Solutions to the 71st William Lowell Putnam Mathematical Competition Saturday, December 4, 2010

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1 Solutons to the 7st Wllam Lowell Putnam Mathematcal Competton Saturday, December 4, 2 Kran Kedlaya and Lenny Ng A The largest such k s n+ 2 n 2. For n even, ths value s acheved by the partton {,n},{2,n },...; for n odd, t s acheved by the partton {n},{,n },{2,n 2},... One way to see that ths s optmal s to note that the common sum can never be less than n, snce n tself belongs to one of the boxes. Ths mples that k + + n/n n + /2. Another argument s that f k > n + /2, then there would have to be two boxes wth one number each by the pgeonhole prncple, but such boxes could not have the same sum. Remark. A much subtler queston would be to fnd the smallest k as a functon of n for whch no such arrangement exsts. A 2 The only such functons are those of the form f x cx + d for some real numbers c,d for whch the property s obvously satsfed. To see ths, suppose that f has the desred property. Then for any x R, 2 f x f x + 2 f x f x + 2 f x + + f x + f x f x + + f x. Consequently, f x + f x. Defne the functon g : R R by gx f x + f x, and put c g, d f. For all x R, g x f x + f x, so gx c dentcally, and f x f x+ f x gx c, so f x cx+d dentcally as desred. A 3 f a b, then the desred result holds trvally, so we assume that at least one of a,b s nonzero. Pck any pont a,b R 2, and let L be the lne gven by the parametrc equaton Lt a,b + a,bt for t R. By the chan rule and the gven equaton, we have d dt h L h L. f we wrte f h L : R R, then f t f t for all t. t follows that f t Ce t for some constant C. Snce f t M for all t, we must have C. t follows that ha,b ; snce a,b was an arbtrary pont, h s dentcally over all of R 2. A 4 Put N n + n + n. Wrte n 2 m k wth m a nonnegatve nteger and k a postve odd nteger. For any nonnegatve nteger j, Snce n n 2 m m +, n s dvsble by 2 n and hence by 2 m+, and smlarly n s dvsble by 2 n and hence by 2 m+. t follows that N mod 2m +. Snce N n > n + 2m +, t follows that N s composte. A 5 We start wth three lemmas. Lemma. f x,y G are nonzero orthogonal vectors, then x x s parallel to y. Proof. Put z x y, so that x,y, and z x y are nonzero and mutually orthogonal. Then w x z, so w x z s nonzero and orthogonal to x and z. However, f x x y, then w x x y x x y x x y s also orthogonal to y, a contradcton. Lemma 2. f x G s nonzero, and there exsts y G nonzero and orthogonal to x, then x x. Proof. Lemma mples that x x s parallel to both y and x y, so t must be zero. Lemma 3. f x,y G commute, then x y. Proof. f x y, then y x s nonzero and dstnct from x y. Consequently, x y x y and y x y x x y. We proceed now to the proof. Assume by way of contradcton that there exst a,b G wth a b. Put c a b a b, so that a,b,c are nonzero and lnearly ndependent. Let e be the dentty element of G. Snce e commutes wth a,b,c, by Lemma 3 we have e a e b e c. Snce a,b,c span R 3, e x for all x R 3, so e. Snce b,c, and b c b c are nonzero and mutually orthogonal, Lemma 2 mples b b c c b c b c e. Hence b c c b, contradctng Lemma 3 because b c. The desred result follows. A 6 Frst soluton. Note that the hypotheses on f mply that f x > for all x [,+, so the ntegrand s a contnuous functon of f and the ntegral makes sense. Rewrte the ntegral as f x +, f x 2m j j mod 2m +.

