ELEMENTARY PROBLEMS AND SOLUTIONS Edited by A. P . Hillman Please send all material for ELEMENTARY PROBLEMS AND SOLUTIONS to A. P. HILLMAN; 709 SOLANO DR., S.E.; ALBUQUERQUE, NM 87108. Each solution should be on a separate sheet (or sheets) and must be received within six months of publication of the problem. Solutions typed in the format used below will be given preference. Proposers of problems should include solutions. Anyone desiring acknowledgment of contributions should enclose a stamped, selfaddressed card (or envelope) . Dr.
BASIC FORMULAS The F i b o n a c c i numbers Fn and t h e Lucas numbers Ln, F
n +2
L
n+Z
= =
F
n+l
+
F
+
n+l
F
n>
F
0
=
°»
F
l
L
n> LQ = 2 , Ll
=
satisfy
l
*
= 1.
A l s o , a = (1 + / 5 ) / 2 , 3 = (1  / 5 ) / 2 , Fn = (an  3 n ) / / 5 , and Ln = an + $n. PROBLEMS PROPOSED IN THIS ISSUE B670
Proposed
Evaluate B671
^
by Russell nF
Euler,
Northwest
Missouri
State
U., Marysville,
MO
n
22 —~n=\ 2
Proposed
by Herta
T. Freitag,
Roanoke,
VA
Show that all even perfect numbers are hexagonal and hence are all triangular. [A perfect number is a positive integer which is the sum of its proper positive integral divisors. The hexagonal numbers are {1, 6, 15, 28, 45, . ..} and the triangular numbers are {1, 3, 6, 10, 15,...}. ] B672
Proposed
by Philip
L. Mana,
Albuquerque,
NM
Let S consist of all positive integers n such that n = lOp and n + 1 = 11^, with p and q primes. What is the largest positive integer d such that every n in S is a term in an arithmetic progression a, a + d, a + 2d, ...? B673
Proposed
by Paul
S. Bruckman,
Edmonds,
WA
Evaluate the infinite product ][ n= 2
B674
Proposed
by Richard
AndreJeannin,
Sfax,
Tunisia
Define t h e sequence {un} by UQ = 0, u\ = 1, un = guni 1
where g i s a r o o t of x deduce t h a t
1990]
(1 + / 5 ) / 2
 un2*
 x  1 =0. = 2 COS(TT/5)
for n in {2, 3 , . . . } ,
Compute un f o r n i n { 2 , 3 , 4 , 5} and t h e n and
(1 
/5)/2
= 2 COS(3TT/5).
277
ELEMENTARY PROBLEMS AND SOLUTIONS
B675
Proposed
by Richard
AndreJeannin,
Sfax,
Tunisia
In a manner analogous t o t h a t f o r t h e p r e v i o u s problem, show t h a t fa + fa = 2 cos £
and
o
fa
 fa = 2 cos ^ o
SOLUTIONS Not T r u e A s y m p t o t i c a l l y B645 1
Proposed ^
by R. Tosic,
(2m 
r
1\
r)(2m
Let £9m =
U. of Novi 
 2
1\
, /
o
+
Sad,
2TT?
\
c
.
f o r
^ • .  D  M •*(„*,) where (£) = 0 f o r k < 0 . Solution
Kwong,
(2m 2 (\m ;>J
2m + l G 2m
and G
2m+z
(2m)I
/2m  1\ \ m /
mlml
(2m + 1\ U + 1/ _
^m +l
/
2/77
\
o
o
««•». ..2
SUNY
College
Let us s t u d y t h e a s y m p t o t i c growth of Gn.
G
i
w = 1, 2, 3 , . . . ,
Prove o r d i s p r o v e t h a t £ n = F n f o r n = 0 , 1, 2 , .
hy Y. //. Harris
Hence,
Yugoslavia
at Fredonia, I t i s evident
Fredonia,
NY
that
ml (m  1 ) ! (2m  1)1
(2m + 1) !
mlml
~ O + l)!/7z! " ( 2 m ) !
