Arithmetic progressions on Pell equations

Transcription

1 Journal of Number Theory 18 (008) Arithmetic progressions on Pell equations A. Pethő a,1,v.ziegler b,, a Department of Computer Science, Number Theory Research Group, Hungarian Acaemy of Sciences an University of Debrecen, H-4010 Debrecen, PO Box 1, Hungary b Institute for Analysis an Computational Number Theory, Graz University of Technology, Steyrergasse 30, A-8010 Graz, Austria Receive 17 May 006; revise 3 August 007 Available online 18 April 008 Communicate by M. Pohst Abstract In this paper we consier arithmetic progressions on Pell equations, i.e. integral solutions (X, Y ) whose X-coorinates or Y -coorinates are in arithmetic progression. 008 Elsevier Inc. All rights reserve. Keywors: Pell equations; Arithmetic progressions; Elliptic curves 1. Introuction In 1999 Bremner [1] consiere arithmetic progressions on elliptic curves. Bremner constructe elliptic curves with arithmetic progressions of length 7, i.e. rational points (X, Y ) whose X-coorinates are in arithmetic progression. In a following paper Bremner, Silverman an Tzanakis [] showe that a subgroup Γ of the elliptic curve E(Q) with E: Y = X(X n ) of rank 1 oes not have non-trivial integral arithmetic progressions, provie n 1. Contrary to the results of Bremner, Silverman an Tzanakis [], Campbell [3] foun an infinite family of elliptic curves with 9 integral points in arithmetic progressions. This result was * Corresponing author. aresses: pethoe@inf.unieb.hu (A. Pethő), ziegler@finanz.math.tugraz.at (V. Ziegler). 1 Research of the author was supporte in part by grants T4985 of the Hungarian National Founation for Scientific Research. The author gratefully acknowleges support from the Austrian Science Fun (FWF) uner project No. P18079-N X/$ see front matter 008 Elsevier Inc. All rights reserve. oi: /j.jnt

2 1390 A. Pethő, V. Ziegler / Journal of Number Theory 18 (008) improve by Ulas [13], where an infinite family was foun with an arithmetic progression consisting of 1 integral points. In this paper we consier curves of genus 0, in particular hyperbola, with integral arithmetic progressions. Inspire by the results of Bremner [1], Bremner, Silverman an Tzanakis [], Campbell [3] an Ulas [13] the aim of this paper is to prove the following theorems. Theorem 1. Let 0 < Z, not a square an 0 m Z. If there are three solutions (X 1,Y 1 ), (X,Y ) an (X 3,Y 3 ) to the Pell equation X Y = m (1) such that X 1 <X <X 3 respectively Y 1 <Y <Y 3 are in arithmetic progression, with X 0 respectively Y 0, then respectively { max Xi } max {1.783 m, m } i For more etails see Table 1. { max Yi } m. i Table 1 Upper bouns for max i=1,,3 { X i } respectively max i=1,,3 { Y i } X m / m B := max{ X i } m / 1 m B m m / < 1 m =1 B m m / < 1 m B m Y m B := max{ Y i } m 1 m B m B m By Theorem 1 we euce an upper boun for the length of an arithmetic progression. Corollary 1. Let,m Z with >0 not a square an m 0. An arithmetic progression on X Y = m has length at most c(,m)τ (m), where τ ( ) is following arithmetic function τ (m) = p α m p splits in Q( ) (α + 1)

