No-three-in-line problem on a torus: periodicity

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1 arxv: v1 [cs.dm] 25 Jan 2019 No-three-n-lne problem on a torus: perodcty Mchael Skotnca skotnca@kam.mff.cun.cz Abstract Let τ m,n denote the maxmal number of ponts on the dscrete torus (dscrete torc grd) of szes m n wth no three collnear ponts. The value τ m,n s known for the case where gcd(m, n) s prme. It s also known that τ m,n 2 gcd(m, n). In ths paper we generalze some of the known tools for determnng τ m,n and also show some new. Usng these tools we prove that the sequence (τ z,n ) n N s perodc for all fxed z > 1. In general, we do not know the perod; however, f z = p a for p prme, then we can bound t. We prove that τ p a,p (a 1)p+2 = 2pa whch mples that the perod for the sequence s p b where b s at most (a 1)p Introducton In 1917, Dudeney ntroduced the No-three-n-lne-problem (see problem 317 n [Dud17]). The goal of ths problem s to place as much as possble ponts on a n n grd so that no three ponts are collnear. Ths problem s stll not solved for all n N. In 2012, Fowler, Groot, Pandya and Snapp ntroduced a modfcaton of ths problem: No-threen-lne-problem on a torus (see [FGPS12]). The purpose s to place as much as possble ponts on a dscrete torus of sze m n where m, n N so that no three of these ponts are collnear. We consder the dscrete torus as a Cartesan product {0,..., m 1} {0,..., n 1} Z 2 and we denote t T m n. A lne on ths torus s an mage of a lne l n Z 2 under a mappng π m,n whch maps a pont (x, y) Z 2 to the pont (x mod m, y mod n). 1 We are nterested n estmatng the quantty τ m,n whch denotes the maxmal number of ponts that can be placed on the torus T m n so that no three of these ponts are collnear. 2 Department of Appled Mathematcs, Charles Unversty n Prague, Malostranské náměstí 25, , Praha 1. 1 As a lne n Z 2 we consder l R 2 Z 2 where l R 2 s a lne n R 2 whch contans at least two ponts of nteger coordnates. 2 From another pont of vew, we may ndentfy T m n wth the group Z m Z n. Then lnes are cosets of cyclc subgroups Z m Z n generated by (x, y) where x, y are relatvely prme. 1

2 Fowler et al. n [FGPS12] proved that τ p,p = p + 1 and τ p,p 2 = 2p where p s an odd prme. They also showed τ m,n = 2 whenever gcd(m, n) = 1 where gcd denotes the greatest common dvsor. Ths fact follows drectly from the Chnese remander theorem. These results were generalzed by Msak, Stȩpeń, A. Szymaszkewcz, L. Szymaszkewcz and Zwerzchowsk [MSS + 16] who determned τ m,n whenever gcd(m, n) = p s a prme. More concretely, they have shown n ths case that τ m,n = 2p f m or n s dvsble by p 2 or f p = 2 and that τ m,n = p + 1 f nether m nor n s dvsble by p 2 and p s an odd prme; see Theorem 1.2. n [MSS + 16]. Msak et al. also provded a useful general upper bound for τ m,n. Theorem 1 (Theorem 1.1 n [MSS + 16]). Let m, n N. Then τ m,n 2 gcd(m, n). Results of the paper: Perodcty. We contrbute to the study of τ m,n by showng that once we fx one of the coordnates, then we get a perodc sequence. Namely, for a postve nteger z, we defne the sequence σ z = (σ z (n)) n=1 by settng σ z (n) := τ z,n. Note that σ z (n) 2n for arbtrary n by Theorem 1. In partcular, σ z attans only fntely many values. We get followng: Theorem 2. The sequence σ z s perodc for all postve ntegers z greater than 1. Theorem 2 n prncple allows to determne the values τ z,n for arbtrary bg n from several ntal values of n (the number of ntal values for n depends on z, of course). Unfortunately, for a general z we do not know what s the perod. Our proof s purely exstental. However, f z s a power of a prme, we can say more. We show that n such case the sequence reaches ts potental maxmum 2z = 2p a for a N and a prme p. Theorem 3. Let T p a p (a 1)p+2 be a torus where p s a prme and a N. Then τ p a,p (a 1)p+2 = 2pa. We can also determne the perod of σ p a. Theorem 4. Let p be a prme, a N. Let us denote m := mn{x; σ p a(x) = 2p a }. Then m = p b for some b a and the sequence σ p a s perodc wth the perod m. Note that Theorem 3 mples that such an m exsts. In fact, we show that each x such that σ p a(x) = 2p a s a perod for σ p a. Therefore, from these two theorems above we get that σ p a s perodc wth the (not necessarly least) perod p (a 1)p+2. Tools. A useful tool for determnng τ m,n s to fnd out for whch values m, n, x, y N we have τ m,n = τ xm,yn. Msak et al. showed τ p,p = τ xp,yp f p s odd prme, x, y are relatvely prme and nether x nor y s dvsble by p. In other words τ p,p = τ a,b f gcd(a, b) = p and nether a nor b s dvsble by p 2 ; see Theorem 4.5 n [MSS + 16]. We generalze ths result to the followng one. Theorem 5. Let m, n, x, y be postve ntegers such that m, n are not both 1 and gcd(x, y) = gcd(m, y) = gcd(n, x) = 1. Then τ xm,yn = τ m,n. 2

