There are only finitely many Diophantine quintuples


 Nathaniel Green
 1 years ago
 Views:
Transcription
1 There are only finitely many Diophantine quintuples Andrej Dujella Department of Mathematis, University of Zagreb, Bijenička esta Zagreb, Croatia Abstrat A set of m positive integers is alled a Diophantine mtuple if the produt of its any two distint elements inreased by 1 is a perfet square. Diophantus found a set of four positive rationals with the above property. The first Diophantine quadruple was found by Fermat (the set {1, 3, 8, 120}). Baker and Davenport proved that this partiular quadruple annot be extended to a Diophantine quintuple. In this paper, we prove that there does not exist a Diophantine sextuple and that there are only finitely many Diophantine quintuples. 1 Introdution A set of m positive integers {a 1, a 2,..., a m } is alled a Diophantine mtuple if a i a j + 1 is a perfet square for all 1 i < j m. Diophantus first studied the problem of finding four numbers suh that the produt of any two of them inreased by unity is a square. He found a set of four positive rationals with the above property: {1/16, 33/16, 17/4, 105/16}. However, the first Diophantine quadruple, {1, 3, 8, 120}, was found by Fermat. Euler was able to add the fifth positive rational, / , to the Fermat s set (see [6], pp ). Reently, Gibbs [13] found examples of sets of six positive rationals with the property of Diophantus. A folklore onjeture is that there does not exist a Diophantine quintuple. The first important result onerning this onjeture was proved in 1969 by Baker and Davenport [3]. They proved that if d is a positive integer suh that {1, 3, 8, d} forms a Diophantine quadruple, then d = 120. This problem was stated in 1967 by Gardner [12] (see also [17]). In 1979 Arkin, Hoggatt and Strauss [1] proved that every Diophantine triple an be extended to a Diophantine quadruple. More preisely, let {a, b, } be a Diophantine triple and ab + 1 = r 2, a + 1 = s 2, b + 1 = t 2, where r, s, t are positive integers. Define d + = a + b + + 2ab + 2rst. 0 Mathematis Subjet Classifiation (2000): Primary 11D09, 11D25, 11D48; Seondary 11B37, 11J68, 11J86. 1
2 There are only finitely many Diophantine quintuples 2 Then {a, b,, d + } is a Diophantine quadruple. Indeed, ad = (at + rs) 2, bd = (bs + rt) 2, d = (r + st) 2. There is a stronger version of the Diophantine quintuple onjeture. Conjeture 1 If {a, b,, d} is a Diophantine quadruple and d > max{a, b, }, then d = d +. Conjeture 1 was proved for ertain Diophantine triples [16, 20] and for some parametri families of Diophantine triples [7, 8, 10]. In partiular, in [10] it was proved that the pair {1, 3} annot be extended to a Diophantine quintuple. A Diophantine quadruple D = {a, b,, d}, where a < b < < d, is alled regular if d = d +. Equivalently, D is regular iff (1) (a + b d) 2 = 4(ab + 1)(d + 1) (see [14]). The equation (1) is a quadrati equation in d. One root of this equation is d +, and other root is d = a + b + + 2ab 2rst. It is easy to hek that all small Diophantine quadruples are regular; e.g. there are exatly 207 quadruples with max{a, b,, d} 10 6 and all of them are regular. Sine the number of integer points on an ellipti urve (2) y 2 = (ax + 1)(bx + 1)(x + 1) is finite, it follows that, for fixed a, b and, there does not exist an infinite set of positive integers d suh that a, b,, d is a Diophantine quadruple. However, bounds for the size [2] and for the number [19] of solutions of (2) depend on a, b, and aordingly they do not immediately yield an absolute bound for the size of suh set. The main result of the present paper is the following theorem. Theorem 1 There are only finitely many Diophantine quintuples. Moreover, this result is effetive. We will prove that all Diophantine quintuples Q satisfy max Q < Hene we almost ompletely solve the problem of the existene of Diophantine quintuples. Furthermore, we prove Theorem 2 There does not exist a Diophantine sextuple. Theorems 1 and 2 improve results from [9] where we proved that there does not exist a Diophantine 9tuple and that there are only finitely many Diophantine 8tuples. As in [9], we prove Conjeture 1 for a large lass of Diophantine triples satisfying some gap onditions. However, in the present paper these gap onditions are muh weaker than in [9]. Aordingly, the lass of Diophantine triples for whih we are able to prove Conjeture 1 is muh larger. In fat, in an arbitrary Diophantine quadruple, we may find a triple for whih we are able to prove Conjeture 1.
3 There are only finitely many Diophantine quintuples 3 In the proof of Conjeture 1 for a triple {a, b, } we first transform the problem into solving systems of simultaneous Pellian equations. This redues to finding intersetions of binary reurrene sequenes. In Setion 5 we almost ompletely determine initial terms of these sequenes, under assumption that they have nonempty intersetion whih indues a solution of our problem. This part is a onsiderable improvement of the orresponding part of [9]. This improvement is due to new gap priniples developed in Setion 4. These gap priniples follow from the areful analysis of the elements of the binary reurrene sequenes with small indies. Let us mention that in a joint paper with A. Pethő [10] we were able to determine initial terms, in a speial ase of triples {1, 3, }, using an indutive argument. Applying some ongruene relations we get lower bounds for solutions. In obtaining these bounds we need to assume that our triple satisfies some gap onditions like b > 4a and > b 2.5. Let us note that these onditions are muh weaker then onditions used in [9], and this is due to more preise determination of the initial terms. Comparing these lower bounds with upper bounds obtained from the Baker s theory on linear forms in logarithms of algebrai numbers (a theorem of Matveev [18]) we prove Theorem 1, and omparing them with upper bounds obtained from a theorem of Bennett [5] on simultaneous approximations of algebrai numbers we prove Theorem 2. In the final steps of the proofs, we use again the above mentioned gap priniples. 2 Systems of Pellian equations Let us fix some notation. Let {a, b, } be a Diophantine triple and a < b <. Furthermore, let positive integers r, s, t be defined by ab + 1 = r 2, a + 1 = s 2, b + 1 = t 2. In order to extend {a, b, } to a Diophantine quadruple {a, b,, d}, we have to solve the system (3) ad + 1 = x 2, bd + 1 = y 2, d + 1 = z 2, with positive integers x, y, z. Eliminating d from (3) we get the following system of Pellian equations (4) (5) az 2 x 2 = a, bz 2 y 2 = b. In [9], Lemma 1, we proved the following lemma whih desribes the sets of solutions of the equations (4) and (5). Lemma 1 There exist positive integers i 0, j 0 and integers z (i) j = 1,..., j 0, with the following properties: (i) (z (i) 0, x(i) 0 ) and (z(j) 1, y(j) 1 0, x(i) 0, z(j) 1, y(j) ) are solutions of (4) and (5), respetively. 1, i = 1,..., i 0,
4 There are only finitely many Diophantine quintuples 4 (ii) (6) (7) (8) (9) z (i) 0, x(i) 0, z(j) 1, y(j) 1 satisfy the following inequalities a( a) s + 1 2(s 1) < 2 1 x (i) 0 < a, 1 z (i) (s 1)( a) 0 < 2a 2 a < 0.421, 1 y (j) 1 b( b) t + 1 2(t 1) < < b, 2 1 z (j) (t 1)( b) 1 < 2b 2 b < (iii) If (z, x) and (z, y) are positive integer solutions of (4) and (5) respetively, then there exist i {1,..., i 0 }, j {1,..., j 0 } and integers m, n 0 suh that (10) (11) z a + x = (z (i) 0 z b + y = (z (j) 1 (i) a + x 0 )(s + a) m, )(t + b) n. b + y (j) 1 Let (x, y, z) be a solution of the system (4) & (5). From (10) it follows that z = v (i) m for some index i and integer m 0, where (12) v (i) 0 = z (i) 0, v(i) 1 = sz (i) 0 + x (i) 0, v(i) m+2 = 2sv(i) m+1 v(i) m, and from (11) we onlude that z = w (j) n for some index j and integer n 0, where (13) w (j) 0 = z (i) 1, w(j) 1 = tz (j) 1 + y (j) 1, w(j) n+2 = 2tw(j) n+1 w(j) n. Let us onsider the sequenes (v m ) and (w n ) modulo 2. From (12) and (13) it is easily seen that (14) (15) v (i) 2m z(i) 0 (mod 2), v (i) 2m+1 sz(i) 0 + x (i) 0 (mod 2), w (j) 2n z(j) 1 (mod 2), w (j) 2n+1 tz(j) 1 + y (j) 1 (mod 2). We are searhing for solutions of the system (4) & (5) suh that d = (z 2 1)/ is an integer. Using (14) and (15), from z = v m = w n we obtain [z (i) 0 ]2 v 2 m z 2 1 (mod ), [z (j) 1 ]2 w 2 n z 2 1 (mod ). Therefore we are interested only in equations v m = w n satisfying [z (i) 0 ]2 [z (j) 1 ]2 1 (mod ).
