a

u

-b

u

)

h

=a

hu

+b

hu

- h(-l)

u

(a

2u

+b

2u

) + 6 .

T h e r e f o r e ,

25


4/4

1979]

ADVANCED PROBLEMS

AND

SOLUTIONS

377

~ 5

F

2n +3m

L

Zn+3m(

F

3m

L

3m ~

^

_ 1

^

F

3m

L

m)

( a p p l y i n g L e m m a 3)

=

J

F

3m

F

kn

+

6

m

L

3m

"

( _ 1 ) L

*)

=

F

m

F

Zm

F

Bm

F

hn

+

6m ^

Lemma

^

T h e r e f o r e :

p^

_

((-l}

m

L

+ l)(F

k

- F

h

)-F

h

=FFFF

S e t t i n g

m - 2v

and

m - 2v

+ 1 y i e l d s ( a ) a n d ( b ) , r e s p e c t i v e l y .

Also solved

by the

proposer.

o

rn

H-280

Proposed

by P.

Bruckman , Concord, CA

(Vol. 16, No. 1, Feb. 1978)

P r o v e

t h e

c o n g r u e n c e s

(1) F

3

.

2

=

2*

+ 2

(mod 2

n + 3

) ;

(2 )

L

3

,

zn

= 2 + 2

2

+ 2

(mod2

2 n +1(

)

, n = 1, 2 , 3

Solution

by

Gregory Wulczyn, Bucknell University, Lewisburg,

PA

(1) n

=

1, F

e

= 8 = 8

(mod 16).

S i n c e L\

k

- 5F\

k

= 4 ,

(L

3k

,F

3k

)

= 2 ,

2

r

\F

3

,

2

n

9

r > 1

9

n >_ 1;

2

t

\L

3

.

z

n

i f ando n l y i f t = 1.

Assume F

3

.

2

n = 2

n+2

(mod 2

3

) =

2

n + 2

(mod 2

n + lf

)

F

3

.

2

n

+

i

= F

3

.

2

n L

3

.

2

n

E

2

n + 3

(mod 2

n + I t

.

(2)

n =

1, L

6

=

18

E 2 +2

h

(mod2

6

)

,

L

2

2k

=

L ^

+

2.

Assume L

3

.

2

-=

2 +

2

2 n

(mod2

2 n f

)

L

3

.

2

n

+

i

=L

2

.

2

- 2 = 2 +2

2 n

>

+

2

4 n+f

(mod2

4 n 5

)

3

.

2

+i E 2 +

2

2 n

f

(mod

2

2 n 6

9

n >1.

Also

solved by the proposer, who noted that this is Corollary 6 in Periodic

Continued Fraction Representations of Fibonacci-type Irrationals / by V. E.

Hoggatt,

Jr.

& Paul

S.

Bruckman,

in The

Fibonacci Quarterly

15,

No .

3

(1977):

225-230.

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