Volume 22, Number 1 July 2018 – October 2018
Olympiad Corner
Below were the problems of the Balkan Mathematical Olympiad which took place in Belgrade, Serbia on May 9, 2018.
Time allowed was 270 minutes. Each problem was worth 10 points
Problem 1. A quadrilateral ABCD is
inscribed in a circle k, where AB > CD and AB is not parallel to CD. Point M is the intersection of the diagonals AC and BD and the perpendicular from M to AB intersects the segment AB at the point E. If EM bisects the angle CED, prove that AB is a diameter of the circle k. (Bulgaria)
Problem 2. Let q be a positive rational
number. Two ants are initially at the same point X in the plane. In the n-th minute (n=1,2,…) each of them chooses whether to walk due north, east, south or west and then walks the distance of qn metres. After a whole number of minutes, they are at the same point in the plane (not necessarily X), but have not taken exactly the same route within that time. Determine all possible values of q. (United Kingdom) (continued on page 4)
Miscellaneous Inequalities
Kin Y. Li
There are many kinds of inequality problems in mathematical Olympiad competitions. Some of these can be solved by applying certain powerful inequalities such as rearrangement or majorization or Muirhead’s inequalities. Some can be solved by techniques like tangent line methods using a bit of differential calculus.In this article, we will be looking at some inequality problems that are not solved by these kinds of powerful tools and techniques.
Example 1. (1983 IMO Shortlisted Problem proposed by Finland) Let p and q be integers with q>0. Show that there exists an interval I of length 1/q and a polynomial P with integral coefficients such that
2 1 ) ( q q p x P for all x∈ I. Solution. Pick P(x) = p((qx−1)2n+1+1)/q
and I = [1/(2q), 3/(2q)]. Then all the coefficients of P are integers and
2 1 2 1 2 1 1 ) ( n n q p qx q p q p x Pfor all x∈ I. Choose n large so that 22n+1
> |pq|. Then we are done.
Example 2 (1994 IMO) Let m and n be positive integers. The set A={a1, a2, … ,
am} is a subset of 1,2,…,n. Whenever ai+aj≤n, 1≤i≤j≤m, ai+aj also belong to A. Prove that . 2 1 2 1 n m a a a m
Solution. We may assume that a1>a2>
⋯ > am. We claim that for i=1,2,…,m, ai+am+1−i≥n+1. (*)
If not, then ai+am+1−i, …, ai+am−1, ai+am are i different elements of A greater than ai, which is impossible. By adding the cases i=1,2,…,m of (*), we get
2(a1+⋯+am) ≥ m(n+1). The result follows.
Example 3 (2001 IMO Shortlisted Problem proposed by Bulgaria). Find all positive integers a1, a2, …, an such that , 100 99 1 2 1 1 0 n n a a a a a a
where a0=1 and (ak+1−1)ak-1≥ak2(ak-1) for k=1,2,…,n-1.
Solution. Let a1, a2, … , an satisfy the conditions of the problem. Then ak >ak-1
and hence ak≥2 for k=1,2,…,n. The inequality (ak+1−1)ak-1≥ak2(ak-1) can be rewritten as . 1 1 1 1 1 k k k k k k a a a a a a
Adding these inequalities for k = i+1,…, n-1 and using an-1/an < an-1/(an-1), we obtain . 1 1 1 1 i i n n i i a a a a a a Then 1 100 99 1 1 1 0 1 i i i i i i a a a a a a a a (*)
for i=1,2,⋯,n-1. Now given a0,a1,…, ai, there is at most one possibility for ai+1. By (*), this yields a1=2, a2=5, a3=56,
a4=78400. These values satisfy the
condition of the problem. So this is a unique solution.
Example 4 (1999 Polish Math Olympiad). Let a1, a2, …, an, b1, b2, …,
bn be integers. Prove that
n j i i j n j i j i j i a b b a b a 1 1 . | | |) | | (| (continued on page 2) Editors: 高 子 眉 (KO Tsz-Mei)梁 達 榮 (LEUNG Tat-Wing)
李 健 賢 (LI Kin-Yin), Dept. of Math., HKUST
吳 鏡 波 (NG Keng-Po Roger), ITC, HKPU Artist: 楊 秀 英 (YEUNG Sau-Ying Camille), MFA, CU Acknowledgment: Thanks to Sindy Ting, Math. Dept., HKUST for general assistance.