5 There are only finitely many Diophantine quintuples 5 We will dedue later more preise information on the initial terms z (i) 0 and z (j) 1 (see Setion 5). Let us mention now a result of Jones [15], Theorem 8, whih says that if < 4b then z (j) 1 = 1. As a onsequene of Lemma 1 and the relations (14) and (15), we obtain the following lemma. From now on, we will omit the supersripts (i) and (j). Lemma 2 1) If the equation v 2m = w 2n has a solution, then z 0 = z 1. 2) If the equation v 2m+1 = w 2n has a solution, then z 0 z 1 < 0 and x 0 s z 0 = z 1. In partiular, if b > 4a and > 100a, then this equation has no solution. 3) If the equation v 2m = w 2n+1 has a solution, then z 0 z 1 < 0 and y 1 t z 1 = z 0. 4) If the equation v 2m+1 = w 2n+1 has a solution, then z 0 z 1 > 0 and x 0 s z 0 = y 1 t z 1. Proof. See [9], Lemma 3. 3 Relationships between m and n In [9], Setion 4, we proved that v m = w n implies m n if b > 4a, > 100b and n 3. We also proved that m 3 2n, provided {a, b, } satisfies some rather strong gap onditions. In this setion we will first prove an unonditional relationship between m and n, and then we will improve that result under various gap assumptions. Lemma 3 If v m = w n, then n 1 m 2n + 1. Proof. We have the following estimates for v 1 : v 1 = sz 0 + x 0 x 0 s z 0 = 2 a z 2 0 x 0 + s z 0 > a 2x 0 > 4x 0 > a, Hene (16) If > 4b, then v 1 < 2x 0 < a a (2s 1)m 1 < v m < a(2s) m 1 for m 1. w 1 2 b z 2 1 y 1 + b z 1 > 4y 1 > If < 4b then, by [15], Theorem 8, z 1 = ±1, y 1 = 1 and w 1 t = a + r = s > a > Furthermore, w 1 < 2y 1 < b and therefore (17) b b b (2t 1)n 1 < w n < b(2t) n 1 for n 1.
6 There are only finitely many Diophantine quintuples 6 We thus get (18) Sine (19) (2s 1) m 1 < ab 2 (2t) n 1. 2s 1 = 2 a > a and 2t = 2 b + 1 < b, it follows that (2s 1) 2 > 3.12a > 2t. It implies and m 2n + 1. On the other hand, we have (2s 1) m 1 < (2t) n < (2t) n+0.43 < (2s 1) 2n+0.86 (2t 1) n 1 < ab 2 (2s) m 1 < ab 2 (2t 1) m 1 < (2t 1) m and n m + 1. Thus we proved the lemma for m, n 0. It remains to hek that v 0 < w 2 and w 0 < v 2. Indeed, 2t w 2 = 2tw 1 w 0 > b ( 4 ) b 2 b > > > v 0, v 2 = 2sv 1 v 0 > 2s a 2 b ( 4 ) a 2 a > > > w 0. 2 a Lemma 4 Assume that > If v m = w n and m, n 2, then 1) > b 4.5 = m 11 9 n + 7 9, 2) > b 2.5 = m 7 5 n + 3 5, 3) > b 2 = m 3 2 n + 1 2, 4) > b 5/3 = m 8 5 n Proof. As in the proof of Lemma 3, assuming > max{b 5/3, }, we have v m > x 0 (2s 1) m 1 > w n < b(2t) n 1, a (2s 1)m 1, for m, n 1. Hene (2s 1) m 1 < ab 2 (2t) n 1 and (20) m 1 a (m 1)/2 (m 1)/2 < n b (n 1)/2+1/4 (n 1)/2+1/4 a 1/4. If m 2 and > b ε, then (20) shows that at least one of the following two inequalities holds: (21) m 1 < ( n ) 1 ε + n 2, (22) m 1 < 1.004n. The inequality (22) implies all statements of the lemma. For ε = 4.5, (21) implies m < 11 9 n and m 11 9 n Similar arguments apply to all other statements of the lemma.
7 There are only finitely many Diophantine quintuples 7 4 Gap priniples In [9], Lemma 14, we proved that if {a, b,, d} is a Diophantine quadruple and a < b < < d, then d 4b. The proof was based on the fat, proved by Jones [15], Lemma 4, that = a + b + 2r or 4ab + a + b. In this setion we will develop a stronger and more preise gap priniple by examining the equality v m = w n for small values of m and n. Let us note that sine (4ab + 1) < d + < 4(ab + 1), we may expet an essential improvement only if we assume that d d +. Lemma 5 Let v m = w n and define d = (v 2 m 1)/. If {0, 1, 2} {m, n}, then d < or d = d +. Proof. From the proof of Lemma 3 we have: v 0 < w 2, w 0 < v 2, v 1 < w 3, w 1 < v 4, v 2 < w 4, w 2 < v 6. Therefore, the ondition {0, 1, 2} {m, n} implies (m, n) {(0, 0), (0, 1), (1, 0), (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (2, 3), (3, 2), (4, 2), (5, 2)}. If 0 {m, n}, then d <. If (m, n) = (1, 1), then d < for z 0 < 0, and d = d + for z 0 > 0 (see [9], proof of Theorem 3). Assume that (m, n) = (1, 2). We have v 1 = sz 0 + x 0, w 2 = z 1 + 2(bz 1 + ty 1 ). By Lemma 2, if z 1 > 0 then z 0 < 0 and x 0 + sz 0 = z 1. Hene w 2 > v 1 = w 0. If z 1 < 0 then z 0 > 0 and (23) x 0 sz 0 = z 1. Inserting (23) into the relation v 1 = w 2 we obtain (24) bz 1 + ty 1 = x 0. From (23), (24) and the system (4) & (5), we obtain (b a)t = z0 2 (b a). Therefore, z 0 = t, x 0 = r, z 1 = st r, y 1 = rt bs. It implies v 1 = st + r and d = v2 1 1 = 1 (ab2 + a + b rst + ab ) = d +. Assume now that (m, n) = (2, 1). In the same manner as in the ase (m, n) = (1, 2), we obtain z 1 = s, y 1 = r, z 0 = st r, x 0 = rs at and d = d +. Let (m, n) = (2, 2). We have v 2 = z 0 + 2(az 0 + sx 0 ), w 2 = z 1 + 2(bz 1 + ty 1 ), and sine z 0 = z 1, we obtain az 0 + sx 0 = bz 0 + ty 1 and (b a)(y 2 1 x b a) = ax x 2 0 by y 2 1 2stx 0 y 1.
8 There are only finitely many Diophantine quintuples 8 Therefore (b a) 2 = (sy 1 tx 0 ) 2 and sine sy 1 = tx 0, we have (25) tx 0 sy 1 = b a. Furthermore, and (a + 1)(bx a b) = a(bx x (b a) 2 2t(b a)x 0 ) (b a)(x atx 0 + a 2 t 2 ) = (b a)(ab + 1)(a + 1) = (b a)r 2 s 2. Finally, x 0 = rs at, y 1 = rt bs, z 0 = st r. Now we obtain and d = d +. v 2 = st r + 2(ast ar + rs 2 ast) = st + r Let (m, n) = (3, 1). We may assume that {a, b, } {1, 3, 8}. Then a 15 and b 48. It implies 2s 1 > a and 2t < b. From (18) we obtain whih implies 6, a ontradition. (1.807 a) 2 < ab 2 < a 3, Let (m, n) = (3, 2). Assume that z 0 > 0, z 1 < 0. We have v 1 > 2sz 0, v 2 > (4s 2 1)z 0, v 3 > (4s 2 1)(2s 1)z 0 > 7(a) 3/2 z 0, and w 1 < 2 2t z 1, w 2 < 2tw 1 < 2 z 1. We have also z 1 > 4x 0 and therefore w 2 < 4x 0. Sine x 0 <, we obtain w 2 < 4 3/2 < 7(a) 3/2 z 0 < v 3. Assume now that z 0 < 0, z 1 > 0. Then z 1 = sz 0 + x 0 and the ondition v 3 = w 2 implies x 0 + 2az 1 = bz 1 + ty 1 > 2bz 1 and x 0 > 4z 1 > x 0, a ontradition. Let (m, n) = (2, 3). If z 0 > 0, z 1 < 0, then v 2 = w 3 implies y 1 + 2bz 0 = az 0 + sx 0 < 2az 0 + z 0. Therefore z0 2 < 4. On the other hand, z2 0 1 (mod ). Hene, z 0 = 1. But, if > 4b then z 0 > 4y 1 > 1. If < 4b then z 1 = 1, y 1 = 1, and Lemma 2 implies = b + 2, whih ontradits the fat that = a + b + 2r. Assume that z 0 < 0, z 1 > 0. As in the ase (m, n) = (3, 2), we have v 2 < 2 z 0 and w 3 > 7.3(b) 3/2 z 1. If > 4b then z 0 > and sine y 1 <, we obtain 4y 1 v 2 < 4 3/2 < 7.3(b) 3/2 z 1 < w 3. If < 4b then w 3 > 7.3b > > 2 > v 2. Let (m, n) = (4, 2). We have v 4 = z 0 + 4(2az 0 + sx 0 ) + 8a 2 (az 0 + sx 0 ), and the relation v 4 = w 2 implies (26) bz 0 + ty 1 = 4az 0 + 2sx 0 + 4a(az 0 + sx 0 ).