3 A. Pethő, V. Ziegler / Journal of Number Theory 18 (008) an the prouct runs over all primes. Furthermore, we have log( m 3 /)+16.9 log + 3 if / m 1, c(,m) = log( m ) log + 3 if / m < 1. Since Theorem 1 we know that there are only finitely many non-trivial arithmetic progressions for fixe, m. This leas to several questions: Are there for fixe only finitely many m respectively for fixe m only finitely many such that (1) amits non-trivial arithmetic progressions. In the secon case the answer is yes (see Theorem ). Theorem. Let 0 m Z be fixe. Then there are only finitely many 0 < Z such that there are three solutions (X 1,Y 1 ), (X,Y ) an (X 3,Y 3 ) to the Pell equation X Y = m such that X 1 <X <X 3 or Y 1 <Y <Y 3 is in arithmetic progression, except the trivial cases (Y 1,Y,Y 3 ) = ( y,0,y), (X 1,X,X 3 ) = ( x,0,x). Moreover if a non-trivial arithmetic progression X 1 <X <X 3 exists then we have 3 m if m is not a perfect square an 9 m otherwise. In the case that Y 1 <Y <Y 3 is in arithmetic progression we obtain 9 m. Note, in the case that Y 1 <Y <Y 3 is in arithmetic progression, we o not get better estimates if we assume m is not a square. In the special case m =±1 we obtain Corollary. Let 0 < Z be not a perfect square. Then there are no three solutions (X 1,Y 1 ), (X,Y ) an (X 3,Y 3 ) to the Pell equation X Y =±1 such that X 1 <X <X 3 or Y 1 <Y <Y 3 is in arithmetic progression, except the trivial case (Y 1,Y,Y 3 ) = ( y,0,y) an the progressions (X 1,X,X 3 ) = ( 3, 1, 1), ( 1, 1, 3) in the case m = 1 an = or = 8. Let us reverse the questions state above. Given an arithmetic progression Y 1 <Y <Y 3 oes there exist a hyperbola such that Y 1, Y, Y 3 are solutions? The answer is given by the following theorem. Theorem 3. For every arithmetic progression Y 1 <Y <Y 3 there exist infinitely many,m Z such that is not a square, m 0 an gc(, m) is square-free such that Y 1, Y an Y 3 are the Y -components of solutions to X Y = m. The next problem we consier is, whether there are,m Z with 0 < not a square an m 0, such that there exists a certain arithmetic progression of length four on X Y = m. In particular we are intereste in the arithmetic progression 1 < 3 < 5 < 7 respectively 0 < 1 < < 3.

4 139 A. Pethő, V. Ziegler / Journal of Number Theory 18 (008) Theorem 4. There are,m Z such that >0 is not a perfect square an m 0, such that 1, 3, 5 an 7 are the Y -components of solutions to X Y = m. We may choose (, m) = (570570, ). In particular there exist arithmetic progressions of length 8. On the other han there are no,m Z with not a perfect square, such that 0, 1, an 3 are the Y -components of solutions to X Y = m. We also prove a converse to Theorem 3: Theorem 5. Let Y 1 <Y <Y 3 <Y 4 <Y 5 be in arithmetic progression such that Y i Y j for any i j. Then there are at most finitely many,m Z such that is not a square, m 0 an gc(, m) is square-free such that Y 1, Y, Y 3, Y 4, Y 5 are the Y -components of solutions to X Y = m. 3 In the next section (Section ) we prove two simple auxiliary results that will help us to prove our theorems. In the following Sections 3 5 an 6 we prove the theorems state above. The proofs of Theorems 1 3 are elementary (but technical). However the proof of Theorem 4 nees some basic knowlege on elliptic curves an the proof of Theorem 5 nees some basic knowlege on algebraic geometry, i.e. we apply a theorem of Faltings [6] (Morell s conjecture). In Section 7 we consier the ual question to Theorems 3 5 an show that the situation is much more simple for the X-component. In the last section we iscuss some open questions that arise by stuying this paper.. Auxiliary results Since with X 1 <X <X 3 also X 3 < X < X 1 respectively with Y 1 <Y <Y 3 also Y 3 < Y < Y 1 is in arithmetic progression, we assume X > 0 respectively Y > 0inthe following. Let us assume X 1 <X <X 3 is in arithmetic progression then the corresponing components Y 1, Y, Y 3 are not uniquely etermine, therefore we choose Y i such that X i Yi X i + Y i,i.e.weassumebothx i an Y i non-negative or non-positive. In particular, this implies that X i (respectively Y i ) is either zero or has the same sign as X i + Y i ; moreover, not all three X 1 Y 1, X3 Y 3, (X Y ) have the same sign. Similarly we choose the corresponing components X i if Y 1 <Y <Y 3 is in arithmetic progression. We call X + Y a solution to (1) if (X, Y ) Z is a solution to (1). We start with the following lemma. Lemma 1. Let α 1 = X 1 + Y 1, α = X + Y an α3 = X 3 + Y 3 be solutions to (1) such that α 1 + α 3 = α. Then α 1 = α = α 3. In other wors not both X 1 <X <X 3 an Y 1 <Y <Y 3 can be arithmetic progressions. Proof. Since 1 an are linearly inepenent over Q we euce, that both (X 1,X,X 3 ) an (Y 1,Y,Y 3 ) form arithmetic progressions. Therefore we write X 1 = x k 1, X = x, X 3 = x + k 1 an similarly Y 1 = y k, Y = y, Y 3 = y + k. Subtracting X Y = m two times from X1 + X 3 (Y 1 + Y 3 ) = m yiels k 1 k = 0. But the Diophantine equation X Y = 0 3 After finishing this paper Dujella, Pethő an Taić [5] prove that apart from ±[0, 1,, 3] each four term arithmetic progression is lying on infinitely many hyperbolas. They also foun some 5, 6 an 7 term arithmetic progressions not containing 0 with the same property.

