Mathematical Excalibur, Volume 2, Number 3

* s -

o

Volume 2, Number 3 May-June, 1996

Olympiad Corner

1996 Canadian Mathematical Olympiad:

Problem 1. If a,

p

and y are the roots of

x3

-

x

-

I = 0, compute l + a

I+p

l + y

-+-+-.

I - a I - p 1-7

Problem 2. Find all real solutions to the following system of equations:

1 + 4 x 2

1 + 4 y 2 - ”

I

*-

Carefully justify your answer.

Problem 3. We denote an arbitrary permutation of the integers 1,2, -.a, n by

a l , a2,

...,

a,. Letfin) be the number of

these permutations such that (i) al = 1;

(continued on page 4 )

Editors: Cheung Pak-Hong, Cum. Studies, HKU

KO, Tsz-Mei, EEE Dept, HKUST

Lmng Tat-Wing Appl. Math Dept HKPU Li, Kin-Yin, Math Dept, HKUST Ng, Keng Po Roger, ITC, HKPU

Artist: Yeung Sau-Ying Camille, MFA, CU The editors welcome contributions from all teachers and students. With your submission, please include your name, address, school, email:

telephone and fax numbers (if available). Electronic submissions, especially in MS Word are encouraged. The deadline for receiving material for the next issue is July 10, 1996. For individual subscription for the five issues foi

the 96-97 academic year, send us five stamped

I

self-addressed envelope. Send all correspondence

I

to:

Dr. Kin-Yin Li Department of Mathematics Hong Kong University of Science and Technology

Clear Water Bay, Kowloon, Hong Kong

Fax: 2358-1643

Fermat‘s Little Theorem and Other Stories

T.

W. Leung

Hong Kong Polytechnic University

Pierre de Fermat (1601-1665), a may rewrite Fermat’s statement as the following which will be referred to as councilor of the provincial High Court

of Judicature in Toulouse, south of Fermat’s Little Theorem: France, practised mathematics during

his spare time. He discussed his findings with his friends via letters. As it turned out, his works significantly influenced the development of modem

If p is a prime number and a is any integer, then

a”

3 a (mod p ) . In

particular, if p does not divide a, then

K1

=

1 (modp).

mathematics. During Fermat’s time, the following ‘‘Chinese hypothesis” was around:

p is a prime if and only if 2p 3 2 (modp).

Now we see how Fermat made use of

his little theorem. He was challenged to determine if there is any even perfect number lying between

lo2’

and 1 02’. (A positive -integer n is called a perfect

number if the sum of all proper factors (i.e., excluding n) of n is equal to n. For

example, 6 = 1 + 2 + 3 and28 = 1 + 2 + 4

+

7

+

14 are perfect numbers.) This problem can be reduced (how?) to check if 237- 1 is prime. Suppose the numberis not prime, and p is an odd prime divisor of that number. then from the third One direction of the hypothesis is not

true. In fact 234’

-

2 is divisible by 341, yet 341 = 1 1 x 31 is composite (not prime). However the other direction is indeed valid. From the manuscripts and letters of Fermat, we conclude that Fermat knew (and most likely could prove) the following facts:

(1) If n is not a prime, then 2” - 1 is not a prime.

(2) If n is a prime, then 2”

-

2 is a multiple of 2n.

(3) If n is a prime, and p is a prime

divisor of 2”

-

1, then p

-

1 is a multiple of n.

The first statement can be proved directly by factoring 2”

-

1. If n = pq (withp > 1 and q > I), then

2 ” - 1 = 2 W - 1

= ( 2 P

-

1)(2p”-” + 2P(‘-*) +

...

+ 1).

The other two statements are variations of the more general statement, indicated

in his other letter:

Given any prime p, and any

geometric progression I , a, a’, - . a .

the number p must divide some number a”

-

1, for which n divides p-1; if then N is any multiple of the smallest number n for which this is

so, p divides also aN

-

1.

With modern mathematical notation, we

statement, p

-

I is a multiple of 37, or p = 37k

+

1, observe thatp is odd, so k is even, or p is of the form 74k‘ + 1. The first few candidates are 149, 223,

....

One then check that

Z3’

-

1 = 137438953471 =223 x 616318177. It is more difficult to check that the second factor is a prime, however Fermat succeeded in showing that

z3’

-

1

is not prime.

