On triangular billiards

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Marilyn Holmes
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1 O triagular billiars Abstract We prove a cojecture of Keyo a Smillie cocerig the oexistece of acute ratioal-agle triagles with the lattice property. MSC-iex: 58F99, 11N25 Keywors: Polygoal billiars, Veech property, Jacobsthal fuctio I a recet paper[?] o Billiars o ratioal-agle triagles, R. Keyo a J. Smillie prove the followig theorem: Theorem 1 Let T be a acute o-isosceles ratioal agle triagle with agles α, β a γ, which ca be writte as p 1 π/q, p 2 π/q a p 3 π/q with q The T is a polygo with the lattice property if a oly if (α, β, γ) is oe of the followig: (π/4, π/3, 5π/12), (π/5, π/3, 7π/15), (2π/9, π/3, 4π/9). They further showe, that the restricitio o q may be roppe, if the followig cojecture was true(see [?], p. 94f): Cojecture 2 Let, s, t be itegers with (, s) = 1, 1 s, t <. Assume that for all p with (p, ) = 1 we have 2 < ps mo + pt mo < 3 2. The oe of the followig coitios hol true: 78, s + t =, s + 2t =, 2s + t =, or is eve, a t s = 2. I this ote we will prove this cojecture: Theorem 3 Cojecture 2 is true. Note that the classificatio of o-obtuse ratioal agle triagles with the lattice-property is complete, sice the cases of isosceles a right agle triagles are completely solve i [?], too. By irect calculatio, R. Keyo a J. Smillie showe, that Theorem 3 is true for We will use this fact at several steps i the proof. The proof will epe o several facts cocerig the istributio of relative prime resiue classes, collecte i the ext Lemma. We write g() for the Jacobsthal fuctio, give by the maximal ifferece of cosecutive itegers relatively prime to, a ω() for the umber of istict prime factors of. Lemma 4 1. We have g() 2 ω(). If ω() 12, we have g() ω() Assume that (a,, ) = 1. The i every iterval [x, x + g()] there is some iteger ν, such that (, ν + a) = 1. 1

2 3. For all > 2 there exists some a with (, a) = 1 a 12 < a < If m is the prouct of the first ω() prime umbers, the g() g(m). 5. We have g(30) = 6, g(210) = 10, g(2310) = 14, g(30030) = 22, g(510510) = 26, g( ) = 34. Proof: The first statemet was prove by Kaol[?]. To prove the seco statemet ote first that it is trivial if (, ) = 1, for if 1 (mo ), the the itegers ν + a are cosecutive (mo ), a oe is coprime to, cotraictig the efiitio of g. Now we may without loss assume that is squarefree. If (, ) = e > 1, the itegers ν + a are coprime to if a oly if they are coprime to /e, thus usig the case (, ) = 1 we get that there is some ν [x, x + g(/e)] such that (ν + a, ) = 1. The thir statemet follows for > 30 from the first oe, for 3 30 by irect ispectio. The fourth statemet was prove by Iwaiec[?]. The fifth statemet ca be checke by irect computatio. Note that the fourth a fifth statemet together greatly improve the first oe for ω() 8. Note further that the asymptotic behaviour of g is much better uerstoo, usig e.g. the result of Iwaiec[?], it is easy to show that there are at most fiitely may exceptios to cojecture 2. The ifficult part of the proof of Theorem 3 is to give a upper bou for a fi properties o the woul-becouterexample which makes it feasible to rule out these fiitely may values. To prove our Theorem, we first ote that we may choose s = 1, sice otherwise we replace p by p ps 1 (mo ). The we have < t < 2. I the first step we exclue o values of. Assume that is a o couterexample to Theorem 3. Defie the iteger k by the relatio 1 1 < t 2 k < 1 1, a a := t (1 2 k ). Sice is 2 k+1 o, 2 k is relatively prime to, hece we get 2 k + 2 k t mo > 2. But we have 2 k t = (2 k 1) + 2 k a, hece 2 k (a + 1) > 2, i.e. a > 1. By the efiitio 2 k+1 of k, we have a <, thus t = [ ( )] ( ) k+1 2. Write t = 1 1 k+1 2 α. k+1 Next we give a upper bou for 2 k. Write t = b. The cases b = 1 a[ b = 2 are] exclue, sice we woul have s + t = resp. 2s + t =. If p 2(b 1), b, we have pt mo +p < 2, thus if there is some p i this iterval relatively prime to, we are oe. Thus we have b 2(b 1) < g() The left ha sie is ecreasig with b, thus if b < the left ha sie is at ( least 2) ( 1), a for > this is > 3. Hece we obtai the bou < 3g(). By Lemma 4 this implies ω() 4, thus g() 10 a < 300. Thus we may suppose b >. Let q < 2 k+1 be a o prime, a efie the iteger l by the relatio 2 l < q < 2 l+1. Assume that q. The (q2 k l, ) = 1, thus we get q2 k l t mo + 2