3 z Table 1: Intal values of τ z,n. gcd(n, ) Table 2: Values of σ 6 (n) accordng to gcd(n, ) = 2 3 j. Msak et al. used the followng dea n [MSS + 16]. Let us consder tor T xm yn and T m n, where m, n, x, y N, a mappng f : T xm yn T m n defned by f((a 1, a 2 )) = (a 1 mod m, a 2 mod n), and the mappngs π xm,yn, π m,n defned above from Z 2 to T xm yn, T m n, respectvely. Then π m,n = f π xm,yn. Hence the mage of every lne on T xm yn under the mappng f s a lne on T m n. Consequently, a set of ponts on T m n such that no three are collnear can be also used on T xm yn. Indeed, f such ponts were collnear on T xm yn they would be collnear also on T m n snce the mage of every lne on T xm yn s a lne on T m n. Ths leads to followng result. Lemma 6 (Lemma 4.1(1) n [MSS + 16]). Let m, n, x, y be postve ntegers. Then τ xm,yn τ m,n. Theorem 5 and Lemma 6 are one of our man tools for the proofs of Theorems 2 and 4. Small values of the sequence σ z (n). For sake of example, we present here few ntal values of τ z,n, our table s analogous to the table n [FGPS12]. See Table 1. The values σ p (n) for a prme number p can be fully determned from Theorem 1.2 n [MSS + 16]. In general, t follows from ther result that the sequence σ p (n) s perodc wth the perod p 2 for a prme number p. Consderng σ 4 (n), we can determne the values σ 4 (n) agan by Theorem 1.2 n [MSS + 16] for n {1, 2, 3, 5, 6, 7}. We also have a computer asssted proof that σ 4 (4) = 6. 3 In partcular, a confguraton of ponts showng σ 4 (4) 6 s shown on Fgure 1. We can also easly deduce that σ 4 (8) = 8: due to Theorem 1, t s suffcent to show that σ 4 (8) 8 whch follows from the example on Fgure 1. Gven that σ 4 (n) = 8, Theorem 4 mples that σ 4 (n) s perodc wth the perod 8. proof. 3 [FGPS12] also clam ths value wthout a detaled proof; wth a moderate effort t s also possble to get a computer-free 3

4 Fgure 1: Example of sx ponts on T 4 4 (left) and eght ponts on T 4 8 (rght) wthout three ponts on a lne. Regardng the values presented n the table (Table 1), we can determne exactly σ 6 (n) for n 0 (mod 6) due to Msak et al. (see [MSS + 16]). For other values we rely on the computer proof. It also gves that σ 6 (36) = σ 6 (54) = 12, whch s the maxmum of σ 6 (n) by Theorem 1, and σ 6 (24) = 8. Unfortunately, we are not able to determne the least perod for σ 6 (n). However, we conjecture σ 6 (2 k 3) = 8 for all k N. 4 If t s true then we can determne all values of σ 6 (n) and the least perod s = 108; See Table 2. Indeed, by Theorem 5, Lemma 6 and by known ntal values we get σ 6 (n) = 2 whenever gcd(n, ) = 1, σ 6 (n) = 4 whenever gcd(n, ) = 3 1, 2 1, 2 2 or 2 3, σ 6 (n) = 6 whenever gcd(n, ) = 3 2 or 3 3, σ 6 (n) = 10 whenever gcd(n, ) = 2 3 2, σ 6 (n) = 12 whenever gcd(n, ) = , or Fnally, σ 6 (n) = 8 whenever gcd(n, ) = 2 3, by our conjecture. 2 Proof of Theorem 5 In ths secton, we prove Theorem 5. For ths purpose we need the followng two well-known theorems. Theorem 7 (Chnese Remander Theorem, see [How02]). Let m, n be postve ntegers. Then two smultaneous congruences x a (mod m), x b (mod n) are solvable f and only f a b (mod gcd(m, n)). Moreover, the soluton s unque modulo lcm(m, n) where lcm denotes the least common multple. Theorem 8 (Drchlet s Theorem, see [Sel49]). Let a, b be postve relatvely prme ntegers. Then there are nfntely many prmes of the form a + nb, where n s a non-negatve nteger. 4 We are able to prove t s true for k 5 usng computer. 4