Another side story comes from the fact that if 2”

+ 1

is prime, then m must be of the form 2”. Fermat conjectured

that all these numbers are prime. Now 22

+

1 = 5,2’ + 1 = 17, 22

+

1 = 257 and 22

+

1 = 65537 are indeed prime numbers. However,

225

+ 1

= 4294967297

is not a prime. In fact, if p is a prime factor of 22” + 1, then 2”+’ is the smallest

(continued on page 2 )

I 2 3

c

Mathematical Excalibur, Vol. 2, No. 3, May-June, 96 Page 2

Fermat's Little Theorem -.

-

(continued from page I )

m satisfying 2"' E 1 (mod p), thus 2"+'

divides p

-

I , or p is of the form k2"+'

+

25

I , hence to look for prime factors of 2

+

1 = 232

+

1, we should consider primes of the form 64k

+

1. The possible candidates are 193, 257, 449, 577, 641, Unfortunately, neither Fermat nor his contemporaries had enough patience to check that 641 indeed divides 2''

+

1. (For readers who are familiar with the law of quadratic reciprocity, one can prove that a prime divisor of 22" + 1 is actually of the form k2m'

+

1 .)

Fermat did not explicitly give any proof of the Fermat's little theorem, and it was Euler who first proved by induction the following fact: if p is a prime then

d

=

a (mod p). Clearly the statement is true if a = 1. Now

=

a

+

1 (modp),

where

(3

= '! E 0 (mod p) for

i ! ( p - i ) !

1 l i s p - 1

There is also another version of the theorem, namely, i f p is a prime and a is relatively prime to p , then

aP-'

E 1

(modp). Euler also gave the first proof by noting that the terms of the series 1, a,

a2,

.--

(modp) must repeat. So for some r 2 0, and some s 2 0, we must have a*' E

a' (mod p), i.e., d'

=

1 (modp). Let s be the smallest positive integer such that d"

=

I (mod p), then one can arrange the p- 1 non-zero congruence classes modulo p into sets { b , ba,

...

,bd"-'}, where each set consists of s elements and the sets are disjoint. Thus s must divide p-1

.

For example, with p = 7 and a = 2, one obtains s = 3 and the numbers 1 to 6 can be grouped into two disjoint sets { 1, 2,4} and {3,6,5). We also observe that p-1 = 6 is divisible by s = 3. Euler generalized this argument to prove the famous Euler's theorem:

If a is relatively prime to n, then a&")

=

1 (mod n),

where $(n) is the Euler totient function that counts the number of integers

between 1 and n that are relatively prime to n. For example, $( 12) = 4 since only 1,5,7, 11 (among the numbers 1-12) are relatively prime to 12.

A formal proof of Euler's theorem goes as follows: Let a be an integer relatively prime to n and let ( a , , u2,

...,

aHfl)} be the set of reduced residues modulo n (i.e., the +(n) positive integers less than n that are relatively prime to n). Then the set {aa,, u q ,

.

. ., aa+(,,)} is also a set of reduced residues modulo n. Hence,

ala2---a+,(,,)

=

a~(")ala~.--a~,) (mod n) or a'(") 3 1 (mod n).

There is however another colouring argument for Fermat's little theorem. Arrange p boxes in a circle and colour them with a colours. There are

d'

possible colouring patterns. Among all these possible colourings, a of them are

such that every box has the same colour. The remaining

d'

-

a colouring patterns

can be grouped into sets of p patterns that are rotations of each other. The p rotations of any one of these colourings are all distinct and thus p divides

d'

- a.

(Where did we use "p is prime"?) Hence,

in essence, the Fermat's little theorem can be proved using the pigeonhole principle.

The following are some applications of Fermat's little theorem and Euler's theorem.

Example 1: If n is an integer > 1, then n does not divide 2"

-

1 .

Solution: If n is even, then the statement is certainly true since 2"

-

1 is an odd integer. For n odd, denote by p the smallest prime divisor of n. Suppose n (and thus also p) divides 2"

-

1. By the Fermat's little theorem, p divides 2"-'

-

1 too. Consequently, p divides 2d

-

1, where d is the greatest common divisor of p

-

1 and n, Since p is the smallest prime divisor of n, d = 1 which leads to the contradiction p divides 1.

Example 2: Let n be an odd number not divisible by 5 , then n divides a number of the form 99..-9.