5 A. Pethő, V. Ziegler / Journal of Number Theory 18 (008) has no solution except (X, Y ) = (0, 0), if is not a perfect square, hence X 1 = X = X 3 an Y 1 = Y = Y 3. The lemma also follows from Bézout s theorem, i.e. a quaratic curve with a line has at most two intersections. Let us assume X 1 <X <X 3 is in arithmetic progression, i.e. X 1 +X 3 = X,onthehyperbola (1). Then the corresponing Y -components cannot fulfill the equation Y 1+Y 3 = Y. Therefore we have Y Y 1+Y 3 = Δ Y with Δ Y 1 Z an Δ Y 0. Similarly we efine Δ X := X X 1+X 3 1 Z,ifY 1 <Y <Y 3 is in arithmetic progression. Because of Lemma 1 we have Δ X 0. In any case we have the lower boun Δ 1/, where Δ = Δ X,Δ Y epening on which component is in arithmetic progression. The next lemma yiels an upper boun for Δ. Lemma. Let X 1 <X <X 3 be in arithmetic progression on the hyperbola (1) such that no Y i = 0 for i = 1,, 3, then Δ Y 3 m. If one (or more) Y i = 0 then we have Δ Y 3 m 3 m if m 1, if m < 1. Now let Y 1 <Y <Y 3 be in arithmetic progression. Then if no Y i = 0(i.e. Y 1 0) an otherwise. Δ X 3 m, m m Δ X + Proof. Let us consier the case where X 1 <X <X 3 is in arithmetic progression. Then (X i Y i )(Xi + Y i )= m, i = 1,, 3. We obtain X i Y i = m X i + Y i, i = 1,, 3, which implies m X i Y i Y i m, ()

6 1394 A. Pethő, V. Ziegler / Journal of Number Theory 18 (008) if Y i 0. Note that we assume that X i an Y i have the same sign. If Y i = 0 we obviously have X i Y i = m. (3) We see that 1 ((X 1 Y 1 ) + (X3 Y 3 ) (X Y )) is equal to ΔY or ΔX accoring as the X i s or the Y i s are in arithmetic progression. Since not all three summans of the left-han sie are of the same sign, it follows, when the X i s are in arithmetic progression, that Δ Y { 1 } max X 1 Y 1 + X3 Y 3, i = 1, 3, X i Y i + X Y, an the max is, clearly, at most 3 m when Y 1Y Y 3 0 an the Y i s are in arithmetic progression we work similarly. 3. Proof of Theorem 1 m + m if some Y i is zero. When We have four ifferent cases. We istinguish whether X 1 <X <X 3 or Y 1 <Y <Y 3 is in arithmetic progression an whether X 1 < X or X < X 1 respectively Y 1 < Y or Y < Y 1. Because the iea for all four cases is the same, we give the etails only for the case X 1 <X <X 3 an X 1 < X an only sketch the proofs for the other cases. We claim that min{ X 1, X } 3 m. Inee, we have X i Y i m an therefore by the proof of Lemma we obtain Δ Y Similarly we obtain min{ Y 1, Y } 3 m 3 m min{ X 1, X } Xi. This yiels immeiately the claim. (Y 1 <Y <Y 3 is in arithmetic progression). Now we claim that min{ Y 1, Y } 3 m We know X Y = m, where (X, Y ) = (X i,y i ) with i {1, } such that X i =min{ X 1, X }, hence Y X + m 9 m + m ( 3 m ) Similarly we obtain min{ X 1, X } 3 m in the other case. Let us consier the equation X Y = m an let us insert the expressions for X an Y. We obtain ( ) X1 + X ( ) 3 Y1 + Y 3 + Δ Y = m. Using the other two equations this implies On the other han X 1 X 3 = Y 1 Y 3 + Δ Y (Y 1 + Y 3 + Δ Y ) + m. (X 1 X 3 ) = ( m + Y 1 )( m + Y 3 ). Inserting the expression for X 1 X 3 we get a Y 3 + a 1Y 3 + a 0 = 0, where

