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ADVANCED PROBLEMS AND SOLUTIONS
Edited by
RAYMO ND E. WHI TNEY
Lock Haven State College, Lock Haven, Pennsylvania
Se nd a l l c om mun ic a t ions c onc e rn ing Adva nc e d Pr ob le m s a nd So lu tions to R a ymond E .
W hi tne y , M a the m a t i c s De p a r tm e n t , Loc k Ha ve n S ta t e C o l l e ge , Loc k Ha ve n , Pe nn s y lva n ia
17745 . Th i s de p a r tm e n t e s pe c ia l ly we lc o me s p rob le ms be l i e ve d to be new o r e x te nd ing o ld
re s u l t s . P r op os e r s s hou ld s ubm i t s o lu t ions o r o the r in fo rma t ion tha t w i l l a s s i s t t he e d i to r .
To fa c i l i t a t e t he i r c ons ide ra t ion , s o lu t ions s hou ld be s ubmi t t e d on s e p a ra t e s igne d s h e e t s
wi th in two mon ths a f t e r pub l i c a t ion of the p rob le m s .
H-195 Proposed by Verner
Hogg att, Jr.,SanJose State College,SanJose, California
Consider the array indicated below:
2
5
3
34
89
2
4
9
22
56
45
3
7
6
38
4
27
65
5
6
6
22
(i) Show tha t the row sum s ar e F , n ^ 2.
( i i) Show tha t the r is i ng diagona l sum s ar e the convolut ion of
(
F
2 n - l ^
= 0
a n d
{ ^ 2 , 2 ) } ;
= 0
,
t he ge n e ra l i z e d nu mb e rs of H a r r i s a nd S ty le s .
H-196 Proposed by J. B. Roberts, Reed College, Portland, Oregon.
(a) Let A
0
be the s e t o f i n t e g ra l pa r t s of t he pos i t i ve in t e g r a l mu l t ip l e s o f r , whe r e
1 + ^ 5
T =
413
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414 ADVANCED PRO BLE MS AND SOLUTIONS [Oct .
a nd l e t A - , m = 0 , 1 , 2 , , be the s e t o f i n t e g ra l pa r t s o f t he num be r s n r
2
for n 61 A . Pr ov e tha t the col lec t ion of Z of a l l pos i t ive in teg ers is the d is jo in t
union of the A..
J
(b) Ge ner a l iz e the pro pos i t ion in (a ) .
H-197 Proposed byLawrence
Somer
Universityof Hi n o s, Urbana, Illinois.
n
s ion re l a t ions h ip :
Le t { i r } _ be the t -F ib ona c c i s e que n c e s w i th pos i t i ve e n t r i e s s a t is fy ing the re c u r -
F ind
(t)
u
n
. l i m
n
t
Vu .
Z
i
n - i
i= l
(t)
u
n+1
u
oo n
SOLUTIONS
HYPER
-TENSION
H-185 Proposedby L
Carlitz
DukeUniversity Durham, North Carolina.
Show that
(1 - 2x)
n
= ( - D
n
"
k
(
n
2
+
k
k
) (
2
k
k
) (1 - v )
n
-
k
2 F l
[-k; n + k + 1; k + 1; x]
k=0 ^ / \ /
w h e r e
2
Fi [a ,b ; c ; x ] de no te s the hyp e rge om e t r i c func tion .
SolutionbytheProposer.
We s tar t wi th the ident i ty
r s
E
(2r + 3s) (y - z) z _
r s ( r + 2s ) ^
+
.2r +3 s+l 1 - y - z
r , s = 0 y*
Now pu t y = u + v, z = v, so tha t
fr
v V
2 r
+
3 s ) L
u v
= 1
1 ;
JL J r l s l ( r + 2s ) l ,-
+
,2r +3 s+l 1 - u - 2v
r . s = 0 *
U v
'
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1972] ADVANCED PRO BLEM S AND SOLUTIONS 415
The right-hand side of (*) is equal to
n=0
while the le f t -h and s ide
oo oo
E
(2r + 3s)l r s V
1
/ -x\
k
f
2r
+ 3 s + k \ , ^
r l s l ( r
+
2s)l
U V
^
{
~
1}
{ k J
( U + V )
r , s = 0 k= 0
It follow s th at
, , . ,n \" ^ , , ,k v ^ .
n
x k / 2 r + 3s + k \ (2r + 3s ) l r s
=
L - < -
1 ) r + t u
s
P +
t
( m o d
P>"
FIBONACCI IS A SQUARE
H-187 Proposed by Ira
Gessel
Harvard University, Cambridge,
Massachusetts.