Solution: If n is odd and not divisible by

5, then n is relatively prime to 10. By the

Euler's theorem, I OH"'= 1 (mod n ) , i.e., n

divides lo+'"' - I , which is a number of the form 99.e-9.

Example 3: Let p be an odd prime number. Then for any set of 2p - 1 integers, there exists a set o f p integers whose sum is divisible by p .

Sketch of Solution: There are

n = -

(

:2

7

distinct sets that each contains p elements. Denote their sums by sI, s2,

.

.

.

, s,,. Suppose none of them is divisible by p. Then, by the Fermat's little theorem,

X S , ~ - '

=

5

1 = n , which is nonzero modulo p. On the other hand, one may use the multinomial expansion to show that $s,"-' is, in fact, divisible byp, and thus lead to a contradiction.

f l

r = l ,=1 1 = I

It is interesting to observe that we use a number theoretic approach to solve a combinatorial problem while using a counting argument to prove Fermat's little theorem.

We have mentioned that the converse of Fermat's little theorem is not true.* That is, there exists composite numbers n such that n divides a""

-

1 . For example, as stated at the beginning of this article, the composite number 341 divides 2340 - 1. Composite numbers n (which must be odd) that divides 2"-' - 1 are called pseudoprimes (in base 2). One may show that there exist infinitely many such pseudoprimes. In fact, if n is a pseudoprime, then m = 2" - 1 will be composite (since n is composite). Also, m - l = 2 " - 2 = n k a n d t h u s 2 " ' ~ ' - 1 =2"k

-

1 is divisible by 2"

-

1 = m. That is, m is another pseudoprime (in base 2).

We may of course try another base. For our example, we find that 341 is no longer a pseudoprime (in base 3), i.e., 341 does not divide 3340

-

1. Well, we may then ask: is it possible to find a composite number n such that for every

a relatively prime to n, a"-' E 1 (mod n).

Such a number is called a Carmichael number. Surprisingly, not only that they exist (with 56 1 being the smallest), there are infinitely many Carmichael numbers, which, in fact, was proved recently!

Mathematical Excalibur, Vo?. 2, No. 3, May-June, 96 Page 3

Problem Corner

We welcome readers to submit solutions to the problems posed below for publication consideration. Solutions should be preceded by the solver’s name, address, school affiliation and grade level. Please send submissions to Dr. Kin-Yin Li, Dept of Mathematics, Hong Kong Wniversiv of Science and Technology, Clear Water Bay, Kowloon. The deadline for submitting solutions is July 10, 1996.

The following problems are selected from the International Mathematics Tournament of the Towns, held in

April 7, 1996.

Problem 36. Let a, b and c be positive numbers such that a2 + b2

-

ab = c2. Prove that (a-c)(b-c) I 0.

Problem 37. Two non-intersecting circles A , and A2 have centres O1 and 0 2

respectively. A l and A 2 are points on A1 and A2 respectively, such that A l A 2 is an external common tangent of the circles. The segment 0102 intersects A I and A2 at B I and B2 respectively. The lines A I B l and A2Bz intersect at C, and the line through C perpendicular to BIB2

intersects A l A 2 at D. Prove that D is the midpoint of A,A2.

Problem 38. Prove that from any sequence of 1996 real numbers, one can choose a block of consecutive terms whose sum differs fiom an integer by at most 0.001.

Problem 39. Eight students took part in a contest with eight problems.

(a) Each problem was solved by 5 students. Prove that there were two students who between them solved all eight problems.

(b) Prove that this is not necessarily the case if 5 is replaced by 4. (A counterexample is enough.)

Problem 40. ABC is an equilateral triangle. For a positive integer n 2 2,

D

is the point on A B such that A D = A B . P I , Pz,

..-,

P,., are points on BC which divide it into n equal segments. Prove that L A P I D + L A P z D

+

. * - + LAPn-ID

= 300

[Hint: Consider Q, such that ADP,Q, is a parallelogram.]

*****************

Solutions

*****************

Problem 31. Show that for any three given odd integers, there is an odd integer such that the sum of the squares

of these four integers is also a square. Solution: Independent solution by William CHEUNG Pok-man (S.T.F.A. Leung Kau Kui College, Form 5), Gary NG Ka Wing (S.T.F.A. Leung Kau Kui College, Form 3), Henry NG Ka Man (S.T.F.A. Leung Kau Kui College, Form 5) and PA1 Hung Ming Tedward (S.K.H. Tang Shiu Kin Secondary School, Form 6 ) .