5 We agan use the dea of Msak et al. Lemma 9 (essentally Lemma 4.1(2) n [MSS + 16]). Let T xm yn, T m n be tor, where m, n, x, y N and m, n are not both 1. Let f : T xm yn T m n be a mappng defned by f((a 1, a 2 )) = (a 1 mod m, a 2 mod n). If the premage of every lne on T m n s a lne on T xm yn, then τ xm,yn = τ m,n. Proof. The Lemma follows from Lemma 6 f τ xm,yn = 2. Suppose τ xm,yn > 2. Let M T xm yn be a maxmal set such that no three ponts from M are collnear. We show that the mage f(m) on T m n also satsfes the no-three-n-lne condton. Frst of all, note that M = f(m). Indeed, f f(a) = f(b) for two dstnct A, B M, then A, B and C would be collnear for some pont C M \ {A, B}. Now, let us consder three dstnct ponts from f(m). If they were collnear, ther premages would be also collnear snce the premage of every lne on T m n s a lne on T xm yn. Consequently, we have τ xm,yn τ m,n and by Lemma 6 we have τ xm,yn τ m,n. Now, we generalze Lemmas 4.3 and 4.4 from [MSS + 16]. Lemma 10. Let x, y, m, n N such that gcd(x, y) = gcd(m, y) = gcd(n, x) = 1. Let T xm yn, T m n be tor and let f : T xm yn T m n be a mappng defned by f((a 1, a 2 )) = (a 1 mod m, a 2 mod n). Then the premage of every lne on T m n s a lne on T xm yn. Proof. Frst, note that t s suffcent to consder only lnes whch contan the orgn. The other lnes are just translatons of such lnes. Let l be a lne on T m n whch contans the orgn. We can express t as l = {π m,n (k (u, v)) ; k Z} where gcd(u, v) = 1. We have to fnd u, v N 0 such that gcd(u, v ) = 1 defnng the lne l = {π xm,yn (k(u, v )) ; k Z} on T xm yn whch t s the premage of l. In other words, for every pont (a 1, a 2 ) from l ts mage f((a 1, a 2 )) belongs to l and for every pont (b 1, b 2 ) from f 1 (l) the followng equatons have soluton. u k b 1 (mod xm), (1) v k b 2 (mod yn). (2) u Let us denote g u,m := gcd(u, m) and g v,n := gcd(v, n). Snce g u,m and m g u,m are relatvely prme, by the Drchlet s Theorem (Theorem 8) there s defnng a prme p u = u g u,m + m g u,m whch s greater than mnxy. At the same way, we can get j defnng another prme p v = v g v,n + j n g v,n greater than mnxy such that p v p u. Ths does not work f ether u or v s zero (one of them s always non-zero). In such case, we set := 1 or j := 1, respectvely, and hence p u = 1 or p v = 1, respectvely. Now, we set ( u u := g u,m p u = g u,m + m ) = u + m, (3) g u,m g u,m ( v v := g v,n p v = g v,n + j n ) = v + jn. (4) g v,n g v,n 5

6 Note that gcd(g u,m, g v,n ) = 1, and therefore gcd(u, v ) = 1. Indeed, g u,m dvdes u, g v,n dvdes v and gcd(u, v) = 1. Snce u = u+m and v +jn, the mage f((a 1, a 2 )) belongs to l for every (a 1, a 2 ) from l. It remans to check whether equaltes (1), (2) have soluton for every pont (b 1, b 2 ) from f 1 (l). Snce (b 1, b 2 ) f 1 (l) there s t N 0 such that b 1 ut b 2 vt (mod m), (mod n). By the defnton of u and v (see (3), (4)) we have b 1 u t b 2 v t v b 1 v u t u b 2 v u t (mod m), (mod n), (mod v m), (mod u n). Therefore v b 1 u b 2 (mod GCD(v m, u n)). Agan, by the defnton of u and v (see (3), (4)) we have ( ) gcd (v m, u m n n) = gcd (g v,n p v m, g u,m p u n) = gcd g v,n p v g u,m, g u,m p u g v,n g u,m g v,n ( ) ( m n = g u,m g v,n gcd, = g u,m g v,n gcd x m, y n ) g u,m g v,n g u,m g v,n ( ) m n = gcd g v,n p v xg u,m, g u,m p u yg v,n = gcd (v xm, u yn). g u,m Hence v b 1 u b 2 (mod gcd(v xm, u yn)). Now, let us consder the followng equatons. g v,n s v b 1 s u b 2 (mod v xm), (mod u yn). Snce v b 1 u b 2 (mod gcd(v xm, u yn)), these equatons have soluton by the Chnese Remander Theorem (Theorem 7). Snce gcd(u, v ) = 1, s = u v k for some k Z. Therefore u k b 1 v k b 2 (mod xm), (mod yn). Consequently, equatons (1), (2) have soluton and the pont (b 1, b 2 ) f 1 (l) les on l. We are done. Now, we are able to prove Theorem 5 usng Lemmas 9 and 10. 6

