Baltic Way 2002 mathematical team contest

Transcription

1 Baltic Way 00 mathematical team cotest Tartu, November, 00 Problems ad solutios 1. Solve the system of equatios a 3 + 3ab + 3ac 6abc = 1 b 3 + 3ba + 3bc 6abc = 1 c 3 + 3ca + 3cb 6abc = 1 i real umbers. Aswer: a = 1, b = 1, c = 1. Solutio. Deotig the left had sides of the give equatios as A, B ad C, the followig equalities ca easily be see to hold: A + B + C = a + b + c) 3, A B + C = a b + c) 3, A + B C = a + b c) 3. Hece, the system of equatios give i the problem is equivalet to the followig oe: a + b + c) 3 = 1 a b + c) 3 = 1, a + b c) 3 = 1 which gives a + b + c = 1 a b + c = 1. a + b c = 1 The uique solutio of this system is a, b, c) = 1, 1, 1).. Let a, b, c, d be real umbers such that a + b + c + d =, ab + ac + ad + bc + bd + cd = 0. Prove that at least oe of the umbers a, b, c, d is ot greater tha 1. Solutio. We ca assume that a is the least amog a, b, c, d or oe of the least, if some of them are equal), there are > 0 egative umbers amog a, b, c, d, ad the sum of the positive oes is x. The we obtai Squarig we get which implies = a + b + c + d a + x. 1) 4 = a + b + c + d 4 a + x ) as the square of the sum of positive umbers is ot less tha the sum of their squares. 1

2 Combiig iequalities 1) ad ) we obtai a + a + ) 4, a + a + 4a 0, a + a + 4a 0. As 3 if all the umbers are egative, the secod coditio of the problem caot be satisfied), we obtai from the last iequality that 4a + 4a 0, aa + 1) 0. As a < 0 it follows that a 1. Alterative solutio. Assume that a, b, c, d > 1. Deotig A = a + 1, B = b + 1, C = c + 1, D = d + 1 we have A, B, C, D > 0. The the first equatio gives A + B + C + D =. 3) We also have ab = A 1)B 1) = AB A B + 1. Addig 5 similar terms to the last oe we get from the secod equatio AB + AC + AD + BC + BD + CD 3A + B + C + D) + 6 = 0. I view of 3) this implies AB + AC + AD + BC + BD + CD = 0, a cotradictio as all the ukows A, B, C, D were supposed to be positive. Aother solutio. Assume that the coditios of the problem hold: a + b + c + d = 4) ab + ac + ad + bc + bd + cd = 0. 5) Suppose that a, b, c, d > 1. 6) If all of a, b, c, d were egative, the 5) could ot be satisfied, so at most three of them are egative. If two or less of them were egative, the 6) would imply that the sum of egative umbers, ad hece also the sum a + b + c + d, is greater tha 1) =, which cotradicts 4). So exactly three of a, b, c, d are egative ad oe is oegative. Let d be the oegative oe. The d = a + b + c) < 1 1 1) = 1. Obviously a, b, c, d < 1. Squarig 4) ad subtractig times 5), we get but a cotradictio. a + b + c + d = 4, a + b + c + d = a + b + c + d < 4,