L e t x = 2 a + l , y = 2 6 + l , z = 2 c + 1 be three given odd integers, then x2 + y 2 + z2

= 2w

+

1, where w = 2(a2 + a

+

b2 t b t c 2 + c ) + 1 isodd. S o x 2 + y 2 + z 2 + w 2 = (w + l)2.

Other commended solver: CHAN Wing Chiu (La Salle College, Form 3), CHENG Wing Kin (S.K.H. Lam Woo Secondary School, Form 4), Calvin CHEUNG Cheuk Lun (S.T.F.A. Leung Kau Kui College, Form 4), W. H. FOK (Homantin Government Secondary School), Alan LEUNG Wing Lun (S.T.F.A. Leung Kau Kui College, Form

4), LIU Wai Kwong (Pui Tak

Canossian College), POON Wing Chi (La Salle College) and YAU Kwan Kiu (Queen’s College, Form 7).

Problem 32. Let a. = 1996 and a,+[ =

an2/(an

+

1) for n = 0, 1,2, .

. .

. Prove that

[a,] = 1996 - n for n = 0, 1,2,

...,

999, where

[XI

is the greatest integer less than or equal to x.

Solution: Independent solution by CHAN Wing Sum (HKUST), W. H. FOK (Homantin Government Secondary School) and KU Yuk Lun (HKUST).

Note that a, > 0 implies anTl > 0 and

> O .

1

a, -a,,+] = 1

- -

a,, + 1

Hence a0 > a l > a2 > ..*. Now a, = a.

+

( a l

-

ao) + ..* + (a, - a,-l)

1 1 = 1996

-

n +-+

+-

a0

+

1 an-l + 1 > 1996-n. For 1 I n I 999, 1 1 n -+

...

+- <- ag

+

1 a,-]

+

1 a,-]

+

1 = 1. 999 999 <-< a998

+

1 1996 - 998

+

1 So [a,] = 1996 - n.

Comments: With 1996 replaced by 1994, 999 replaced by 998, this was a problem proposed by USA in the 1994 IMO.

Other commended solver: William CHEUNG Pok-man (S.T.F.A. Leung Kau Kui College, Form 5), Henry NG Ka Man (S.T.F.A. Leung Kau Kui College, Form 5 ) , POON Wing Chi (La Salle College) and YAU Kwan Kiu (Queen‘s College, Form 7).

Problem 33. Let A , B, C be noncollinear points. Prove that there is a unique point X in the plane of ABC such that XA2

+

XA2

+

CA2. (A problem proposed by Germany in the last IMO.)

xg2 + A B ~ = X B ~

+

xC2

-t

gc2

=

xc2

+

Solution: Henry NG Ka Man

(S.T.F.A. Leung Kau Kui College, Form 5 ) .

Without loss of generality, we may ~

assume A, B, C have coordinates (a,O),

(b,O), (O,c), (where a#b and cf0) respectively. Let

X

be a point in the plane ofABC with coordinates (xy). For

X

to satisfy the given conditions, the equations on x and y are M

-

cy = a’

-

c2

-

ab, bx - cy = b2

-

c2

-

ab andx = a + b

(after simplification), which has a unique solution (xy) = (a+b, c+2ab/c). Other commended solvers: Calvin CHEUNG Cheuk Lun (S.T.F.A. Leung Kau Kui College, Form 4), William CHEUNG Pok-man (S.T.F.A. Leung Kau Kui College, Form 5 ) , W. H. FOK (Homantin Government Secondary School), Alan LEUNG Wing Lun (S.T.F.A. Leung Kau Kui College, Form

4), LIU Wai Kwong (Pui Tak

Canossian College) and Gary NG Ka Wing (S.T.F.A. Leung Kau Kui College, Form 3).

Problem 34. Let n > 2 be an integer, c be a nonzero real number and z be a nonreal (continued on page 4 )

-

Mathematical Excalibur, VoI. 2, No. 3, May-June, 96 Page 4

Problem Corner

(continuedfrom page 3) root ofX"

+

c X + 1. Show that

Solution 1: W. H. FOK (Homantin Government Secondary School). Write z = r(cose

+

isine) with sine f 0.

