Solutions to the 62nd William Lowell Putnam Mathematical Competition Saturday, December 1, 2001 Manjul Bhargava, Kiran Kedlaya, and Lenny Ng ✁ ✄✂✆☎✞✝✠✟✡☎☛✂☞✟✡☎✌ ✍✂✎☎☛✝✠✟✑✏✒✂ A–1 The hypothesis implies ✝✔✓✕✂✗✖✙✘ ✝ ✂✚☎✛✝ for all (by replacing by ✍✂✤☎✣✝✠),✟★and hence ✝✡☎✜ ✄✂✢☎✣✝✠✟✤✏✥✂ ✝✔✓✕✂✦✖✧✘ ☎✣✂✩✏✪ ✝ for all (using ) A–2 Let ✫★✬ denote✮ the desired probability Then and, for ✴✗✵ , ✫ ✬ ✏✷✶ ✸ ✴ ✸ ✴✺✹ ✏✷✶ ✸ ✴ ✮✠✻ ✫ ✬✽✼✾✭ ✹ ❀❂✮ ✮✠✻ ✸ ✴✺✹ ✫ ✬✽✼✾✭ ✹ ✏ ✮ ✶ ✁✮❁❀ ✮✿✻ ✸ ✆ ✴ ✹ ✮ ✫✣✭ ✏✯✮✱✰✳✲ ✫ ✬✽✼✾✭ , ✟ ✓✕♦❈✓❚♣ ✮❄❃ ✸ ✺ ✴ ✹ ✰❈❇ ✏❊✲✳✰✳❋ ✸ The recurrence yields ✫❆❅ , ✫★❉ , and by a simple induction, one then checks that for general ✴ one ✏ ✰✿ ✮✱✟ ✸ ✴✺✹ has ✫ ✬ ✴ Note: Richard Stanley points out the following nonin ✄❍✔✟✤✏❏■ ✬❑▼▲ ✄❍ ✟✁✰✿ ✮✱✟ ✸✳◆ ✹ ductive argument Put ● ❍❖ ; ✹ ✸✳◆ ✭ ✍❍✟ then the coefficient of in ● is the probability of getting exactly ◗ heads Thus the desired number is ❘✮✱✟❙❀ ❚❀❯✮✱✟✁✟❘✰ ✸ , and both values of ● can be com● ● ✁✮✱✟✤✏❱✮ puted directly: ● , and ● ❚❀❯✮✱✟✩✏ ✮ ✲ ✲❳❲ ❇❳❲❩❨✱❨❬❨❈❲ ❀✑✮ ✸ ✴ ✸ ✴✺✹ ✮ ✏ ✍❍✟❫✏❵❴ ✮ ✸ ✴✺✹ ✮ ❃ ❍ ❅ Note: a more sophisticated interpretation of this argument can be given using Galois theory Namely, if ❛ is neither a square nor twice a square, then the number ❧♠ ❛ ✟ ❧♠ ✟ ❞ ✸ are distinct quadratic fields, fields and ❞ so their compositum is a number field❢ of degree 4, whose Galois group acts transitively on ❜ ❞ ❛❂❜ ❞ ✸✠✐ Thus ✫ ❭ is irreducible ✏ A–3 By the quadratic formula, if ✫❪❭ , then ❛❝❜ ✸❈❞ ✸ ❛ ✸ , and hence the four roots of ✫ ❭ are ✹ ✘❡✏❣❢ ❜ ❛❤❜ ❞ ✸✠✐ given by If ✫★❭ factors into two ❞ nonconstant polynomials over the integers, then some ✘ subset of consisting of one or two elements form the roots of a polynomial with integer coefficients