7 Theorem 5. Let m, n, x, y be postve ntegers such that m, n are not both 1 and gcd(x, y) = gcd(m, y) = gcd(n, x) = 1. Then τ xm,yn = τ m,n. Proof. Consderng mappng f : T xm yn T m n defned by f((a 1, a 2 )) = (a 1 mod m, a 2 mod n), the premage of every lne on T m n s a lne on T xm yn by Lemma 10. Therefore τ xm,yn = τ m,n by Lemma 9. 3 Perodcty In ths secton, we prove that the sequence σ z s perodc for all z N (Theorem 4 and Theorem 2). Frst, we menton two lemmas whch are used n the proofs of Theorem 4 and 2. They are specal cases of Theorem 5 and Lemma 6, respectvely. Lemma 11. Let z, x, m N such that z > 1 and gcd(x, z) = 1. Then σ z (m) = σ z (xm). Proof. We may use Theorem 5 for the tor T m z and T xm z. Lemma 12. Let z, x, m N. Then σ z (m) σ z (xm). Proof. We use Lemma 6 for the tor T m z and T xm z. Theorem 4. Let p be a prme, a N. Let us denote m := mn{x; σ p a(x) = 2p a }. Then m = p b for some b a and the sequence σ p a s perodc wth the perod m. Proof. Note that the exstence of such m follows from Theorem 3 whch s proven n the followng secton (Secton 4). More precsely, m p (a 1)p+2 by Theorem 3. Frst, we observe that m = p b for some b N. Indeed, f m were hp b for some h > 1 such that p does not dvde h, then σ p a(p b ) = σ p a(hp b ) by Theorem 11. Ths would contradct the mnmalty of m. Let us consder an arbtrary x {1,..., p b }. We show that σ p a(x) = σ p a(x+αp b ) for any α N. Snce x p b we can express t as x = rp l where 0 l b and gcd(r, p) = 1. By Theorem 11 σ p a(x) = σ p a(p l ). We consder two cases: 1. x < p b. In ths case x + αp b = p l (r + αp b l ). Snce b l 1, we get gcd(r + αp b l, p) = 1. Therefore σ p a(p l (r + αp b l )) = σ p a(p l ) = σ p a(x) by Lemma x = p b. In ths case x + αp b = p b (1 + α). Snce σ p a(p b ) = 2p a = max σ p a, Lemma 12 mples σ p a(p b ) = σ p a(hp b ) for any h > 0 and hence also for h = (1 + α). We proved that σ p a(x) = σ p a(x + αp b ) for any x {1,..., p b }. Now, we prove that the sequence σ z (x) s perodc for all z N. For ths purpose, we use the followng observaton and lemma. 7

8 Observaton 13. Every nfnte sequence {C () } N of n tuples of natural numbers C () N n contans an nfnte non-decreasng subsequence, that s, {C (t) } N such that C (t ) j C (t +1) j for all, j N. Proof. It holds trvally for n = 1. For n > 1 we may proceed by nducton. Lemma 14. Let z N and z = I pa be ts prme factorzaton. There exsts m z = I pb where b a for each I whch satsfes the followng condton. J I : σ z J p b J for arbtrary 0 c < b, d b and where J := I \ J. p c = σ z J p d J p c (5) Proof. Frst, note that the left expresson n (5) s always less than or equal to the rght expresson by Lemma 12. The same lemma also mples that f (5) does not hold for some J I, (c ) J, (d ) J (that s, the rght sde of (5) s strctly greater than the left sde) then t does not hold also for J I, (c ) J, (δ ) J where all δ = max{d j ; j J}. For a contradcton, let us assume there s no m z whch satsfes condton (5). Ths mples there s an nfnte sequence C of counterexamples C (b) = ( J (b), C (b), δ (b)), where J (b) I, C (b) = (c (b) ) J (b), δ (b) b, for all b N such that σ z J (b) p b J (b) p c (b) < σ z (b) p δ(b) p c J (b) J (b). (6) Snce I s fnte, there exsts an nfnte subsequence C of counterexamples whose ndex sets J (b) I are equal to some J I. Snce σ z attans only fntely many values, there exsts an nfnte subsequence C of counterexamples from C such that the rght sdes of (6) for them are equal to some S N. Fnally, Observaton 13 mples there exsts an nfnte subsequence C = {C (tb) } b N of counterexamples from C such that {C (tb) } b N s a non-decreasng sequence. Now, let t b δ (t1). Then we get σ z J p t b J p c(t b ) < σ z J p δ(t b ) J p c(t b ) = S = σ z J p δ(t 1 ) J p c(t 1 ) σ z J A contradcton. Note that the frst nequalty holds by (6). The second nequalty holds by Lemma 12 snce t b δ (t1) and c (t 1) c (t k) for all J. Fnally, we prove Theorem 2 usng the lemma above. 8 p t b J p c(t b ).

9 Theorem 2. The sequence σ z s perodc for all postve ntegers z greater than 1. Proof. Let z = I pa be the prme factorzaton of z. We show that σ z s perodc wth the perod m z = I pb gven by Lemma 14 satsfyng condton (5). We am to show that σ z (x) = σ z (x + αm z ) for all x m z and for all α N 0. We can express x as x = r I pc such that gcd(r, p ) = 1 for all I. Lemma 11 mples σ z (x) = σ z ( I pc ). We consder two cases: 1. c < b for all I. In ths case x + αm z = I pc (r + α I pb c ). Therefore gcd(r + α I pb c, p j ) = 1 for all j I snce gcd(r, p j ) = 1. Lemma 11 mples σ z (x + αm z ) = σ z ( I pc ) = σ z(x). 2. There s I such that c b. Let J := { I; c < b}, K := { I; c = b} and L := { I; c > b}. Then we get x + αm z = ( p c p b r p c b + α ) p b c. J K L L J The expresson ( r L pc b + α ) J pb c s not dvsble by p for J L. However, t could be dvsble by p for K. Therefore ( ) x + αm z = h J p c K L for some d b where K L 5 and h N such that gcd(h, p ) = 1 for all I = J K L. Now, we use the property of m z (see (5)) and Lemma 11: ( σ z (x + αm z ) = σ z h ) ( ) ( ) p c p d = σ z p c p d = σ z p c p b J K L J K L J K L ( ) ( ) = σz p c p c = σ z p c = σ z (x), J K L I where holds by Lemma 11, and hold by the property of m z (see (5)) snce c < b for J, d b for K L = J ( ) and c b for K L = J ( ). That s, σ z s perodc for all z > 1 wth the perod m z. Note, that t s Lemma 14 what makes the proof of ths theorem exstence. 5 In fact, d = b for L. p d 9