2 2 and suppose by way of contradcton that t converges to a fnte lmt L. For n, defne the Lebesgue measurable set n {x [,] : f x + n + f x + n /2}. Then L n 2 µ n, so the latter sum converges. n partcular, there exsts a nonnegatve nteger N for whch nn µ n < ; the ntersecton n [,] [,] n nn nn then has postve Lebesgue measure. By Taylor s theorem wth remander, for t [,/2], { } log t t +t 2 sup t 2 t [,/2] t t2 5 3 t. For each nonnegatve nteger n N, we then have n L N n N n N 3 5 n N f x + f x log 5 log n log N f x + + f x + f x + + f x + f x + f x + + f x + f x + + f x + N f x + n. For each x, log f x + N/ f x + n s a strctly ncreasng unbounded functon of n. By the monotone convergence theorem, the ntegral log f x + N/ f x + n grows wthout bound as n +, a contradcton. Thus the orgnal ntegral dverges, as desred. Remark. Ths soluton s motvated by the commonlyused fact that an nfnte product + x + x 2 converges absolutely f and only f the sum x +x 2 + converges absolutely. The addtonal measure-theoretc argument at the begnnng s needed because one cannot bound log t by a fxed multple of t unformly for all t [,. Greg Martn suggests a varant soluton that avods use of Lebesgue measure. Note frst that f f y > 2 f y +, then ether f y > 2 f y + /2 or f y + /2 > 2 f y +, and n ether case we deduce that b a f there exst arbtrarly large values of y for whch f y > 2 f y+, we deduce that the orgnal ntegral s greater than any multple of /7, and so dverges. Otherwse, for x large we may argue that f x f x + f x > 3 5 log f x f x + as n the above soluton, and agan get dvergence usng a telescopng sum. Second soluton. Communcated by Paul Allen. Let b > a be nonnegatve ntegers. Then f x f x + f x b ka b ka b ka f x + k f x + k + f x + k f x + k f x + k + f x + k f x + k f x + k + f x + a f x + a f x + b. f x + a Now snce f x, gven a, we can choose an nteger la > a for whch f la < f a + /2; then f x+a f x+la f x+a f la f a+ > /2 for all x [,]. Thus f we defne a sequence of ntegers a n by a, a n+ la n, then f x f x + f x > an+ n n and the fnal sum clearly dverges. f x f x + a n f x /2, Thrd soluton. By Joshua Rosenberg, communcated by Cataln Zara. f the orgnal ntegral converges, then on one hand the ntegrand f x f x+/ f x f x + / f x cannot tend to as x. On the other hand, for any a, f a + < f a < a+ f x f a a f x f x + f a a f x f x +, f x a and the last expresson tends to as a. Hence by the squeeze theorem, f a + / f a as a, a contradcton. y+/2 y /2 f x f x + > 2 > f x 2 7.

3 3 B Frst soluton. No such sequence exsts. f t dd, then the Cauchy-Schwartz nequalty would mply contradcton. 8 a 2 + a a 4 + a a 3 + a , Second soluton. Communcated by Cataln Zara. Suppose that such a sequence exsts. f a 2 k [,] for all k, then a 4 k a2 k for all k, and so 4 a 4 + a a 2 + a , contradcton. There thus exsts a postve nteger k for whch a 2 k. However, n ths case, for m large, a2m k > 2m and so a 2m + 2m. + a2m 2 Thrd soluton. We generalze the second soluton to show that for any postve nteger k, t s mpossble for a sequence a,a 2,... of complex numbers to satsfy the gven condtons n case the seres a k + ak 2 + converges absolutely. Ths ncludes the orgnal problem by takng k 2, n whch case the seres a 2 + a2 2 + conssts of nonnegatve real numbers and so converges absolutely f t converges at all. Snce the sum a k converges by hypothess, we can fnd a postve nteger n such that n+ a k <. For each postve nteger d, we then have n kd a kd n+ a kd <. We thus cannot have a,..., a n, or else the sum n akd would be bounded n absolute value by n ndependently of d. But f we put r max{ a,..., a n } >, we obtan another contradcton because for any ε >, lm supr ε kd n d a kd >. For nstance, ths follows from applyng the root test to the ratonal functon z d, n a k z d n a kd whch has a pole wthn the crcle z r /k. An elementary proof s also possble. Fourth soluton. Communcated by Noam Elkes. Snce k a 2 k 2, for each postve nteger k we have a 2 k 2 and so a4 k 2a2 k, wth equalty only for a2 k {,2}. Thus to have k a 4 k 4, there must be a sngle ndex k for whch a 2 k 2, and the other a k must all equal. But then k a 2m k 2 m 2m for any postve nteger m > 2. Remark. Manjul Bhargava ponts out t s easy to construct sequences of complex numbers wth the desred property f we drop the condton of absolute convergence. Here s an nductve constructon of whch several varants are possble. For n,2,... and z C, defne the fnte sequence s n,z z e2π j/n : j,...,n. Ths sequence has the property that for any postve nteger j, the sum of the j-th powers of the terms of s n,z equals /z j f j s dvsble by n and otherwse. Moreover, any partal sum of j-th powers s bounded n absolute value by n/ z j. The desred sequence wll be constructed as follows. Suppose that we have a fnte sequence whch has the correct sum of j-th powers for j,...,m. For nstance, for m, we may start wth the sngleton sequence. We may then extend t to a new sequence whch has the correct sum of j-th powers for j,...,m +, by appendng k copes of s m+,z for sutable choces of