9 There are only finitely many Diophantine quintuples 9 It holds ty 1 b z 0 < z 0. Hene, if z 0 > 0 then the left hand side of (26) is 2bz 0 + z 0 < 3z 0, while the right hand side is > 4a 2 z 0 > 3z 0. If z 0 < 0, then the left hand side of (26) is z 0, while the right hand side is 4a 4a z 0 > a. Let (m, n) = (5, 2). It follows from (18) that and 10.66a 2 2 < 10.65b, a ontradition. (1.807 a) 4 < b b We now an prove the following gap prinipal, whih we will improve again in Proposition 1 below. Lemma 6 If {a, b,, d} is a Diophantine quadruple and a < b < < d, then d = d + or d b 1.5. Proof. By Lemma 5, if d d + then m 3 and n 3. From (17) it follows that and w 3 > (2t 1) b > 81 4 b d 1.161b > b 1.5. Using the gap priniple from Lemma 6 we an prove the following lemma. Lemma 7 Under the notation from above, we have v 3 w 3. Proof. If z 0, z 1 > 0 then define z := x 0 sz 0 = y 1 tz 1, and if z 0, z 1 < 0 then define z := x 0 + sz 0 = y 1 + tz 1. Define also d 0 = (z 2 1)/. Then d 0 is an integer. Furthermore, d = z 2, ad = 1 (a2 x 2 0 2asx 0 z 0 + a 2 z az 2 0 a + ) = (ax 2 0 2asx 0 z 0 + a 2 z x 2 0) = (sx 0 az 0 ) 2, bd = 1 (b2 y 2 1 2bty 1 z 1 + b 2 z 2 1 b + ) = (ty 1 bz 1 ) 2. We have Therefore z = 2 a z 2 0 x 0 + s z 0 > d 0 > a a and z <. > > 0, a
10 There are only finitely many Diophantine quintuples 10 and thus the set {a, b,, d 0 } forms a Diophantine quadruple. Sine d 0 <, by Lemma 6, we have two possibilities: the quadruple {a, b,, d 0 } is regular or (27) > 1.16d b 1.5. Assume that {a, b,, d 0 } is regular, i.e. d 0 = d. Then z = r st. From (x 0 r) = s( z 0 t) and gd(, s) = 1 it follows that z 0 t (mod ), and sine z 0 <, t <, we onlude that z 0 = t and x 0 = r. We an proeed analogously to prove that z 1 = s and y 1 = r. The ondition v 3 = w 3 implies sz 0 + 3x 0 + 4a(x 0 + sz 0 ) = tz 1 + 3y 1 + 4b(y 1 + tz 1 ) and we obtain a = b, a ontradition. Hene we may assume that the quadruple {a, b,, d 0 } is not regular. But it means that > 10 6, whih implies and z = 2 a z 2 0 x 0 + s z 0 > x 0 > a > a d 0 > a 1 > a Thus from (27) we obtain > a 1.25 b 1.5 > a 0.25 and a , whih ontradits the assumption that > Now we an prove the following strong gap priniple, whih is the main improvement to [9] and whih we will use several times later. Observe that espeially the dependene on is muh better than in the gap priniple in [9]. Proposition 1 If {a, b,, d} is a Diophantine quadruple and a < b < < d, then d = d + or d > a 2.5. Proof. From Lemmas 5 and 7 it follows that m 4 or n 4. By (16), we have v 4 Therefore, if m 4 then a (2s 1) a 5 5. Similarly, from (17) it follows w 4 and for n 4 we obtain d 2.696a b (2t 1)3 d 3.919b > a b 5 5, > b 2.5 > a 2.5.
11 There are only finitely many Diophantine quintuples 11 Corollary 1 If {a, b,, d, e} is a Diophantine quintuple and a < b < < d < e, then e > d 3.5 b 2.5. Proof. Assume that {b,, d, e} is a regular Diophantine quadruple. Then e 4d(b + 1) < d 3. The quadruple {a,, d, e} is not regular and, by Proposition 1, we have e > 2.695d 3.5 a 2.5 > d 3. Therefore the quadruple {b,, d, e} is not regular and hene e > 2.695d 3.5 b Determination of the initial terms Using the gap priniples developed in the previous setion we will improve Lemma 2 and obtain more speifi information on the initial terms of the sequenes (v m ) and (w n ). Lemma 8 1) If the equation v 2m = w 2n has a solution, then z 0 = z 1. Furthermore, z 0 = 1 or z 0 = r st or z 0 < min{0.869 a 5/14 9/14, b }. 2) If the equation v 2m+1 = w 2n has a solution, then z 0 = t, z 1 = r st and z 0 z 1 < 0. 3) If the equation v 2m = w 2n+1 has a solution, then z 0 = r st, z 1 = s and z 0 z 1 < 0. 4) If the equation v 2m+1 = w 2n+1 has a solution, then z 0 = t, z 1 = s and z 0 z 1 > 0. Proof. 1) From Lemma 2 we have z 0 = z 1. Define d 0 = (z 2 0 1)/. Then d 0 is an integer and d = z 2 0, ad = 1 (az2 0 a + ) = x 2 0, bd = 1 (bz2 1 b + ) = y 2 1. Hene, we have three possibilities: d 0 = 0 or {a, b,, d} is a regular Diophantine quadruple or {a, b,, d} is an irregular Diophantine quadruple. If d 0 = 0 then z 0 = 1. If the quadruple {a, b,, d} is regular, then (sine d 0 < ) we have d 0 = d and z 0 = r st. Otherwise we may apply Proposition 1 to obtain 2.695d a2.5. Sine z 0 1, we have z We may assume that > 106 and therefore d 0 = z2 0 1 z2 ( ) > z Hene we obtain 4.5 > z 7 0 a2.5 and z 0 < 0.869a 5/14 9/14. Analogously, from Lemma 6, we obtain z 0 < 0.972b ) Let z = z 1 = x 0 + sz 0 if z 1 > 0, and z = z 1 = x 0 sz 0 if z 1 < 0. Define d 0 = (z 2 1)/. Then d 0 is an integer and d = z 2, ad = (sx 0 ± az 0 ) 2, bd = y 2 1. In the proof of Lemma 7 it is shown that d 0 > 0 and d 0 <. Therefore {a, b,, d} is a Diophantine quadruple, and by the proof of Lemma 7, it must be regular. It means that d 0 = d and z = r st. It implies z 1 = r st and z 0 = t.
12 There are only finitely many Diophantine quintuples 12 3) Let z = z 0 = y 1 + tz 1 if z 0 > 0, and z = z 0 = y 1 tz 1 if z 0 < 0, and define d 0 = (z 2 1)/. Then d = z 2, ad = x 2 0, bd = (ty 1 ± bz 1 ) 2 and 0 < d 0 <. If the quadruple {a, b, d 0, } is not regular, then from Lemma 6 we have (28) We an assume that > If > 4b then and if < 4b then Therefore, z = 2 b z 2 1 y 1 + t z d b 1.5. > y 1 > b > b, z > a > b. d 0 > b 1 > b Now (28) implies > /4 b 1/4 and b , a ontradition. It follows that the quadruple {a, b,, d} is regular, i.e. d 0 = d and z = r st. Hene z 0 = r st and (y 1 r) = t( z 1 s). Sine gd(t, ) = 1 we have z 1 s (mod ), whih implies z 1 = s. 4) Let z = x 0 s z 0 = y 1 t z 1 and d 0 = (z 2 1)/. In the proof of Lemma 7 we have shown that {a, b, d 0, } is a regular Diophantine quadruple, and that this fat implies z 0 = t and z 1 = s. 6 Standard Diophantine triples In [9] we proved Conjeture 1 for triples satisfying some gap onditions like b > 4a and > max{b 13, }. In Setion 7 we will prove Conjeture 1 under ertain weaker assumptions. This results will suffie for proving Theorem 1 sine we will show that every Diophantine quadruple ontains a triple whih satisfies some of our gap assumptions. Definition 1 Let {a, b, } be a Diophantine triple and a < b <. Diophantine triple of We all {a, b, } a the first kind if > b 4.5, the seond kind if b > 4a and > b 2.5, the third kind if b > 12a and b 5/3 < < b 2, the fourth kind if b > 4a and b 2 < 6ab 2. A triple {a, b, } is alled standard if it is a Diophantine triple of the first, seond, third or fourth kind.