7 A. Pethő, V. Ziegler / Journal of Number Theory 18 (008) a = 4Δ + 4Y 1 Δ m, a 1 = 8Δ 3 + 4Δm + 4Y1 Δ + 1Y 1Δ + Y 1 m = a (Y 1 + Δ) + 3m(Y 1 + Δ), a 0 = Y1 m + 4Δ4 + 4Δ m + 4Y 1 Δm + 8Y 1 Δ 3 + 4Y1 Δ = a 4 + 3a m + 5m 4 my 1, an we write Δ for Δ Y. In the case that X < X 1 we use the equation X 1 Y 1 = m an insert Y 1 = Y Y 3 Δ respectively X 1 = X X 3. Similarly as above we obtain a Y 3 + a 1 Y 3 + a 0 = 0 with a = m + Y Δ Δ, a 1 = my + mδ 4Y Δ + 6Y Δ Δ 3 = a ( Y + Δ), a 0 = my 4mY Δ + mδ 4Y Δ + 4Y Δ 3 Δ 4 = a + m + my. In the case that Y 1 <Y <Y 3 is in arithmetic progression we obtain a X 3 + a 1X 3 + a 0 = 0, where a = m + 4X 1 Δ + 4Δ, a 1 = mx 1 4mΔ + 4X1 Δ + 1X 1Δ + 8Δ 3 = a (X 1 + Δ) 3m(X 1 + Δ), a 0 = mx1 4mX 1Δ 4mΔ + 4X1 Δ + 8X 1 Δ 3 + 4Δ 4 = a 4 3a m + 5m 4 + mx 1, provie Y 1 Y respectively a = m X Δ + Δ, a 1 = mx + mδ + 4X Δ 6X Δ + Δ 3 = a ( X + Δ), a 0 = mx 4mX Δ + mδ + 4X Δ 4X Δ 3 + Δ 4 = a m + mx, in the case that Y < Y 1. We have a = 0 in the cases iscusse above if an only if m = 4Δ(Y 1 + Δ), Y Δ + Δ, 4X 1 Δ 4Δ an X Δ Δ respectively. Then

8 1396 A. Pethő, V. Ziegler / Journal of Number Theory 18 (008) X 1 = m + Y 1 = (Y 1 + Δ), X = m + Y = (Y Δ), Y1 = X 1 m = (X 1 + Δ) an Y = X m = (X Δ) respectively. These relations are absur, as is not a square, thus a 0. Now we solve the quaratic equation a Y3 + a 1Y 3 + a 0 = 0 an istinguish whether m =1 or m an whether m / 1or m / < 1. Taking care of all these ifferent cases we can estimate Y 3 by using the formula Y 3 = a 1 arithmetic progression with X 1 < X we have a 1 4a a ± a 1 4a a 0 a = Y mY 1 + 1mY 1Δ + 1mΔ a + 9m Y 1 + 4Y 1Δ + 4Δ 4a Y my 1 +9m Y 1 4 a 0 a. In the case that X 1 <X <X 3 is in 5m 4 5m 4a + 6 my 1 Δ +9 m Y 1 Δ + 6 mδ + 9m Δ + 5 m +5 m, 4 because 1 a m +4 Y 1 Δ +4Δ. By previous computations we know that Y an Δ 3 m / if m / 1 an Δ 3/ m / otherwise. Therefore we obtain 3 m a 1 4a a 0 a ( ) m m 16m + 7 m m 4 c 1 m 4 in the case that m / 1 an a 1 4a a ( 0 81 a m 4 m m + 59m + 81m + 5 m 4m ( m 7 + m + 81 )) c m otherwise, where c = if we assume m =1 respectively c 1 = an c = 114 if we assume m. Note that in the case that m / < 1 we may assume Y 1 3. These computations yiel

9 A. Pethő, V. Ziegler / Journal of Number Theory 18 (008) Y 3 = a 1 ± a a 1 4a a 0 a Y my 1 +6 mδ + Δ + m a ( m c m ) 4 m m c 1 c1 respectively Y 3 = a 1 ± a a 1 4a a 0 a Y my 1 + Δ + 3 mδ + m c a a ( c m m ( ) ) 1 + m / m m c in the case that m / 1 respectively m / < 1, where if we assume m =1 respectively c = 6 + c 1 = an c = if we assume m. Since max{ X 1, X, X 3 } = X 3 = X 3 m + m4 c 1 = c m 1 + m m Y3 we obtain c 3 m 1 respectively X 3 m +m c = m c + 1 m m c, where c if we assume m = 1 an c an c if we assume m. The estimations for the other cases run analogously (see also Table 1).

