Zsigmondy's Theorem - Department of Mathematics - The Hong ...

Mathematical Excalibur, Vol. 16, No. 4, Feb. 12 Page 4
Other commended solvers: Alumni
2011 (Carmel Alison Lam Foundation
Secondary School), CHAN Chun Wai
and LEE Chi Man (Statistics and
Actuarial Science Society SS HKUSU),
Andrew KIRK (Mearns Castle High
School, Glasgow, Scotland), Andy
LOO (St. Paul’s Co-educational
College), MANOLOUDIS Apostolos
(4° Lyk. Korydallos, Piraeus, Greece),
NGUYEN van Thien (Luong The
Vinh High School, Dong Nai, Vietnam),
Mihai STOENESCU (Bischwiller,
France) and ZOLBAYAR Shagdar
(Orchlon School, Ulaanbaatar,
Mongolia).
Problem 384. For all positive real
numbers a,b,c satisfying a + b + c = 3,
prove that
2 2 2 2 2 2
a + 3b
b + 3c
c + 3a
+ + ≥4.
2
2
2
ab (4−ab)
bc (4−bc)
ca (4−ca)
Solution. William PENG.
Let
a b c
A=
+ + ,
2
2
2
b (4−ab)
c (4−bc)
a (4−ca)
1 1 1
B=
+ + ,
a(4
−ab)
b(4
−bc)
c(4
−ca)
4−ab
4−bc
4−ca
C=
+ +
a b c
and D 1 1 1
= + .
a b
+ Then A+3B is the
c
left side of the desired inequality. Now
since a + b + c = 3, we have C = 4D −3.
By the Cauchy-Schwarz inequality, we
have (a+b+c)D ≥ 3 2 , AC ≥ D 2 and BC
≥ D 2 . The first of these gives us D ≥ 3
so that (D−3)(D−1)≥0, which implies
D 2 ≥ 4D−3. The second and third imply
4D
A + 3B
≥
C
2
2
4D
= ≥ 4.
4D
− 3
Other commended solvers: Alumni
2011 (Carmel Alison Lam Foundation
Secondary School), AN-anduud
Problem Solving Group (Ulaanbaatar,
Mongolia), Andrew KIRK (Mearns
Castle High School, Glasgow,
Scotland), LKL Excalibur (Madam
Lau Kam Lung Secondary School of
MFBM), Andy LOO (St. Paul’s
Co-educational College), NGUYEN
van Thien (Luong The Vinh High
School, Dong Nai, Vietnam) and Paolo
PERFETTI (Math Dept, Università
degli studi di Tor Vergata Roma, via
della ricerca scientifica, Roma, Italy).
Problem 385. To prepare for the IMO,
in everyday of the next 11 weeks, Jack
will solve at least one problem. If every
week he can solve at most 12 problems,
then prove that for some positive integer n,
there are n consecutive days in which he
can solve a total of 21 problems.
Solution. AN-anduud Problem Solving
Group (Ulaanbaatar, Mongolia), CHAN
Chun Wai and LEE Chi Man (Statistics
and Actuarial Science Society SS
HKUSU), Andrew KIRK (Mearns
Castle High School, Glasgow, Scotland),
Andy LOO (St. Paul’s Co-educational
College) and Yan Yin WANG (City
University of Hong Kong, Computing
Math, Year 2).
Let S i be the total number of problems
Jack solved from the first day to the end of
the i-th day. Since he solves at least one
problem everyday, we have 0 < S 1 < S 2 < S 3
< ⋯ < S 77 . Since he can solve at most 12
problems every week, we have
S 77 ≤12×11=132.
Consider the two strictly increasing
sequences S 1 , S 2 , ⋯ , S 77 and S 1 +21, S 2 +21,
⋯, S 77 +21. Now these 154 integers are at
least 1 and at most 132+21=153. By the
pigeonhole principle, since the two
sequences are strictly increasing, there
must be m < k such that S k =S m +21.
Therefore, Jack solved a total of 21
problems from the (m+1)-st day to the end
of the k-th day.
Olympiad Corner
(continued from page 1)
Problem 4 (cont). Is it possible to empty
both bags after a finite sequence of
operations?
Operation b) is now replaced with
b') Triple the number of balls in one bag.
Is it now possible to empty both bags after
a finite sequence of operations?
Problem 5. Prove that the product of four
consecutive positive integers cannot be
equal to the product of two consecutive
positive integers.
Problem 6. Let ABC be an acute-angled
triangle. The feet of all the altitudes from
A, B and C are D, E and F respectively.
Prove that DE+DF ≤ BC and determine
the triangles for which equality holds.
Zsigmondy’s Theorem
(continued from page 2)
Example 8. Find all positive integral
solutions to
(a+1)(a 2 +a+1)⋯(a n +a n–1 +⋯+1)
= a m +a m-1 +⋯+1.
Solution. Note that n = m = 1 is a trivial
solution. Other than that, we must have
m>n. Write the equation as
2 3 n+ 1 m+
1
a −1 a −1 a −1
a −1
⋅ ⋅...
⋅ = ,
a−1 a−1 a−1 a−1
then rearranging we get
( )( ) (
2 3 n+
1
a −1 a −1 ... a −1
m+
1
( a )( a )
n−1
= −1 −1 .
By Zsigmondy’s theorem, we must
have a = 2 and m + 1 = 6, i.e. m = 5
(otherwise, a m –1 has a prime factor that
does not divide a 2 –1, a 3 –1, …, a n+1 –1,
a contradiction), which however does
not yield a solution for n.
The above examples show that
Zsigmondy’s theorem can instantly
reduce many number theoretic
problems to a handful of small cases.
We should bear in mind the exceptions
stated in Zsigmondy’s theorem in order
not to miss out any solutions.
Below are some exercises for the
readers.
Exercise 1 (1994 Romanian Team
Selection Test). Prove that the
sequence a n =3 n –2 n contains no three
terms in geometric progression.
Exercise 2. Fermat’s last theorem
asserts that for a positive integer n ≥ 3,
the equation x n +y n =z n has no integral
solution with xyz≠0. Prove this
statement when z is a prime.
Exercise 3 (1996 British Math
Olympiad Round 2). Determine all
sets of non-negative integers x, y and z
which satisfy the equation 2 x +3 y =z 2 .
Reference
[1] PISOLVE, The Zsigmondy
Theorem,
http://www.artofproblemsolving.com/
Forum/viewtopic.php?f=721&t=42233
0
[2] Lola Thompson, Zsigmondy’s
Theorem,
http://www.math.dartmouth.edu/~tho
mpson/Zsigmondy's%20Theorem.pdf
)