a postve nteger k and a complex number z wth z < m 2. Ths last restrcton ensures that the resultng nfnte sequence a,a 2,... s such that for each postve nteger m, the seres a m + am 2 + s convergent though not absolutely convergent. ts partal sums nclude a subsequence equal to the constant value m, so the sum of the seres must equal m as desred. B 2 The smallest dstance s 3, acheved by A,, B 3,, C,4. To check ths, t suffces to check that AB cannot equal or 2. t cannot equal because f two of the ponts were to concde, the three ponts would be collnear. The trangle nequalty mples that AC BC AB, wth equalty f and only f A,B,C are collnear. f AB, we may assume wthout loss of generalty that A,, B,. To avod collnearty, we must have AC BC, but ths forces C /2,y for some y R, a contradcton. One can also treat ths case by scalng by a factor of 2 to reduce to the case AB 2, treated n the next paragraph. f AB 2, then we may assume wthout loss of generalty that A,,B 2,. The trangle nequalty mples AC BC {,}. Also, for C x,y, AC 2 x 2 + y 2 and BC 2 2 x 2 + y 2 have the same party; t follows that AC BC. Hence c,y for some y R, so y 2 and y 2 + BC 2 are consecutve perfect squares. Ths can only happen for y, but then A,B,C are collnear, a contradcton agan. Remark. Manjul Bhargava ponts out that more generally, a Heronan trangle a trangle wth nteger sdes and ratonal area cannot have a sde of length or 2 and agan t s enough to treat the case of length 2. The orgnal problem follows from ths because a trangle whose vertces have nteger coordnates has area

4 4 equal to half an nteger by Pck s formula or the explct formula for the area as a determnant. B 3 t s possble f and only f n 5. Snce , for n 4, we can start wth an ntal dstrbuton n whch each box B starts wth at most balls so n partcular B s empty. From such a dstrbuton, no moves are possble, so we cannot reach the desred fnal dstrbuton. Suppose now that n 5. By the pgeonhole prncple, at any tme, there exsts at least one ndex for whch the box B contans at least balls. We wll descrbe any such ndex as beng elgble. The followng sequence of operatons then has the desred effect. a Fnd the largest elgble ndex. f, proceed to b. Otherwse, move balls from B to B, then repeat a. b At ths pont, only the ndex can be elgble so t must be. Fnd the largest ndex j for whch B j s nonempty. f j, proceed to c. Otherwse, move ball from B to B j ; n case ths makes j elgble, move j balls from B j to B. Then repeat b. c At ths pont, all of the balls are n B. For 2,...,2, move one ball from B to B n tmes. After these operatons, we have the desred dstrbuton. B 4 Frst soluton. The pars p, q satsfyng the gven equaton are those of the form px ax + b,qx cx + d for a,b,c,d R such that bc ad. We wll see later that these ndeed gve solutons. Suppose p and q satsfy the gven equaton; note that nether p nor q can be dentcally zero. By subtractng the equatons pxqx + px + qx px qx pxqx, we obtan the equaton pxqx + + qx qxpx + + px. The orgnal equaton mples that px and qx have no common nonconstant factor, so px dvdes px + + px. Snce each of px + and px has the same degree and leadng coeffcent as p, we must have px + + px 2px. f we defne the polynomals rx px + px, sx qx + qx, we have rx + rx, and smlarly sx + sx. Put Then rx a,sx c for all x Z, and hence dentcally; consequently, px ax + b,qx cx + d for all x Z, and hence dentcally. For p and q of ths form, pxqx + px + qx bc ad, so we get a soluton f and only f bc ad, as clamed. Second soluton. Communcated by Cataln Zara. Agan, note that p and q must be nonzero. Wrte px p + p x + + p m x m qx q + q x + + q n x n wth p m,q n, so that m degp,n degq. t s enough to derve a contradcton assumng that max{m, n} >, the remanng cases beng treated as n the frst soluton. Put Rx pxqx+ px+qx. Snce m+n 2 by assumpton, the coeffcent of x m+n n Rx must vansh. By easy algebra, ths coeffcent equals m np m q n, so we must have m n >. For k,...,2m 2, the coeffcent of x k n Rx s j p q j p j q k k j + j>k, j> and must vansh. For k 2m 2, the only summand s for, j m,m, so p m q m p m q m. Suppose now that h and that p q j p j q s known to vansh whenever j > h. By the prevous paragraph, we ntally have ths for h m. Take k m+h 2 and note that the condtons + j > h, j m force h. Usng the hypothess, we see that the only possble nonzero contrbuton to the coeffcent of x k n Rx s from, j h,m. Hence p h q m p m q h ; snce p m,q m, ths mples p h q j p j q h whenever j > h. By descendng nducton, we deduce that p q j p j q whenever j >. Consequently, px and qx are scalar multples of each other, forcng Rx, a contradcton. Thrd soluton. Communcated by Davd Feldman. As n the second soluton, we note that there are no solutons where m degp, n degq are dstnct and m+n 2. Suppose p,q form a soluton wth m n 2. The desred dentty asserts that the matrx px px + qx qx + has determnant. Ths condton s preserved by replacng qx wth qx t px for any real number t. n partcular, we can choose t so that degqx t px < m; we then obtan a contradcton. a r,b p,c s,d q.