On-line: http://www.math.ust.hk/excalibur/
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 December 1, 2018.
For individual subscription for the next five issues for the 17-18 academic year, send us five stamped self-addressed envelopes. Send all correspondence to:
Dr. Kin-Yin LI, Math Dept., Hong Kong Univ. of Science and Technology, Clear Water Bay, Kowloon, Hong Kong
Fax: (852) 2358 1643 Email: [email protected]
© Department of Mathematics, The Hong Kong University of Science and Technology
Mathematical Excalibur, Vol. 22, No. 1, Jul. 18 – Oct. 18 Page 2 Solution. For integer x, let f{a,b}(x)=1 if
either a≤x<b or b≤x<a and f{a,b}(x)=0
otherwise. Observe that when a,b are integers, |a-b| equals the sum of f{a,b}(x)
over all integers x. Now fix an integer x and suppose a≤ is the number of values
of i for which ai≤x.
Define a>, b≤, b< analogously. We have
(a≤-b≤) + (a>-b>)
= (a≤+a<) – (b≤+b>)
= n – n = 0,
which implies (a≤-b≤)(a>-b>)≤0.Thus
a≤a> + b≤b>≤ a≤b> + a> b≤. Now
n j i a b x f a a i j 1 { , } ). (because both sides count the same set of pairs and the other terms reduce similarly, yielding
n j i b a n j i b b a a x f x f x f i j i j i j 1 } , { 1 } , { } , { ( ) ( ) ( ).Because x was an arbitrary integer, this last inequality holds for all integers x. Summing over all integers x and using our first observation, we get the desired inequality. Equality holds if and only if the above inequality is an inequality for all x, which is true precisely when the ai equal the bi in some order.
Example 5 (2007 Chinese Math Olympiad). Let a,b,c be complex numbers. Let |a+b|=m, |a-b|=n and mn≠0. Prove that . |} | |, max{| 2 2 n m mn bc a b ac Solution. Since ) | | | (| 2 | | | | | | | | | | | | | | | ) ( ) ( | | | | | | | | | | | | | max{| |,| |} 2 2 2 2 b a a b a b a b a b b a bc a a b ac b a b bc a a b ac b ac b a bc
and m2+n2=|a-b|2+|a+b|2=2(|a|2+|b|2),
so . |} | |, max{| 2 2 n m mn bc a b ac
Example 6 (1999 Balkan Math Olympiad). Let x0,x1,x2,… be a non-
decreasing sequence of nonnegative integers such that for every k≥0, the number of terms of the sequence which
are less than or equal to k is finite; let this number be yk. Prove that for all positive integers m and n, ). 1 )( 1 ( 0 0
m n y x m j j n i iSolution. Under the given construction, ys≤t if and only if xt>s. Thus the sequences x0,x1,x2,… and y0,y1,y2,… are
dual, meaning that applying the given algorithm to y0,y1,y2,… will restore the
original x0,x1,x2,….
To find x0+x1+⋯+xn, observe that among the numbers x0,x1,⋯,xn , there are exactly y0 terms equal to 0, y1-y0 terms equal to
1, … and 2 1 n n x x y y terms equal to xn-1,
while the remaining n+1-xn-1 terms equal
to xn. Hence, x0+x1+⋯+ xn equals . ) 1 ( ) 1 ( ) ( 1 1 0 1 1 1 1 n x x n x i i i x n y y y y n x y y i n n n
First suppose that xn-1≥m. Write xn-1=m+k for k≥0. Because xn>m+k, from our initial observations we have ym+k≤n. Then n+1 ≥ ym+k ≥ ym+k-1 ≥ ⋯ ≥ ym. So ). 1 )( 1 ( ) 1 ( ) 1 )( 1 ( ) 1 ( ) 1 ( 1 0 0 0 0 1
m n n k k m n y x n y y x n y x k m m i i n x j m i i j n m j j n i i nNext suppose that xn-1<m. Then xn≤m implies ym>n, which implies ym-1≥n. Because x0,x1,x2,… and y0,y1,y2,… are
dual, we may apply the same argument with the roles of the two sequences reversed. This completes the proof. Example 7 (2007 Chinese Girls’ Math Olympiad). Let m,n be integers, m>n≥2, S={1,2,…,m} and T={a1,a2,…,an} be a subset of S. Suppose every two elements of T are not both the divisors of any element of S. Prove that
. 1 1 1 2 1 m n m a a a n
Solution. For i=1,2,…,n, let ki be the integer such that ki≤m/ai<ki+1. Let Ti = {kai : k = 1,…, ki}.Then |Ti| = ki Since every two elements of T are not both the divisors of any element of S, so if i≠i’, then Ti∩Ti’ is empty. Hence,
. | | | | | | 1 1 m S T T k n i i n i i
Since m/ai < ki +1, we have . ) 1 ( / 1 1 1 n m k a m n i i n i i
Dividing by m, we get the desired conclusion.