3 3. Fid all sequeces a 0 a 1 a... of real umbers such that a m + = a m + a for all itegers m, 0. Aswer: a 0, a 1 ad a =. Solutio. Deotig f) = a we have fm + ) = f m) + f ). 7) Substitutig m = = 0 ito 7) we get f0) = f 0), hece either f0) = 1 these cases separately. or f0) = 0. We cosider 1) If f0) = 1 the substitutig m = 1 ad = 0 ito 7) we obtai f1) = f 1) + 1 4, whece f1) 1 ) = 0 ad f1) = 1. Now, f) = f1 + 1 ) = f 1) = 1, f8) = f + ) = f ) = 1, etc, implyig that f i ) = 1 for arbitrarily large atural i ad, due to mootoity, f) = 1 for every atural. ) If f0) = 0 the by substitutig m = 1, = 0 ito 7) we obtai f1) = f 1) ad hece, f1) = 0 or f1) = 1. This gives two subcases. a) If f0) = 0 ad f1) = 0 the by the same techique as above we see that f i ) = 0 for arbitrarily large atural i ad, due to mootoity, f) = 0 for every atural. b) If f0) = 0 ad f1) = 1 the we compute Now, f) = f1 + 1 ) = f 1) =, f4) = f + 0 ) = f ) = 4, f5) = f + 1 ) = f ) + f 1) = 5. f 3) + f 4) = f5) = f 5) + f 0) = 5, hece f 3) = 5 16 = 9 ad f3) = 3. Further, f8) = f + ) = f ) = 8, f9) = f3 + 0 ) = f 3) = 9, f10) = f3 + 1 ) = f 3) + f 1) = 10. From the equalities f 6) + f 8) = f 10) + f 0), f 7) + f 1) = f 5) + f 5) we also coclude that f6) = 6 ad f7) = 7. It remais to ote that k + 1) + k ) = k 1) + k + ), k + ) + k 4) = k ) + k + 4), ad by iductio it follows that f) = for every atural. 3

4 4. Let be a positive iteger. Prove that x i 1 x i ) i=1 1 1 ) for all oegative real umbers x 1, x,..., x such that x 1 + x + + x = 1. Solutio. Expadig the expressios at both sides we obtai the equivalet iequality i x 3 i + i x i It is easy to check that the left had side is equal to i ) x i x i 1 ) ad hece is oegative. Alterative solutio. First ote that for = 1 the required coditio holds trivially, ad for = we have ) x + 1 x) x1 x) + 1 x)x = x1 x) = 1 4 = 1 ) 1. So we may further cosider the case 3. Assume first that for each idex i the iequality x i < 3 holds. Let fx) = x1 x) = x x + x 3, [ the f x) = 6x 4. Hece, the fuctio f is cocave i the iterval 0, ]. Thus, from Jese s 3 iequality we have x i 1 x i ) = i=1 ) x x fx i ) f = f i=1 1 ) 1 = 1 ) 1. = 1 ) 1 = If some x i 3 the we have x i 1 x i ) 1 1 ) = For the rest of the terms we have j i x j 1 x j ) j i x j = 1 x i 1 3. Hece, i=1 x i 1 x i ) = ) as Fid all pairs a, b) of positive ratioal umbers such that a + b =

5 1 Aswer: a, b) =, 3 ) 3 or a, b) =, 1 ). Solutio. Squarig both sides of the equatio gives a + b + ab = + 3 8) so ab = r + 3 for some ratioal umber r. Squarig both sides of this gives 4ab = r r 3, so r 3 is ratioal, which implies r = 0. Hece ab = 3/4 ad substitutig this ito 8) gives a + b =. 1 Solvig for a ad b gives a, b) =, 3 ) 3 or a, b) =, 1 ). 6. The followig solitaire game is played o a m rectagular board, m,, divided ito uit squares. First, a rook is placed o some square. At each move, the rook ca be moved a arbitrary umber of squares horizotally or vertically, with the extra coditio that each move has to be made i the 90 clockwise directio compared to the previous oe e.g. after a move to the left, the ext oe has to be doe upwards, the ext oe to the right etc). For which values of m ad is it possible that the rook visits every square of the board exactly oce ad returs to the first square? The rook is cosidered to visit oly those squares it stops o, ad ot the oes it steps over.) Aswer: m, 0 mod. Solutio. First, cosider ay row that is ot the row where the rook starts from. The rook has to visit all the squares of that row exactly oce, ad o its tour aroud the board, every time it visits this row, exactly two squares get visited. Hece, m must be eve; a similar argumet for the colums shows that must also be eve. It remais to prove that for ay eve m ad such a tour is possible. We will show it by a iductiolike argumet. Labellig the squares with pairs of itegers i, j), where 1 i m ad 1 j, we start movig from the square m/ + 1, 1) ad first cover all the squares of the top ad bottom rows i the order show i the figure below, except for the squares m/ 1, ) ad m/ + 1, ); ote that we fiish o the square m/ 1, 1) The ext square to visit will be m/ 1, 1) ad ow we will cover the rows umbered ad 1, except for the two middle squares i row. Cotiuig i this way we ca visit all the squares except for the two middle squares i every secod row ote that here we eed the assumptio that m ad are eve):

