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ADVANCED PROBLEMS AND SOLUTIONS
EDITED BY
FLORIAN LUCA
Please send all communications concerning ADVANCED PROBLEMS AND SOLUTIONS
to FLORIAN LUCA, MATHEMATICAL INSTITUTE, UNAM, CP 04510, MEXICO DF,
MEXICO or by e-mail at [email protected] as files of the type tex, dvi, ps, doc, html,
pdf, etc. 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, all solutions sent by regular mail should be submitted on separate
signed sheets within two months after publication of the problems.
PROBLEMS PROPOSED IN THIS ISSUE
H-743 Proposed by Romeo Meštrović, Kotor, Montenegro.
Let p ≥ 5 be a prime and qp (2) = (2p−1 − 1)/p be the Fermat quotient of p to base 2. Prove
that
(p−1)/2
1 X (−3)k
qp (2) ≡ −
(mod p).
2
k
k=1
H-744 Proposed by D. M. Bătineţu-Giurgiu, Bucharest and Neculai Stanciu,
Buzău, Romania.
Prove that
n
(1)
(3)
n+3−Ln+2
e
≤
en+1−Fn+2 ≤
1X 1
n
Lk
k=1
n
1X 1
n
Fk
k=1
!n
;
(2)
e
!n
;
(4)
en−Fn Fn+1
!n
n
1X 1
≤
;
n
L2k
k=1
!n
n
X
1
1
≤
.
n
Fk2
n+2−Ln Ln+1
k=1
H-745 Proposed by Kenneth B. Davenport, SCI-Dallas, PA.
√
Prove that (a2 − 1) cos(n + 3)θ − 2 a cos nθ = (a − 1)2 cos(n +
√ 1)θ, where a is the real
number satisfying a3 = a2 + a + 1 and θ is given by cos θ = (1 − a) a/2.
H-746 Proposed by H. Ohtsuka, Saitama,Japan.
n
Define the generalized Fibonomial coefficient
by
k F ;m
Fmn Fm(n−1) · · · Fm(n−k+1)
n
=
for 0 ≤ k ≤ n
k F ;m
Fmk Fm(k−1) · · · Fm
374
VOLUME 51, NUMBER 4
ADVANCED PROBLEMS AND SOLUTIONS
n
n
with
= 1 and
= 0 (otherwise). Let εi = (−1)(m+1)i . For positive integers
0 F ;m
k F ;m
n, m and s prove that
X
n
n
n
εi
= εs
.
i F ;m j F ;m
s F ;2m
i+j=2s
SOLUTIONS
Fibonacci Numbers and Derivatives of Polynomials
H-717
Proposed by Samuel G. Moreno, Jaén, Spain,
(Vol. 50, No. 2, May 2012)
Prove that if p is a polynomial such that p(x) > 0 for all x ∈ R, then
deg(p)
X
Fk+1 y k p(k) (x) > 0
for all
k=0
x, y ∈ R.
Solution by the proposer.
For a fixed y ∈ R, y 6= 0, we consider the second-order linear differential equation with
constant coefficients
(I − yD − y 2 D 2 )q(x) = q(x) − yq 0 (x) − y 2 q 00 (x) = p(x),
(1)
in which I stands for the identity operator, and D = d/dx stands for the derivative. If α
denotes the golden ratio, the two distinct roots of the auxiliary equation of (1) are λ1 = −α/y
and λ2 = −(1 − α)/y. Moreover, a particular solution of (1) is
∞
X
2 2 −1
q0 (x) = I − yD − y D
p(x) =
Fk+1 y k D k p(x)
k=0
deg(p)
=
X
Fk+1 y k p(k) (x).
k=0
Thus, the general solution of (1) reads q(x) = q0 (x) + C1 eλ1 x + C2 eλ2 x . Therefore, the unique
polynomial solution of the differential equation considered is q0 .
Taking into account that p must be a polynomial of even degree, and also that the asymptotic
behavior of q0 is governed by F1 y 0 p(0) (x) = p(x), we observe that q0 tends to infinity as |x|
does, so there exists (at least) one absolute minimum m0 of q0 on R. Using that q00 (m0 ) = 0
and q000 (m0 ) ≥ 0, and using also (1), we conclude
q0 (x) ≥ q0 (m0 ) = p(m0 ) + (yq00 (m0 ) + y 2 q000 (m0 )) = p(m0 ) + y 2 q000 (m0 ) > 0,
for all reals x.