10 1398 A. Pethő, V. Ziegler / Journal of Number Theory 18 (008) The iea for the proof of Corollary 1 is to compute the number of solutions to X Y = m with X M respectively Y M. Therefore we have to iscuss properties of the solutions to X Y = m. We prove the following lemma. Lemma 3. Let M be fixe. Then the equation X Y = m has at most c(ɛ,m,m)τ (m) solutions with X M respectively Y M, where log( Mɛ (ɛ ) m ) c(ɛ,m,m) = log ɛ an ɛ>1 is the funamental solution to X Y = 1. log( Mɛ (ɛ 1) m ) log ɛ + 1 We start with some general remarks on Diophantine equation (1). If α = u+v is a solution to (1) an ɛ = x + y is a solution to X Y = 1, (4) then also αɛ = (ux + vy) + (uy + vx) is a solution to (1). We say that two solutions α 1 = u 1 + v 1 an α = u + v belong to the same class of solutions if there exists a solution ɛ = x + y to (4), such that α 1 = ɛα.letα 1,...,α l be representatives for each class (note that there are only finitely many classes, see [11]). Then we obtain by the formula αɛ k, where 0 k Z, ɛ is a funamental solution to (4) an α runs through the representative system α 1,...,α l, a set of solutions L which can be extene to the set of all solutions by ajoining (±x,±y) to L for all (x, y) L. Let α = u + v be a solution to (1) in a certain class C, such that v is non-negative an least possible, then we call α the funamental solution of the class C (if u + v is in the same class with u + v then choose u positive). Note that every solution can be written uniquely as αɛ k where α is a funamental solution an k Z. We call k the exponent of the solution. We use following notation: Let α = x + y be a solution to (1), then we call ᾱ := x y its conjugate solution. Warning. Do not confuse this notation with the complex conjugation! A theorem ue to Nagell states (see [11] or [10]). Theorem 6 (Nagell). Let α = u + v be the funamental solution to (1) an ɛ = x + y a funamental solution to (4) with x,y > 0. Then 0 < u 1 (x + 1)m, 0 v y (x + 1) m, if m>0 an 0 u 1 (x 1)m, 0 <v y (x 1) m, if m<0.

11 A. Pethő, V. Ziegler / Journal of Number Theory 18 (008) As a corollary we prove the following lemma. Lemma 4. Let α be a funamental solution to (1) an ɛ the funamental solution to (4). Then m ɛ α, ᾱ ɛ m. (5) Proof. We write α = u + v so that ᾱ = u v u an ɛ = x + y as in Nagell s Theorem 6. By this theorem we have u 1 (x + 1) m, v y (x 1) m. Obviously ɛ = x +y x +1. Since x y = 1wehave y yiel x 1 = x+1 α u +v 1 ɛ m ɛ m + = ɛ m. ɛ. These estimations On the other han ᾱ = u v u +v ɛ m as shown above. Having prove so far the upper boun α, ᾱ ɛ m, we obtain from m = α ᾱ the lower boun for both α an ᾱ. m ɛ Proof of Corollary 1. First, let us count the possibilities for the exponents. Let X + Y = αɛ k be a solution, where α is a funamental solution an k Z. Then X + Y = ᾱɛ k.in view of the relation α ᾱ = m an Lemma 4, we may assume m ɛ α m (if not interchange the roles of α an ᾱ). We claim ᾱɛ k ɛ αɛk if k 1, an αɛ k ɛ ᾱɛ k if k 1. These statements are easily verifie using Lemma 4 an the relation αᾱ = m. Now we consier for k 1 M max { αɛ k +ᾱɛ k, αɛ k ᾱɛ k } (1 αɛk ) ɛ an for k 1 M max { αɛ k +ᾱɛ k, αɛ k ᾱɛ k } ᾱɛ k( 1 ɛ ).