Example 8 (1987 IMO Shortlisted Problem proposed by Netherland). Given five real numbers u0, u1, u2, u3,
u4, prove that it is always possible to
find five real numbers v0, v1, v2, v3, v4
that satisfy the following conditions: (i) ui-vi∈ℕ.
(ii) ∑0≤i<j≤4 (vi-vj)2 < 4. Solution. Observe that
4 0 2 4 0 2 4 0 2 4 0 2 4 0 2 . ) ( 5 ) ( ) ( 5 )] ( ) [( ) ( i i i i i i j i j i j i j i v v v v v v v v v v v vLet us take vi’s satisfying the last line with v0≤ v1≤ v2≤ v3≤ v4≤1+v0. Define
v5=1+v0. We see that one of the
differences vi+1-vi, i=0,…,4, is at most 1/5. Let v=(vi+1+vi)/2. Then place the other three vi’s in [v-1/2,v+1/2]. Now we have |v-vi|≤1/10, |v-vi+1|≤1/10 and |v-vk|≤1/2 for any k other than i and i+1. Finally, we have
4 0 2 2 2 5(2(1/10) 3(1/2) ) 4. ) ( j i j i v vExample 9 (2000 Romanian Math Olympiad). Let n≥1 be an odd positive integer and x1, x2, …, xn be real numbers such that |xk+1-xk|≤1 for k=1,2,…,n-1. Show that
n k n k k k n x x 1 2 1 . 4 1 | |Solution. Let P, N be the sets of positive, negative numbers among x1, x2, …, xn respectively. Without loss of generality, assume that there are more k such that xk is negative than there are k such that xk is positive. Let (a1,…,an) be a permutation of (x1,…,xn) such that a1,…,an is a nondecreasing sequence. By construction, |P|≤(n-1)/2.
Problem Corner
We welcome readers to submit their solutions to the problems posed below for publication consideration. The solutions should be preceded by the solver’s name, home (or email) address and school affiliation. Please send submissions to Dr. Kin Y. Li, Department of Mathematics, The Hong Kong University of Science & Technology, Clear Water Bay, Kowloon, Hong Kong. The deadline for sending solutions is December 1, 2018.
Problem 521. Given 20 points in space so that no three of them are collinear, prove that the number of planes determined by these points is not equal to 1111.
Problem 522. Determine all functions f:ℝ→ℝ such that for all real x and y, (x-2) f(y) + f (y + 2 f(x)) = f (x + y f(x)).
Problem 523. Find all positive integers n for which there exists a polynomial P(x) with integer coefficients such that P(d) = (n/d)2 for each positive divisor d
of n.
Problem 524. (proposed by Andrew WU, St. Albans School, Mc Lean, VA, USA) In ∆ABC with centroid G, M and N are the midpoints of AB and AC, and the tangents from M and N to the circumcircle of ∆AMN meet BC at R and S, respectively. Point X lies on side BC satisfying ∠CAG =∠BAX. Show that GX is the radical axis of the circumcircles of ∆BMS and ∆CNR.
Problem 525. Find all positive integer n such that n(n+2)(n+4) has at most 15 positive divisors.
*****************
Solutions
****************
Problem 516. Determine all triples (p,m,n) of positive integers such that p is prime and 2mp2+1=n5 holds.
Solution. CHUI Tsz Fung (Ma Tau Chung Government Primary School) and ZHANG Yupei (HKUST).
Let q=n4+n3+n2+n+1. Then 2mp2 =
(n-1)q and gcd(n-1,q)=gcd(n-1,5) = 1 or 5. Now q>1 is odd and so p is an odd prime. Let p=2k+1. Then gcd(2m,p2)=1.