Also solved by Paul S. Bruckman.
NOVEMBER 2013
375
THE FIBONACCI QUARTERLY
Inequalities with Fibonacci Numbers and Radicals
H-718
Proposed by Hideyuki Ohtsuka, Saitama, Japan.
(Vol. 50, No. 2, May 2012)
2n−2m−3
4
4
Let An,m = Fn+m
(Fn+m
− Fn−m
). Prove that
(1)
2n
Y
Fk ≤ An,m for n ≥ m ≥ 1;
k=2m
s
n
3
Y
p
F2m−1
F2n−1 F2n
F2k < An,m 4
for n ≥ m ≥ 2.
(2)
F2m−3 F2m−2 F2n+1
k=m
Solution by the proposer.
(1) Let n ≥ m ≥ 1. If n = m, then LHS = RHS = F2n .
Let n > m. We have
Q2n
2n
n−m
Y
Y Fn+m−j Fn+m+j
Fk
k=2m Fk
=
=
·
2n−2m+1
Fn+m
Fn+m
Fn+m
Fn+m
j=0
k=2m
=
n−m
Y
2
Fn+m
j=0
=
n−m
Y
1−
j=0
− (−1)n+m−j Fj2
2
Fn+m
(−1)n+m−j Fj2
2
Fn+m
!
(By Catalan’s Identity)
.
If n − m is odd,
n−m
Y
j=0
<
≤
1−
(−1)n+m−j Fj2
2
Fn+m
(n−m−1)/2 Y
r=0
F4
1 − n−m
.
4
Fn+m
F2
1 + 2r+1
2
Fn+m
!
=
(n−m−1)/2 Y
r=0
F2
1 − 2r+1
2
Fn+m
F2
1 + 22r
Fn+m
≤
(n−m−1)/2 Y
r=0
F2
1 − 2r+1
2
Fn+m
F4
1 − 2r+1
4
Fn+m
If n − m is even,
n−m
Y
j=0
≤
1−
(−1)n+m−j Fj2
2
Fn+m
(n−m)/2 Y
r=0
F2
1 + 22r
Fn+m
F4
≤ 1 − n−m
.
4
Fn+m
376
!
=
(n−m)/2 Y
r=0
F2
1 − 22r
Fn+m
F2
1 + 2r−1
2
Fn+m
≤
2 F2r
1− 2
Fn+m
(n−m)/2 Y
r=0
F4
1 − 42r
Fn+m
VOLUME 51, NUMBER 4
ADVANCED PROBLEMS AND SOLUTIONS
Therefore, we obtain
2n
Y
k=2m
Fk ≤
2n−2m+1
Fn+m
F4
1 − n−m
4
Fn+m
= An,m .
(2) First, we have Ft−2 Ft−1 Ft+1 Ft+2 < Ft4 by the Gelin–Cesàre Identity. Therefore, for
t ≥ 3, we have
Ft2
Ft−1 Ft+1
<
.
(1)
Ft−2 Ft+2
Ft2
Let n ≥ m ≥ 2. We have
n
n
n
2
2
Y
F2k−1
F2k
F2n Y F2k−2 F2k
F2n Y
=
<
(by (1))
2
2
F2m−2
F2m−2
F2k−3 F2k+1
F2k−1
F2k−1
k=m
k=m
k=m
F2m−1 F2n−1 F2n
=
.
F2m−3 F2m−2 F2n+1
Thus, we have
s
n
n
Y
F2m−1 F2n−1 F2n Y
F2k <
F2k−1 .
F2m−3 F2m−2 F2n+1
k=m
k=m
Q
Multiplying both sides of this inequality by nk=m F2k , we get
s
n
n
Y
F2m−1 F2n−1 F2n Y
2
F2k <
F2k−1 F2k .
F2m−3 F2m−2 F2n+1
k=m
k=m
Here, we have
n
Y
F2k−1 F2k =
k=m
Thus, we have
2n
Y
Fk = F2m−1
k=2m−1
n
Y
2
F2k
< An,m
k=m
2n
Y
k=2m
s
Fk ≤ F2m−1 An,m
(by (1)).