12 1400 A. Pethő, V. Ziegler / Journal of Number Theory 18 (008) Solving these inequalities with respect to k an integer an taking into account the possibility of k = 0. We obtain that there are at most log( Mɛ (ɛ ) m ) c(ɛ,m,m) := log ɛ log( Mɛ (ɛ 1) m ) log ɛ + 1 possible k s. Next we want to count the possible classes of solutions. Therefore we write m =± l i=1 p α i i such that the prime ieals (p i ) = p i p i split, the (q i ) = q i ramify an the (r i ) = r i are inert. The following relations are easy to see: m i=1 q β i i n i=1 p N i (α) pn i (ᾱ), q N i (α) qn i (ᾱ), r N i (α) rn i (ᾱ), where N is an integer. This means that the exponents of the primes r i an q i in (α) respectively (ᾱ) are fixe. Only the exponents of the p i can vary. This means that we have l i=1 (α i + 1) possibilities for α, which proves Lemma 3. Now using ɛ 1 + Corollary 1 follows from Lemma 3, Theorem 1 an some estimations. 4. Proof of Theorem The proof of Theorem is rather easy in view of Lemma. If m is not a square then Y = 0is impossible an we obtain Δ Y 3 m respectively Δ X 3 m. By Lemma 1 we know that at least one of Δ X an Δ Y is non-zero an therefore their absolute value is at least 1/. This yiels immeiately the theorem in this case. If m is a square then the case Y = 0 is possible. Let us consier the case that X 1 <X <X 3 is in arithmetic progression. Then by Lemma we have Δ Y 3 m. The same argument as above yiels Theorem in this case. Now we investigate the case that Y 1 <Y <Y 3 is in arithmetic progression, Y 1 = 0 an m = c, with c Z. This yiels X 1 = c, Y 3 = Y an δ:= r γ i i { Δ X = c ( }} ){ + m 1. X 3 + Y X + Y Since Δ X 1 Z also δ 1 Z. Let us exclue the case δ = 0. In this case we woul obtain X 3 + 4Y = X + Y,

13 A. Pethő, V. Ziegler / Journal of Number Theory 18 (008) hence 4Y = Y,i.e.Y = 0 which is a contraiction. On the other han, as Y 3 >Y 1 an the sign of X, Y an X 3, Y 3 are the same, we get δ m ( + 1 ) = 5 m 4. By a similar argument as above we obtain 6.5 m < 9 m. Now we can prove Corollary. We only have to check for =, 3, 5, 6, 7, 8 if there are any solutions with absolute value at most that form an arithmetic progression. This can be one easily by a computer. 5. Proof of Theorem 3 Let Y 1 = a, Y = a + k, Y 3 = a + k with a,k Z be the given arithmetic progression. Since 0 X ( 0 )Y = m 0 is equivalent to the equation X Y = m we may assume a, k are coprime. If there are,m Z that fulfill Theorem 3 then the system X 1 a = m, X (a + k) = m, of Diophantine equations has a solution. But then also the system X 3 (a + k) = m, (6) X X 1 = k(a + k), X 3 X = k(a + 3k), has the same solution an also the equation C: X (4a + 4k) = X 1 (a + 3k) + X 3 (a + k) (7) has this solution. It is not har to see that this projective curve C has genus 0 an can be parameterize by a line. The projective point P = (1, 1, 1) P lies on C an let Q = (p, q, 0) lie on the line L: X 3 = 0. By Bézout s theorem the straight line from P to Q has only two intersections (with multiplicities) with C. One intersection is P an let the other intersection be R. Because the genus of C is 0 the point R must be rational if Q was rational. Let us compute R in epenence of Q to obtain all rational points. Of course an m epen on the representative of the projective solution (X 1,X,X 3 ) since (λx 1,λX,λX 3 ) = λ (X 1,X,X 3 ) an similarly m(λx 1,λX,λX 3 ) = λ m(x 1,X,X 3 ). We can erive for each rational projective solution to (7) exactly one pair (, m) such that gc(, m) is square-free. Therefore every pair (, m) that fulfills the properties state in Theorem 3 correspons to exactly one rational point on the projective curve C. The line from P to Q is given by the equation Inserting this in Eq. (7) yiels qx 1 px + (p q)x 3 = 0.

