There are only finitely many Diophantine quintuples

Size: px
Start display at page:

Download "There are only finitely many Diophantine quintuples"

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 m-tuple 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 m-tuple 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 9-tuple and that there are only finitely many Diophantine 8-tuples. 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*

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 information

5.2 The Master Theorem

5.2 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 information

Sequences CheatSheet

Sequences CheatSheet Sequenes CheatSheet Intro This heatsheet ontains everything you should know about real sequenes. It s not meant to be exhaustive, but it ontains more material than the textbook. Definitions and properties

More information

Set Theory and Logic: Fundamental Concepts (Notes by Dr. J. Santos)

Set 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 information

Chapter 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 information

DIOPHANTINE m-tuples FOR LINEAR POLYNOMIALS. II. EQUAL DEGREES

DIOPHANTINE m-tuples FOR LINEAR POLYNOMIALS. II. EQUAL DEGREES DIOPHANTINE m-tuples 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 information

Lecture 5. Gaussian and Gauss-Jordan elimination

Lecture 5. Gaussian and Gauss-Jordan elimination International College of Eonomis and Finane (State University Higher Shool of Eonomis) Letures on Linear Algebra by Vladimir Chernyak, Leture. Gaussian and Gauss-Jordan elimination To be read to the musi

More information

Diameter growth and bounded topology of complete manifolds with nonnegative Ricci curvature

Diameter growth and bounded topology of complete manifolds with nonnegative Ricci curvature Diameter growth and bounded topology of omplete manifolds with nonnegative Rii urvature Huihong Jiang and Yi-Hu Yang Abstrat In this Note, we show that a omplete n-dim Riemannian manifold with nonnegative

More information

Solving a family of quartic Thue inequalities using continued fractions

Solving a family of quartic Thue inequalities using continued fractions Solving a family of quartic Thue inequalities using continued fractions Andrej Dujella, Bernadin Ibrahimpašić and Borka Jadrijević Abstract In this paper we find all primitive solutions of the Thue inequality

More information

Channel Assignment Strategies for Cellular Phone Systems

Channel 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 information

Restricted Least Squares, Hypothesis Testing, and Prediction in the Classical Linear Regression Model

Restricted Least Squares, Hypothesis Testing, and Prediction in the Classical Linear Regression Model Restrited Least Squares, Hypothesis Testing, and Predition in the Classial Linear Regression Model A. Introdution and assumptions The lassial linear regression model an be written as (1) or (2) where xt

More information

arxiv: v2 [math.co] 10 Oct 2016

arxiv: v2 [math.co] 10 Oct 2016 arxiv:1610.00360v [math.co] 10 Ot 016 Improving the bound for maximum degree on Murty-Simon Conjeture A. Bahjati 1, A. Jabalameli 1, M. Ferdosi 1, M. M. Shokri 1, M. Saghafian and S. Bahariyan 1 Department

More information

A 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 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 information

chapter > Make the Connection Factoring CHAPTER 4 OUTLINE Chapter 4 :: Pretest 374

chapter > 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 information

Applications of Fermat s Little Theorem and Congruences

Applications 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 information

USA Mathematical Talent Search. PROBLEMS / SOLUTIONS / COMMENTS Round 3 - Year 12 - Academic Year 2000-2001

USA Mathematical Talent Search. PROBLEMS / SOLUTIONS / COMMENTS Round 3 - Year 12 - Academic Year 2000-2001 USA Mathematial Talent Searh PROBLEMS / SOLUTIONS / COMMENTS Round 3 - Year - Aademi Year 000-00 Gene A. Berg, Editor /3/. Find the smallest positive integer with the property that it has divisors ending

More information

Chapter 1 Microeconomics of Consumer Theory

Chapter 1 Microeconomics of Consumer Theory Chapter 1 Miroeonomis of Consumer Theory The two broad ategories of deision-makers in an eonomy are onsumers and firms. Eah individual in eah of these groups makes its deisions in order to ahieve some

More information

Convergence of c k f(kx) and the Lip α class

Convergence 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 information

GENERALIZED PYTHAGOREAN TRIPLES AND PYTHAGOREAN TRIPLE PRESERVING MATRICES. Mohan Tikoo and Haohao Wang