Taking the real and imaginary parts of

z n + cz

+

1 = 0 using De Moivre's theorem, we have r"cosn0

+

crcose

+

1 = 0 r"sinn8

+

crsine = 0. and Then

r"sin(n-1)O = Ssinnecose

-

r"cosn8sine

= -crsinecosQ

+

(crcos0 + 1)sine

= sine. Since

Isin(k+l)BI = lsinkecose

+

coskesinel I lsinkel

+

Isinel,

induction gives Isinkel I klsinel for every positive integer k. So

lzl" = r" = IsinO/sin(n-1)81 L ll(n-I). Solution 2: LEUNG Hoi-Ming (SKH Lui Ming Choi Secondary School). Let r = (zI and w = z/r. Then Iw1= 1 and

w

-

= 1 . Since (rw)"

+

crw

+

1 = 0, multiplying by W , then conjugating, we get

+ c r + E = O rnWn-l

and r"W"-' + c r

+

w = 0 .

Subtracting these equations and solving for r", we get

Since r is real and IwI = 1, by the triangle inequality,

I =o

Other commended solvers: William CHEUNG Pok-man (S.T.F.A. Leung Kau Kui College, Form 5 ) .

Problem 35. On a blackboard, nine 0's

and one 1 are written. If any two of the numbers on the board may both be

replaced by their average in one operation, what is the least positive number that can appear on the board after a finite number of such operations? Solution: POON Wing Chi (La Salle College).

Let m be the least positive number on the board and n be the number of zeros on the board after an operation. Consider the number c = m12". If two positive numbers are both replaced by their average, then n does not change, but m (and c) may increase. If a 0 is averaged with a positive number r, then n decreases by one and m remains unchanged or becomes r12 (2 m12.) The new c value will be greater than or equal to (m12)12"-' = m/2", which is the old c value. In the beginning, c = 11512. After a finite number of operations, c 2 1 6 1 2 and m 2 2"/512 2 11512. To obtain exactly 11512, start with 1 and average with each of the nine 0's.

Comments: This problem comes from an article in the MarchIApril 1994 issue of Quanrum, published by Springer Verlag. The article dealt with the concept of monoinvariant, which is an expression like c in the problem that increases after each operation. Studying such expression often solves the problem.

Olympiad Corner

(continuedfi.om page 1 )

(ii)

I

a,

-

a,+'

I

I 2, i = I, ---, n-I. Determine whether

A

1996) is divisible by 3.

Problem 4. In AABC, AB = AC.

Suppose that the bisector of L B meets

AC at D and that BC = BD

+

AD. Determine LA.

Problem 5. Let r l , r2, . .., r,,, be m given positive rational numbers such that

m

k = l

crk

= 1.

Define the hnction f by

for each positive integer n. Determine the minimum and maximum values of

An).

From the Editors' Desk:

Thanks to our readers for another year of support, especially the submission of articles and problem solutions. If you would like to receive your personal copy for the five issues for the 96-97 academic year, send five stamped self- addressed envelopes to Dr. Kin-Yin Li, Hong Kong University of Science and Mathematics, Clear Water Bay, Kowloon, Hong Kong.

Technology, Department of

*****************

APMO and IMO: The Eighth APMO took place on March 16th. The Hong Kong students had a very strong (record setting) performance. The top 8 scorers are as follow. (Note the maximum is

7x5=35 points.)

1. %@?t$l, (Bobby POON Wai Hoi), St. Paul's College, 35 points (Perfect score! First time for Hong Kong) 2.

&E@

(YU Chun Ling), Ying Wa

College, 33 points

3.m$%% (HO Wing Yip), Clementi Secondary School, 32 points

4.

SF@)

(MOK Tsz Tao), Queen's College, 3 1 points

5 . %*3#f (TSE Shan Shan), Tuen Mun Government Secondary Schod, 29 points

6.

RZ@$%

(LAW Siu Lung), Diocesan Boy's School, 26 points

7.

$$is)&

(YUNG Hon Wai), Heep Woh College, 26 points

8.

!&x@

(CHU Tim Kin), King's College, 24 points

The first 6 students are invited to be the Hong Kong team members to participate in the 37th International Mathematical Olympiad to be held in India this summer. The selection was based on their outstanding performance in the APMO and throughout the Hong Kong Math Olympiad training program.