13 There are only finitely many Diophantine quintuples 13 The Diophantine triples of the first and the seond kind appear naturally when we try to modify results from [9] using Lemma 8. They orrespond to triples with properties b > 4a, > b 13 and b > 4a, > b 5, onsidered in [9]. On the other hand, triples of the third and the fourth kind ome from the analysis what kind of triples a regular Diophantine quadruple may ontain. Note also that these four ases are not mutually exlusive. We now use the improved gap priniple (Proposition 1) again to show that the set of all standard Diophantine triples is large. Proposition 2 Every Diophantine quadruple ontains a standard triple. Proof. Let {a, b,, d} be a Diophantine quadruple. If it is not regular, then by Proposition 1 and [9], Lemma 14, we have d > 3.5 and > 4a. Hene {a,, d} is a triple of the seond kind. Assume that {a, b,, d} is a regular quadruple. Then (4ab + 1) < d < 4(ab + 1). If b > 4a and b 1.5, then d > b 2.5 and we see that {a, b, d} is a triple of the seond kind. If b > 4a and < b 1.5, then we have two possibilities: if = a + b + 2r then < 4b, 2 < d < 4b 2 and therefore {b,, d} is a triple of the fourth kind; if 4ab + a + b then d < 2, d > b > 5/3 and it follows that {a,, d} is a triple of the third kind. We may now assume that b < 4a. By [15], Theorem 8, we have = k (k 1) or = k (k 2), where the sequenes ( k ) and ( k ) are defined by 0 = 0, 1 = a + b + 2r, k = (4ab + 2) k 1 k 2 + 2(a + b), 0 = 0, 1 = a + b 2r, k = (4ab + 2) k 1 k 2 + 2(a + b). If > b 2.5 then d > 4ab > b 4.5 and we see that {a, b, d} is a triple of the first kind. Sine 2 = 4r(a + r)(b + r) > 4ab 2 > b 3, the ondition b 2.5 implies = 1 or = 2. If = 1 then a + b ab = ab ( a b + b a ) ab( ) < 4.81 ab and 4r. Hene d > 4ab > > 2 and d < 4r 2 2 r < 2a 2. Therefore the triple {a,, d} is of the fourth kind. If b 2.5 and = 2 4ab + 2a + 2b (we may assume b > a + 2, sine otherwise 2 = 1 ), then d < 2 and d > 4ab > b 2 > 1.8 > 5/3. Therefore the triple {a,, d} is of the third kind. 7 Lower bounds for solutions The main tool in obtaining lower bounds for m and n satisfying v m = w n (m, n > 2) is the ongruene method introdued in the joint paper of the author with A. Pethő [10].
14 There are only finitely many Diophantine quintuples 14 Lemma 9 1) v 2m z 0 + 2(az 0 m 2 + sx 0 m) (mod 8 2 ) 2) v 2m+1 sz 0 + [2asz 0 m(m + 1) + x 0 (2m + 1)] (mod 4 2 ) 3) w 2n z 1 + 2(bz 1 n 2 + ty 1 n) (mod 8 2 ) 4) w 2n+1 tz 1 + [2btz 1 n(n + 1) + y 1 (2n + 1)] (mod 4 2 ) Proof. See [9], Lemma 4. If v m = w n then, of ourse, v m w n (mod 4 2 ) and we an use Lemma 9 to obtain some ongruenes modulo. However, if a, b, m and n are small ompared with, then these ongruenes are atually equations. It should be possible to prove that these new equations are in ontradition with the starting equations v m = w n. This will imply that m and n annot be too small. We will prove a lower bound for n (and therefore also for m by Lemma 3) depending on and we will do this separately for Diophantine triples of the first, seond, third and fourth kind in the following four lemmas. Lemma 10 Let {a, b, } be a Diophantine triple of the first kind and > If v m = w n and n > 2, then n > Proof. Assume that n By Lemma 17, we have max{ m/2 +1, n/2 +1} < n Aording to Lemma 8, we will onsider six ases. 1.1) v 2m = w 2n, z 0 = 1 From Lemma 9 we have (29) ±am 2 + sm ±bn 2 + tn (mod 4). Sine > b 4.5, we have am 2 < <, sm < <, bn 2 < <, tn < <. Therefore we may replae by = in (29): (30) ±am 2 + sm = ±bn 2 + tn. From (30), squaring twie, we obtain [(am 2 bn 2 ) 2 m 2 n 2 ] 2 4m 2 n 2 (mod ). Sine 4m 2 n 2 < < and [(am 2 bn 2 ) 2 m 2 n 2 ] 2 < <, we therefore have [(am 2 bn 2 ) 2 m 2 n 2 ] 2 = 4m 2 n 2, and (31) am 2 bn 2 = ±m ± n. From (30) and (31) we obtain (32) m(s ± 1) = n(t ± 1). Inserting (32) into (30) we obtain (33) n = (s ± 1)[t(s ± 1) (t ± 1)s] ±[a(t ± 1) 2 b(s ± 1) 2 ] = (s ± 1)(±t s) ±(±2at + 2a ± 2bs 2b).
15 There are only finitely many Diophantine quintuples 15 Sine (s ± 1)(±t s) (s 1)(t s) = (s 1)(b a) t + s > 2(s 1) 2 b a > 2 b and ± 2at + 2a 2bs 2b 4bs + 4b < 6b a, (33) implies that n > ontradition. 1.2) v 2m = w 2n, z 0 = r st We may assume that b 4. Then we have 12 b > 0.377, a z 0 = z 1 = 2 a b 1 r + st > r > ab, and from z 1 < 2 it follows < b 5.4a2 b < 5.4b 3 < b 4.5, a ontradition. 1.3) v 2m = w 2n, z 0 1, r st From Lemma 9 we obtain (34) az 0 m 2 + sx 0 m bz 0 n 2 + ty 1 n (mod 4). By Lemma 8, we have Therefore, (35) az 0 m 2 < <, sx 0 m < (a z 0 + z 0 )m < 0.8 <, bz 0 n 2 < <, ty 1 n < (b z 1 + z 1 )n < 0.87 <. az 0 m 2 + sx 0 m = bz 0 n 2 + ty 1 n. Assume now that b > 4a. Sine z 0 1, we have z0 2 max{ + 1, 3/a}. This implies 0 sx 0 a z 0 1 = x2 0 + a a2 a z 0 (sx 0 + a z 0 ) 1.001a 2a 2 z 2 0 < , 0 ty 1 b z 1 1 = y2 1 + b b2 b z 1 (ty 1 + b z 1 ) 1.001b 2b 2 z 2 0 < If z 0 > 1 then, by (35), we have az 0 m(m ) bz 0 n(n + 1), and sine m, n 2, we obtain a m 2 > bn 2 and m n But now Lemma 4 implies n 1, a ontradition. If z 0 < 1 then (35) implies a z 0 m(m 1)) b z 0 n(n ). Hene, by Lemma 4, we onlude that 4n(n ) < ( 11 9 n + 7 )( n 11 ) < n n , 18 whih implies n n < 0 and n 1, a ontradition.
16 There are only finitely many Diophantine quintuples 16 (36) Assume now that b < 4a. Squaring twie the relation (35) we obtain By Proposition 1, we have [(am 2 bn 2 ) 2 x 2 0m 2 y 2 1n 2 ] 2 4x 2 0y 2 1m 2 n 2 (mod ). y 2 1 < b z < 4a a 5/7 9/ < 3.018(a) 2/7 < This implies that the both sides of the ongruene (36) are less than. Indeed, the left hand side is bounded above by max{ , , } < <, while 4x 2 0 y2 1 m2 n 2 < = <. Therefore we have an equality in (36), and this implies (37) am 2 bn 2 = ±x 0 m ± y 1 n. From (35) and (37) we obtain x 0 m(s ± 1) = y 1 n(t ± 1) and n = A/B, where A = x 2 0y 1 (s ± 1)(±t s), B = z 0 [ab(y 2 1 x 2 0) + 2(a b) ± 2aty 2 1 2bsx 2 0]. We have the following estimates These estimates yield A x 2 0y 1 (s + 1)(t + s) < 2.005x 2 0y 1 ab, B > z 0 [ab(2y 1 1) 2b 4aty1 2 2s(b a)] [ > z 0 y 1 ab ty 1 y 1 ay 1 b 2s ] > z 0 y 1 ab. ay 1 n < 2.005x2 0 y 1 ab z 0 y 1 ab < x 2 0 z 0 ab < az z 0 ab < 1.28 z 0 < 1/4 < 1, a ontradition. 2) v 2m+1 = w 2n The impossibility of this ase is proven in 1.2). 3) v 2m = w 2n+1 By Lemma 8 and 1.2), z 0 > 5.4b 3 < b 4.5, a ontradition. 4) v 2m+1 = w 2n+1 From Lemmas 8 and 9 we obtain (38) 2.155, and from z ab 0 < 2 a it follows < 5.4ab2 < ±astm(m + 1) + rm ±bstn(n + 1) + rn (mod 2), where m = m, n = n if z 0 < 0, and m = m + 1, n = n + 1 if z 0 > 0. Multiplying (38) by s and t, respetively, we obtain (39) (40) ±atm(m + 1) + rsm ±btn(n + 1) + rsn (mod 2), ±asm(m + 1) + rtm ±bsn(n + 1) + rtn (mod 2). We have btn(n + 1) < <, rtn < <. Therefore we have equalities in (39) and (40). This implies rm = rn and am(m + 1) = bn(n + 1), and finally m = n = 0.