So n-1=2m, q=p2. Then n=2m+1. So n4+n3+n2+n=p2-1 can be expressed as
(22m+2m+1+2)(22m+3·2m+2)=4k(k+1).
If m≥2, then the left side is 4 (mod 8) and the right side is 0 (mod 8). Hence, m=1. Then p=11 and n=3. So (p,m,n)=(11,1,3) only.
Other commended solvers: Ioan Viorel
CODREANU (Satulung, Maramures,
Romania), Akash Singha ROY (West Bengal, India), Ioannis D. SFIKAS (Athens, Greece), Toshihiro SHIMIZU (Kawasaki, Japan) and Nicuşor ZLOTA (“Traian Vuia” Technical College, Focşani, Romania).
Problem 517. For all positive x and y, prove that
x2y2(x2+y2-2) ≥ (xy-1)(x+y).
Solution. CHUI Tsz Fung (Ma Tau Chung Government Primary School). Let k=xy. We have
0 ) 1 2 2 ( ) 1 ( 1 2 2 2 2 2 2 2 k k k k k k k k k k k Since 2 k 2 xyxy, so y x xy y x y x y x 2 2 2 2 2 . 1 1 2 2 2 2 2 2 2 y x xy k k k k k Then x2y2(x2+y2-2) ≥ (xy-1)(x+y).
Other commended solvers: LEUNG Hei
Chun, Paolo PERFETTI (Math Dept,
Università degli studi di Tor Vergata Roma, via della ricerca scientifica, Roma, Italy), Ioannis D. SFIKAS (Athens, Greece), Nicuşor ZLOTA (“Traian Vuia” Technical College, Focşani, Romania).
Problem 518. Let I be the incenter and AD be a diameter of the circumcircle of ∆ABC. Let point E be on the ray BA and point F be on the ray CA. If the lengths of BE and CF are both equal to the semiperimeter of ∆ABC, then prove that lines EF and DI are perpendicular.
Solution. ZHANG Yupei (HKUST).
S J O B C E F I L A D R K
Let circle ABC intersect line DI at S. Let K, J, L be the feet of the perpendiculars from I to sides AC, CB, BA of ∆ABC respectively. Since AD is a diameter of the circumcircle of ∆ABC, we get ∠ASD =∠AKI =∠ALI = 90º. So A,S,K,I,L are concyclic. Next, ∠BLS = 180º - ∠ALS = 180º - ∠AKS =∠CKS and ∠LBS =∠ KCS. So ∆BLS, ∆CKS are similar. Since BE=CF, AF/AE = BL/CK = SB/SC. We get ∠EAF = ∠CAB =∠CSB. So ∆EAF ≅ ∆CSB. Then ∠SBC =∠SAC =∠EFA. We get EF || AS. Then DI ⊥ EF.
Other commended solvers: Andrea
FANCHINI (Cantù, Italy), William
KAHN (Sidney, Australia), Akash
Singha ROY (West Bengal, India),
Ioannis D. SFIKAS (Athens, Greece), and Toshihiro SHIMIZU (Kawasaki, Japan).
Problem 519. Let A and B be subsets of the positive integers with 10 and 9 elements respectively. Suppose for every x,y,u,v∈A satisfying x+y=u+v, we have {x,y}={u,v}. Prove that the set A+B={a+b: a∈A, b∈B} has at least 50 elements.
Solution. CHUI Tsz Fung (Ma Tau Chung Government Primary School). If a1, a2∈A and b1, b2∈B such that
a1+b1=a2+b2, then a1-a2=b2-b1 (with
a1≠a2 and b1≠b2). Assume the
equation x+b1=y+b2 has two distinct
solutions (x,y) = (a3, a4) and (a5, a6)
such that a3, a4, a5, a6 ∈A. Then we
have a3-a4 = b2-b1 = a5-a6, which
implies a3+a6=a4+a5. By the condition
of A, we have {a3, a6}={a4, a5}. Then
we have 2 cases.
Case 1: a3=a4 and a5=a6. From
a3+b1=a4+b2, we get b1=b2. Then
|a3-a4| + |b1-b2|=0, contradiction.
Case 2: a3=a5 and a4=a6.Then (a3,
a4)=(a5, a6), contradiction.
So x+b1=y+b2 has at most one solution.
Since there are 36 choices of b1≠b2∈B,
so there must be 36 solutions of (a1, a2,
b1, b2) such that a1≠a2∈A, b1≠b2∈B
and a1+b1=a2+b2.