3
F2m−1
F2n−1 F2n
,
F2m−3 F2m−2 F2n+1
which leads to the desired inequality.
Note. We obtain the following inequality in the same manner as (2):
s
n
3
Y
p
F2m−1
F2n−1 F2n
F2k−1 < An,m 4
(for n ≥ m ≥ 2).
F2m−2 F2n+1 F2n+2
k=m
Also solved by Paul S. Bruckman and Dmitry Fleischman.
NOVEMBER 2013
377
THE FIBONACCI QUARTERLY
Alternating Sums of High Powers of Fibonacci Numbers
H-719
Proposed by Hideyuki Ohtsuka, Saitama, Japan.
(Vol. 50, No. 2, May 2012)
Let Tj (n) = (−1)n(j+1) (Fn Fn+1 )j . Given a positive integer m prove that there are rational
numbers λ1 , . . . , λm such that
n
m
X
X
(−1)k(m+1) Fk2m =
λj Tj (n).
j=1
k=1
Show the identities
n
X
2
1
(1)
(−1)k Fk4 = − T1 (n) + T2 (n);
3
3
k=1
n
X
1
1
1
Fk6 = T1 (n) − T2 (n) + T3 (n);
(2)
2
4
4
k=1
n
X
8
4
2
1
(3)
(−1)k Fk8 = − T1 (n) + T2 (n) − T3 (n) + T4 (n).
21
21
7
7
k=1
Solution by Harris Kwong, SUNY Fredonia, NY.
Lemma. For any integer i ≥ 1, there exist rational numbers ai,` such that
bi/2c
Fki
=
i
Fk+1
i
+ (−1)
i
Fk−1
+
X
ai,` (−1)k` Fki−2` .
`=0
Equivalently, we can write
bi/2c
i
i
Fk+1
+ (−1)i Fk−1
=
X
bi,` (−1)k` Fki−2`
`=0
for some rational numbers bi,` .
Proof. Induct on i. The result is obviously true when i = 1, because Fk+1 − Fk−1 = Fk . For
i ≥ 2,
i−1
X
i
i
i
i i
r i
Fk = (Fk+1 − Fk−1 ) = Fk+1 + (−1) Fk−1 +
(−1)
F i−r F r .
r k+1 k−1
Fk2
r=1
k
+ (−1) implies
When i is even, Casini’s identity Fk+1 Fk−1 =
that the middle term in the
summation, where r = i/2, is
i
i
(−1)i/2
(Fk+1 Fk−1 )i/2 = (−1)i/2
[Fk2 + (−1)k ] i/2
i/2
i/2
X
i/2 i
i/2
2(i/2−`)
i/2
= (−1)
Fk
(−1)k`
i/2
`
`=0
i/2
=
X
ci/2,` (−1)k` Fki−2` ,
`=0
378
VOLUME 51, NUMBER 4
ADVANCED PROBLEMS AND SOLUTIONS
i i/2
where ci/2,` = (−1)i/2 i/2
` .
In general, for 1 ≤ r ≤ b(i − 1)/2c, due to symmetry, we can group the rth term with the
(i − r)th term; and it follows from the induction hypothesis that
i
i−r r
r i
i−r
(−1)
+ (−1)
F F
F r F i−r
r k+1 k−1
i − r k+1 k−1
r i
i−2r
i−2r
= (−1)
(Fk+1 Fk−1 )r [Fk+1
+ (−1)i−2r Fk−1
]
r
i−2r
i−2r
r i
= (−1)
(Fk2 + (−1)k ]r [Fk+1
+ (−1)i−2r Fk−1
]
r


# b(i−2r)/2c
"X
r X
i
r
2(r−s)
(−1)ks 
bi−2r,t (−1)kt Fki−2r−2t 
= (−1)r
Fk
r
s
s=0
t=0
bi/2c
=
X
cr,` (−1)k` Fki−2` ,
`=0
where cr,` =
P
r i
s+t=` (−1) r
rsbi−2r,t is a rational number. The result follows immediately.