3 q2 k l > 2. Usig the relatio t = ( k+1 ) α with 0 < α < 1, this becomes Sice q 2 l + 1, this implies q2 k l t mo + q2 k l > 2 q 2 l+1 q2k l α + q2 k l > 2 2 q 2 l+1 + q2k l > 0 0 < 2 l+1 + q2k l 2 l+1 + 2k+1 2 l+1 + hece 2 l+1. Thus is ivisible by all o primes. Usig the elemetary bou θ() > /2, where θ(x) = p x log p, this implies 2 > e /2, which i tur implies < 121. However, Theorem 3 is true for all < 10000, thus we coclue that it is true for all o. Thus assume that (, t) is a couterexample to Theorem 3 with eve. We show that t caot be too close to /2 or to. The proofs for these two cases ru parallel, a we will oly give the first oe. Set t = 2 + b. Let p be ay iteger relatively prime to, i particular, p is o. The we have pt = p 2 + bp + bp (mo ) 2 thus if is a couterexample to our Theorem, we coclue that bp [/2, 3/2 p], i.e. p [ 2b, 3 2b p ] b. The case b = 1 is exclue, thus the upper bou of this iterval is b, thus i particular we have p [ 2b, ] b. But the oly coitios impose o p were that p is o a coprime to. Sice all eve itegers are ot coprime to, we get that the iterval [ 2b, ] b cotais o iteger relatively prime to. Hece g() > 2b, thus b > 2g(), i.e. t > /2 + 2g(). I the same way we have t < 2g(). Set w = (t, ). As p rus over all itegers relatively prime to, pt rus over all itegers with (pt, ) = w, a pt mo has perio /w. Hece there is some p < /w, relatively prime to with pt w (mo ). But the pt mo + p w + /w, a this is /2, uless w = 1, 2, /2 or. The last two cases are trivially exclue. Thus we are left with the cases w = 1, 2. Now t is a ratioal umber with eomiator >, thus applyig Dirichlet s Theorem we fi a iteger a some e, such that t e < 1. Assume that = 1. The t e < 1, a because /2 < t <, we coclue t >. Together with the bou prove above we obtai the iequality > 2g(), i.e. 2g() >. Usig the first statemet of Lemma 4, this yiels ω() 4, thus < 1156, but for < the Theorem is alreay prove. I the same way we exclue the case = 2. Now assume > 2. The by Lemma 4, statemet 3, we fi some a relatively prime to with 12 < a < Let p be a iteger relatively prime to which also satisfies p ae 1 (mo ). 3

4 Note that the right ha sie exists, sice (e, ) = 1. Write p = k + a. The we have pt = pe + θ p = ke + a e + θ p a + θ p (mo ) where θ is some real umber of absolute value < 1. But pt mo is > 2 p, thus either the right ha sie is > 2 p, which yiels a + p > 2 p or the right ha sie is egative, which yiels a p < 0 From ow o, we will oly cosier the first iequality, because the seco oe ca be ealt with similarly, but gives a little stroger bous. By the choice of a we have a/ 5/12, thus we get p( + 1) > /12. By Lemma 4, statemet 2, p ca be chose to be (g()+1). Thus we obtai the iequality ( +)(g()+1) > /12. Sice, we fially coclue g() > /24 1. The bou g() < 2 ω() shows that this is oly possible for ω() 9. Now the improve bou g() ω() 2 lowers the bou to 7, a we ca use the fifth statemet from Lemma 4 to coclue < (24 27) 2, thus ω() 6 a < (24 23) 2 = Assume that p is some prime umber, such that the least positive resiue of ep (mo ) is i the iterval [/12, 5/12]. The by the argumet above, we get p( + 1) > /12 or p. Hece all primes p which satisfy this cogruece coitio, have to ivie. By the bous give above, it suffices to fi 7 such primes to exclue the pair (, ). To fiish the proof of Theorem 3, ote first that = 552. Choose some, a compute p max = /+1. Cout the umber of resiue classes a relatively prime to, with /12 < a < 5/12, a call this umber N.Cout the prime umbers up to p max i all reuce resiue classes (mo ), a choose those N sequeces with the least umber of primes i it. If is a couterexample to Theorem 3, a is correspoig i the sese escribe above, the is ivisible by all these prime umbers, i particular there are at most 6 such primes. Doig this for all 552, we fou o such that there coul correspo some givig a couterexample to Theorem 3. All computatios were performe o a Silico Graphics Iy workstatio usig Mathematica 3.0. Refereces [1] H. Iwaiec, O the problem of Jacobsthal Demostr. Math. 11, (1978) 4

5 [2] W. Iwaiec, A ew form of the error term i the liear sieve Acta Arith. 37, (1980). [3] H.-J. Kaol, Ueber eie zahletheoretische Fuktio vo Jacobsthal Math. A. 170, (1967) [4] R. Keyo, J. Smillie, Billiars o ratioal-agle triagles, Commet. Math. Helv. 75, (2000) [5] W. A. Veech, Teichmueller curves i mouli space, Eisestei series a a applicatio to triagular billiars Ivet. Math. 97, No.3, (1989). 5

ALGEBRA HW 7 CLAY SHONKWILER

ALGEBRA HW 7 CLAY SHONKWILER Prove, or isprove a salvage: If K is a fiel, a f(x) K[x] has o roots, the K[x]/(f(x)) is a fiel. Couter-example: Cosier the fiel K = Q a the polyomial f(x) = x 4 + 3x 2 + 2.