However, we have a1+b1, a2+b2∈A+B.
Since A+B has 90 not necessary distinct elements, so A+B has at least 54 distinct elements. In particular, A+B has at least 50 distinct elements.
Mathematical Excalibur, Vol. 22, No. 1, Jul. 18 – Oct. 18 Page 4 Other commended solvers: William
KAHN (Sidney, Australia), Akash
Singha ROY (West Bengal, India),
George SHEN, Toshihiro SHIMIZU
(Kawasaki, Japan) and ZHANG
Yupei (HKUST).
Problem 520. Let P be the set of all polynomials f(x)=ax2+bx,where a,b are
nonnegative integers less than 201018.
Find the number of polynomials f in P for which there is a polynomial g in P such that g(f(k))≡k (mod 201018) for all
integers k.
Solution. William KAHN (Sidney, Australia) and George SHEN.
We will show that there exists Q(x) = cx2+dx for P(x) = ax2+bx if and only if
2810059|a and gcd(2010,b)=1. Then it
follows that the answer is 2·20109·
201018(1-1/2)(1-1/3)(1-1/5)(1-1/67)
= 253·11·201026.
Assume that Q(P(n))≡n (mod 201018)
for all n. Then n→P(n) is one-to-one (mod 201018) and using the Chinese
remainder theorem we deduce that n→P(n) is one-to-one (mod p18) for p
in {2,3,5,67}.
Let p ∈{2,3,5,67}. If p|b, then P(p17) ≡
P(0) (mod p18) gives a contradiction.
Hence, p∤ b. If p∤a, then P(-a−1b) ≡
P(0) (mod p18) gives a contradiction.
So p | a. Hence 2010 | a and gcd(2010,b) = 1. In particular, (b(a2-b2))−1 (mod
201018) exists. Since
Q(P(1)) ≡ 1 (mod 201018)
⇒ c(a+b)2+d(a+b)≡1 (mod 201018)
⇒ 2b(a2-b2)c ≡2a (mod 201018)
and
Q(P(-1)) ≡ -1 (mod 201018)
⇒ c(a-b)2+d(a-b) ≡ -1 (mod 201018)
⇒ 2b(a2-b2)d≡-2(a2+b2) (mod 201018)
we have
c≡(b(a2-b2))−1a+201018e (mod 201018)
and d ≡ -(b(a2-b2))−1(a2+b2) + 201018e (mod 201018), where e = 0 or ½. Therefore, Q(P(x))-x ≡ -(b(a2-b2))−1a2x(x-1)(x+1)(ax+b) +201018ex(x-1) ≡ -(b(a2-b2))−1a2x(x-1)(x+1)(ax+2b) (mod 201018).
Now if x=2, we get 201018 | 223a2, hence
2810059 | a.
Conversely, if 2810059 | a and gcd(2010,b)
=1, then we can define c and d as above. Since 2 | n(n-1) and 2 | an+2b for all n, Q(P(n))≡n (mod 201018) follows.
Other commended solvers: Toshihiro
SHIMIZU (Kawasaki, Japan) and
ZHANG Yupei (HKUST).
Olympiad Corner
(Continued from page 1)
Problem 3. Alice and Bob play the
following game: They start with two non-empty piles of coins. Taking turns, with Alice playing first, each player choose a pile with an even number of coins and moves half of the coins of this pile to the other piles. The game ends if a player cannot move, in which case the other player wins. (Cyprus)
Problem 4. Find all primes p and q such that 3pq-1+1 divides 11p+17p. (Bulgaria)
Miscellaneous Inequalities
(Continued from page 2)
Suppose that 1≤i≤n-1. In the sequence
x1,…,xn, there must be two adjacent terms xk and xk+1 which are separated by the interval (ai, ai+1), i.e. such that either xk≤ai≤ai+1≤xk+1 or xk+1≤ai≤ai+1≤xk. So ai+1-ai≤|xk-xk+1| ≤1. That is a1,…,an is a nondecreasing sequence of terms, such that any two adjacent terms differ by at most 1.
Let σP denote the sum of the numbers in P. We claim that σP≤(n2-1)/8. This is certainly true if P is empty.