10 4 Proof of Theorem 3 In the last part, we prove Theorem 3. Frst of all, we need a tool for determnng whether three ponts are collnear. Let A = (a 1, a 2 ), B = (b 1, b 2 ), C = (c 1, c 2 ) be the ponts on the torus T m n. Let D(A, B, C) denote the followng determnant a 1 b 1 c 1 a 2 b 2 c 2. Lemma 15 (see Lemma 3.2. n [MSS + 16]). Let m, n be postve ntegers and let A = (a 1, a 2 ), B = (b 1, b 2 ), C = (c 1, c 2 ) be the ponts on the torus T m n. 1. A, B, C are not collnear f and only f there exst α 1, α 2, β 1, β 2, δ 1, δ 2, γ 1, γ 2 Z such that D((a 1 + α 1 m, a 2 + α 2 n), (b 1 + β 1 m, b 2 + β 2 n), (c 1 + γ 1 m, c 2 + γ 2 n)) = If A, B, C are collnear then D((a 1, a 2 ), (b 1, b 2 ), (c 1, c 2 )) 0 (mod gcd(m, n)). In our proof of Theorem 3 we wll also need to be able to determne the length of a lne on a torus. Lemma 16. Let m, n be postve ntegers and l = {π m,n (A + k(u, v)); ( k Z} be a lne ) on T m n where m A, (u, v) T m n such that gcd(u, v) = 1. Then the length of l s lcm. gcd(m,u), n gcd(n,v) Proof. The lne l s gven by the vector (u, v) and ts length s equal to the order of the element (u, v) n Z m Z n. Whch s The order of an element h n the group Z x s ord Zm Z n (u, v) = lcm(ord Zm (u), ord Zn (v)). ord Zx (h) = lcm(x, h) h = xh gcd(x,h) h = x gcd(x, h). We also need lemmas about propertes of lnes on a torus. Lemma 17. Let m, n be postve ntegers such that m s dvsble by p f and only f n s dvsble by p for every prme p. Let (a 1, a 2 ) be a pont on T m n such that gcd(a 1, a 2 ) = 1. Then there s exactly one lne contanng the orgn and (a 1, a 2 ). Proof. Let l denote the lne {π m,n (k(a 1, a 2 )); k Z}. Ths lne contans the orgn and (a 1, a 2 ). Let us assume that there s another lne l = {π m,n (k(b 1, b 2 )); k Z} contanng the orgn and (a 1, a 2 ). That means there exsts k Z such that kb 1 a 1 kb 2 a 2 (mod m), (mod n). 10

11 Therefore l l. Moreover, a 1 = kb 1 s 1 m and a 2 = kb 2 s 2 n for some s 1, s 2 Z. Snce gcd (m, ( kb 1 s 1 m) = ) gcd (m, kb 1 ) and gcd (n, kb 2 s 2 n) = gcd (n, kb 2 ), the length ( of the lne l) s m lcm by Lemma 16. By the same lemma the length of l m s lcm. gcd(m,kb 1 ), n gcd(n,kb 2 ) gcd(m,b 1 ), n gcd(n,b 2 ) Now, we show that gcd (m, kb 1 ) = gcd (m, b 1 ). For a contradcton, let us assume gcd (m, kb 1 ) > gcd (m, b 1 ). Then there s a prme p whch dvdes both k and m. By the assumpton p also dvdes n and thus gcd(a 1, a 2 ) = gcd(kb 1 s 1 m, kb 2 s 2 n) p 1. A contradcton. Analogously we get gcd (n, kb 2 ) = gcd (n, b 2 ). Consequently, l has the same length as l. Moreover, l = l snce l l. Lemma 18. Let T p a p be a torus where p s a prme, a b, and let (x, y) be a pont on T b p a pb. Let g denote gcd(x, y). If x or y s not dvsble by p, then there s exactly one lne between the orgn and (x, y). Proof. Let l = {π p a,p b(k(v 1, v 2 )); k Z} be a lne whch contans (x, y). Then for some k Z kv 1 x (mod p a ), kv 2 y (mod p b ). Let g denote gcd(x, y). Snce gcd(g, p) = 1, g has an nverse element modulo p b. Let us denote t g 1. Then g 1 kv 1 x g g 1 kv 2 y g (mod p a ), (mod p b ). Hence every lne on T p a p contans (x, y) f and only f t contans ( x, y ). Snce gcd( x, y ) = 1, Lemma 17 b g g g g mples that there s exactly one lne contanng the orgn and ( x, y ). Consequently, there s exactly one g g lne contanng the orgn and the pont (x, y). Corollary 19. Let T p a p be a torus where p s a prme and let A = (a 1, a b 2 ), B = (b 1, b 2 ) T p a p be b ponts on that torus such { that a 1 b 1 } s not dvsble by p. Then there s exactly one lne between A, B and ts length s max p a,. p b gcd(p b,a 2 b 2 ) Proof. The lnes between (a 1, a 2 ), (b 1, b 2 ) are n one-to-one correspondence wth the lnes between the orgn and P := π p a,p b(a 1 b 1, a 2 b 2 ). Snce a 1 b 1 s not dvsble by p there s exactly one lne between the orgn and P by Lemma 18. Such lne s gven by the vector π p a,p b( a 1 b 1, a 2 b 2 ). We can express t as (v, gcd(a 1 b 1,a 2 b 2 ) gcd(a 1 b 2,a 2 b 2 ) wpr ) where v, w s not dvsble by p and r < b. Moreover, p r = gcd(p b a, 2 b 2 )) = gcd(a 1 b 1,a 2 b 2 ) gcd(pb, a 2 b 2 ) snce a 1 b 1 s not dvsble by p. Lemma 16 mples the length of the lne between the orgn and P s ( ) ) } { } p a lcm gcd(p a, v), p b = lcm (p a, pb = max {p a, pb = max p a p b,. gcd(p b, wp r ) p r p r gcd(p b, a 2 b 2 ) 11