6 The rest of the squares ca be visited easily: We draw covex quadrilaterals i the plae. They divide the plae ito regios oe of the regios is ifiite). Determie the maximal possible umber of these regios. Aswer: The maximal umber of regios is Solutio. Oe quadrilateral produces two regios. Suppose we have draw k quadrilaterals Q 1,..., Q k ad produced a k regios. We draw aother quadrilateral Q k+1 ad try to evaluate the umber of regios a k+1 ow produced. Our task is to make a k+1 as large as possible. Note that i a maximal cofiguratio, o vertex of ay Q i ca be located o the edge of aother quadrilateral as otherwise we could move this vertex a little bit to produce a extra regio. Because of this fact ad the covexity of the Q j s, ay oe of the four sides of Q k+1 meets at most two sides of ay Q j. So the sides of Q k+1 are divided ito at most k +1 segmets, each of which potetially grows the umber of regios by oe beig part of the commo boudary of two parts, oe of which is couted i a k ). But if a side of Q k+1 itersects the boudary of each Q j, 1 j k twice, the its edpoits vertices of Q k+1 ) are i the regio outside of all the Q j -s, ad the the segmets meetig at such a vertex are o the boudary of a sigle ew part recall that it makes o sese to put vertices o edges of aother quadrilaterals). This meas that a k+1 a k 4k + 1) 4 = 8k. By cosiderig squares iscribed i a circle oe easily sees that the situatio where a k+1 a k = 8k ca be reached. It remais to determie the expressio for the maximal a k. Sice the differece a k+1 a k is liear i k, a k is a quadratic polyomial i k, ad a 0 =. So a k = Ak + Bk +. We have 8k = a k+1 a k = Ak + 1) + B for all k. This implies A = 4, B = 4, ad a = Let P be a set of 3 poits i ) the plae, o three of which are o a lie. How may possibilities are 1 there to choose a set T of triagles, whose vertices are all i P, such that each triagle i T has a side that is ot a side of ay other triagle i T? Aswer: There is oe possibility for = 3 ad possibilities for 4. Solutio. For a fixed poit x P ), let T x be the set of all triagles with vertices i P which have x as 1 a vertex. Clearly, T x =, ad each triagle i T x has a side which is ot a side of ay other triagle i T x. For ay x, y P such that x y, we have T x T y if ad oly if 4. We will show that ay possible set T is equal to T x for some x P, i.e. that the aswer is 1 for = 3 ad for 4. Let { )} 1 T = t i : i = 1,,...,, S = { s i : i = 1,,..., )} 1 such that T is a set of triagles whose vertices are all i ) P, ad s i is a side of t i but ot of ay t j, j i. Furthermore, let C be the collectio of all the triagles whose vertices are i P. Note that 3 C \ T = ) 3 ) ) 1 1 =. 3 6