_ 2m + 1 m + 1
2,
\ 777 /
so t h a t Gn/Gni
~ 2.
However, i t i s w e l l known t h a t
Thus, Gn * Fn f o r s u f f i c i e n t l y l a r g e n. I n f a c t , from n u m e r i c a l c o m p u t a t i o n s , we have Gn = Fn f o r 0 < n < 14, and £ n > Fn f o r rc > 1 5 . AZso solved by Charles Ashbacher, Piero Filipponi, L. Kuipers, and the
Paul S. Bruckman, proposer.
James
E.
Desmond,
Triangular Number Analogue B646
Proposed
by A. P. Hillman
in memory
of Gloria
C.
Padilla
We know t h a t F2n = FnLn = Fn(Fn_l + Fn + l) . Find m a s a f u n c t i o n of n so a s t o have t h e a n a l o g o u s formula Tm = Tn(Tn.l + Tn+1), where Tn i s t h e t r i a n g u l a r number n(n + l ) / 2 . Solution
by H.J.
Seiffert,
We have: ^ ( T n _ x + Tn+l)
Berlin, = Tn(Tn
Germany n
= n(n + l)(n(n 278
+ Tn+n+l)=
Tn(2Tn
+ 1)
+ 1) + l ) / 2 = ^ ( n + i ) . [Aug.
ELEMENTARY PROBLEMS AND SOLUTIONS
Also solved by Richard AndreJeannin, Wray G. Brady, Paul S. Bruckman, Nicos D. Diamantis, Russell Euler, Piero Filipponi, Herta T. Freitag, Russell Jay Hendel, L. Kuipers, Jack Lee, Carl Libis, Bob Prielipp, Jesse Nemoyer & Joseph J. Kostal & Durbha Subramanyam, Sahib Singh, Lawrence Somer, Gregory Wulczyn, and the proposer. Much Ado a b o u t Zero B647
Proposed
by L.
Kuipers,
Serre,
Switzerland
Simplify [£2n + 7(l)»][L3„ Solution
by
Sahib
Singh,
+3
 2 (  l ) » £ „ ]  3(l)nLn_2L2n
Clarion
U. of Pennsylvania,
The given e x p r e s s i o n s i m p l i f i e s t o z e r o . numbers, i t f o l l o w s t h a t L2 + 7 (  l ) n = Ln_2L expression is L
oL +0[L, n + ZL
n2
Mo
 2(l)nL v
3n + 3
/
+2
(LZ.0L

v
n
 V2Lnl^+2
Clarion,
PA
By u s i n g t h e B i n e t form of Lucas «• ^n v i  e w °f t h i s s t h e g i v e n
, + 3(l)nL
,9)].
n + 2 n~\
n+
Z/J
Again, a p p l y i n g t h e B i n e t form of Lucas numbers, we see t h a t L20L , + 3(l)nL ^ 9 = Lq .  2 (  l ) n L „ . v
n + 2 n\
J
n+2
Also Carl the
solved by Paul S. Libis, Bob Prielipp, proposer.
v
3n + 3
Hence, t h e r e q u i r e d c o n c l u s i o n
'
n
follows.
Bruckman, Herta H.J. Seiffert,
T. M.
Freitag, Wachtel,
Pell P r i m i t i v e P y t h a g o r e a n B648
Proposed
by M. Wachtel,
Zurich,
Y. H. Gregory
Harris Kwong, Wulczyn, and
Triples
Switzerland
The P e l l numbers Pn and Qn a r e d e f i n e d by P
n+2
2P
=
n+l 2
Show t h a t (P. , P D kn> P2n { 1 , 2, . . . } . Solution
by Paul
S.