17 There are only finitely many Diophantine quintuples 17 Lemma 11 Let {a, b, } be a Diophantine triple of the seond kind and > v m = w n and n > 2, then n > If Proof. Assume that n Then max{ m/2 + 1, n/2 + 1} < n ) v 2m = w 2n, z 0 = 1 Sine am 2 < 0.408, sm < 0.701, bn 2 < and tn < 0.701, (29) implies (30). Furthermore am 2 sm < m a < 0.06 < 0.001, bn 2 tn < n b < 0.06 < Hene m n t s > b a > 1.99, ontrary to Lemma ) v 2m = w 2n, z 0 = r st We have z 0 = z 1 = 2 a b 1 > r + st 2r > ab, 2 b it follows < 5.25a2 b < 1.32ab 2. Hene {a, b, } is a Diophantine and from z 1 < triple of the fourth kind, and this ase will be treated in Lemma ) v 2m = w 2n, z 0 1, r st By Proposition 1, we have > b 3.5, and Lemma 8 implies az 0 m 2 < <, sx 0 m < <, bz 0 n 2 < 0.98 <, ty 1 n < <. Therefore, equation (35) holds again. As in Lemma 10, we obtain a ontradition with Lemma 4. 2) v 2m+1 = w 2n This ase is impossible by Lemma 2. Namely, if 100a then a and a 1.03, a ontradition. 3) v 2m = w 2n+1 As in 1.2), we obtain < 5.25ab 2 and again {a, b, } is a triple of the fourth kind. 4) v 2m+1 = w 2n+1 Relation (38) implies [am(m + 1) bn(n + 1)] 2 r 2 (n m) 2 (mod 2). Sine a 2 m 2 (m + 1) 2 < 0.96, b 2 n 2 (n + 1) 2 < 0.96, and r 2 m 2 < 0.48, we obtain (41) bn(n + 1) am(m + 1) = ±r(m n), whih implies (42) 2n + 1 a 2m + 1 b = ±4r(m n) + b a [(2n + 1) b + (2m + 1) a] b(2m + 1).
18 There are only finitely many Diophantine quintuples 18 By Lemma 4, the right hand side of (42) is 4r(m n) b(2n + 1)(2m + 1) + Therefore, (43) b b(2n + 1)(2m + 1) < 2n + 1 2m + 1 < < b 2 4n+5 10 b(2n + 1)(2m + 1) < = If n = 1 then m = 1 (by Lemma 4), ontrary to (43). If n = 2 then m = 2 or m = 3, whih both ontradit the relation (43). Hene we may assume that n 3, m 3, and now (42) implies 2n+1 2m+1 < On the other hand, from Lemma 4 we see that 2n+1 2m (2m+1) 0.653, a ontradition. Lemma 12 Let {a, b, } be a Diophantine triple of the third kind and > v m = w n and n > 2, then n > If Proof. Assume that n Then max{ m/2 + 1, n/2 + 1} < n ) v 2m = w 2n, z 0 = 1 Sine am 2 < 0.9, sm < 0.951, bn 2 < 0.9 and tn < <, the proof is idential to that of Lemma ) v 2m = w 2n, z 0 = r st From (34) we have (44) ±astm(m 1) + rm ±bstn(n 1) + rn (mod ). Multiplying (44) by 2st we obtain (45) ±2[am(m 1) bn(n 1)] 2rst(n m) (mod 2). Let α be the absolutely least residue of 2rst modulo 2, and let A = (2rst 2r 2 +)(st+r). Then α (st + r) A and A = 2ar + 2br + 2r + st 2 r < 2ar + 2br + 2r. Sine r st = 2 b a 1 r+st < 2 2 < b b < br, we obtain A < 2r(a+b+1) < 13 ab 6 br and α < 13br 12 < 1.15b. We have 2am(m 1) < ab 0.904, 2bn(n 1) < 0.904, α(m n) < Therefore ±2[am(m 1) bn(n ± 1)] = α(n m), and it implies (46) 2n 1 a 2m 1 b = ±2α(m n) + b a [(2n 1) b + (2m 1) a] b(2m 1). By Lemma 4, the right hand side of (46) is 1.15b 6n+3 5 b(2n 1)(2m 1) + 1 (2n 1)(2m 1) < 0.495,
19 There are only finitely many Diophantine quintuples 19 and therefore (47) 2n 1 2m 1 < If n = 2 then Lemma 4 implies m = 2 or 3. The ase m = 2 ontradits the inequality (47), while for m = 3 relation (46) implies 2n 1 2m 1 < But for n = 2, m = 3 we have 2n 1 2m 1 { 3 5, 5 7 }, a ontradition. Hene n 3, and sine n = m = 3 ontradits (47), we have also m 4. Now, from (46) we obtain 2n 1 2m+1 2m 1 < But for m 4 we have 2n+1 > 2n 1 2m 1 > (2m 1) > 0.517, a ontradition. 1.3) v 2m = w 2n, z 0 1, r st Proposition 1 implies > b 3.5 and therefore this ase is impossible. 2) v 2m+1 = w 2n The impossibility of this ase is shown in Lemma 11. 3) v 2m = w 2n+1 From Lemmas 8 and 9 it follows that and (48) ±2astm(m 1) + r(2m ± 1) ±2bstn(n + 1) + r(2n + 1) (mod 4) ±2[am(m 1) bn(n + 1)] 2rst(n m + δ) (mod 2), where δ {0, 1}. Let α be defined as in 1.2). As in 1.2), we obtain and (49) ±2[am(m 1) bn(n + 1)] = α(n m + δ) 2n + 1 a 2m 1 b = The right hand side of (49) is and therefore (50) ±2α(m n δ) + b a [(2n + 1) b + (2m 1) a] b(2m 1). 1.15b 6n+11 5 b(2n 1)(2m 1) + 1 (2n 1)(2m 1), 2n + 1 2m 1 < 0.861, 2n + 1 2m + 1 < 0.617, respetively. If n = 1, then by Lemmas 4 and 7 we have m = 2. This is learly impossible if we have 2n+1 2n+1 2m 1 on the left hand side of (49), while for 2m+1 we obtain 3 5 < 0.509, a ontradition. If n = 2 then, by Lemma 17, we have m = 2, 3 or 4, and (50) implies m = 4. Sine m = n = 3 is also impossible by (50), we onlude that n 2 and m 4. This implies 2n+1 2m 1 < 0.469, 2n+1 2m+1 < 0.429, respetively. On the other hand, we have 2n+1 2m 1 > 2n+1 2m m+1 > 0.513, a ontradition. 4) v 2m+1 = w 2n+1
20 There are only finitely many Diophantine quintuples 20 Let α be defined as in 1.2). From (38) we obtain whih yields ±2[am(m + 1) bn(n + 1)] = α(n m), (51) 2n + 1 a 2m + 1 b = ±2α(m n) + b a [(2n + 1) b + (2m + 1) a] b(2m + 1). From (51) it follows that 2n+1 2m+1 a ontradition. < But Lemma 4 implies 2n+1 2m (2m+1) 0.55, Lemma 13 Let {a, b, } be a Diophantine triple of the fourth kind and > v m = w n and n > 2, then n > 0.2. If Proof. Assume that n 0.2. Then max{ m/2 + 1, n/2 + 1} < n ) v 2m = w 2n, z 0 = 1 The proof is idential to that of Lemma ) v 2m = w 2n, z 0 = r st The first part of the proof is idential to that of Lemma 12. In partiular, the relation (45) is valid. Estimating r st, we get r st < 2 ab < 6ab2 2 ab < 3b ab < 3br. Therefore A < 2r(a + b + 1) < 2.5br and α < 2.5br 2 < 1.33b. Note that ab α 2 4r 2 s 2 t 2 4r 2 (mod 2), 4r 2 4.5ab < 1.125b 2 < 2 and α 2 < 1.77b 2 < 2. Hene, α 2 = 4r 2 and α = ±2r. Sine am(m + 1) < 0.9, bn(n + 1) < 0.9, r(m n) < 0.7, we have (52) bn(n + 1) am(m + 1) = ±r(m n). Inserting α = ±2r in (46) and using Lemma 4 and the estimate α = 2r < b < 1.001b, we obtain (53) 2n 1 2m 1 < n + 1 (2n 1)(2m 1) < If (m, n) = (3, 2), then (52) implies 6b 12a = ±r and (4b 9a)(9b 16a) = 1, whih is learly impossible for b > 4a. Hene n 3, m 4 and (53) yields 2n 1 2m 1 < On the other hand, 2n+1 2m+1 > 2n 1 2m (2m 1) > 0.619, a ontradition. 1.3) v 2m = w 2n, z 0 1, r st Proposition 1 implies > b 3.5, and we have < 6ab 2 < 1.5b 3. Therefore, this ase is impossible. 2) v 2m+1 = w 2n
1.3 Complex Numbers; Quadratic Equations in the Complex Number System*
04 CHAPTER Equations and Inequalities Explaining Conepts: Disussion and Writing 7. Whih of the following pairs of equations are equivalent? Explain. x 2 9; x 3 (b) x 29; x 3 () x  2x  22 x  2 2 ; x
More information5.2 The Master Theorem
170 CHAPTER 5. RECURSION AND RECURRENCES 5.2 The Master Theorem Master Theorem In the last setion, we saw three different kinds of behavior for reurrenes of the form at (n/2) + n These behaviors depended
More informationSet Theory and Logic: Fundamental Concepts (Notes by Dr. J. Santos)
A.1 Set Theory and Logi: Fundamental Conepts (Notes by Dr. J. Santos) A.1. Primitive Conepts. In mathematis, the notion of a set is a primitive notion. That is, we admit, as a starting point, the existene
More informationChapter 6 A N ovel Solution Of Linear Congruenes Proeedings NCUR IX. (1995), Vol. II, pp. 708{712 Jerey F. Gold Department of Mathematis, Department of Physis University of Utah Salt Lake City, Utah 84112
More informationDIOPHANTINE mtuples FOR LINEAR POLYNOMIALS. II. EQUAL DEGREES
DIOPHANTINE mtuples FOR LINEAR POLYNOMIALS. II. EQUAL DEGREES ANDREJ DUJELLA, CLEMENS FUCHS, AND GARY WALSH Abstract. In this paper we prove the best possible upper bounds for the number of elements in
More informationChannel Assignment Strategies for Cellular Phone Systems
Channel Assignment Strategies for Cellular Phone Systems Wei Liu Yiping Han Hang Yu Zhejiang University Hangzhou, P. R. China Contat: wliu5@ie.uhk.edu.hk 000 Mathematial Contest in Modeling (MCM) Meritorious