First suppose this subset has a single element, say ❛❥❜ ❞ ✸ ; this element must be a rational number ❞ ✟ ✏ ✸ ✹ ❛❊❜ ✸❈❞ ✸ ❛ Then ❞ ❛❤❜ ❞ ✸ ❅ is an integer, ✏ ❅ ❛ ✸ ✴ But then so ❛ is twice a perfect square, say ✏ ❜ ✮ ❛✑❜ ❞ ✸ ✏❱ ❜ ✮✱✟ ❞ ✸ ✴ is only rational if ✴ , i.e., ❞ ✏ ✸ if ❛ Next, suppose that the subset contains two elements; ❢ ❢ then we can take it to be one of ❞ ❛✯❜ ❞ ✸❦✐ , ❞ ✸ ❜ ❢ ✟ ❛ ✐ or ❜ ❞ ❛ ✹ ❞ ✸ ✐ In all cases, the sum and the ❞ product of the elements of the subset must be a ratio✖❱❧ , nal number In the first case, this means ✸ ❞ ❛ ❛ so is a perfect square In the second case, we have ✖✙❧ ✸ ❞ ✸ , ✟ contradiction In the third case, ❛ ✖✪❧ we have ❅ ✖❡❧ ✹ ❞ ✸ , or ❛ ✹ ✸ ✹ ✸❈❞ ✸ ❛ , which ❞ means that ❛ is twice a perfect square ✄❍✔✟ We conclude that ✫ ❭ factors into two nonconstant polynomials over the integers if and only if ❛ is either a square or twice a square ✏ ✓✕t✡✉✥✏❡♦ ✉✆✓✕✈ ✏ A–4 Choose so that q❯r ♥ ♥▼s❯r r s ♣ r❯s , and let ✇ ①✌②✆③✩④ denote the area of triangle ✉ ✏ ✉⑤t Then ✇ s✎q✚④ ①❫②☛③ ✇ q✚④ since the triangles have the same altitude and base t Also ✉ ✏⑥ ✰ ✟ ✉ ✏✷✮✚❀ ✏ ✇ s❯q⑤④ s✎q s❯r ✇ s❯r⑤④ ♥ , and ✇⑦q❯r ④ ✰ ✟☞ t❁✰ ✉⑤✟ ✉ ✏ ✁✮✚❀⑧♦❬✟ (e.g., by the q✎r s❯r r r ✇ s❯r✚④ ♥ law of sines) Adding this all up yields ✮❁✏ ✏ or ♥ ❘✮ ✹ ✉ t t s❯q✚④✿✹❂✇ s ④❦✹❂✇⑦q❯r ④ ❘✮✩❀ ✟ ❘✮✩❀⑩♦✱✟✢✏ ❀ ❀ ✸ ✸ ♥ ✹⑨♥ ♥ ✇ ✉ ♦✱✟✤✏❱✮ Similarly ♦❈ ❘✮ ✹ ♣✕✟✤✏⑧♣✱ ❘✮ ♥ ✹⑨♥ ♦ ✟✤✏❶✮ ❴ ✓✕❺✑✟❩❻ ✽ ❴✿✓❼❺✑✟ given by Let ●❸ ❷❹✮❬✇ ✰✿ ✁✮ ✇ ✄❍✔✟❽ ✏❵ ❍✟ be the ✟✁✟❘function ✟♠✏ ● ✹ ● ● ● ♥ ♥ However, ; then ✄❍✔✟ ❍ ✍❍✟✁✟ is strictly decreasing in , so ● ● is increas● ✄❍✔✟❘✟✁✟ ing and is decreasing Thus there is at most ● ● ● ❍ ✄❍✔✟❘✟✁✟✜✏✥❍ one such that ● ● ● ; in fact, since the equa ✄❾❈✟✤✏✥❾ ❾✎✏❶ ❿❀✛✮ ❇✳✟✁✰ ✸ , we tion ● has a positive root ✹ ❞ ✏✥♦✜✏⑧♣✢✏➀❾ must have ♥ ✉ t ✏➁ ➂✉⑤t❁✰☞✉ ✟ ✉ ✏➃❾ We now ✏ compute , ✇ s ④ r ✇ s❯r✚④ ✉ ✰ t❯✟ ✉ t ✏ ❾✳✰ ✸ , analogously ✇ s❯➄✦④ s❯➄ s ✇ s ④ ✘ ✏ ✉✜➅ ✏➆❾✳✰ ✘✾➅ ✏➁➇ ✉ ❀ ✸ , and ✇⑦➄ ✇⑦s❯r ④ ✇⑦r ④ ④ ✇ s❯r✚④ ✉ ❀ ✘ ❀ ✉✚➅ ➇❬✏✯➇➈✮✤❀⑩✲✳❾❈✰ ➇✱✏❊➉ ✼ ❉❼➊ ➋ ➌ ✸ ✇ s❯➄✦④ ✇⑦s❯r ④ ✇⑦r ④ ✁✮ ♦❬✟❫✏➆✮ Note: the key relation ♥ can also be de✹ rived by computing using homogeneous coordinates or vectors ✝ ✬➎➍★✭ ❀✷ ✄✝ ✮❬✟ ✬ ✏ ❴✳❴✽✮ ✸ that ✹ ❬ ✮ ✟ ⑩ ❀ ✮ ✝ Notice ✝ ✬➎➍★✭ ✬ ✹❂✇ ✹ ④ is a multiple of ; thus divides ❴❈❴ ✏ ❋ ✮❈✮ ✮ ✲ ❬ ✸ ✸ ✸ ❲ ❲ ❲ ❴❈❴✿✮ ✝❤➏➐✮ ✸ Since is divisible by✝ 3, we must have ➒➑❹➓✠➔❽✲✳✟ ✄✝ ✮ ✟ ✬ ✱ ✬➎➍★✭ , otherwise one of and ✹ is a mul- A–5 Suppose ✝ ✄✝ tiple of and the other is not, so their difference cannot ✝ ✬✠➍★✭ ➏❶✮❳ ➒➑❹➓✠➔✌✲✳✟ be divisible by Now , so we must ✄✝ ✮❬✟ ✬ ➏❱✮❳ ➒➑❽➓✠➔❫✲✳✟ have ✹ , which forces ✴ to be even, and in particular at least ✝ ✝ ✬➎➍★✭ ❀❯ ✄✝ ✮✱✟ ✬ ➏→❀❯ ✄✝ ✮✱✟ ✬ ➒➑❹➓✠➔❽➣➎✟ ✹ ✹ ❀❯ ✄✝ ✮✱✟ ✬ ➏↔❀❯✮❳ ➒➑❹➓➎➔❹➣➎✟ ✹ Since If is even, then Since ✴ is even, ❴✳❴✽✮❯➏❊✮↕ ➂➑❽➓✠➔❽➣✠✟ ✝ ,✮✱this is impossible Thus is odd, ❴❈❴✿✮♠✏✷❋ ✮✳✮ ✮❬✲ ❲ ❲ and so must divide Moreover, ✝ ✬✠➍★✭ ❀➙ ✍✝ ✮✱✟ ✬ ➏✥✝✑ ➒➑❹➓✠➔❽➣➎✟ ✝❯➏❱✮↕ ➂➑❽➓✠➔❽➣✠✟ , so ✹ ✸ ❋ ✮❈✮ every row contains exactly elements of s , ➘ ✝↕✏❥✮ ✮✚❀ ❴✳❴✽✮ ✬✥✏ ➛ ✸ ✸ ✝✯✏➜✮❬✲ We cannot have , since for any ✴ Thus the only possibility is One eas✝❂✏↔✮✱✲✽✓ ✏ ✸ is a solution; all that ✴ ily checks that ✴ ✮❩ remains is to check no ❴❈other works ✮❬✲ ✬➎➍★that ❴✿✮⑩➏➁ ➂➑❽➓✠➔❽➞✳In ✟ fact, ✭ ➏ ✸ , then ✸ ✴➝✵ But if ✮✱✲ ✬✠➍★✭ ➏➃✮❬✲⑩ ➂➑❽➓✠➔✌➞❈✟ ✴ is even, contradiction since ✝✺✏❡✮❬✲✿✓ ✏ ✸ is the unique solution Thus ✴ Note: once one has that ❴✳❴ ✏➟✝ ✮✎❀✙ ✍✝ ✸ ✬➎➍★✭ ✸ ✹ ➊ ❅✁➦ ✼★✭➨➧ ➦ ✮ ➩ ✹ ✉ ✸ ❅ ➊ ❅✁➦ ✼★✭ ➥ ✸ ✹ ✮ ✸ ❅ ➊ ❅✁➦ ✼✾✭ ➥ ✉➲➯ ➴❼➷▼➮ ✮ ➫ where we have artificially introduced ❍❹➸❂❴ grand in the last step Now, for , ✮ ❞ since ✉ ➻❪➩ ➼ ✮ ❍ ❅ ❀✦❍❹✏ ✹ ➫ ❞ ✮ ➻➽➩ ➼ ✹ , we have ✸ ✮ ✹ ❅ ➊ ❅✁➦ ✼★✭ ✵ ✹ ❍ ➭ ❍✔✰✿ ✸ ✄❍ ✹ ❍ ❅ ❀❶❍✔✟ ➭ ❍ ➻ ➩ ❀✦❍ ✸ ❞ ✹ ❃ ✴✞✹❐➶ ✍♦✱✟ ✮✜✏ ✹ ✄♦❬✟ ● ➴❼➷▼✃ ✴❹✹❳➶ ✄♦❬✟ ✮✳✓ ✹ ❴❤✮ ✮✪❴ ✻ ❻ ✶ ✮✥❴ ❴➡✮ ✻ or vice versa, which transforms into s If we identify and with red and black, then the given coloring and the checkerboard coloring both satisfy the sum condition Since the desired result is clearly true for the checkerboard coloring, and performing the matrix operations does not affect this, the desired result follows in general ✮ ✓ ✸➵➳ ✹ ❍ ❅ B–2 By adding and subtracting the two given equations, we obtain the equivalent pair of equations ➸ ✮ ✸ ✄❍ ✹ ✮✱✟✠➺ ✸ ✮✱✟✁✟ ✮ ➴❼➷▼✃ ✸ ➘ ✶ diverges, so does Hence, for sufficiently large ❞ ✄♦❬✟ ➶ ✰ Note: Richard Stanley points out a theorem of Ryser (see Ryser, Combinatorial Mathematics, Theorem❴✚3.1) ✉ ❀❂✮ that can also be applied Namely, if and s are matrices with the same row and column sums, then there is a sequence of operations on ✸ ❲ ✸ matrices of the form into the inte- ✮ ❍ ❅ ✄♦❬✟✣✏ ❃ ✉ ❍ ❅ ❀⑩❍✟ ➭ ❍✎❀ ✹ ✸ as desired ❍ ❅ ➭ ❍ ➩ ✄♦❬✟ ● ✄♦❬✟ ● ➴❼➷▼✃ ➶ ➄ ➘ ➘ It follows that ✉⑤❍✔✟ ❅❄➭ ❍ ✸ ✹ ➫ ➩ ✮ ✏ ✹ ➫ ✮ ✏ ✴ ✸ ✄♦❬✟✣✏ ● ➴➬➷▼➮ A–6 The answer is yes Consider the arc of the parabola ❍ ❅ ➠ ✏➡✉⑤❍ ❅ ➠ ❀✪✮✱✟ ❅ ✏❥✮ inside the circle ✹ , where ✉ ✮❬✰ ✸ This intersects the we initially assume that ✵ ✉✑❀❂✮❬✰➢✉✆✓✱ ✉⑨❀ ✄❴✿✓❼❴✳✟ ❜ ✸ ❞ ✸ circle in three points, and ✮✱✟❘✰☞✉⑤✟ ✉ We claim that for sufficiently ✍large, the ❴✿✓❼❴✳✟ ➤ length of the parabolic arc between and ✉✑❀✑✮✱✰☞✉✺✓❬ ✉↕❀❳✮✱✟❘✰☞✉⑤✟ ✸ ❞ ✸ is greater than ✸ , which implies