If P is nonempty, then the elements of P are ai≤ai+1≤⋯≤an for some 2≤i≤n. Because ai-1≤0 by assumption and ai≤ ai-1+1 from the previous paragraph, we
have ai≤1.
Similarly, ai+1≤ ai+1≤2 and so on up to an≤|P|. Hence, σP≤1+2+⋯+|P|. From |P|≤(n-1)/2, we get σP≤(n2-1)/8, as claimed.
Let σN denote the sum of the numbers in N. The left-hand side of the required inequality then equals
4 1 8 1 2 | 2 | | | | | 2 2 n n P N P N P as needed.
Example 10 (2000 Asia Pacific Math Olympiad). Let n, k be positive integers with n > k. Prove that
. ) ( )! ( ! ! ) ( 1 1 k n k n k n k n k n k n k n k n k n k n n
Solution. By the binomial theorem, we have nn=(k+(n-k))n=a 0+⋯ +an, where for i=0,1,…,n, . 0 ) ( i ni i k n k i n a We claim that . 1 n i n n a n n
The right inequality holds because nn = a0+⋯+an>ai. To prove the left inequality, it suffices to prove that ai is larger than a0,…, ai-1, ai+1, …, an because then
n m i i n m m n a a n a n 0 0 . ) 1 (Next, we will show ai is increasing for i≤k and decreasing for i≥k. Observe that . 1 1 i n i n i i n Hence . 1 ) ( 1 ) ( 1 1 1 k i i n k n k n k i n k n k i n a a i n i i n i i i
This expression is less than 1 when i<k and it is greater than 1 when i≥k. In other words, a0<⋯<ak and ak>⋯>an as desired.
1. Denote by N the intersection of the lines AD and BC and by P the intersection of the lines AB and CD. The line MN is the polar of point P with respect to circle
k, so MN .1 OP, where O is the center of circle k.
Let the lines EM and CD meet at Kand the lines MN and AB meet at E'. Since
EK and EP are the interior and exterior bisector of angle CED, the quadruple
(C,D;P,K) is harmonic. By projecting from M we get that (A,B;P,E) is also a
harmonic quadruple. On the other hand, it follows from the complete quadrilateral
ABCD that the quadruple (A, B; P, E') N
is harmonic, so we must have E'
=
Eand hence MN ..L AB. Thus O lies on the line AB, i.e. AB is a diameter of circle k.
Second solution. Suppose that AB is not a diameter and take points A' on the ray
CA and B' on the ray DB such that
<r.A'DB
=
<r.B'CA=
90°.~ E ' I -..,,.,B'
ThequadrilateralA'B'CD is cyclic, so <r.B' A'C = <r.B'DC = <r.BAC, which implies that A' B'
II
AB and ME ..L A' B'. Let the line ME meet A' B' at point E'. From the cyclic quadrilaterals A' E' MD and B' E' MC we obtain <r.D E' E=
<r.D A' M =<r.CB'M
=
<r.CE'E, which together with <r.DEE' = <r.CEE' implies that the triangles GEE' and DEE' are congruent and hence CE=
DE. Finally, the bisector EM of angle CED is perpendicular to CD, so ABII
CD, contrary to the problem condition.2. Denote by a,. the sum of x-and y-coordinates of the first ant, and by bn that of the second ant, after n minutes. Then we have
Ian -
an-11=
jb,. - bn-11=
q", soa,.= e:1q + e:2q2 + · · · + €nqn. and bn
=
1]1q + 1]2q2 + · · · + T/nq" for some coeffciients€i, 1/i E {-1, 1}. If the ants meet after n minutes, then
O _ £in - bn _ €1 - 'T/1 + €2 -1J2 2 + + €n - 'T/n n _ P( )
- 2 - --2-q --2-q . . . 2 q - q '
where the polynomial P has all coefficients Ei
2
!1• in the set { -1, 0, 1}. Therefore, if q=
%
(a, b EN), then aI
1 and bI
1, so we must have q=
1.The value q
=
1 is obviously possible, e.g. if the first ant goes east and west, and the second one goes north and south., '
Second solution. Consider the positions O!k and f3k of the ants in the complex plane after k steps, assuming that their starting point is at the origin and the steps are parallel to the axes. We know that O:k - O:k-1
=
akqk and f3k - f3k-I=
bkqk, where ak, bk E {l, -1, i, -i}. So, if O:n=
f3n for some n > 0, thenn
~(ak - bk)l
=
0, where ak - bk E {O, ±1 ± i, ±2, ±2i}.k=l
Observe that the coefficient ak - bk is always divisible by 1 + i in the Gaussian integers Z[i]: indeed,
Ck=
a1
~!k
E {0,±1,±i,±l ±i}.Upon cancelling 1 + i we obtain c1q + c2q2 + · · · + c,.q"
=
0. Assuming withoutloss of generality that c1, Cn =/= 0, we know that if q
=
%
(a, b EN) then aI
c1 andb
I
Cn in Z[i], which is possible only if a= b=
1.3. By v2 ( n) we denote the largest nonnegative integer r such that 2r
I
n.A position (a,b) (i.e. two piles of sizes a and b) is said to be k-happy if v2(a)
=
·112(1,) = k for some integer k ;;., 0, and k-unhappy if min{·112(a), 112(b)}
=
k <max{v2(a),v2(b)}. We shall prove that Bob has a winning strategy if and only if the initial position is k-happy for some even k.