7 Let m be the umber of pairs s, t) such that s S is a side of t C \ T. Sice every s S is a side of exactly 3 triagles from C \ T, we have ) ) 1 1 m = S 3) = 3) = 3 = 3 C \ T. 3 O the other had, every t C \ T has at most three sides from S. By the above equality, for every t C \ T, all its sides must be i S. Assume that for p P there is a side s S such that p is a edpoit of s. The p is also a vertex of each of the 3 triagles i C \ T which have s as a side. Cosequetly, p is a edpoit of sides i S. Sice every side i S has exactly edpoits, the umber of poits p P which occur as a vertex of some s S is S = ) 1 = 1. Cosequetly, there is a x P which is ot a edpoit of ay s S, ad hece T must be equal to T x. 9. Two magicias show the followig trick. The first magicia goes out of the room. The secod magicia takes a deck of 100 cards labelled by umbers 1,,..., 100 ad asks three spectators to choose i tur oe card each. The secod magicia sees what card each spectator has take. The he adds oe more card from the rest of the deck. Spectators shuffle these 4 cards, call the first magicia ad give him these 4 cards. The first magicia looks at the 4 cards ad guesses what card was chose by the first spectator, what card by the secod ad what card by the third. Prove that the magicias ca perform this trick. Solutio. We will idetify ourselves with the secod magicia. The we eed to choose a card i such a maer that aother magicia will be able to uderstad which of the 4 cards we have chose ad what iformatio it gives about the order of the other cards. We will reach these two goals idepedetly. Let a, b, c be remaiders of the labels of the spectators three cards modulo 5. There are three possible cases. 1) All the three remaiders coicide. The choose a card with a remaider ot equal to the remaider of spectators cards. Deote this remaider d. Note that we ow have differet remaiders, oe of them i 3 copies this will be used by the first magicia to distiguish betwwe the three cases). To determie which of the cards is chose by us is ow a simple exercise i divisio by 5. But we must also ecode the orderig of the spectators cards. These cards have a atural orderig by their labels, ad they are also ordered by their belogig to the spectators. Thus, we have to ecode a permutatio of 3 elemets. There are 6 permutatios of 3 elemets, let us eumerate them somehow. The, if we wat to iform the first magicia that spectators form a permutatio umber k with respect to the atural orderig, we choose the card umber 5k + d. ) The remaiders a, b, c are pairwise differet. The it is clear that exactly oe of the followig possibilities takes place: either b a = a c, or a b = b c, or a c = c b 9) the equalities are cosidered modulo 5). It is ot hard to prove it by a case study, but oe could also imagie choosig three vertices of a regular petago these vertices always form a isosceles, but ot a equilateral triagle. Each of these possibilities has oe of the remaiders distiguished from the other two remaiders these distiguished remaiders are a, b, c, respectively). Now, choose a card from the rest of the deck havig the distiguished remaider modulo 5. Hece, we have three differet remaiders, oe of them distiguished by 9) ad preseted i two copies. Let d be the distiguished remaider ad s = 5m + d be the spectator s card with this remaider. Now we have to choose a card r with the remaider d such that the first magicia would be able to uderstad which of the cards s ad r was chose by us ad what permutatio of spectators it implies. This ca be doe easily: if we wat to iform the first magicia that spectators form a permutatio umber k with respect to the atural orderig, we choose the card umber s + 5k mod 100). 7

