Solutions to the Fifty-Sixth William Lowell Putnam Mathematical Competition Saturday, December 2, 1995 Kiran Kedlaya ✂✁☎✄✆ ✞✝✠✟☛✡ A–1 Suppose on the contrary that there exist with ✂✁☞ ✞✝✌✟✎✍ ✁☎✄ ✝✑✟✎✍ ✁ ✝✌✟✒✡ ✏ ✏ ✏ ✏ and with Then ✓ ✂✁☞ ✞✝✕✔ ✁ ✝✖✟✗✍ ✂✁✘ ✞✝ ✓ ✁ ✝✙✔✚✟☛✡ ✏ ✏ ✏ ✏ while , contradiction ★✩✝ easiest A–2 The integral converges iff ✛✑✜✣✢ The✓✆✤✖ ✥✧✦ ✔ ✁✆proof ✝ ✝ ✜ uses “big-O” notation and the fact that ✤✠✥✑✦✫✪✭✬✮✥✑✯✮✓✰✦ ✔ ✦ ✤ ✯✮✓✴✦ ✔ for ✱ ✱✳✲ ✦ ✝ (Here means bounded by a constant times ) So ✵ ✵ ✦✶✥ ✦ ✛✳✷ ✦ ✜ ✦ ✜ hence✿ ✵ ✁✂★✂✝ ✵ ✦✶✥ and similarly ✿ ✵ ✦ ✦ ✛❀✷ ✦ ✜ ✵ ✦ ✷ ✷❆✢❇✜ ✓✆✸ ✁✂★✂✝ ✦ ✤✹✥ ✓✞✤✹✥ ✁✆★✂❁ ✁✂★✂❁ ✛ ✪☎✦ ✤ ✔ ✛ ✷ ✝ ✪ ✬✭✦✺✥✻✯✮✓✰✦✽✼ ✔✆✔✾✄ ✭ ✓✞✤✹✥ ✓✆✤✚✥ ✛ ✢ ✪❃❂❄✦✺✥❅✯✮✓✴✦ ✼ ✝ ✪❃❂❈✦✳✥❅✯✮✓✴✦ ✼ ✝ ✔ ✔✘❉ Hence the integral we’re looking at is ❊✻❋ ✦ ✁✆★✂❁ ✓✂✓ ● ✛✺✷❍✢ ✔ ✪❃❂❈✦✺✥✻✯✮✓✰✦ ✼ ✝ ✔❏■ ✦ ❉ The term is bounded by a constant times ✦ ✼▲❑ ★✂❁ , whose integral converges Thus we only have to ✆ ★ ❁ ✦ ✼▲▼ ✓ ✦ ✼✫▼ ★✂❁ ✔ ✪☎❂ decide whether converges But ✛◆✷❅✢ has divergent integral, so we get convergence if and only if ✛✶✜❖✢ (in which case the integral telescopes anyway) ✁ ❉✙❉❘❉ ❙ A–3 Let and ◗ be the numbers ✓ and❙ ❯ ❯ , re❳ ■ ✔ ✤✕❳ ✼ ❱ ✥ spectively We are given that ❯❃❱❲✷ ❱ ❩❨ ❙ ✓✴❬✮❭❫❪❵❴ ✔ ✓❜❛ ❳❅✓✰❬✮❭❝❪❞❴ ✔ ✔ ✤✕❳ ✼ ❱ ✥ and ❱❏✷❵❯❃❱ ◗✌❨ for ❡❢✜ ✤ ✄❘❉✙❉❘❉✾✄✩❣ ✤ ✄✙❉❘❉✙❉✾✄✂❣ Sum the first relation over ❡❤✜ and we ✥ ❣ ❳❦✓✰❬✮❭❝❪❞❴ ✔ ✥ ❳♠✓✴❬✮❭❫❪❞❴ ✔ ❥❨ ❧❨ , or ◗ get ◗✐✷✠ Now add the first and second relations for any particu❙ ✓♥❛ ✥ ❳ ■ ✔ ✤✕❳ ✼ ❱ ✥ ◗ ♦❨ lar value ❱ ✓✴❬✮❭❫ ❪❵❴ ✔ of ❡ and we get ✥ ❱ ✷ is divisible by 7, and 10 But we know ◗ ❛ ❳❦✓✴❬✮❭❫❪❞❴ ✔ ■ is coprime to 7, so ❱♣✷ ❱♣❨ ✦ ✁ ✥❵s✙s❘s❜✥◆✦ ✓✴✈ ✤ ✔ ✪☎✈ ⑥ q❃r✠✜ r❲✷✉t ✷ , so that q✕✇✮✜❖q✙①②✜ A–4 Let ❳ These form a cyclic sequence that doesn’t change when you rotate the necklace, except that the entire sequence gets translated ✦ by a constant In particular, it makes sense to✦ choose ❱ for which q✕❱ is maximum and ❳ ✇ ; this way q✕❱❤③ make that one for all ❡ , which gives ✦ ✁ ✥✧s✙s❘s✭✥④✦ ✓✴✈ ✤ ✔ ✪❃✈ , but the right side may be ❱✚③⑤ ✷ ✤ ❡ replaced by ❡♣✷ since the left side is an integer ⑥ ✝✭✄✙❉❘❉❘❉✙✄❄⑥ ❷ ⑦ ✇ but nots ✦ to Choose a ✦❸ vector to ⑦ ❳ orthogonal ⑥ ✓ ✆✔❺❹ ❻❹❽❼ ⑥ ⑥ ⑦⑥ ✁ Since as , the same is true of ❷ ; ✓ ❷ ⑥ s ⑦ ⑥ ✁ ✔ ❶ ✁ ❛ ✁ ✓ ✆✔ In other words, if but that is simply ❳ ❛ ✓ ✆✔ ❶ , then ❱ must also go to ❱❿❾✜ However, it is easy to exhibit a solution which does not go✓ to The sum of the eigenvalues of the matrix ➀ ➀ ✔ ✜ ✛ ❱➂➁ , also known as the trace of , being the sum ➀ ➀ of the diagonal entries of , is nonnegative, so has an eigenvalue ➃ with nonnegative real part, and a cor⑥ ⑥ responding eigenvector ⑦ Then ❯☎➄❘➅ ⑦ is a solution that does not go to (If ➃ is not real, add this solution to its complex conjugate to get a real solution, which still doesn’t go to 0.) Hence one ❳ of the ❶ ❱ , say ✦❸ ⑥ ✓ ✆✔ s ❷ ⑥ ✜ for all ✦ ✁✂★✆❁ ✯✮✓✴✦ ✼ ✝ ✔ ■ ✁ ❉❘❉✙❉✂■❚❙ A–5 Everyone (presumably) knows that the set of solutions of a system of linear first-order differential equations ✈ with constant coefficients❛ is -dimensional, with ba✓ ✆✔☎⑥ sis vectors of the form ❱ ⑦ ❱ (i.e a function times ⑥ a constant vector), where the ⑦ ❱ ✦❤are linearly indepen⑥ ✓ ✆✔ dent In particular, our solution can be written as ⑧ ✇ ✁✽❶ ❛ ✓ ✆✔☎⑦ ⑥ ✁ ❱ ❱ ❱⑩⑨ ❶ ✁ , is zero, in which case A–6 View as a random walk/Markov process with states ✓ ✄♥➆❈✄ this ✔ ❡ t the triples of integers with sum 0, corresponding to the difference between the first, second and third rows with their average (twice the number of columns) Adding a new✓✞✤ column adds on a random permutation ✤ ✔ ✄ ❳ ✄ of the vector I prefer to identify the triple ✷ ✝ ✓ ✄♥➆❈✄ ✔ ✓ ➆❝✔ ✥❆✓ ➆ ✔➈➇ ✥❆✓ ✔➈➇ in ❡ t with the point ❡❏✷ ✷☛t t❇✷➉❡ ➇ the plane, where is a cube root of unity Then adding a new column corresponds to moving to one of the six neighbors of the current position in a triangular lattice ✈ What we’d like to argue is that for large enough , the ratio of the probabilities of being in any two particular states goes to Then in fact, we’ll see that eventually, ✤ ✄ about✤ six times as many matrices have ✛❵✜➊✢❇✷ ✢✳✜ ❶ than ✛☛✜➋✢✖✜ ❶ This is a pain to prove, though, ✷ and in fact is way more than we actually need ➀ Let ➌✹✇ and ✇ be the probability that we are at the origin, or at a particular point➀ adjacent to the origin, ✁ ✁ ✇ (In fact, ➌ ✇✭➍ respectively Then ➌ ✇❄➍ is ✜ ✤❃✪✭➎ times the sum of the