12 Lemma 20. Let T p a p be a torus where p s a prme and let A, B, C be ponts on t such that π b p a 1,p b(a) = π p a 1,p b(b). If each lne l contanng the ponts A, C on T p a p has the same length as ts mage b l := π p a 1,pb(l) then A, B, C are not collnear. (See Fgure 2 for an example.) Proof. If there s a lne l whch contans A, B, C then ts mage l has smaller length snce π p a 1,p b(a) = π p a 1,pb(B). However, t s mpossble by our assumpton. Lemma 21. Let T p a p be a torus where p s a prme and a b. Let (v 1p c, v b 2 p d ) be a pont on T p a p b such that p does not dvde v 1 or v 2, c < a, d < b and c d. Then each lne whch contans the orgn and (v 1 p c, v 2 p d ) also contans some pont (w 1, w 2 ) such that w 1, w 2 satsfy the followng equatons. w 1 v 1 (mod p a c ), w 2 v 2 p d c (mod p b c ). Proof. Let l be a lne whch contans the orgn and (v 1 p c, v 2 p d ). Then we can express t as l = {π p a,p b(k(u 1, u 2 )); k Z} such that gcd(u 1, u 2 ) = 1 and there exsts k Z whch satsfes the followng equatons. ku 1 v 1 p c (mod p a ), ku 2 v 2 p d (mod p b ). Snce gcd(u 1, u 2 ) = 1, p does not dvde u 1 or u 2. Therefore p c has to dvde k snce c d. Consequently, w 1 := k p c u 1 v 1 (mod p a c ), w 2 := k p c u 2 v 2 p d c (mod p b c ). Now, we are able to prove Theorem 3. Theorem 3. Let T p a p (a 1)p+2 be a torus where p s a prme and a N. Then τ p a,p (a 1)p+2 = 2pa. Proof. Theorem 1 mples τ p a,p(a 1)p+2 2p. Therefore, t s suffcent to show a constructon of 2p ponts where no three of them are collnear. Let X := {(, 2 p); P } and Y := {(, 2 p + 1); P } where P = {0,..., p 1}. Msak et al. (see Theorem 1.2(3a) n [MSS + 16]) proved that no three ponts from the set π p,p 2(X Y ) are collnear on the torus T p p 2. In ths proof, we nductvely construct a set X a Y a Z 2 by takng multple copes of X Y and prove by nducton that no three ponts from π p a,p (a 1)p+2(X a Y a ) are collnear on the torus T p a p (a 1)p+2. Frst, let us defne the set X a Y a. 12

13 B C A A C Fgure 2: An applcaton of Lemma 20. Let A = (0, 0), B = (3, 0) and C = (1, 2) be ponts on T 9,9. There s exactly one lne between A and C by Theorem 17 and has the length 9 (see the upper pcture). Now, let us consder ponts A = π 3,9 (A) and C = π 3,9 (C) on T 9,9. The lne contanng A and C also has the length 9 (see the lower pcture). Snce A = π 3,9 (B), we conclude that A, B, C are not collnear. 13

14 1. For a = 1 we defne X a := X and Y a := Y. 2. For a > 1 we take p copes of the set X a 1 Y a 1 on Z 2 so that X a := X a 1 p 1 =1 p 1 X a 1 + ( p a 1, p (a 2)p++3), Y a := Y a 1 Y a 1 + ( p a 1, p (a 2)p++3). =1 As an example of the constructon see Fgures 3 and 4. For a pont A π p a,p (a 1)p+2(X a Y a ) on T p a p we call the pont  X (a 1)p+2 a Y a such that π p a,p(a 1)p+2(Â) = A the premage of A (see Fgure 4). Now, we nductvely prove that π p a,p (a 1)p+2(X a Y a ) does not contan three collnear ponts. Case a = 1 was proved by Msak et al. (see Theorem 1.2(3a) n [MSS + 16]). Let us assume a > 1 and let A, B, C π p a,p (a 1)p+2(X a Y a ). We have to consder several cases. Frst, let us assume that π p a 1,p (a 2)p+2(A), π p a 1,p (a 2)p+2(B), π p a 1,p (a 2)p+2(C) X a 1 Y a 1 are three dstnct ponts. In other words, the premages of A, B, C n X a Y a are copes of three dstnct ponts from X a 1 Y a 1. If A, B, C are collnear on T p a p (a 1)p+2, then π p a 1,p (a 2)p+2(A), π p a 1,p (a 2)p+2(B), π p a 1,p (a 2)p+2(C) are collnear on T p a 1 p as well whch contradcts the nducton hypothess. (a 2)p+2 Now, we assume A, B, C satsfy π p a 1,p (a 2)p+2(A) = π p a 1,p(a 2)p+2(B). In other words, the premages of A and B are copes of the same pont from X a 1 Y a 1. In the frst case, we assume that the premages of A, B are from X a and the premage of C s from Y a. Therefore A = (k + pt 1, k 2 p + pt 2 ), B = (k + pt 1 + up a 1, k 2 p + pt 2 + xp q ), C = (m + pv 1, m 2 p + pv 2 + 1) for some k, m {0,..., p 1}, u {1,..., p 1}, for sutable t 1, t 2, v 1, v 2, x N 0, q (a 2)p + 4 a and where p does not dvde x. Note that u > 0; otherwse A = B. We may use the translaton gven by the vector ( pt 1, pt 2 ) and we get A t = (k, k 2 p), B t = (k + up a 1, k 2 p + xp q ), C t = (m + ps 1, m 2 p + ps 2 + 1), where s 1 = v 1 t 1 mod p a and s 2 = v 2 t 2 mod p (a 1)p+2. The ponts A t, B t, C t are collnear f and only f A, B, C are collnear. We compute D(A t, B t, C t ). D(A t, B t, C t ) = D((k, k 2 p), (k + up a 1, k 2 p + xp q ), (m + ps 1, m 2 p + ps 2 + 1)) = kxp q + up a + up a 1 + up a s 2 mxp q xp q+1 s 1 up a = p a 1 (u + pr) 14