8 The decodig procedure is easy: if we have two umbers p ad q that have the same remaider modulo 5, calculate p q mod 100) ad q p mod 100). If p q mod 100) > q p mod 100) the r = q is our card ad s = p is the spectator s card. The case p q mod 100) = q p mod 100) is impossible sice the sum of these umbers is equal to 100, ad oe of them is ot greater tha 6 5 = 30.) 3) Two remaiders say, a ad b) coicide. Let us choose a card with the remaider d = a+c)/ mod 5. The a d = d c mod 5, so the remaider d is distiguished by 9). Hece we have three differet remaiders, oe of them distiguished by 9) ad oe of the o-distiguished remaiders preseted i two copies. The first magicia will easily determie our card, ad the rule to choose the card i order to eable him also determie the order of spectators is similar to the oe i the 1-st case. Alterative solutio. This solutio gives a o-costructive proof that the trick is possible. For this, we eed to show there is a ijective mappig from the set of ordered triples to the set of uordered quadruples that additioally respects iclusio. To prove that the desired mappig exists, let s cosides a bipartite graph such that the set of ordered triples T ad the set of uordered quadruples Q form the two disjoit sets of vertices ad there is a edge betwee a triple ad a quadruple if ad oly if the triple is a subset of the quadruple. For each triple t T, we ca add ay of the remaiig 97 cards to it, ad thus we have 97 differet quadruples coected to each triple i the graph. Coversely, for each quadruple q Q, we ca remove ay of the 4 cards from it, ad reorder the remaiig 3 cards i 3! = 6 differet ways, ad thus we have 4 differet triples coected to each quadruple i the graph. Accordig to the Hall s theorem, a bipartite graph G = T, Q, E) has a perfect matchig if ad oly if for each subset T T the set of eighbours of T, deoted NT ), satisfies NT ) T. To prove that this coditio holds for our graph, cosider ay subset T T. Because we have 97 quadruples for each triple, ad there ca be at most 4 copies of each of them i the multiset of eighbours, we have NT ) 97 4 T > 4 T, which is eve much more tha we eed. Thus, the desired mappig is guarateed to exist. Aother solutio. Let the three chose umbers be x 1, x, x 3 ). At least oe of the sets {1,,..., 4}, {5, 6,..., 48}, {49, 50,..., 7} ad {73, 74,..., 96} should cotai oe of x 1, x ad x 3, let S be such set. Next we split S ito 6 parts: S = S 1 S... S 6 so that 4 first elemets of S are i S 1, four ext i S, etc. Now we choose i {1,,..., 6} correspodig to the order of umbers x 1, x ad x 3 if x 1 < x < x 3 the i = 1, if x 1 < x 3 < x the i =,...,if x 3 < x < x 1 the i = 6). At last let j be the umber of elemets i {x 1, x, x 3 } that are greater tha elemets of S ote that ay x k, k {1,, 3}, is either greater or smaller tha all the elemets of S ). Now we choose x 4 S i so that x 1 + x + x 3 + x 4 j mod 4 ad add the card umber x 4 to those three cards. Decodig of {a, b, c, d} is straightforward. We first put the umbers ito icreasig order ad the calculate a+b+c+d mod 4 showig the added card. The added card belogs to some S i i {1,,..., 6}) for some S ad i shows us the iitial orderig of cards. 10. Let N be a positive iteger. Two persos play the followig game. The first player writes a list of positive itegers ot greater tha 5, ot ecessarily differet, such that their sum is at least 00. The secod player wis if he ca select some of these umbers so that their sum S satisfies the coditio 00 N S 00 + N. What is the smallest value of N for which the secod player has a wiig strategy? Aswer: N = 11. Solutio. If N = 11, the the secod player ca simply remove umbers from the list, startig with the smallest umber, util the sum of the remaiig umbers is less tha 1. If the last umber removed was ot 4 or 5, the the sum of the remaiig umbers is at least 1 3 = 189. If the last umber removed was 4 or 5, the oly 4-s ad 5-s remai, ad there must be exactly 8 of them sice their sum must be less tha 1 ad ot less tha 1 4 = 188. Hece their sum S satisfies 8 4 = 19 S 8 5 = 00. I ay case the secod player wis. O the other had, if N 10, the the first player ca write 5 two times ad 3 seve times. The the sum of all umbers is 11, but if at least oe umber is removed, the the sum of the remaiig oes is at most 188 so the secod player caot wi. 8

9 11. Let be a positive iteger. Cosider poits i the plae such that o three of them are colliear ad o two of the distaces betwee them are equal. Oe by oe, we coect each poit to the two poits earest to it by lie segmets if there are already other lie segmets draw to this poit, we do ot erase these). Prove that there is o poit from which lie segmets will be draw to more tha 11 poits. Solutio. Suppose there exists a poit A such that A is coected to twelve poits. The there exist three poits B, C ad D such that BAC 60, BAD 60 ad CAD 60. We ca assume that AD > AB ad AD > AC. By the cosie law we have BD = AD + AB AD AB cos BAD < AD + AB AB cos BAD = AD + AB 1 cos BAD) AD sice 1 cos BAD). Hece BD < AD. Similarly we get CD < AD. Hece A ad D should ot be coected which is a cotradictio. Commet. It would be iterestig to kow whether 11 ca be achieved or the actual boud is lower. 1. A set S of four distict poits is give i the plae. It is kow that for ay poit X S the remaiig poits ca be deoted by Y, Z ad W so that XY = XZ + XW. Prove that all the four poits lie o a lie. Solutio. Let S = {A, B, C, D} ad let AB be the logest of the six segmets formed by these four poits if there are several logest segmets, choose ay of them). If we choose X = A the we must also choose Y = B. Ideed, if we would, for example, choose Y = C, we should have AC = AB + AD cotradictig the maximality of AB. Hece we get AB = AC + AD. 10) Similarly, choosig X = B we must choose Y = A ad we obtai AB = BC + BD. 11) O the other had, from the triagle iequality we kow that AB AC + BC, AB AD + BD, where at least oe of the iequalities is strict if all the four poits are ot o the same lie. Hece, addig the two last iequalities we get AB < AC + BC + AD + BD. O the other had, addig 10) ad 11) we get a cotradictio. AB = AC + AD + BC + BD, 13. Let ABC be a acute triagle with BAC > BCA, ad let D be a poit o side AC such that AB = BD. Furthermore, let F be a poit o the circumcircle of triagle ABC such that lie F D is perpedicular to side BC ad poits F, B lie o differet sides of lie AC. Prove that lie F B is perpedicular to side AC. Solutio. Let E be the other poit o the circumcircle of triagle ABC such that AB = EB. Let D be the poit of itersectio of side AC ad the lie perpedicular to side BC, passig through E. The ECB = BCA ad the triagle ECD is isosceles. As ED BC, the triagle BED is also isosceles ad BE = BD implyig D = D. Hece, the poits E, D, F lie o oe lie. We ow have EF B + F DA = BCA + EDC = 90. 9