More informationchapter > Make the Connection Factoring CHAPTER 4 OUTLINE Chapter 4 :: Pretest 374
CHAPTER hapter 4 > Make the Connetion 4 INTRODUCTION Developing seret odes is big business beause of the widespread use of omputers and the Internet. Corporations all over the world sell enryption systems
More informationA DIOPHANTINE EQUATION RELATED TO THE SUM OF POWERS OF TWO CONSECUTIVE GENERALIZED FIBONACCI NUMBERS
A DIOPHANTINE EQUATION RELATED TO THE SUM OF POWERS OF TWO CONSECUTIVE GENERALIZED FIBONACCI NUMBERS ANA PAULA CHAVES AND DIEGO MARQUES Abstract. Let F n) n 0 be the Fibonacci sequence given by F m+ =
More informationApplications of Fermat s Little Theorem and Congruences
Applications of Fermat s Little Theorem and Congruences Definition: Let m be a positive integer. Then integers a and b are congruent modulo m, denoted by a b mod m, if m (a b). Example: 3 1 mod 2, 6 4
More informationUSA Mathematical Talent Search. PROBLEMS / SOLUTIONS / COMMENTS Round 3  Year 12  Academic Year 20002001
USA Mathematial Talent Searh PROBLEMS / SOLUTIONS / COMMENTS Round 3  Year  Aademi Year 00000 Gene A. Berg, Editor /3/. Find the smallest positive integer with the property that it has divisors ending
More informationChapter 1 Microeconomics of Consumer Theory
Chapter 1 Miroeonomis of Consumer Theory The two broad ategories of deisionmakers in an eonomy are onsumers and firms. Eah individual in eah of these groups makes its deisions in order to ahieve some
More informationConvergence of c k f(kx) and the Lip α class
Convergene of and the Lip α lass Christoph Aistleitner Abstrat By Carleson s theorem a trigonometri series k osπkx or k sin πkx is ae onvergent if < (1) Gaposhkin generalized this result to series of the
More informationFrom (2) follows, if z0 = 0, then z = vt, thus a2 =?va (2.3) Then 2:3 beomes z0 = z (z? vt) (2.4) t0 = bt + b2z Consider the onsequenes of (3). A ligh
Chapter 2 Lorentz Transformations 2. Elementary Considerations We assume we have two oordinate systems S and S0 with oordinates x; y; z; t and x0; y0; z0; t0, respetively. Physial events an be measured
More informationComputer Networks Framing
Computer Networks Framing Saad Mneimneh Computer Siene Hunter College of CUNY New York Introdution Who framed Roger rabbit? A detetive, a woman, and a rabbit in a network of trouble We will skip the physial
More informationAPPLICATIONS OF THE ORDER FUNCTION
APPLICATIONS OF THE ORDER FUNCTION LECTURE NOTES: MATH 432, CSUSM, SPRING 2009. PROF. WAYNE AITKEN In this lecture we will explore several applications of order functions including formulas for GCDs and
More informationMOP 2007 Black Group Integer Polynomials Yufei Zhao. Integer Polynomials. June 29, 2007 Yufei Zhao yufeiz@mit.edu
Integer Polynomials June 9, 007 Yufei Zhao yufeiz@mit.edu We will use Z[x] to denote the ring of polynomials with integer coefficients. We begin by summarizing some of the common approaches used in dealing
More informationDSPI DSPI DSPI DSPI
DSPI DSPI DSPI DSPI Digital Signal Proessing I (879) Fall Semester, 005 IIR FILER DESIG EXAMPLE hese notes summarize the design proedure for IIR filters as disussed in lass on ovember. Introdution:
More informationSebastián Bravo López
Transfinite Turing mahines Sebastián Bravo López 1 Introdution With the rise of omputers with high omputational power the idea of developing more powerful models of omputation has appeared. Suppose that
More informationcos t sin t sin t cos t
Exerise 7 Suppose that t 0 0andthat os t sin t At sin t os t Compute Bt t As ds,andshowthata and B ommute 0 Exerise 8 Suppose A is the oeffiient matrix of the ompanion equation Y AY assoiated with the
More informationStatic Fairness Criteria in Telecommunications
Teknillinen Korkeakoulu ERIKOISTYÖ Teknillisen fysiikan koulutusohjelma 92002 Mat208 Sovelletun matematiikan erikoistyöt Stati Fairness Criteria in Teleommuniations Vesa Timonen, email: vesatimonen@hutfi
More informationContinued Fractions and the Euclidean Algorithm
Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction
More informationPowers of Two in Generalized Fibonacci Sequences
Revista Colombiana de Matemáticas Volumen 462012)1, páginas 6779 Powers of Two in Generalized Fibonacci Sequences Potencias de dos en sucesiones generalizadas de Fibonacci Jhon J. Bravo 1,a,B, Florian
More informationWeighting Methods in Survey Sampling
Setion on Survey Researh Methods JSM 01 Weighting Methods in Survey Sampling Chiaohih Chang Ferry Butar Butar Abstrat It is said that a welldesigned survey an best prevent nonresponse. However, no matter
More informationIRREDUCIBLE OPERATOR SEMIGROUPS SUCH THAT AB AND BA ARE PROPORTIONAL. 1. Introduction
IRREDUCIBLE OPERATOR SEMIGROUPS SUCH THAT AB AND BA ARE PROPORTIONAL R. DRNOVŠEK, T. KOŠIR Dedicated to Prof. Heydar Radjavi on the occasion of his seventieth birthday. Abstract. Let S be an irreducible
More informationEvery Positive Integer is the Sum of Four Squares! (and other exciting problems)
Every Positive Integer is the Sum of Four Squares! (and other exciting problems) Sophex University of Texas at Austin October 18th, 00 Matilde N. Lalín 1. Lagrange s Theorem Theorem 1 Every positive integer
More informationOptimal Sales Force Compensation
Optimal Sales Fore Compensation Matthias Kräkel Anja Shöttner Abstrat We analyze a dynami moralhazard model to derive optimal sales fore ompensation plans without imposing any ad ho restritions on the
More informationTHREE DIMENSIONAL GEOMETRY
Chapter 8 THREE DIMENSIONAL GEOMETRY 8.1 Introduction In this chapter we present a vector algebra approach to three dimensional geometry. The aim is to present standard properties of lines and planes,
More informationU.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009. Notes on Algebra
U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009 Notes on Algebra These notes contain as little theory as possible, and most results are stated without proof. Any introductory
More informationAUDITING COST OVERRUN CLAIMS *
AUDITING COST OVERRUN CLAIMS * David PérezCastrillo # University of Copenhagen & Universitat Autònoma de Barelona Niolas Riedinger ENSAE, Paris Abstrat: We onsider a ostreimbursement or a ostsharing
More informationMA257: INTRODUCTION TO NUMBER THEORY LECTURE NOTES
MA257: INTRODUCTION TO NUMBER THEORY LECTURE NOTES 2016 47 4. Diophantine Equations A Diophantine Equation is simply an equation in one or more variables for which integer (or sometimes rational) solutions
More informationCapacity at Unsignalized TwoStage Priority Intersections
Capaity at Unsignalized TwoStage Priority Intersetions by Werner Brilon and Ning Wu Abstrat The subjet of this paper is the apaity of minorstreet traffi movements aross major divided fourlane roadways
More informationSymmetric Subgame Perfect Equilibria in Resource Allocation
Proeedings of the TwentySixth AAAI Conferene on Artifiial Intelligene Symmetri Subgame Perfet Equilibria in Resoure Alloation Ludek Cigler Eole Polytehnique Fédérale de Lausanne Artifiial Intelligene
More informationInverses and powers: Rules of Matrix Arithmetic
Contents 1 Inverses and powers: Rules of Matrix Arithmetic 1.1 What about division of matrices? 1.2 Properties of the Inverse of a Matrix 1.2.1 Theorem (Uniqueness of Inverse) 1.2.2 Inverse Test 1.2.3
More informationQuotient Rings and Field Extensions
Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F.