GENERALIZED PYTHAGOREAN TRIPLES AND PYTHAGOREAN TRIPLE PRESERVING MATRICES. Mohan Tikoo and Haohao Wang VOLUME 1, NUMBER 1, 009 3 GENERALIZED PYTHAGOREAN TRIPLES AND PYTHAGOREAN TRIPLE PRESERVING MATRICES Mohan Tikoo and Haohao Wang Abstract. Traditionally, Pythagorean triples (PT) consist of three positive

More information

From (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

From (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 information

Computer Networks Framing

Computer 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 information

APPLICATIONS OF THE ORDER FUNCTION

APPLICATIONS 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 information

MOP 2007 Black Group Integer Polynomials Yufei Zhao. Integer Polynomials. June 29, 2007 Yufei Zhao yufeiz@mit.edu

MOP 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 information

DSP-I DSP-I DSP-I DSP-I

DSP-I DSP-I DSP-I DSP-I DSP-I DSP-I DSP-I DSP-I Digital Signal Proessing I (8-79) Fall Semester, 005 IIR FILER DESIG EXAMPLE hese notes summarize the design proedure for IIR filters as disussed in lass on ovember. Introdution:

More information

Static Fairness Criteria in Telecommunications

Static Fairness Criteria in Telecommunications Teknillinen Korkeakoulu ERIKOISTYÖ Teknillisen fysiikan koulutusohjelma 92002 Mat-208 Sovelletun matematiikan erikoistyöt Stati Fairness Criteria in Teleommuniations Vesa Timonen, e-mail: vesatimonen@hutfi

More information

Sebastián Bravo López

Sebastián 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 information

Javier Cilleruelo. Departamento de Matemáticas. Universidad Autónoma de Madrid MADRID ESPAÑA

Javier Cilleruelo. Departamento de Matemáticas. Universidad Autónoma de Madrid MADRID ESPAÑA B [g]sequences WHOSE TERMS ARE SQUARES Javier Cilleruelo Departamento de Matemátias Universidad Autónoma de Madrid MADRID-8049. ESPAÑA INTRODUCTION Sixty years ago Sidon [7] asked, in the ourse of some

More information

cos t sin t sin t cos t

cos 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 information

Continued Fractions and the Euclidean Algorithm

Continued 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 information

PYTHAGOREAN NUMBERS. Bowling Green State University, Bowling Green, OH (Submitted February 1988)

PYTHAGOREAN NUMBERS. Bowling Green State University, Bowling Green, OH (Submitted February 1988) S u p r i y a Mohanty a n d S. P. Mohanty Bowling Green State University, Bowling Green, OH 43403-0221 (Submitted February 1988) Let M be a right angled triangle with legs x and y and hypotenuse z. Then

More information

The Quantum Theory of Radiation

The Quantum Theory of Radiation A. Einstein, Physikalishe Zeitshrift 18, 121 1917 The Quantum Theory of Radiation A. Einstein (Reeived Marh, 1917) The formal similarity of the spetral distribution urve of temperature radiation to Maxwell

More information

1. R In this and the next section we are going to study the properties of sequences of real numbers.

1. R In this and the next section we are going to study the properties of sequences of real numbers. +a 1. R In this and the next section we are going to study the properties of sequences of real numbers. Definition 1.1. (Sequence) A sequence is a function with domain N. Example 1.2. A sequence of real

More information

Exponential Maps and Symmetric Transformations in Cluster-Spin. You-Gang Feng

Exponential Maps and Symmetric Transformations in Cluster-Spin. You-Gang Feng Exponential Maps and Symmetri Transformations in Cluster-Spin System for Lattie- Ising Models You-Gang Feng Department of Basi Sienes, College of Siene, Guizhou University, Cai-jia Guan, Guiyang, 550003

More information

Weighting Methods in Survey Sampling

Weighting Methods in Survey Sampling Setion on Survey Researh Methods JSM 01 Weighting Methods in Survey Sampling Chiao-hih Chang Ferry Butar Butar Abstrat It is said that a well-designed survey an best prevent nonresponse. However, no matter

More information

Every 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) 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 information