Pro b le m: Show tha t
a
pos i t i ve in t e ge r
n is a
F i b o n a c c i n u m b e r
if
and only
if
e i the r
5n
2
+
4 o r
5n
2
- 4 is a
s q u a r e .
Solution by the Proposer.
L e t
F
0
= 0, F
4
= 1, F - = F
+
F _-
be the F ibona c c i s e r i e s a nd L
0
= 2 , =
1 ,
L
,- = L
+
L
n
be the Lu cas ser ie s . I t i s wel l known tha t
r+ 1
r r - 1
(1) ( -I)"
+
F*
r
=
F
r + 1
F
r
_
1
(2)
L
r
=
F
r + 1
+
F
r
_
x
Sub t ra c t ing fou r t ime s the f i r s t f rom the s qua re of the secon d equat io n, we have
whe nc e
L
2
_
4 ( - l )
r
-
4 F
2
= (F _,, - F -
)
2
= F
2
,
r
r r+1 r - 1 r
5 F
2
+
4 ( - l )
r
= L
2
.
r
r
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418 ADVANCED PR OB L EM S
AND
SOLUTIONS
[Oct.
T h u s if n is a F i b o n ac c i n u m b e r , e i t h e r 5n
2
+ 4 o r 5n
2
- 4 is a s q u a r e .
I have
two
p r o o f s
of the
c o n v e r s e .
F i r s t
Proof. We us e the
t h e o r e m ( H a r dy
and
W r i g h t ,
An
In t roduc t ion
to the
T h e o r y
of
N u m b e r s ,
p. 153)
tha t
if p and q ar e
i n t e g e r s ,
x is a
r e a l n u m b e r ,
and
| (p /q)
- x | 2, so by the two t h e o r e m s
quoted above , k /n = F - / F for s o m e r, and s inc e bo th f ra c t ions a re r e d u c e d , n = F .
Second Proof. A s s u m e 5n
2
4 = m
2
. T h e n m
2
- 5n
2
= 4, so
m + \/ 5n m - N/5n _
n
"
2 * 2
l
'
and s ince
m and n
ha ve
the
s a m e p a r i t y ,
m
+ \
r
5n , m - V5n
_ _ and
a r e i n t e g e r s
in Q ( N / 5 ) ,
w h e r e
Q is the
r a t i o n a l s ,
and
s inc e the i r p roduc t
is 1,
the y
are
u n i t s . It is
wel l known (Hardy
and
W r igh t ,
p. 221)
t h a t
the
on ly in t e g ra l un i t s
of
Q(\/I>)
are
of the fo rm
x~
, w h e r e x = (1 + \/"5)/2.
T h e n
we
have
(m
+
^ 5 n ) / 2
= x
r
= \ (x
r
+ y
r
) + f L ^ ^ L N/5 ,
2
L ^ J
w h e r e y = -1 /x . Now x. + y = L and
8/12/2019 fibonacci advanced
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1972] ADVANCED PROB LEM S AND SOLUTIONS 419
(
X
r
- y
r
) / ^ 5 = F
r
(Hardy and Wri ght , p . 148) . Thu s
| ( m + N/5n) = | ( L
r
+ ^ 5 F
r
> ,
so n = F .
S U M S E R IE S
H-189 Proposedby L Carlitz DukeUniversity Durham, NorthCarolina Corrected).
Show that
(2r + 3s ) l (a - b y )
r
b
S
y
r + 2 s
^ r l s ( r + 2 s ) l
H
_, ,2r +3 s+l , ,
2
r 9 S = 0
' (1 + ay) 1 - ay - by
2
Solution
b y
theProposer.
P u t
i - ar
so tha t
1 - ax - bx
2
A
m=0
m
x
m
1 = E *
(1 - ax - bx
2
)(l - y)
n
J
m , n = 0
v
ni n
x
y
R e p la c ing y by x" y th i s be c o me s
1 V ^
m
~
n n
z /
J
G x y
m , n = 0
1 - ax - bx
2
)( l - x y)
Hence tha t par t of the expans ion of
(1 - ax - bx
2
) ( l - x V )
tha t i s independen t of x is equal to
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420 ADVANCED PRO BLE MS AND SOLUTIONS [Oct .
(*)
1 - ay - by
2
On the oth er hand, s ince
(1 - ax - bx
2
)( l - x~ y) = (1 + ay) - x(a - by) - bx
2
- x " y ,
we have
(1 - a x - b x ^ U - x - V ) "
1
= [ ^ a - b ^ + b x ^ + x - V l
v T S t
E
(r + s + t ) l (a - by) b y r+ 2s -t
r ls l t l , - ^
x
r + s + t + l
x
r,s,t=0
( 1 + a
^
The pa rt of th is sum tha t i s independent of x is obta ined by taking t = r + 2s . We ge t
E
( 2 r + 3 s) l ( a - b y ) V y
r 2 s
r
T
.sI(r + 2s)l ,.. , v2 r+3 s+i
r , s = 0
( 1 + a y )
Since th is i s equal to (*) , we have proved the s ta ted ident i ty .