14 140 A. Pethő, V. Ziegler / Journal of Number Theory 18 (008) ( p (a + 3k) 4q (a + k) ) X + ( p(q p)(a + 3k) ) X X 3 + ( 4q (a + k) + p(p q)(a + 3k) ) X 3 = 0. This equation has two solutions for X. The first solution is X = X 3 an the secon solution is X = a(p pq + q ) + k(3p 6pq + 4q ) p (a + 3k) 4q X 3. (a + k) The first solution yiels X 1 = X = X 3, which implies Y 1 = Y = Y 3. But this has been exclue. In the other case we get X 1 = a(p 4pq + q ) + k(3p 8pq + 4q ) p (a + 3k) 4q X 3. (a + k) Let X 3 = k(p (a + 3k) 4q (a + k)), then we obtain by system (6) = 4kp(p q)q ( (a + 3k)p 4(a + k)q ), m = k(p q) ( (a + 3k)p kq )( kp (a + k)q )( (a + 3k)p (a + k)q ). Let us keep p fixe, non-zero an choose q square-free such that gc(q, ) = gc(q, p) = gc(q, a) = gc(q, k) = gc(q, a + 3k) = 1. Let us note that for such q s is not a perfect square. Moreover, also gc(q, m) = 1. Note that there are infinitely many q s with the properties state above, hence we can construct infinitely many pairs (, m) that satisfy the conitions of Theorem 3. The conition >0 is fulfille whenever we choose p>0 an q large enough. 6. Proofs of Theorems 4 an 5 The first part of the proof of Theorem 4 is rather easy. One has to check that 1, 3, 5, 7 are part of the solutions of the Pell equations given in Theorem 4. In this section we want to show how to fin a pair (, m) that fulfills the conitions of Theorem 4 or to prove that there oes not exist such a pair. Similarly as in the proof of Theorem 3 we are le to projective curves lying in P 3 respectively P 4 (Theorem 5). The proof of Theorem 4 will lea us to an elliptic curve for which we will fin some rational points. One of these rational points will yiel a pair (, m) such that is positive an square-free. On the other han a genus 5 curve will be establishe in the proof of Theorem 5. This will result in only finitely many pairs (, m). Let us start with the proof of Theorem 5. As mentione in the section above we may assume that a an k are relative prime. Similarly as in the proof of Theorem 3 we obtain from X 1 a = m, X (a + k) = m, X 3 (a + k) = m, X 4 (a + 3k) = m, X 5 (a + 4k) = m, (8)

15 A. Pethő, V. Ziegler / Journal of Number Theory 18 (008) the system X X 1 = k(a + k), X 3 X = k(a + 3k), X4 X 3 = k(a + 5k), X 5 X 4 = k(a + 7k) (9) respectively X (4a + 4k) = X 1 (a + 3k) + X 3 (a + k), X3 (4a + 8k) = X (a + 5k) + X 4 (a + 3k), X4 (4a + 1k) = X 3 (a + 7k) + X 5 (a + 5k), (10) which efines a curve X in the projective space P 4. By the conitions of Theorem 5 we have to exclue the cases a = k, k, 3k, k/, 3k/, 5k/, 7k/, i.e. none of the coefficients in (10) is zero. Lemma 5. Let a i,j be non-zero integers, an let the curve X be efine by X1 a 1,1 + X a 1, + X3 a 1,3 = 0, X a,1 + X3 a, + X4 a,3 = 0, X3 a 3,1 + X4 a 3, + X5 a 3,3 = 0. (11) Let F 1 = a, a 3, a,3 a 3,1, F = a 1, a, a 1,3 a,1, If F 1 F F 3 0 then the genus of X is 5. F 3 = a, a 3, a 1, a,3 a 1, a 3,1 a 3, a 1,3 a,1. Proof. We use Hurwitz s formula (see [7, Corollary IV..4] or any other book on algebraic geometry) in orer to prove Lemma 5. Let X be the curve efine by (11) an Y the curve given by (11) where the last equation is replace by X 5 = 0. Let f : X Y be the morphism, given by (X 1,X,X 3,X 4,X 5 ) (X 1,X,X 3,X 4, 0). We see that a point P = (X 1,X,X 3,X 4, 0) Y has two istinct points as preimage, if an only if X 5 is not zero. Otherwise P has only one point as preimage an is therefore ramifie with ramification inex e P =. Let p 1 be the number of ramification points. Then we obtain g X = (g Y ) + p 1.