IRREDUCIBLE OPERATOR SEMIGROUPS SUCH THAT AB AND BA ARE PROPORTIONAL. 1. Introduction

IRREDUCIBLE 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 information

The square-free kernel of x 2n a 2n

The square-free kernel of x 2n a 2n ACTA ARITHMETICA 101.2 (2002) The square-free kernel of x 2n a 2n by Paulo Ribenboim (Kingston, Ont.) Dedicated to my long-time friend and collaborator Wayne McDaniel, at the occasion of his retirement

More information

Real quadratic fields with class number divisible by 5 or 7

Real quadratic fields with class number divisible by 5 or 7 Real quadratic fields with class number divisible by 5 or 7 Dongho Byeon Department of Mathematics, Seoul National University, Seoul 151-747, Korea E-mail: dhbyeon@math.snu.ac.kr Abstract. We shall show

More information

THREE DIMENSIONAL GEOMETRY

THREE 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 information

SUM OF TWO SQUARES JAHNAVI BHASKAR

SUM 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 information

Powers of Two in Generalized Fibonacci Sequences

Powers of Two in Generalized Fibonacci Sequences Revista Colombiana de Matemáticas Volumen 462012)1, páginas 67-79 Powers of Two in Generalized Fibonacci Sequences Potencias de dos en sucesiones generalizadas de Fibonacci Jhon J. Bravo 1,a,B, Florian

More information

U.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 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 information

Further linear algebra. Chapter I. Integers.

Further linear algebra. Chapter I. Integers. Further linear algebra. Chapter I. Integers. Andrei Yafaev Number theory is the theory of Z = {0, ±1, ±2,...}. 1 Euclid s algorithm, Bézout s identity and the greatest common divisor. We say that a Z divides

More information

MA257: INTRODUCTION TO NUMBER THEORY LECTURE NOTES

MA257: 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 information

ON INTEGERS EXPRESSIBLE BY SOME SPECIAL LINEAR FORM

ON INTEGERS EXPRESSIBLE BY SOME SPECIAL LINEAR FORM Acta Math. Univ. Comenianae Vol. LXXXI, (01), pp. 03 09 03 ON INTEGERS EXPRESSIBLE BY SOME SPECIAL LINEAR FORM A. DUBICKAS and A. NOVIKAS Abstract. Let E(4) be the set of positive integers expressible

More information

AUDITING COST OVERRUN CLAIMS *

AUDITING COST OVERRUN CLAIMS * AUDITING COST OVERRUN CLAIMS * David Pérez-Castrillo # University of Copenhagen & Universitat Autònoma de Barelona Niolas Riedinger ENSAE, Paris Abstrat: We onsider a ost-reimbursement or a ost-sharing

More information

Concentration of points on two and three dimensional modular hyperbolas and applications

Concentration 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 (CSIC-UAM-UC3M-UCM) and Departamento de Matemáticas Universidad

More information

Capacity at Unsignalized Two-Stage Priority Intersections

Capacity at Unsignalized Two-Stage Priority Intersections Capaity at Unsignalized Two-Stage Priority Intersetions by Werner Brilon and Ning Wu Abstrat The subjet of this paper is the apaity of minor-street traffi movements aross major divided four-lane roadways

More information

Optimal Sales Force Compensation

Optimal Sales Force Compensation Optimal Sales Fore Compensation Matthias Kräkel Anja Shöttner Abstrat We analyze a dynami moral-hazard model to derive optimal sales fore ompensation plans without imposing any ad ho restritions on the

More information

Symmetric Subgame Perfect Equilibria in Resource Allocation

Symmetric Subgame Perfect Equilibria in Resource Allocation Proeedings of the Twenty-Sixth AAAI Conferene on Artifiial Intelligene Symmetri Subgame Perfet Equilibria in Resoure Alloation Ludek Cigler Eole Polytehnique Fédérale de Lausanne Artifiial Intelligene

More information

Phase Velocity and Group Velocity for Beginners

Phase Velocity and Group Velocity for Beginners Phase Veloity and Group Veloity for Beginners Rodolfo A. Frino January 015 (v1) - February 015 (v3) Eletronis Engineer Degree from the National University of Mar del Plata - Argentina Copyright 014-015