15 Fgure 3: An example of the constructon of π p a,p (a 1)p+2(X a Y a ) from Theorem 3 for p = 3, a = 1 (see the upper pcture) and a = 2 (see the lower pcture). 15

16 X a 1 Y a 1 + (p a 1, p (a 2)p+5 ) A π p a,p (a 1)p+2 π p a,p (a 1)p+2(X a 1 Y a 1 + (p a 1, p (a 2)p+5 ))  X a 1 Y a 1 + (p a 1, p (a 2)p+4 ) = π p a,p (a 1)p+2(X a 1 Y a 1 + (p a 1, p (a 2)p+4 )) X a 1 Y a 1 = π p a,p (a 1)p+2(X a 1 Y a 1 ) T p a,p (a 1)p+2 Z 2 Fgure 4: A schematc pcture of the constructon of π p a,p (a 1)p+2(X a Y a ) from Theorem 3 for p = 3 and an example of the premage  of a pont A π p a,p (a 1)p+2(X a Y a ). for some R Z. Therefore D(A t, B t, C t ) 0 (mod p a ) snce u > 0. Consequently, A, B, C are not collnear on T p a p (a 1)p+2 by Lemma 15 (2). Smlarly, we check the case when the premages of A, B are from Y a and the premage of C from X a. Hence and we get A = (k, k 2 p + 1), B = (k + up a 1, k 2 p + xp q + 1), C = (l + ps 1, l 2 p + ps 2 ), D(A, B, C) = k 2 up a + up a 1 + lxp q + s 1 k 2 p a + s 1 xp q+1 up a l 2 up a s 2 xp q k = p a 1 (u + pr). Therefore A, B, C are not collnear on T p a p (a 1)p+2 by the same lemma. Now, we check the case when the premages of all three ponts are from X a. The case where they are all from Y a s just a translaton. We can express the ponts as follows: A = (k + s 1, k 2 p + s 2 ), B = (k + s 1 + up a 1, k 2 p + s 2 + xp q ), C = (m + s 3, m 2 p + s 4 ), where p dvdes s 1, s 3 and p 4 dvdes s 2, s 4 by the defnton of the constructon (the frst teraton for a = 2), q {p (a 2)p+4,..., p (a 1)p+2 }, u 1 s not dvsble by p, x {0, 1} and k, m {0,..., p 1}. 16

17 1. Frst, we assume m k. Let us map these ponts to the smaller torus T p a p (a 2)p+4 usng π p a,p (a 2)p+4 and let A, B, C be the mages. Then A = (k + s 1, k 2 p + s 2), B = (k + s 1 + up a 1, k 2 p + s 2), C = (m + s 3, m 2 p + s 4), where p 4 dvdes s 2, s 4. Let us check the lne between A and C. Snce ((k m) + s 1 s 3 ) s not dvsble by p, Corollary 19 mples there s exactly one lne between them and ts length s d = max{p a p, (a 2)p+4 gcd(p (a 2)p+4,p(k m)(k+m)+s 2 s )} = max{pa, p (a 2)p+4 h } where h {1, 2} (the sum of k 4 and m may be p). Now, let us map A, B, C to the torus T p a 1 p and let (a 2)p+4 A, B, C be the mages. We can express them as A = (k + s 1, k 2 p + s 2), B = A, C = (m + s 3, m 2 p + s 4), where p dvdes s 1 a s 3. Agan, by Corollary 19 there s exactly one lne between A and C and ts length s d = max{p a 1, p (a 2)p+4 h }. Snce (a 2)p + 4 h a for a > 1, d = d and A, B, C are not collnear by Lemma 20. Therefore A, B, C are not collnear on T p a p (a 1)p Let us check the case when k = m. That means π p,p 2(A) = π p,p 2(C). Frst, let us assume that π p a 1,p (a 2)p+2(A) π p a 1,p(a 2)p+2(C). In other words, the premages of all three ponts are copes of the same pont from X but they are not copes of the same pont from X a 1. We have A = (k + s 1, k 2 p + s 2 ), B = (k + s 1 + up a 1, k 2 p + s 2 + xp q ), C = (k + s 3, k 2 p + s 4 ), where p dvdes s 1, s 3 and p 4 dvdes s 2, s 4 by the defnton of the constructon (the frst teraton for a = 2), q {p (a 2)p+4,..., p (a 1)p+2 }, u 1 s not dvsble by p, x {0, 1} and k, m {0,..., p 1}. We can see that p a 1 does not dvde s 1 and p (a 2)p+4 does not dvde s 2 or p a 1 does not dvde s 3 and p (a 2)p+4 does not dvde s 4 ; otherwse π p a 1,p (a 2)p+2(A) = π p a 1,p (a 2)p+2(C). Now, we use the translaton gven by the vector ( k s 1, k 2 p s 2 ) and we get A t = (0, 0), B t = (up a 1, xp q ), C t = (vp t, yp r ), 17