5 5 B 5 Frst soluton. The answer s no. Suppose otherwse. For the condton to make sense, f must be dfferentable. Snce f s strctly ncreasng, we must have f x for all x. Also, the functon f x s strctly ncreasng: f y > x then f y f f y > f f x f x. n partcular, f y > for all y R. For any x, f f x b and f x a >, then f x > a for x > x and thus f x ax x + b for x x. Then ether b < x or a f x f f x f b ab x +b. n the latter case, b ax +/a+ x +. We conclude n ether case that f x x + for all x. t must then be the case that f f x f x for all x, snce otherwse f x > x + for large x. Now by the above reasonng, f f b and f a >, then f x > a x + b for x >. Thus for x > max{, b /a }, we have f x > and f f x > a x + b. But then f f x > for suffcently large x, a contradcton. Second soluton. Communcated by Cataln Zara. Suppose such a functon exsts. Snce f s strctly ncreasng and dfferentable, so s f f f. n partcular, f s twce dfferentable; also, f x f f x f x s the product of two strctly ncreasng nonnegatve functons, so t s also strctly ncreasng and nonnegatve. n partcular, we can choose α > and M R such that f x > 4α for all x M. Then for all x M, f x f M + f Mx M + 2αx M 2. n partcular, for some M > M, we have f x αx 2 for all x M. Pck T > so that αt 2 > M. Then for x T, f x > M and so f x f f x α f x 2. Now f T 2T f 2T T f t 2T f t 2 dt α dt; T however, as T, the left sde of ths nequalty tends to whle the rght sde tends to +, a contradcton. Thrd soluton. Communcated by Noam Elkes. Snce f s strctly ncreasng, for some y, we can defne the nverse functon gy of f for y y. Then x g f x, and we may dfferentate to fnd that g f x f x g f x f f x. t follows that g y / f y for y y ; snce g takes arbtrarly large values, the ntegral y dy/ f y must dverge. One then gets a contradcton from any reasonable lower bound on f y for y large, e.g., the bound f x αx 2 from the second soluton. One can also start wth a lnear lower bound f x βx, then use the ntegral expresson for g to deduce that gx γ logx, whch n turn forces f x to grow exponentally. B 6 For any polynomal px, let [px]a denote the n n matrx obtaned by replacng each entry A j of A by pa j ; thus A [k] [x k ]A. Let Px x n + a n x n + + a denote the characterstc polynomal of A. By the Cayley-Hamlton theorem, A PA A n+ + a n A n + + a A A [n+] + a n A [n] + + a A [] [xpx]a. Thus each entry of A s a root of the polynomal xpx. Now suppose m n +. Then [x m+ n px]a A [m+] + a n A [m] + + a A [m+ n] snce each entry of A s a root of x m+ n px. On the other hand, A m+ n PA A m+ + a n A m + + a A m+ n. Therefore f A k A [k] for m + n k m, then A m+ A [m+]. The desred result follows by nducton on m. Remark. Davd Feldman ponts out that the result s best possble n the followng sense: there exst examples of n n matrces A for whch A k A [k] for k,...,n but A n+ A [n+].