More information

Inverses and powers: Rules of Matrix Arithmetic

Inverses 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 information

Quotient Rings and Field Extensions

Quotient 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 information

On the recoil and Doppler shifts

On the recoil and Doppler shifts Journal of Modern Optis Vol. 00, No. 00, DD Month 200x, 1 5 On the reoil and Doppler shifts Stephen M. Barnett Department of Physis, SUPA, University of Strathlyde, Glasgow G4 0NG, UK and Department of

More information

Homework until Test #2

Homework 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 information

Math 319 Problem Set #3 Solution 21 February 2002

Math 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 information

Ideal Class Group and Units

Ideal 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 information

arxiv:astro-ph/0304006v2 10 Jun 2003 Theory Group, MS 50A-5101 Lawrence Berkeley National Laboratory One Cyclotron Road Berkeley, CA 94720 USA

arxiv:astro-ph/0304006v2 10 Jun 2003 Theory Group, MS 50A-5101 Lawrence Berkeley National Laboratory One Cyclotron Road Berkeley, CA 94720 USA LBNL-52402 Marh 2003 On the Speed of Gravity and the v/ Corretions to the Shapiro Time Delay Stuart Samuel 1 arxiv:astro-ph/0304006v2 10 Jun 2003 Theory Group, MS 50A-5101 Lawrene Berkeley National Laboratory

More information

An Iterated Beam Search Algorithm for Scheduling Television Commercials. Mesut Yavuz. Shenandoah University

An Iterated Beam Search Algorithm for Scheduling Television Commercials. Mesut Yavuz. Shenandoah University 008-0569 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 information

Supply chain coordination; A Game Theory approach

Supply 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 Jean-Claude Hennet x and Yasemin Arda xx x LI CNR-UMR 668 Université

More information

Perpendicular Bisectors of Triangle Sides

Perpendicular Bisectors of Triangle Sides Forum Geometriorum Volume 13 (013) 3 9. FORUM GEOM ISSN 13-1178 Perpendiular isetors of Triangle Sides Douglas W. Mithell strat. Formulas, in terms of the sidelengths and area, are given for the lengths

More information

Trigonometry & Pythagoras Theorem

Trigonometry & 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 information

Practice Problems for First Test

Practice Problems for First Test Mathematicians have tried in vain to this day to discover some order in the sequence of prime numbers, and we have reason to believe that it is a mystery into which the human mind will never penetrate.-

More information

FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z

FACTORING 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 information

MATH2001 Development of Mathematical Ideas History of Solving Polynomial Equations

MATH2001 Development of Mathematical Ideas History of Solving Polynomial Equations MATH2001 Development of Mathematical Ideas History of Solving Polynomial Equations 19/24 April 2012 Lagrange s work on general solution formulae for polynomial equations The formulae for the cubic and

More information

Elementary Number Theory We begin with a bit of elementary number theory, which is concerned

Elementary 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

3 Game Theory: Basic Concepts

3 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 information

Chemical Equilibrium. Chemical Equilibrium. Chemical Equilibrium. Chemical Equilibriu m. Chapter 14

Chemical Equilibrium. Chemical Equilibrium. Chemical Equilibrium. Chemical Equilibriu m. Chapter 14 Chapter 14 Chemial Equilibrium Chemial Equilibriu m Muh like water in a U-shaped tube, there is onstant mixing bak and forth through the lower portion of the tube. reatants produts It s as if the forward

More information

Algebraic and Transcendental Numbers

Algebraic 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 information

Unique Factorization

Unique 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 information

The Dirichlet Unit Theorem

The 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 information

Isaac Newton. Translated into English by

Isaac 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 information

Chapter 2 Michelson Morley Experiment and Velocity of Light

Chapter 2 Michelson Morley Experiment and Velocity of Light Chapter 2 Mihelson Morley Experiment and Veloity of Light 2. Attempts to Loate Speial Privileged Frame Sientists tried to determine relative veloity of light with respet to earth. For this purpose, the

More information

6.3 Point of View: Integrating along the y-axis

6.3 Point of View: Integrating along the y-axis math appliation: area between urves 7 EXAMPLE 66 Find the area of the region in the first quadrant enlosed by the graphs of y =, y = ln x, and the x- and y-axes SOLUTION It is easy to sketh the region

More information

Special Relativity and Linear Algebra

Special 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 information

k, then n = p2α 1 1 pα k

k, 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 information

) ( )( ) ( ) ( )( ) ( ) ( ) (1)

) ( )( ) ( ) ( )( ) ( ) ( ) (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 information

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.