More informationarxiv:astroph/0304006v2 10 Jun 2003 Theory Group, MS 50A5101 Lawrence Berkeley National Laboratory One Cyclotron Road Berkeley, CA 94720 USA
LBNL52402 Marh 2003 On the Speed of Gravity and the v/ Corretions to the Shapiro Time Delay Stuart Samuel 1 arxiv:astroph/0304006v2 10 Jun 2003 Theory Group, MS 50A5101 Lawrene Berkeley National Laboratory
More informationSupply chain coordination; A Game Theory approach
aepted for publiation in the journal "Engineering Appliations of Artifiial Intelligene" 2008 upply hain oordination; A Game Theory approah JeanClaude Hennet x and Yasemin Arda xx x LI CNRUMR 668 Université
More informationAn Iterated Beam Search Algorithm for Scheduling Television Commercials. Mesut Yavuz. Shenandoah University
0080569 An Iterated Beam Searh Algorithm for Sheduling Television Commerials Mesut Yavuz Shenandoah University The Harry F. Byrd, Jr. Shool of Business Winhester, Virginia, U.S.A. myavuz@su.edu POMS 19
More informationIdeal Class Group and Units
Chapter 4 Ideal Class Group and Units We are now interested in understanding two aspects of ring of integers of number fields: how principal they are (that is, what is the proportion of principal ideals
More informationConcentration of points on two and three dimensional modular hyperbolas and applications
Concentration of points on two and three dimensional modular hyperbolas and applications J. Cilleruelo Instituto de Ciencias Matemáticas (CSICUAMUC3MUCM) and Departamento de Matemáticas Universidad
More informationHomework until Test #2
MATH31: Number Theory Homework until Test # Philipp BRAUN Section 3.1 page 43, 1. It has been conjectured that there are infinitely many primes of the form n. Exhibit five such primes. Solution. Five such
More informationSUM OF TWO SQUARES JAHNAVI BHASKAR
SUM OF TWO SQUARES JAHNAVI BHASKAR Abstract. I will investigate which numbers can be written as the sum of two squares and in how many ways, providing enough basic number theory so even the unacquainted
More informationTrigonometry & Pythagoras Theorem
Trigonometry & Pythagoras Theorem Mathematis Skills Guide This is one of a series of guides designed to help you inrease your onfidene in handling Mathematis. This guide ontains oth theory and exerises
More informationMath 319 Problem Set #3 Solution 21 February 2002
Math 319 Problem Set #3 Solution 21 February 2002 1. ( 2.1, problem 15) Find integers a 1, a 2, a 3, a 4, a 5 such that every integer x satisfies at least one of the congruences x a 1 (mod 2), x a 2 (mod
More informationFACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z
FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Abstract We consider polynomials with integer coefficients and discuss their factorization
More informationChemical Equilibrium. Chemical Equilibrium. Chemical Equilibrium. Chemical Equilibriu m. Chapter 14
Chapter 14 Chemial Equilibrium Chemial Equilibriu m Muh like water in a Ushaped tube, there is onstant mixing bak and forth through the lower portion of the tube. reatants produts It s as if the forward
More informationOn the largest prime factor of x 2 1
On the largest prime factor of x 2 1 Florian Luca and Filip Najman Abstract In this paper, we find all integers x such that x 2 1 has only prime factors smaller than 100. This gives some interesting numerical
More information3 Game Theory: Basic Concepts
3 Game Theory: Basi Conepts Eah disipline of the soial sienes rules omfortably ithin its on hosen domain: : : so long as it stays largely oblivious of the others. Edard O. Wilson (1998):191 3.1 and and
More informationThe Dirichlet Unit Theorem
Chapter 6 The Dirichlet Unit Theorem As usual, we will be working in the ring B of algebraic integers of a number field L. Two factorizations of an element of B are regarded as essentially the same if
More informationUnique Factorization
Unique Factorization Waffle Mathcamp 2010 Throughout these notes, all rings will be assumed to be commutative. 1 Factorization in domains: definitions and examples In this class, we will study the phenomenon
More informationIsaac Newton. Translated into English by
THE MATHEMATICAL PRINCIPLES OF NATURAL PHILOSOPHY (BOOK 1, SECTION 1) By Isaa Newton Translated into English by Andrew Motte Edited by David R. Wilkins 2002 NOTE ON THE TEXT Setion I in Book I of Isaa
More informationAlgebraic and Transcendental Numbers
Pondicherry University July 2000 Algebraic and Transcendental Numbers Stéphane Fischler This text is meant to be an introduction to algebraic and transcendental numbers. For a detailed (though elementary)
More informationElementary Number Theory We begin with a bit of elementary number theory, which is concerned
CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,
More information) ( )( ) ( ) ( )( ) ( ) ( ) (1)
OPEN CHANNEL FLOW Open hannel flow is haraterized by a surfae in ontat with a gas phase, allowing the fluid to take on shapes and undergo behavior that is impossible in a pipe or other filled onduit. Examples
More informationSpecial Relativity and Linear Algebra
peial Relativity and Linear Algebra Corey Adams May 7, Introdution Before Einstein s publiation in 95 of his theory of speial relativity, the mathematial manipulations that were a produt of his theory
More informationFundamentals of Chemical Reactor Theory
UNIVERSITY OF CALIFORNIA, LOS ANGELES Civil & Environmental Engineering Department Fundamentals of Chemial Reator Theory Mihael K. Stenstrom Professor Diego Rosso Teahing Assistant Los Angeles, 3 Introdution
More informationChapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.
Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize
More informationk, then n = p2α 1 1 pα k
Powers of Integers An integer n is a perfect square if n = m for some integer m. Taking into account the prime factorization, if m = p α 1 1 pα k k, then n = pα 1 1 p α k k. That is, n is a perfect square
More informationNumber Theory Hungarian Style. Cameron Byerley s interpretation of Csaba Szabó s lectures
Number Theory Hungarian Style Cameron Byerley s interpretation of Csaba Szabó s lectures August 20, 2005 2 0.1 introduction Number theory is a beautiful subject and even cooler when you learn about it
More informationClassical Electromagnetic Doppler Effect Redefined. Copyright 2014 Joseph A. Rybczyk
Classial Eletromagneti Doppler Effet Redefined Copyright 04 Joseph A. Rybzyk Abstrat The lassial Doppler Effet formula for eletromagneti waves is redefined to agree with the fundamental sientifi priniples
More informationEE 201 ELECTRIC CIRCUITS LECTURE 25. Natural, Forced and Complete Responses of First Order Circuits
EE 201 EECTRIC CIUITS ECTURE 25 The material overed in this leture will be as follows: Natural, Fored and Complete Responses of First Order Ciruits Step Response of First Order Ciruits Step Response of
More informationIntelligent Measurement Processes in 3D Optical Metrology: Producing More Accurate Point Clouds
Intelligent Measurement Proesses in 3D Optial Metrology: Produing More Aurate Point Clouds Charles Mony, Ph.D. 1 President Creaform in. mony@reaform3d.om Daniel Brown, Eng. 1 Produt Manager Creaform in.
More informationPYTHAGOREAN TRIPLES PETE L. CLARK
PYTHAGOREAN TRIPLES PETE L. CLARK 1. Parameterization of Pythagorean Triples 1.1. Introduction to Pythagorean triples. By a Pythagorean triple we mean an ordered triple (x, y, z) Z 3 such that x + y =
More informationChapter 5 Single Phase Systems
Chapter 5 Single Phase Systems Chemial engineering alulations rely heavily on the availability of physial properties of materials. There are three ommon methods used to find these properties. These inlude
More informationCollinear Points in Permutations
Collinear Points in Permutations Joshua N. Cooper Courant Institute of Mathematics New York University, New York, NY József Solymosi Department of Mathematics University of British Columbia, Vancouver,
More informationLectures on Number Theory. LarsÅke Lindahl
Lectures on Number Theory LarsÅke Lindahl 2002 Contents 1 Divisibility 1 2 Prime Numbers 7 3 The Linear Diophantine Equation ax+by=c 12 4 Congruences 15 5 Linear Congruences 19 6 The Chinese Remainder
More informationNodal domains on graphs  How to count them and why?
Proeedings of Symposia in Pure Mathematis Nodal domains on graphs  How to ount them and why? Ram Band, Idan Oren and Uzy Smilansky, Abstrat. The purpose of the present manusript is to ollet known results
More informationOn the generation of elliptic curves with 16 rational torsion points by Pythagorean triples
On the generation of elliptic curves with 16 rational torsion points by Pythagorean triples Brian Hilley Boston College MT695 Honors Seminar March 3, 2006 1 Introduction 1.1 Mazur s Theorem Let C be a
More informationTheory of linear elasticity. I assume you have learned the elements of linear
Physial Metallurgy Frature mehanis leture 1 In the next two letures (Ot.16, Ot.18), we will disuss some basis of frature mehanis using ontinuum theories. The method of ontinuum mehanis is to view a solid
More informationREDUCTION FACTOR OF FEEDING LINES THAT HAVE A CABLE AND AN OVERHEAD SECTION
C I E 17 th International Conferene on Eletriity istriution Barelona, 115 May 003 EUCTION FACTO OF FEEING LINES THAT HAVE A CABLE AN AN OVEHEA SECTION Ljuivoje opovi J.. Elektrodistriuija  Belgrade 
More informationSUBGROUPS OF CYCLIC GROUPS. 1. Introduction In a group G, we denote the (cyclic) group of powers of some g G by
SUBGROUPS OF CYCLIC GROUPS KEITH CONRAD 1. Introduction In a group G, we denote the (cyclic) group of powers of some g G by g = {g k : k Z}. If G = g, then G itself is cyclic, with g as a generator. Examples
More informationMODULAR ARITHMETIC. a smallest member. It is equivalent to the Principle of Mathematical Induction.