We now prove the original problem. The case of m = 1 is valid:
n
X
Fk2 = Fn Fn+1 = T1 (n).
k=1
Since
Fk2m
=
Fkm
· Fkm ,
n
X
(−1)k(m+1) Fk2m
k=1
the lemma asserts that


bm/2c
n
X
X
m
m
=
(−1)k(m+1) Fkm Fk+1
+ (−1)m Fk−1
+
am,` (−1)k` Fkm−2` 
k=1
`=0
bm/2c
m
= (−1)n(m+1) Fnm Fn+1
+
X
`=0
am,`
n
X
2(m−`)
(−1)k(m−`+1) Fk
.
k=1
P
Solving for nk=1 (−1)k(m+1) Fk2m yields the desired result from induction.
In practice, it is easier to compute the coefficients λj directly. For example, when m = 2,
2
2
2
2
Fk2 = (Fk+1 − Fk−1 )2 = Fk+1
+ Fk−1
− 2Fk+1 Fk−1 = Fk+1
+ Fk−1
− 2Fk2 − 2(−1)k .
This leads to
n
n
n
n
X
X
X
X
2
2
(−1)k Fk4 =
(−1)k Fk2 (Fk+1
+ Fk−1
)−2
(−1)k Fk4 − 2
Fk2
k=1
k=1
2
= (−1)n Fn2 Fn+1
−2
Pn
k 4
k=1 (−1) Fk
k=1
n
X
k=1
k=1
(−1)k Fk4 − 2T1 (n).
Thus, 3
= T2 (n) − 2T1 (n), which proves (1).
In a similar manner, we find
3
3
3
3
Fk3 = Fk+1
− Fk−1
− 3Fk+1 Fk−1 (Fk+1 − Fk−1 ) = Fk+1
− Fk−1
− 3[Fk2 + (−1)k ]Fk .
NOVEMBER 2013
379
THE FIBONACCI QUARTERLY
Hence,
n
X
Fk6 =
k=1
n
X
k=1
3
3
Fk3 (Fk+1
− Fk−1
)−3
3
= Fn3 Fn+1
−3
This yields 4
Pn
6
k=1 Fk
n
X
k=1
Fk6 − 3
n
X
k=1
Fk6 − 3
n
X
(−1)k Fk4
k=1
1
2
T2 (n) − T1 (n) .
3
3
= T3 (n) − T2 (n) + 2T1 (n), thereby proving (2).
The case of m = 4 is slightly more complicated. First we obtain
4
4
2
2
2
2
Fk4 = Fk+1
+ Fk−1
− 4Fk+1 Fk−1 (Fk+1
+ Fk−1
) + 6Fk+1
Fk−1
4
4
2
2
= Fk+1
+ Fk−1
− 4Fk+1 Fk−1 [(Fk+1 − Fk−1 )2 + 2Fk+1 Fk−1 ] + 6Fk+1
Fk−1
4
4
2
2
= Fk+1
+ Fk−1
− 4Fk+1 Fk−1 Fk2 − 2Fk+1
Fk−1
4
4
= Fk+1
+ Fk−1
− 4[Fk2 + (−1)k ]Fk2 − 2[Fk2 + (−1)k ]2
4
4
= Fk+1
+ Fk−1
− 6Fk4 − 8(−1)k Fk2 − 2.
Therefore,
n
n
n
n
n
X
X
X
X
X
k 8
k 4
4
2
k 8
6
(−1) Fk =
(−1) Fk (Fk+1 + Fk−1 ) − 6
(−1) Fk − 8
Fk − 2
(−1)k Fk4 .
k=1
k=1
k=1
k=1
k=1
We conclude that
n
X
1
1
1
1
1
2
n 8
(−1) Fk =
T4 (n) − 8
T3 (n) − T2 (n) + T1 (n) − 2
T2 (n) − T1 (n)
7
4
4
2
3
3
k=1
1
2
4
8
T4 (n) + T3 (n) +
T2 (n) −
T1 (n),
7
7
21
21
which establishes (3).
=
Also solved by Paul S. Bruckman, Kenneth B. Davenport, Dmitry Fleischman
and Zbigniew Jakubczyk.
Late Acknowledgements. Kenneth B. Davenport, M. N. Deshpande, Harris Kwong, and
Anastasios Kotronis all solved H-716.
380
VOLUME 51, NUMBER 4