Chapter 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 information

Collinear Points in Permutations

Collinear 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 information

Fundamentals of Chemical Reactor Theory

Fundamentals 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 information

Notes on Coaxial Transmission Lines

Notes on Coaxial Transmission Lines ELEC344 Mirowave Engineering Handout#4 Kevin Chen Notes on Coaxial Transmission Lines TEM Fields in Coaxial Transmission Lines Beause it is a two-wire transmission line, oaxial line an support TEM waves.

More information

On the largest prime factor of x 2 1

On 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 information

Chapter 6. Number Theory. 6.1 The Division Algorithm

Chapter 6. Number Theory. 6.1 The Division Algorithm Chapter 6 Number Theory The material in this chapter offers a small glimpse of why a lot of facts that you ve probably nown and used for a long time are true. It also offers some exposure to generalization,

More information

DISTRIBUTIVE PROPERTIES OF ADDITION OVER MULTIPLICATION OF IDEMPOTENT MATRICES

DISTRIBUTIVE PROPERTIES OF ADDITION OVER MULTIPLICATION OF IDEMPOTENT MATRICES J. Appl. Math. & Informatics Vol. 29(2011), No. 5-6, pp. 1603-1608 Website: http://www.kcam.biz DISTRIBUTIVE PROPERTIES OF ADDITION OVER MULTIPLICATION OF IDEMPOTENT MATRICES WIWAT WANICHARPICHAT Abstract.

More information

Classical Electromagnetic Doppler Effect Redefined. Copyright 2014 Joseph A. Rybczyk

Classical 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 information

Number 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 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 information

2 The Euclidean algorithm

2 The Euclidean algorithm 2 The Euclidean algorithm Do you understand the number 5? 6? 7? At some point our level of comfort with individual numbers goes down as the numbers get large For some it may be at 43, for others, 4 In

More information

PYTHAGOREAN TRIPLES PETE L. CLARK

PYTHAGOREAN 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 information

A Non-Existence Property of Pythagorean Triangles with a 3-D Application

A Non-Existence Property of Pythagorean Triangles with a 3-D Application A Non-Existence Property of Pythagorean Triangles with a 3-D Application Konstantine Zelator Department of Mathematics and Computer Science Rhode Island College 600 Mount Pleasant Avenue Providence, RI

More information

MODULAR ARITHMETIC. a smallest member. It is equivalent to the Principle of Mathematical Induction.

MODULAR 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 information

EE 201 ELECTRIC CIRCUITS LECTURE 25. Natural, Forced and Complete Responses of First Order Circuits

EE 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 information

Chapter 5 Single Phase Systems

Chapter 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 information

Intelligent Measurement Processes in 3D Optical Metrology: Producing More Accurate Point Clouds

Intelligent 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 information

Lectures on Number Theory. Lars-Åke Lindahl

Lectures on Number Theory. Lars-Åke Lindahl Lectures on Number Theory Lars-Åke 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 information

On 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 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 information

Prices and Heterogeneous Search Costs

Prices and Heterogeneous Search Costs Pries and Heterogeneous Searh Costs José L. Moraga-González 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 information

Theory of linear elasticity. I assume you have learned the elements of linear

Theory 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 information

10 k + pm pm. 10 n p q = 2n 5 n p 2 a 5 b q = p

10 k + pm pm. 10 n p q = 2n 5 n p 2 a 5 b q = p Week 7 Summary Lecture 13 Suppose that p and q are integers with gcd(p, q) = 1 (so that the fraction p/q is in its lowest terms) and 0 < p < q (so that 0 < p/q < 1), and suppose that q is not divisible

More information

SUBGROUPS 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. 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 information

ORDERS OF ELEMENTS IN A GROUP

ORDERS OF ELEMENTS IN A GROUP ORDERS OF ELEMENTS IN A GROUP KEITH CONRAD 1. Introduction Let G be a group and g G. We say g has finite order if g n = e for some positive integer n. For example, 1 and i have finite order in C, since

More information