18 where u, x, v, y are not dvsble by p, t a 2, r (a 2)p + 2. Let us have a look at the mages of these ponts on the torus T p a p (a 2)p+4. We get A = (0, 0), B = (up a 1, 0), C = (vp t, y p r ) for sutable y whch s not dvsble by p. Let us compute the length of the lnes between A a C. By Lemma 21, every such lne passng through some pont (w 1, w 2 ) satsfes w 1 v (mod p a t ), w 2 y p r t (mod p (a 2)p+2 t ). By Lemma 19, there s exactly one lne between the orgn and such (w 1, w 2 ) and ts length s max{p a, p(a 2)p+4 }. Therefore each lne between A and C has the length d = max{p a, p(a 2)p+4 }. p r t p r t Let us map A, B, C to the torus T p a 2 p and let (a 2)p+4 A, B, C be the mages. We get A = A, B = A C = (v p t, y p r ) for sutable v whch s not dvsble by p. Smlarly, the length d of each lne between A and C s d = max{p a 1, p(a 2)p+4 } by Corollary 19. Note that the maxmal r t satsfyng the defnton of p r t the constructon equals max x {1,...,a 2} (xp+2 x) = (a 2)(p 1)+2. Hence r t (a 2)p a+4 and (a 2)p + 4 (r t) a. Consequently, d = d and Lemma 20 mples that A, B, C are not collnear and, therefore, A, B, C are not collnear on T p a p (a 1)p+2. Fnally, we check the case when all three A, B, C ponts are copes of one pont from X a 1 whch s wthout loss of generalty the orgn. Otherwse, we use a translaton smlarly to the prevous cases. Therefore A = (up a 1, p q ), B = (vp a 1, p r ), C = (wp a 1, p s ) for dstnct q, r, s {p (a 2)p+4,..., p (a 1)p+2 } and for dstnct u, v, w {0,..., p 1} satsfyng the defnton of the constructon. Let us denote b := (a 1)p + 2 and D := D((up a 1 + α 1 p a, p q + α 2 p b ), (vp a 1 + β 1 p a, p r + β 2 p b ), (wp a 1 + γ 1 p a, p s + γ 1 p b )) the determnant for A, B, C 18

19 from Lemma 15 (1) where α 1, α 2, β 1, β 2, γ 1, γ 2 Z. Then D = D(((up a 1, p q ), (vp a 1, p r ), (wp a 1, p s )) + p a D((α 1, p q ), (β 1, p r ), (γ 1, p s )) + p b D((up a 1, α 2 ), (vp a 1, β 2 ), (wp a 1, γ 2 )) + p a+b ((α 1, α 2 ), (β 1, β 2 ), (γ 1, γ 2 )) = p a+r 1 (u w) + p a+q 1 (w v) + p a+s 1 (v u) + p a (p r (α 1 γ 1 ) + p q (γ 1 β 1 ) + q s (β 1 α 1 )) + p a+b 1 Q + p a+b W for sutable Q, W Z. Wthout loss of generalty let q be the smallest number of {q, r, s}. Then ndeed q < b and thus D s dvsble by p a+q 1. We get D = p a+q 1 ((w v) + ph) for sutable H Z. Therefore D 0 and A, B, C are not collnear on T p a p by Lemma 15 (1) and we (a 1)p+2 are done. Acknowledgement I would lke to thank to my supervsor Martn Tancer for all hs support and advce. I would also lke to thank to the anonymous revewers for valuable advce and remarks. References [Dud17] H. E. Dudeney. Amusements n mathematcs. Nelson, Ednburgh, pp. 94, 222. [FGPS12] J. Fowler, A. Groot, D. Pandya, and B. Snapp. The no-three-n-lne problem on a torus arxv: [How02] F. T. Howard. A generalzed Chnese remander theorem. The College Mathematcs Journal, 33(4): , [MSS + 16] A. Msak, Z. Stȩpeń, A. Szymaszkewcz, L. Szymaszkewcz, and M. Zwerzchowsk. A note on the no-three-n-lne problem on a torus. Dscrete Mathematcs, 339(1): , [Sel49] A. Selberg. An elementary proof of Drchlet s theorem about prmes n an arthmetc progresson. Annals of Mathematcs, 50(2): ,