I T S A M O D W O R L D
H-190 ProposedbyH. H.Ferns Victoria, ritish Columbia.
P r o v e t h e f o l l o w i n g
2
r
F = n (mod 5)
2
r
L = 1 (mod 5) ,
whe re F a nd L a r e the n F ibona c c i a nd n Luc a s nu m be rs , r e s p e c t iv e ly , a nd r i s
the lea s t res idu e of n - 1 (mod 4) .
Solutionb ytheProposer.
In an unpub l ished pape r by the pr op os er , i t i s shown tha t
^\ E
2k
n
+1
)
5
k
k=0
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1 9 7
2 ] ADVANCED PROB LEM S AND SOLUTIONS 421
Hence
IT]
k = l
x
'
Thus
5
k
.
2
n 1
F
n
= n (mod 5)
Let n - 1 = 4m + r , wh ere 0 < r < 4 . Then
But
Hence
2 F = 2 F = 2 2 F = n (mod 5) .
2
4 m
= ( 2
4
}
m ^
1 ( m o d 5 )
^
To p rove
use
2
r
F = n (mod 5) .
2
r
L = 1 (mo d 5) ,
M
k= 0
%
'
(which is der ived in the same paper) and proceed as above*
JU S T S O MA N Y T W O S
H-192 Proposed byRonaldAlter UniversityofKentucky Lexington, Kentucky.
I f
3n+l
c
n
3=0
M1+J.
. s( J H
prove tha t
c
=
2
6 n + 3
. N
s
(N odd , n >0 ) .
n
olution
by
the
Proposer
In the s eque nce
b
k
= b
k - l "
3 b
k - 2 '
( k
"
l j b
i =
b
2 = D >
it is ea sy to show tha t 2 is the hi gh est pow er of 2 tha t div ide s b. if and only if k = 3
(mod 6) , Als o , by der iv in g the ap pro pr ia te Bine t for mu la , i t fo l lows tha t
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422 ADVANCED PROB LEM S AND SOLUTIONS [Oct .
Thus
^ i m - c - ^ i - -
k ^ 1
2
j=o
The de s i re d re s u l t fo l lows by obs e rv ing
6n+3 6n+2 n *
Ed i t o r i a l No te : P l e a s e s ubmi t s o lu t ions fo r a ny o f t he p rob le m p r opo s a l s . W e ne e d f re s h
blood
A GOLDEN SECTION SEARCH PROBLEM
R E X H . S H U D D E
Waval Post Graduate School, Monterey, California
Afte r t i r i ng of u s ing num e rou s qua d ra t i c func t ions a s ob je c t ive func t ions fo r e xa m ple s
in my ma the ma t i c a l p rog ra m min g c ou rs e , I pos e d the fo llowing p ro b le m fo r
myself:
De s ign
a un imoda l funct ion ove r t he (0,1) i n t e rva l wh ic h i s c onc a ve , h a s a ma x im um in the in t e r io r
o f (0 ,1 ) , a nd i s no t a qua d ra t i c func t ion . The pu rp os e w a s to de m ons t ra t e num e r i c a l ly the
golden sec t ion search.*
My f i r s t thoughts we re to add two funct ions which ar e concav e o ver the (0 ,1) in te rva l
with the pr op er ty tha t one goes to -o a t 0 and the o th er goes to -o a t 1 . My two in i t ia l
c ho ic e s we r e log x a nd l / (x - 1 ) . The go lde n s e c t ion s e a rc h s t a r t s a t t he two po in t s x
t
=
1 - (1/0) and x
2
= l/(p wh ere 0 = (1 +
N / 5 ) / 2 .
Af t e r s e a rc h ing w i th 8 po in t s , I no t i c e d tha t
the in te rva l of un cer ta in ty s t i l l conta in ed the f i r s t se ar ch point so I thought i t abou t t im e to
f ind the loca t io n of the ma xim um an alyt i ca l ly . I wa s dumfounded to d is co ve r tha t i f I con t in-
ued indefin i te ly wi th the se ar ch m y in ter va l of un cer ta in ty would s t i l l conta in the in i t ia l sea rc h
point .
* Doug la s J . W i lde , Op t imum Se e k ing M e tho ds , P re n t i c e Ha l l , Inc . (1964) .
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