probabilities of being at ➀each ✈ neighbor of the origin at time , but these are all ✇ ) ✪ ➀ So the desired result, which is➀ that ✪ ➌✹➀ ✇ ✈ ✁ ✇➉➏ some large , is equivalent to ✇✭➍ ✬❈✪✭➐ ✇♠➏ ✬❚✪☎➐ for B–4 The infinite continued fraction as❴ the✤❃✪limit ✬❄✬❄❳❚❴ ✄ is defined ✬❈✬✭❳❚ ✁ ➭❺✇✭➍ ✜ ✷ ➭❺✇ of the sequence ➭❺①✻✜ Notice that the sequence is strictly decreasing (by induction) and thus indeed has a limit ➭ , which satisfies ✝ ✬❄✬❄❳❚❴ ✤☎✪ ✬❄✬✭❳❝❴ ✥❖✤ ❳ ➭❖✜ ✷ ➭ , or rewriting, ➭ ✷ ➭ ✜ Moreover, we want the greater of the two roots Suppose on the contrary that this is not the case; then ✈ ➀ ❶ ✓❜✬❈✪✭➐ ✔ ✇ ✇ for some constant However, if ✲ ✈ ➎✭➑ ✜ , the probability that we chose of the six ➑ ✓❜➎✭➑ each ✔✘➒ ✪❫➓ ➑ ➒ ➔ ➎ ➔✂→↔➣ types of moves times is already , which by Stirling’s approximation is asymptotic to a ✩ ★ ✝ ➑ ✼✫↕ This➀ term✪ ➀ alone is✬❚✪☎bigger than constant times ❜ ✓ ❚ ✬ ☎ ✪ ➐ ➐ ✔ ✁ ✇ ❶ ✇✭➍ ✇♠➏ , so we must have for some ✈ ✤ ➀ ✁ ✪ ➀ ✇♠➏ ✷♠➙ for any (In ❳ fact, we must have ✇❄➍ ➙❇➛ ) Now how to compute the eighth root of✤ ➭ ? Notice that ✦ ✦ ✝ ✦☛✥➯ ❳ if satisfies the quadratic ✷❅✛ ✜ , then we have ❳ ✜ B–1 For a given ➜ , no more than three different values of ✓✴✦ ✔ ➜ are possible (four would require one part each of size at least 1,2,3,4, and that’s already more than 9✓✴✦ el✦ ✄✆➝ ✓ ✓✴✦ ✔✘✄ ✔✂✔ ➜ ➜✽➞ ements) If no such exist, each pair ✦ occurs for at most element of , and since there are ➐↔➟✺➐ only possible pairs, each must occur exactly once ✓✴✦ ✔ In particular, each value of ➜ must ✓✰✦ ✔ occur times ✓✴✦ ✔ However, clearly any given value of ➜ occurs t❝➜ times, where t is✓✴✦ the number of distinct partitions of ✔ that size Thus ➜ can occur times only if it equals or 3, but we have three distinct values for which it occurs, contradiction ❊ ✓ ✸ ① ✷➡✛✹➢✂➤⑩➥❇➦ ✝ ✥✧✓ ✔ ❊ ✜ ✝✂➠❄➨ ✢✫➧ ✸ ❭ ➢❫➦ ✔ ✝ ■ ✤✹✥⑤✓ ❶ ✪ B–5 This problem is dumb if you know the SpragueGrundy theory of normal impartial games (see Conway, Berlekamp and Guy, Winning Ways, for details) I’ll describe how it applies here To each position you assign a nim-value as follows A position with no moves (in which case the person to move has just lost) takes value Any other position is