the desired result by symmetry We express ➤ using the usual formula for arclength: ✏➁➥ Similarly, because every column contains exactly ✰ elements of ➄ and➘ ✴ ✸ elements of s , ➘ is even, one can use✝ that✮ ✮✱✟ ✬ is divisible by ✹ ✴ to rule out cases ➤ ✴ ➴➬➷➱➮ elements of ✰ and ✮✱✲ ❲ Of the divisors of ❲ , those congruent to mod are precisely those not divisible by 11 (since 7❋ and ✮❬13 ✝ ✲ ❲ are both congruent to✟ mod 3) Thus divides ✝✎➏❶ ✮❳ ➒➑❹➓➎➔❹➣➎ ✝ ✮✱✲ Now is only possible if divides ✰ ✴ ❍ ❅ ❀✌❍✔✟ ➭ ❍ ✵ ✸ ✰✳❍✆✏✪❍ ➌ ✮✱✰ ➠ ✏✪❇❈❍ ➌ ✮❬❴✳❍ ❅ ➠ ❅ ✹ ✮✱❴❈❍ ❅ ➠ ❅ ✹ ❇ ➠ ➌ ✹ ➌ ➠ ✹ ❃ ❍ Multiplying the former by and the latter by ➠ , then adding and subtracting the two resulting equations, we obtain another pair of equations equivalent to the given ones, , and hence ➤⑨✵ Note:✉ a numerical computation shows that one must ✲▼➣ ❋ ❃ to obtain ➤⑧✵ ✸ , and that the maximum take ✵ ➣ ❴❈❴ ❋ ✉✪➾✪➚➵➣ ✮ ❃ value of ➤ is about ❃ ✸ , achieved for ✲⑤✏❱ ✄❍ B–1 Let ➄ (resp s ) denote the set of red (resp black) ♦✥✖ squares in such a coloring, and for ➄❶➪✑s , let ✍♦✱✟ ✄♦❬✟ ✮ ♦ ● ✴✣✹☛➶ ✹ denote the number written in square , ❴✆➹ ✍♦✱✟☞✓ ✄♦❬✟✚➹ ❀✑✮ where Then it is clear that the ● ➶ ✴ ✍♦✱✟ ♦ value of ● ✍♦✱✟ depends only on the row of , while the ♦ value of ➶ depends only on the column of Since ➠ ✟ ➋ ✓ ✹ ❍❹✏❶ ✍✲ ✭❿➧ ✮✜✏→ ✄❍❒❀ ✮✱✟✁✰ ➠ ✟ ➋ ❃ ✏❶ ✍✲ ✭❘➧ ❀❂✮✱✟❘✰ ➋ ➋ ✸ and ➠ It follows that ✹ is the unique solution satisfying the given equations B–3 Since ◆ ❅ ✹ ◆ ◆ ✹ ❀✪✮✱✰ ✟ ❅ ✸ ✮❬✰▼➣ ✏ ◆ ❅ ❀ ◆ ✹ , we have that ✮❬✰▼➣ ❮❚✴❄❰ ✏ and ◆ ◆ ✹ ✮✱✰ ✸ ✟ ❅ ✸ ✏ if and only if ❅ ❀ ◆➘ ✬➎Ð ✸❈Ï ➼ ▲ ✬ ✮✚➹ ◆ ✹ ✹ ✭ Ĩ Alternate Ư ✼ ✸ ✸ ✬ ➹ ✴ ✬➎Ð Ï ❅ ◆ Hence ➘ ◆ ✹➘ ✏ ➼ ✬ ▲ ✬➎➍ solution: Ö ✼✬ ❀ ➼ ✼❭ ✹ ✼ ✸ ✸ ✬ ❑ ✸ ✼ ➍★✭ ❑ ✹ ✟☞ ❅❼Õ ❀ ➘ ❅✕Õ ❀ ✸ ➼ ❑ ▲ ➱ ❍ ❑ Ò ✼ ✸ ❑ ✼ ✸ ✸ ✬ ❑ ➍ ❑❈ĨƠ❑ ✼ ✬➎Ð Ï ✸ ✼ ✼ ❑ ✟ ➍ ❅❼Õ ✟ ✼ ✸ ❀ ❑ Ị ❑✳ĨƠ❑ ✭ ✮✸ ✸ ✭ ❭ ✬✠Ð Õ Ï ▲ ✸ ✹ ✬➼ Ò ➼ ✹ ✭ ✬ ✭ ✸ ✏ ▲ ✸ ➩ ✼ ✼✬ ✬✿✼ ❉ ✬➎Ð Õ Ï ✹ ➼ ❭ ▲ ✸ ✼❭ Ò ✏ ✮✜✏❂✲ ✸ ✹ ✘ ✴ Let B–6 Yes, there must exist infinitely many such be ✓✕✝ ✟ ➸✙❴ ✬ the convex hull ✘ of the set of points ✴ for ✴ Geometrically, is the intersection of all convex sets ✓❼✝ ✟ ✬ ; ✴ (or even all halfplanes) containing the ✘ ✄❍★points ✓ ➠ ✟ algebraically, is ✓❼the set of points which can ❑ ❑ ✓❼✝ ✝ ✟ ✟ ✬➎æ for some be✓ written as â ✭ ✴ ✭ ✬✿å ✹ ❨✱❨❬❨ ✹⑩â ✴ ✓ ❑ â❬✭ ❃❬❃✱❃ â which are nonnegative of sum ❃ ✭ ✰✳Ø × B–4 For a rational number expressed in➇ Ø✿lowest terms, ➇ ➇ ➇ ✰✳Ø▼✟ × ✹ Ù × define ✰✳its height to be Then for Ø❥✖✒✘ any × ✰✳Ø▼✟❘✟❫✏➃➇ Ø expressed in lowest terms, we have ➇ Ø✿➇ ❅ ❀ ❅ ➇ Ù ● × × ✹ × ; ➇ since by assumption Ø ➇ ✏✯ ➛ ➇ Ø✿➇ , we have × and are nonzero integers with × Ù ● × ✰❈Ø▼✟❘✟✢❀ Ù Ĩ ➇ Ø✿➇☞❀➀➇ ☞ ➇ ❀⑧➇ Ø✿➇ ❅ ➇ × ✹ × × ➇ ✿ Ø ➇☞❀❂➇ ➇❬❀❂➇ Ø✿➇ ✹ × × ✏❶ ❼➇ ➇❬❀✑✮✱✟➢ ❼➇ Ø✿➇☞❀⑧✮❬✟ ➸ ✸ ❃ × ✹ ✸ ➸⑧✲ ✄✘➽✟ ✬ It follows that ● Õ consists solely of numbers of height strictly larger than ✸ ✴✆✹ ✸ , and hence Ú ✬➼ ▲ Ó ✭ ● ✬ Õ ✍✘➽✟✣✏✥Û ✝ ❑ ❃ ✄❍✔✟❁✏ ❍✌✏ ➠ ➠ ✟ ✝ ✬ ✹ ❛ ◆ ❀ ✴ ỗ (1) ọ Next, we show that given one ✴ satisfying (1), there exists a✝ larger one❴ also satisfying (1) Again, the ✍✝ con❑ ✰ ❑ ❀ ❻ ❻ ❺ ◆ ◆ dition as implies that ✝ ✟❘✰✿ ❀ ✟❩❻ ❴ ❻ ❺ ◆ ✬ ✴ as ◆ Thus the sequence ❢➎ ✍✝ ❑ ❀❒✝ ✟✁✰✿ ❀ ✟ ❑✳è ◆ ✐ ✬ ✬ has a maximum element; sup✴ ✏ ♥ is the largest value of ◆ that achieves this pose ◆ ✏❤ ✍✝❈é✦❀➙✝ ✟✁✰✿ ❀ ✟ ✬ maximum, and put ❛ ♥ ✴ Then the ✓✕✝➎é✱✟ ✓✕✝ ❑ ✟ ❛ ◆ line through ♥ of slope lies strictly above ✓✕✝ ❑ ✟ for ◆ ✵✯♥ and passes through or lies above ◆ for ✟❘✟ ➠ implies that ➶ ➶ ➶ ➶ from the given equation That is, ➶ is ➶ ✟ ✴ We first show that satisfies ✄✝ (1) condition ❑ ❀❽✝ The ✝ ❑ ✰ ❻❵❴ ❻ä❺ ✟✁✰✿ ❀♠✮✱✟✤❻ ◆ ◆ ◆ ✭ as implies that ❴ ❢➎ ✍✝ ❑ ❀✺✝ ✟❘✰✽ ❀❽✮✱✟ ◆ ✐ has an upper ✭ as well Thus the set ✝ ❑ ➹❂✝ ❛ ◆ ❀✑✮✱✟ ✭ ✹ bound ❛ , and now , as desired Note: many choices for the height function are possible: ✰✳Ø▼✟✡✏✯➑❫Ü▼Ý⑤➇ ➇Ơ✓▼➇ Ø✿➇ ✰✳Ø▼✟ one can take Ù × × , or Ù × Ø equal to the total number of prime factors of × and , and so on The key properties of the height function are that on one hand, there are only finitely many rationals with height below any finite bound, and on the other hand, the height function is a sufficiently “algebraic” function ✰❈Ø of its argument that one can relate the heights of × ✰✳Ø▼✟ and ● × B–5 Note that ➶ and hence ✓❼✝ ✬ We prove that for infinitely many ✴ , ✴ is a vertex ✘ on the upper boundary of , and that these ✴ satisfy the ✓❼✝ ✟ ✬ given condition The condition that ✴ is a vertex ✘ to the exison the upper boundary of is equivalent ✓✕✝ ✟ ✬ tence of a✘ line passing through ✴ with all other ❴ points of below it That is, there should exist ❛ ✵ such that ỉ ỉ ì ò Suppose is strictly increasing If â✱❅ for some ❍ ✬ choice of ➩ , then ✬ is dominated by ♥ ❅ for ✴ suffi❍ ❍ ✴ sufficiently negative But taking ✬ and ✬➎➍ ❅ for ❴ ❍ ✬❂✏ ciently negative of the right parity, we get ❍ ✄❍ ✟ ✄❍ ✟ ✬ ✵✪➶ ✬➎➍ ❅ , contradiction Thus â ❅ ✬➎➍ ❅ but ➶ ❴ ❍ ✏ ❍ ✏ ✄❍✔✟✜✏ ❍ â➱✭ and ✭ â➱✭✕♥▼✭ , we have ➶ ♥➵✭ ; since❍ ➩ for ✏❡ all❴ Analogously, if ➶ is strictly decreasing, then ❍ ✬ or else ✬ is ❍ dominated by ♥ ✭ for ✴ sufficiently â➢❅ ❍ positive But taking ✬ and ✬➎➍ ❅ ❴ for ✴ ❍ sufficiently pos❍ ✬➎➍ ❅❐à ✬ but itive of the right parity, we get ✍❍ ✟ ✄❍ ✟ ✬✠➍ ❅ ✬ , contradiction Thus in that case, ➶ à❸➶ ❍ ✍❍✟➽✏ ❍ for all ➶ ♥✱❅ Ó rewrite ▲ ❍ ✬ for all ✴ Pick ➩ arbitrary, and ✍❍define recur❍ ✏ ✟ ❴ ❍ ✏ ➶ ✬ sively by ✬➎➍★✭ for ✴❸✵ , and ✬✿✼✾✭ ✍ ❍ ✟ ❴ Þ ✏ ✄ ✝ ✝ ✠ ➣ ❼ ✂ ❘ ✟ ✰ ✼★✭ ❅ ✸ ➶ ✬ ✹ ❞ ✹ for ✴❥à Let ♥➵✭ ✏➁ ✄✝✧❀ ✝ ❅ ➣✠✂➢✟❘✰ ❞ ✸ and ♥ ❅ be the roots of ♥ ❅ ✹ and ❍ ❅ ❀❶✝✠❍❳❀❡✂➙✏↔❴ ✮ ❴ , so that ♥ ✭ ✵ ✓ ✵ ✵á ➇ ➇ ➇ ➇ ✖❣♥✱ã ❅ and ♥▼✭ ✵ ♥ ❅ Then there exist â➱✭ â ❅ such that ❍ ✏ ✖❫ß ✬ ✬ ✬ â ✭ ♥ ✭ ✹⑩â☞❅✱♥ ❅ for all ✴ ✏ä ❴ ➛ ✼ ❅❼Õ the sum as ✏➛ Note that ❮➨✴❄❰ ❛ ❅ ❛ ❛ for some ✹ ❰ if and only if ✴ ❀ Thus ✴✞✹❂❮❚✴❄❰ and ✴ ❮➨✴❄❰ each increase by except at ✏ ❛ ❅ ❛ , where the former skips from ❛ ❅ ✹ ✸ ❛ to ✴ ✹ ❛ ❅ ❛ ❛ ❅ Thus ✹ ✸ ✹ ✸ and the latter repeats the value the sums are ➘ ➘ ➘ ➘ ✬➼ ❮➨✴❩✹ ▲ ✼ ✬➎Ð ✸❈Ï ➼ ❑ ▲ ➱ ▲ ❑ ✔ ✬ Ñ ✬➎➘ Ð ➘ ✭ ✿ Ï➱ ❑ Ò ❑ ➍ ✏ ➼ ❑➱▲ ▲✔❑ Ò ❑ ➘ ✭ ✬ ✼ ❑ ✏ ➼ ✸ ✹ ✸ ❑➱▲ ➘ ✭ ❑❈ĨƠ❑ ✏ ✼ ➼ ✼ ✸ ❑➱▲ ➘ ✭ ❑✳Ó➈❑ ✏ ➼ ✼ ✼ ✸ ❑➱▲ ✭ ✏➁✲ ❃ injective Since ➶ is also continuous, ➶ is either strictly increasing or strictly decreasing Moreover, ➶ cannot ❍✗❻ ❺ tend to a finite limit , or else we’d have ➤ as ✹ ✍❍✟✁✟✤❀✑✝ ✄❍✔✟❁✏✯✂➢❍ , with the left side bounded and ➶ ➶ ➶ the right side unbounded Similarly, ➶ cannot tend to a ❍⑨❻Þ❀✜❺ finite limit as Together with monotonicity, this yields that ➶ is also surjective ◆ ❛ à❤♥ ❀❐ê Thus (1) holds for ê for suitably small ✵ ✴ ✏ ❴ ♥ with ❛ ✏ë✮✳✓ ✓ ❀❤✮ ❀ ✓❼✝ ❃✱❃❬❃ ✴ ◗ , the points ✴ ✕✓ ◗ ✝ ✓✕✝ ✟ ✴✚✹♠◗ ✬➎➍ ❖ lie below the line through ✴ ✬ ✝ ✝ ✝ ❛ ❛ ✬ ✹ That means ✬➎➍ ❖ ◗ and ✬✽✼ ❖ ✝ ✝ adding these together gives ✬✿✼ ❖ ✹ ✬➎➍ ❖ ✸ replaced by for By induction, we have that (1) holds for infinitely many ❴ ❛ such that ✴ For any such ✴ there exists ✵ sired ✬✽✼ ❖ ✟ ✝ ✝ ✟ and of❀ slope ❛ ✬ ◗; ✬ , as de-