+
P
n>
P
0
= °>
P
l
= ^
®n + 2 =
2
®n+l
+ l 1, 3Pl + 1) i s a p r i m i t i v e s 3P 2n
+
Bruckman,
Edmonds,
+ ^n'^0
=
l
= «1 •
P y t h a g o r e a n t r i p l e for n i n
WA
The Pell numbers satisfy the following identities: (1) (2) Hence,
2P
2n®2n
=
2P
P
hn;
«L  L = !•
P p + (3) I t i s known t h a t p r i m i t i v e Pythagoren t r i p l e s a r e g e n e r a t e d by
«L  L  L !•
(4)
a2  b2,
{lab,
a2 + £ 2 ) , where g . c . d .
We may l e t a = Q2 , b = P2n. lab = P ^ 2
a 1990]
~ b
2
( a , 2?) = 1.
We see from (2) t h a t g . c . d .
( a , 2?) = 1.
Also
[using ( 1 ) ] ;
= P\n + 1 [using (3)]; 279
ELEMENTARY PROBLEMS AND SOLUTIONS
and
a2 + b2
+ 1
3Po
[adding 2b2 to both sides of (3)],
This proves the assertion. Also solved by Nicos D. Diamantis, Ernest J. Eckert, Russell Euler, Piero Filipponi, Herta T. Freitag, Russell Jay Hendel, L. Kuipers, Jesse Nemoyer & Joseph J. Kostal & Durbha Subramanyam, Bob Prielipp, H.J. Seiffert, Sahib Singh, Lawrence Somer, Gregory Wulczyn, and the proposer. Sides Differing b y 17 B649
Proposed
by M. Wachtel,
Zurich,
Switzerland
Give a r u l e for c o n s t r u c t i n g a sequence of p r i m i t i v e P y t h a g o r e a n ( a „ , bn, cn) whose f i r s t few t r i p l e s a r e i n t h e t a b l e
n
bn Cn
and which
\an  bn\ c
6 1248 1265 1777
5 572 555 797
4 224 207 305
3 88 105 137
8 7332 7315 10357
7 3276 3293 4645
satisfy
a< 2 n  l and
2 28 45 53
1 24 7 25
CLyl
triples
l n
_
+
= 17,
a
=
2n cln
l +
26P
2n = h2n~l 26Q2n.
=
+
h
2n>
[Pn and Qn are the Pell numbers of B648.] Rule
by Paul S. (a2ni,
b2n\,
= (10P
2
(a2n,
b2n,
by Ernest
Let matrix
(ai,
Edmonds,
2
+ 26PnQn + 5Q2,
26P2  lhPnQn
+ 13^),
+ 26PnQn + 12«2, 24P2 + 26PnQn  5Q2,
26P2 + UPnQn
+ 13^).
J.
Eckert,
bzn1
c
b2n
c2n]
[a2n
2kP2
c2n)
U. of South
Z?i, C i ) = ( 2 4 ,
Then [a2ni
WA
oln\)
+ 26PnQn  12Q2,
= (10P n Rule
Bruckman,
7,
25),
is tne
2nl^ =
Carolina,
(a2,
b2,
o2)
Aiken,
SC
= ( 2 8 , 4 5 , 53) a n d A d e n o t e
matrix product
[a\
n 1
Gl]A

the
and
o2]A n\
[a2
EditorTs note: The derivations and proofs given by Bruckman and Eckert are not included because of space limitations; however, since each term in the required equations satisfies the same 3 rd order linear homogeneous recursion w
n+3
=
5(wn
+2
+ Wn + l )

W n,
it suffices to verify the rules for n = 1, 2, and 3. Also 280
solved
by
Gregory
Wulczyn
and
the
proposer. [Aug.