assigned the smallest number not assigned to a valid move from that position For a single pile, one sees that an empty pile has value 0, a pile of has value 1, a pile of has value 2, a pile of has value 0, a pile of has value 1, and a pile of has value ➦ ✛✚➧ ❭ ➢ ✦✫✪ ✛ ✔ ✝ You add piles just like in standard Nim: the nim-value of the composite of two games (where at every turn you pick a game and make a move there) is the “base addition without carries” (i.e exclusive OR) of the nimvalues of the constituents So our✬➡starting position, ➺✻❳↔➺➸✤✹ ➺✻❳ ➐ with ✜ piles of 3, 4, 5, 6, has nim-value ■ ✦ ❉ ① ✦➩✪ Let✝ ➦❧✥ ✜ ❭ ✝ ✛ in the second integral and write ➢✆➤⑩➥ ➦ ➧ ➢ ➦ and you get ✝✩➠ ❊ ✝ ✝ ✝ ❭ ✝ ✸ ✥ ■ ✛ ➢✂➤⑩➥ ➦ ✢ ➧ ➢ ➦ ➦ ✿ ① ✝✂➠ ❊ ✝ ✝ ✥✧✓ ✝ ✥ ❶ ✝ ✔ ❭ ✝ ■ ❉ ✜ ✛ ➢✆➤⑩➥ ➦ ✛ ➧ ➢ ➦ ➦ ① as A position is a win for the player to move if and only if it has a nonzero value, in which case the winning strategy is to always move to a position (This is always possible from a nonzero position and never from a zero position, which is precisely the condition that defines the set of winning positions.) In this case, the winning move is to reduce the pile of down to 2, and you can easily describe the entire strategy if you so desire ✝ ✝ Since the left side is increasing as✝ a✥ function of ✢ , we ❶ have equality if and only if ✢ ✜⑤✛ ✈ ✤ ✈ ✝ Clearly, of ✦ ✝ ✦✶then, ✥✧✤ the positive square roots ✦ ✝ ✓ ✝ the ✥❦quadratic ✬ ✔ ✁✆★✩✝ ✦✮✥ ✷❆✢ satisfy the quadratic ✁✂✷ ★✩✝ ✢ ✤ ❳ ✜ ✁✂★✂❁ ➭ ✦ ✝ Thus compute is the greater ❂❝❴❃we ✦✗✥➳ ✤ ❳ that ✜ root✦ ✝ of ❴❃✦❞✷ ✥✌✤ , ✁✂➭ ★✂➵ is the greater root ✵ ❳ ✝ ✷ ✜ ➭ of , and is the greater root of ✦ ➐❄✦✺✥➸✤ ❳ ✓✰➐②✥ ➫ ✔ ✪❄✬ ✷ ✜ , otherwise known as B–2 For those who haven’t taken enough physics, “rolling without slipping” means that the perimeter of the ellipse and the curve pass at the same rate, so all we’re saying is that the perimeter of the ellipse equals the length of one period of the sine curve So set up the integrals: ✝✩➠ ✝ ✦✶✥⑤✤ ✔ ✓✴✦ ✥ ✦✶✥⑤✤ ✔ ✷➲✛ ✛ ❁ ✝ ✝ ✦ ✓ ✬ ✔ ✦ ✥ ✤ ❉ ⑤ ✷ ✛ ✷ ✓✴✦ ✜ ➐ B–3 For ✜ we obviously get 45, while for ✜ the answer is because it both changes sign (because determinants are alternating) and remains unchanged (by symmetry) when you switch ✈ ✬ any two rows other than the first one So only ✜ is left By the multilinearity of the determinant, the answer is the determinant of the matrix whose first (resp second) row is the sum of all possible first (resp second) ✓✴rows There are 90 ❂❚➫❄❳ ✄ ❂❈❳❈➫ ✔ first rows whose sum is the✓✴❂❚vector , and 100 ➫❄❳ ✄ ❂❚➫✭❳ ✔ second rows whose sum is ❂❝➫✺➟➉❂❝➫✭.❳ Thus the answer ❂❚➫❄❳✺➟☛❂❚➫❄❳ ❂❝➫✭❳✶➟➉❂❚❳❈➫ ✬❄❳❈✬❈➫✭❳ ❉ is ✷ ✜ ✜ ✄✆➼❻✄✆➽ B–6 Obviously ➻ have to be greater than 1, and no two can both be rational, so without loss of✦✽generality as➚ ✦ ✦❏➶ ➼ sume that ➻ and are irrational Let ➾ ✜ ✷➳➪ ✦ ➑ ✟❆➹ ✓ ✔ ➻ denote ❛❤ the Then if and ✓✴➑❞fractional ✪ ✔✚✟ ✓✞✤ part ✤❃✪ of✄ ✤ ✔❚ ❳❏➚ ➘ ➻ ✷ ➻ ➾ only if In particular, this✤ ✤ ✄✙❉❘❉✙❉✾✄ ✈❤➚ ✓✴✈➬✥➲✤ ✔ ✪ ➹ ✓ ✔❈➴ ➾ ➻✈ ✽➮❤✷ means that ➻ contains ➷ elements, and similarly Hence for every integer , ✈ ✈❐✥⑤✤ ✜✃➱ ➻ ❒ ✥ ➱ ✈✉✥⑤✤ ➼ ❒ ✥ ➱ ✈✉✥⑤✤ ➽ ❒ ✷ ➐ ❉ ✈ Dividing through ✤❃✪ ✥⑤by ✤❃✪ ➼ shows that ✈ ➻ that for all , ❰ ✷ ✈◆✥➸✤ ➻ Ï ✥ ❰ ✈ ✥ and ⑤ ✤❃✪ taking ✤ the limit as ➽ ✜ That in turn ✷ ✈✮✥⑤✤ ➼ Ï ✥ ❰ ✷ ✈◆✥⑤✤ ➽ Ï ✜ ✓ ❹❮❼ ✈ Ð ➚ ✄ ✈ ➚ ➾ q of points ➾ is dense (and in fact equidistributed) in the unit square In✪ particular, our claim def✪ ✥ ➼ ❶ for some integers initely holds unless ✛ ➻ ✢ ✜ ✄ ✄ ❶ ✛ ✢ implies ✬ ❉ On the other hand, suppose that such a relation ➼ does hold Since ➻ and are irrational, by the one-dimensional Weil theorem, the set✓✴✦ of points ✓ ✈❲✪ ➚ ✄ ✈❲✪ ➼ ➚ ✄✆➝❏✔ is ✦✮ dense in the set of in the ➾❚✷ ➻ ➾❚✷ ✥ ➝ unit square such that ✛ is an integer It is✓✰✦ simple ✢ ✄✂➝❫✔➡✟ enough to show that this set meets the region ➾ ➓ ❳ ✄ ✤ ➣ ✝❀Ñ ✦✮✥ ➝ ✤✭➚ ✥ unless ✲ ✛ ✢ is an integer, and that ✤❃✪ ✥➸✤☎✪ ➼ would imply that ➻ , a quantity between and 1, is an integer We have our desired contradiction Our desired contradiction is equivalent to showing that ✈ the left side actually takes the value for some Since the left side is an integer, it suffices to show that ✓✴✈✉✥➸✤ ✔ ✪ ➚✚✥ ✓✰✈✉✥⑤✤ ✔ ✪ ➼ ➚ ✤ ✈ ➾❈✷ ➻ ➾❈✷ ✲ for some A result in ergodic theory (the two-dimensional version ✤ ✄✆Ð☎✄ of the Weil equidistribution theorem) states that if q are linearly independent over the rationals, then the set