MODULAR ARITHMETIC 1 Working With Integers The usual arithmetic operations of addition, subtraction and multiplication can be performed on integers, and the result is always another integer Division, on
More informationPrices and Heterogeneous Search Costs
Pries and Heterogeneous Searh Costs José L. MoragaGonzález Zsolt Sándor Matthijs R. Wildenbeest First draft: July 2014 Revised: June 2015 Abstrat We study prie formation in a model of onsumer searh for
More informationThe Notebook Series. The solution of cubic and quartic equations. R.S. Johnson. Professor of Applied Mathematics
The Notebook Series The solution of cubic and quartic equations by R.S. Johnson Professor of Applied Mathematics School of Mathematics & Statistics University of Newcastle upon Tyne R.S.Johnson 006 CONTENTS
More informationPUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.
PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include
More informationON GALOIS REALIZATIONS OF THE 2COVERABLE SYMMETRIC AND ALTERNATING GROUPS
ON GALOIS REALIZATIONS OF THE 2COVERABLE SYMMETRIC AND ALTERNATING GROUPS DANIEL RABAYEV AND JACK SONN Abstract. Let f(x) be a monic polynomial in Z[x] with no rational roots but with roots in Q p for
More informationOptimal Online Buffer Scheduling for Block Devices *
Optimal Online Buffer Sheduling for Blok Devies * ABSTRACT Anna Adamaszek Department of Computer Siene and Centre for Disrete Mathematis and its Appliations (DIMAP) University of Warwik, Coventry, UK A.M.Adamaszek@warwik.a.uk
More informationThe van Hoeij Algorithm for Factoring Polynomials
The van Hoeij Algorithm for Factoring Polynomials Jürgen Klüners Abstract In this survey we report about a new algorithm for factoring polynomials due to Mark van Hoeij. The main idea is that the combinatorial
More informationThe sum of digits of polynomial values in arithmetic progressions
The sum of digits of polynomial values in arithmetic progressions Thomas Stoll Institut de Mathématiques de Luminy, Université de la Méditerranée, 13288 Marseille Cedex 9, France Email: stoll@iml.univmrs.fr
More informationWORKFLOW CONTROLFLOW PATTERNS A Revised View
WORKFLOW CONTROLFLOW PATTERNS A Revised View Nik Russell 1, Arthur H.M. ter Hofstede 1, 1 BPM Group, Queensland University of Tehnology GPO Box 2434, Brisbane QLD 4001, Australia {n.russell,a.terhofstede}@qut.edu.au
More information5. Factoring by the QF method
5. Factoring by the QF method 5.0 Preliminaries 5.1 The QF view of factorability 5.2 Illustration of the QF view of factorability 5.3 The QF approach to factorization 5.4 Alternative factorization by the
More informationA Comparison of Default and Reduced Bandwidth MR Imaging of the Spine at 1.5 T
9 A Comparison of efault and Redued Bandwidth MR Imaging of the Spine at 1.5 T L. Ketonen 1 S. Totterman 1 J. H. Simon 1 T. H. Foster 2. K. Kido 1 J. Szumowski 1 S. E. Joy1 The value of a redued bandwidth
More informationAn example of a computable
An example of a computable absolutely normal number Verónica Becher Santiago Figueira Abstract The first example of an absolutely normal number was given by Sierpinski in 96, twenty years before the concept
More informationDerivation of Einstein s Equation, E = mc 2, from the Classical Force Laws
Apeiron, Vol. 14, No. 4, Otober 7 435 Derivation of Einstein s Equation, E = m, from the Classial Fore Laws N. Hamdan, A.K. Hariri Department of Physis, University of Aleppo, Syria nhamdan59@hotmail.om,
More informationCHAPTER 14 Chemical Equilibrium: Equal but Opposite Reaction Rates
CHATER 14 Chemial Equilibrium: Equal but Opposite Reation Rates 14.1. Collet and Organize For two reversible reations, we are given the reation profiles (Figure 14.1). The profile for the onversion of
More informationa 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2.
Chapter 1 LINEAR EQUATIONS 1.1 Introduction to linear equations A linear equation in n unknowns x 1, x,, x n is an equation of the form a 1 x 1 + a x + + a n x n = b, where a 1, a,..., a n, b are given
More informationRole of the Reference Frame in Angular Photon Distribution at ElectronPositron Annihilation
Journal of Modern Physis,, 5, 353358 Published Online April in SiRes. http://www.sirp.org/journal/jmp http://dx.doi.org/.36/jmp..565 Role of the Referene Frame in Angular Photon Distribution at EletronPositron
More informationON DEGREES IN THE HASSE DIAGRAM OF THE STRONG BRUHAT ORDER
Séminaire Lotharingien de Combinatoire 53 (2006), Article B53g ON DEGREES IN THE HASSE DIAGRAM OF THE STRONG BRUHAT ORDER RON M. ADIN AND YUVAL ROICHMAN Abstract. For a permutation π in the symmetric group
More informationPROOFS BY DESCENT KEITH CONRAD
PROOFS BY DESCENT KEITH CONRAD As ordinary methods, such as are found in the books, are inadequate to proving such difficult propositions, I discovered at last a most singular method... that I called the
More informationPrime Numbers and Irreducible Polynomials
Prime Numbers and Irreducible Polynomials M. Ram Murty The similarity between prime numbers and irreducible polynomials has been a dominant theme in the development of number theory and algebraic geometry.
More informationInteger roots of quadratic and cubic polynomials with integer coefficients
Integer roots of quadratic and cubic polynomials with integer coefficients Konstantine Zelator Mathematics, Computer Science and Statistics 212 Ben Franklin Hall Bloomsburg University 400 East Second Street
More informationFIRST YEAR CALCULUS. Chapter 7 CONTINUITY. It is a parabola, and we can draw this parabola without lifting our pencil from the paper.
FIRST YEAR CALCULUS WWLCHENW L c WWWL W L Chen, 1982, 2008. 2006. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It It is is
More informationFACTORING CERTAIN INFINITE ABELIAN GROUPS BY DISTORTED CYCLIC SUBSETS
International Electronic Journal of Algebra Volume 6 (2009) 95106 FACTORING CERTAIN INFINITE ABELIAN GROUPS BY DISTORTED CYCLIC SUBSETS Sándor Szabó Received: 11 November 2008; Revised: 13 March 2009
More information(x + a) n = x n + a Z n [x]. Proof. If n is prime then the map
22. A quick primality test Prime numbers are one of the most basic objects in mathematics and one of the most basic questions is to decide which numbers are prime (a clearly related problem is to find
More informationGranular Problem Solving and Software Engineering
Granular Problem Solving and Software Engineering Haibin Zhu, Senior Member, IEEE Department of Computer Siene and Mathematis, Nipissing University, 100 College Drive, North Bay, Ontario, P1B 8L7, Canada
More informationAbout the inverse football pool problem for 9 games 1
Seventh International Workshop on Optimal Codes and Related Topics September 61, 013, Albena, Bulgaria pp. 15133 About the inverse football pool problem for 9 games 1 Emil Kolev Tsonka Baicheva Institute
More informationSrinivas Bollapragada GE Global Research Center. Abstract
Sheduling Commerial Videotapes in Broadast Television Srinivas Bollapragada GE Global Researh Center Mihael Bussiek GAMS Development Corporation Suman Mallik University of Illinois at Urbana Champaign
More informationA matrix over a field F is a rectangular array of elements from F. The symbol
Chapter MATRICES Matrix arithmetic A matrix over a field F is a rectangular array of elements from F The symbol M m n (F) denotes the collection of all m n matrices over F Matrices will usually be denoted
More informationSMT 2014 Algebra Test Solutions February 15, 2014
1. Alice and Bob are painting a house. If Alice and Bob do not take any breaks, they will finish painting the house in 20 hours. If, however, Bob stops painting once the house is halffinished, then the
More informationCHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY
January 10, 2010 CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY The set of polynomials over a field F is a ring, whose structure shares with the ring of integers many characteristics.
More informationProcurement auctions are sometimes plagued with a chosen supplier s failing to accomplish a project successfully.
Deision Analysis Vol. 7, No. 1, Marh 2010, pp. 23 39 issn 15458490 eissn 15458504 10 0701 0023 informs doi 10.1287/dea.1090.0155 2010 INFORMS Managing Projet Failure Risk Through Contingent Contrats
More information