10 The required result ow follows. B PSfrag replacemets E A D C F 14. Let L, M ad N be poits o sides AC, AB ad BC of triagle ABC, respectively, such that BL is the bisector of agle ABC ad segmets AN, BL ad CM have a commo poit. Prove that if ALB = MNB the LNM = 90. Solutio. Let P be the itersectio poit of lies M N ad AC. The P LB = P N B ad the quadragle P LNB is cyclic. Let ω be its circumcircle. It is sufficiet to prove that P L is a diameter of ω. Let Q deote the secod itersectio poit of the lie AB ad ω. The P QB = P LB ad QP L = QBL = LBN = LP N, ad the triagles P AQ ad BAL are similar. Therefore, P Q P A = BL BA. 1) We see that the lie P L is a bisector of the iscribed agle NP Q. Now i order to prove that P L is a diameter of ω it is sufficiet to check that P N = P Q. The triagles N P C ad LBC are similar, hece P N P C = BL BC. 13) Note also that AB BC = AL CL. 14) by the properties of a bisector. Combiig 1), 13) ad 14) we have P N P Q = AL AP CP CL. We wat to prove that the left had side of this equality equals 1. This follows from the fact that the quadruple of poits P, A, L, C) is harmoic, as ca be prove usig stadard methods e.g. cosiderig the quadrilateral MBNS, where S = MC AN ). 10

11 PSfrag replacemets A B C D E F N B M C L A P Q 15. A spider ad a fly are sittig o a cube. The fly wats to maximize the shortest path to the spider alog the surface of the cube. Is it ecessarily best for the fly to be at the poit opposite to the spider? Opposite meas symmetric with respect to the ceter of the cube.) Aswer: o. Solutio. Suppose that the side of the cube is 1 ad the spider sits at the middle of oe of the edges. The the shortest path to the middle of the opposite edge has legth. However, if the fly goes to a poit o this edge at distace s from the middle, the the legth of the shortest path is mi ) 4 + s 9 3, 4 + s. If 0 < s < 3 7)/ the this expressio is greater tha. 16. Fid all oegative itegers m such that a m = m+1) + 1 is divisible by at most two differet primes. Aswer: m = 0, 1, are the oly solutios. Solutio. Obviously m = 0, 1, are solutios as a 0 = 5, a 1 = 65 = 5 13, ad a = 105 = We show that these are the oly solutios. Assume that m 3 ad that a m cotais at most two differet prime factors. Clearly, a m = 4 m is divisible by 5, ad a m = m+1 + m ) m+1 m ). The two above factors are relatively prime as they are both odd ad their differece is a power of. Sice both factors are larger tha 1, oe of them must be a power of 5. Hece, m+1 m ± 1) = 5 t 1 = 5 1) t 1 ) for some positive iteger t, where ± reads as either plus or mius. For odd t the right had side is ot divisible by 8, cotradictig m 3. Therefore, t must be eve ad m+1 m ± 1) = 5 t/ 1) 5 t/ + 1). Clearly, 5 t/ + 1 mod 4). Cosequetly, 5 t/ 1 = m k for some odd k, ad 5 t/ + 1 = m k + divides m ± 1), i.e. m 1 k + 1 m ± 1. This implies k = 1, fially leadig to a cotradictio sice for m 3. m < m ± 1 < m 1 + 1) 11