ELEMENTARY PROBLEMS AND SOLUTIONS
A v e r a g e Age of G e n e r a l i z e d B650
Proposed by Piero & David Singmaster,
Rabbits
Filipponi, Fond. U. Bor.doni, Rome Italy Polytechnic of the South Bank, London,
UK
Let us i n t r o d u c e a p a i r of 1monthold r a b b i t s i n t o an e n c l o s u r e on t h e f i r s t day of a c e r t a i n month. At t h e end of one month, r a b b i t s a r e mature and each p a i r produces k  1 p a i r s of o f f s p r i n g . Thus, a t t h e b e g i n n i n g of t h e second month t h e r e i s 1 p a i r of 2monthold r a b b i t s and k  1 p a i r s of 0montholds. At t h e b e g i n n i n g of t h e t h i r d month, t h e r e i s 1 p a i r of 3  m o n t h  o l d s , k  1 p a i r s of 1montholds, and k(k  1) p a i r s of 0  m o n t h  o l d s . Assuming t h a t t h e r a b b i t s a r e i m m o r t a l , what i s t h e i r a v e r a g e age An a t t h e end of t h e nth month? S p e c i a l i z e t o t h e f i r s t few v a l u e s of k. What happens as n > °°? Solution
by Sahib
Singh,
Clarion
U. of Pennsylvania,
Clarion,
I f Ai d e n o t e s t h e a v e r a g e age a t t h e end of t h e i t h following recurrence r e l a t i o n : Ai
=  ( 1 + A^,
+l
where A1 =  ;
PA
month, t h e n we have t h e
k > 1.
Using t h i s , we conclude t h a t kn \ Thus,
i=o
k + 2 A2 =  ^  ;
A A*.3 =
Limit A^n = „ n.oo
Also
solved
 1)
k2 + k + 2  r — ,
etc.
1
k 
by Paul
kn(k
I
S.
1
Bruckman
and
the
proposers.
Multiples of a Prime p B651
Proposed
by L.
Van Hamme,
Vrije
Universiteit,
Brussels,
Belgium
Let UQ, ui, . . . be d e f i n e d by u0 = 0, U\ = 1, and un+2 = un+i  un. Also l e t p be a prime g r e a t e r t h a n 3 , and f o r n i n X = { 1 , 2, . . . , p  l } 5 l e t n d e n o t e t h e y i n I w i t h nv E 1 (mod p ) . Prove t h a t pi
(n~lun + k)
£
E
° ( m ° d P)
n= 1
for all nonnegative integers k. Solution
by
the
proposer.
Let p be a zero of 1 + X + X2.
Hence, p 3 = 1.
2P
P
(1 + p )  1  P P =  Q ~ 1  P
= (p2 + 1 + p) = 0 = (p p is also a zero of (1 + X) 1990]
p
2
1
+ 1 + p ) =0
 1  X.
Since
P
if p = 1
(mod 3)
if p = 1 (mod 3 ) ,
Hence, 281
ELEMENTARY PROBLEMS AND SOLUTIONS
g o .   o . :?!«;> = »• Multiplying the first equation with p^3 the second with p~k, easily verified formula ^
J
r
and using the
Yl\
y
we get
Dividing by p and using tP)
p\n/ p\n
=^^— n
(mod rp ) ,
1 < n < pr ~ 1 3
we set the assertion
Also solved by Paul S.
Bruckman.
(continued from page 288) Z^(t) represents the number of zeros of ft which are eclose to n^. By invariance of the complex integral, the functions Z^(t) are constant since the functions ft vary continuously and do not vanish on the path of integration. Hence., 2^(0) = 2^(1) for each i. This says that in a small neighborhood of each zero of f±, there is a onetoone correspondence of zeros of f± with zeros of f 0 , in the required manner. Q In the case of our given functions, we find that the zeros of the polynomial fn{z) are close to the zeros of g (z), which lie on the circle \z\ = as as requireds and the zeros of fn get closer to the circle as n •> °°. H
Also solved proposer.
282
by P. Bruckman,
O. Brugia
& P. Filipponi,
L. Kuipers,
and
the
[Aug.