• Given a 0-happy position, the play.er in turn is unable to play and loses. • Given a k-happy position (a, b) with k;;., 1, the player in turn will transform
it into one of the positions (a+ !b, !b) and (b + !a, !a), both of which are
(k -1)-bappy because v2(a +kb)= v2(kb)
=
v2(b +!a)= v2(!a) = k -1. Therefore, if the starting position is k-happy, after k moves they will get stuck at a 0-happy position, so Bob will win if and only if k is even.• Given a k-unhappy position (a, b) with k odd and ·112(a) = k < ·112(b)
=
f.,Alice can move to position (!a, b + !a). Since v2(!a)
=
v2(b +fa)= k - 1, this position is (k -1)-happy with21
k - 1, so Alice will win.• Given a k-unhappy position (a,b) with k even and v2(a) = k < v2(b) = f.,
Alice must not play to position (fa, b + fa), because the new position is (k-1)-happy and will lead to Bob's victory. Thus she must play to position (a+ }b,
}b),
which is also k-unhappy. Indeed, if f. > k+ 1, then v2(a+ }11)=
k
<
v2(2b) = £ - 1, whereas if£=
k + 1, then v2(a + !b)>
v2(!b) = k.Similarly, Bob will have to play to another k-unhappy position and so on. Hence a k-unhappy position is winning for Alice if 2
f
k, and drawing if 2I
k.4. The only solution is (p, q)
=
(3, 3).It is directly verified that there are no solut,ions with p
=
2. From now on we assume that p > 2.Since N
=
1IP+
17P=
4 (mod 8), we see that 8f
3pq-l+
1 > 4. Consider an odd prime divisor r of 3pq-l+
1. Clearly, r\t
{3, 11, 17}. If b E Z is such that 17b=
-1 (mod r), then rI
bPN=
aP - I (mod r), where a= llb. Hence the order of a modulo r divides p, i.e. ordr(a) E {l,p}. In the case ordr(a)=
1, we haverI
a - I= (11+
l 7)b (mod r), so the only possibility is r=
7. On the other hand, if ordr(a)=
p, then p Ir - 1. Thus the prime factorization of 3pq-l+
1 has the form3pq - l
+
1 _ 2u7/Jp,1 p'k- ' 1 . . . k '
(*)
where Pi fi {2, 7} are primes such that Pi= 1 (mod p).
We have already obtained (.I'. ~ 2. Furthermore, if p
=
7 then /3=
0, and otherwise HP+ 17P28
=
l -llp-217
+
llp-3172 - • · ·+
17p-l=
p · 4P-l (mod 7),so HP
+
17P is not divisible by 72• In either case, (3 ::;;; 1.For q
=
2 ( *) becomes 3p+
1=
2u7fJPi' · - -Pk\ but since Pi ~ 2p+
l, this is only possible if 'Yi=
0 for all i, i.e. 3p+
l=
2°'7/3 E {2, 4, 14, 28}. In this case we get no solutions.In the remaining case q > 2 we have 4
I
3p9 - 1+
1, i.e. a=
2. Now the right hand side in ( *) is congruent to 4 or 28 modulo p, which implies that p=
3. Consequently, 3q+
1I
6244, which holds only for q=
3. The pair (p, q)=
(3, 3) is indeed a solution.Remark. If we allow q to be composite, it remains to check the case a