12 17. Show that the sequece ) ) ) ,,,..., cosidered modulo 00, is periodic. Solutio. Defie ) x k = k ad ote that + 1 x k +1 x k = k ) ) ) = = x k 1. k k 1 Let m be ay positive iteger. We will prove by iductio o k that the sequece {x k } =k is periodic modulo m. For k = 1 it is obvious that x k = is periodic modulo m with period m. Therefore it will suffice to show that the followig is true: the sequece {x } is periodic modulo m if its differece sequece, d = x +1 x, is periodic modulo m. Furthermore, if t the the period of {x } is equal to ht where h is the smallest positive iteger such that hx t x 0 ) 0 modulo m. Ideed, let t be the period of {d } ad h be the smallest positive iteger such that hx t x 0 ) 0 modulo m. The +ht 1 1 t 1 x +ht = x 0 + d j = x 0 + d j + h = j=0 j=0 = x + hx t x 0 ) x mod m) for all, so the sequece {x } is i fact periodic modulo m with a period dividig ht). 18. Fid all itegers > 1 such that ay prime divisor of 6 1 is a divisor of 3 1) 1). Solutio. Clearly = is such a iteger. We will show that there are o others. Cosider the equality 6 1 = + 1) + 1) 3 1). The iteger + 1 = 1) + 1 clearly has a odd divisor p. The p Therefore, p does ot divide 3 1 ad cosequetly p 1. This implies that p divides 3 +1)+ 1) = +1). As p does ot divide, we obtai p + 1. Also, p 1) + 1) =. From p + 1 ad p it follows that p = 3, so + 1 = 3 r for some positive iteger r. The discrimiat of the quadratic r ) must be a square of a iteger, hece r ) = 34 3 r 1 1) must be a squareof a iteger. Sice for r the umber 4 3 r 1 1 is ot divisible by 3, this is possible oly if r = 1. So = 0 ad =. 19. Let be a positive iteger. Prove that the equatio x + y + 1 x + 1 y = 3 does ot have solutios i positive ratioal umbers. Solutio. Suppose x = p q ad y = r satisfy the give equatio, where p, q, r, s are positive itegers s ad gcdp, q) = 1, gcdr, s) = 1. We have p q + r s + q p + s r = 3, 1 j=0 d j

13 or p + q )rs + r + s )pq = 3pqrs, so rs r + s )pq. Sice gcdr, s) = 1, we have gcdr + s, rs) = 1 ad rs pq. Aalogously pq rs, so rs = pq ad hece there are either two or zero itegers divisible by 3 amog p, q, r, s. Now we have p + q )rs + r + s )rs = 3rs), p + q + r + s = 3rs, but 3rs 0 mod 3) ad p + q + r + s is cogruet to either 1 or modulo 3, a cotradictio. 0. Does there exist a ifiite o-costat arithmetic progressio, each term of which is of the form a b, where a ad b are positive itegers with b? Aswer: o. Solutio. For a arithmetic progressio a 1, a,... with differece d the followig holds: S = = a 1 a a +1 a 1 a 1 + d a 1 + d 1 1 m ), + 1 where m = maxa 1, d). Therefore S teds to ifiity whe icreases. O the other had, the sum of reciprocals of the powers of a atural umber x 1 is 1 x + 1 x = 1 x 1 1 x = 1 xx 1). Hece, the sum of reciprocals of the terms of the progressio required i the problem caot exceed = ) =, a cotradictio. Alterative solutio. Let a k = a 0 + dk, k = 0, 1,.... Choose a prime umber p > d ad set k p a 0 )d 1 mod p. The a k = a 0 + k d p mod p ad hece, a k ca ot be a power of a atural umber. Aother solutio. There ca be at most squares i the set {1,,..., }, at most 3 cubes i the same set, etc. The greatest power that ca occur i the set {1,,..., } is log ad thus there are o more tha log powers amog the umbers 1,,...,. Now we ca estimate this sum above: log log 1) < < log = o). This meas that every arithmetic progressio grows faster tha the share of powers. 13