16 1404 A. Pethő, V. Ziegler / Journal of Number Theory 18 (008) We apply this reuction also to the curves Y P 3 an Z P 3, where Y is given by an Z is given by X 1 a 1,1 + X a 1, + X 3 a 1,3 = 0, X a,1 + X 3 a, + X 4 a,3 = 0, (1) X 1 a 1,1 + X a 1, + X 3 a 1,3 = 0 an X 4 = 0. Note that the curve Y in this paragraph is the same as in the paragraph above just imbee into P 3, where P 3 P 4 an P = (X 1,X,X 3,X 4,X 5 ) P 3 if an only if X 5 = 0. Obviously the curve Z has genus 0. Let now f : Y Z be the morphism given by (X 1,X,X 3,X 4 ) (X 1,X,X 3, 0). In this case P is ramifie if an only if X 4 = 0. Let p be the number of points on Y that are ramifie. By Hurwitz s formula we obtain g Y = (0 ) + p, hence g X = p + p 1 / 3. In orer to prove the lemma we have to compute p 1 an p. Let us compute p 1. Note, if X 5 = 0 then X 4 0, since otherwise X 3 = X = X 1 = 0 = X 4 = X 5, which oes not yiel an element of P 4. We obtain X 3 X 4 = a 3, a 3,1. Since a 3, an a 3,1 are not zero we have for fixe X 4 exactly two possibilities to choose X 3.Now we insert this relation into the secon line of (11) an obtain X X 4 = a,a 3, a,3 a 3,1 a,1 a 3,1 = F 1 a,1 a 3,1. Therefore we have exactly two possibilities for X if F 1 0 for fixe X 4. We compute by the formulae above an the first equation of (11) X 1 X 4 = a,a 3, a 1, + a,3 a 1, a 3,1 + a 3, a 1,3 a,1 =. a,1 a 3,1 a 1,1 a,1 a 3,1 a 1,1 F 3 Since this formula there are exactly two possibilities for X 1 if F 3 0 for fixe X 4. In particular we have p 1 = 8, if F 1 F 3 0. Similarly we obtain for the curve Y efine by (1), that X 4 = 0 yiels exactly two possibilities for X if X 3 is fixe. Furthermore, there are exactly two possibilities for X 1 if F 0. Therefore p = 4 provie F 0. Combining these results yiels the lemma. Corollary 3. The curve efine by (10) uner the assumption a k, k, 3k, k/, 3k/, 5k/, 7k/ has genus g = 5.

17 A. Pethő, V. Ziegler / Journal of Number Theory 18 (008) Proof. We have to show F 1 F F 3 0. Some computations yiel F 1 = 3(a + 5k) 0, F = 3(a + 3k) 3 0, F 3 = 8(a + k)(a + 3k)(a + 5k) 0. Because of Falting s theorem [6] (Morell s conjecture) the curve given by (10) has at most finitely many rational points, hence there are only finitely many pairs (, m), which satisfy the conitions of Theorem 5. Therefore Theorem 5 is prove. Proof of Theorem 4. The proof of Lemma 5 shows that one has to eal with curves of genus 1, i.e. elliptic curves, in orer to fin the examples given in Theorem 4. First, we want to transform the curve X given by X (4a + 4k) = X 1 (a + 3k) + X 3 (a + k), X3 (4a + 8k) = X (a + 5k) + X 4 (a + 3k), (13) into a plane curve. Similarly as in Section 5 we project X onto the plane X 4 = 0 from the point P = (1, 1, 1, 1) X. The line from P to Q = (x,y,z,0) is given by the system If we solve the system zx yx 3 + (y z)x 4 = 0, zx 1 xx 3 + (x z)x 4 = 0. we obtain X3 (4a + 8k) X (a + 5k) X 4 (a + 3k) = 0, zx yx 3 + (y z)x 4 = 0, zx 1 xx 3 + (x z)x 4 = 0, (14) X 1 = k( 10xy + 5y + 16xz 8z ) + a( 4xy + y + 8xz 4z ) (a + 5k)y 4(a + k)z X 4, X = (a + 5k)y + 8(a + k)yz 4(a + k)z (a + 5k)y 4(a + k)z X 4, X 3 = (a + 5k)y (a + 5k)yz + 4(a + k)z (a + 5k)y 4(a + k)z X 4. (15) Since we are working with projective coorinates we may choose

18 1406 A. Pethő, V. Ziegler / Journal of Number Theory 18 (008) X 1 = k ( 10xy + 5y + 16xz 8z ) + a ( 4xy + y + 8xz 4z ), X = (a + 5k)y + 8(a + k)yz 4(a + k)z, X 3 = (a + 5k)y (a + 5k)yz + 4(a + k)z, X 4 = (a + 5k)y 4(a + k)z. Note that X 4 = 0 implies a + k an a + 5k are both squares, which is a contraiction for a = 1 an k = respectively a = 0 an k = 1. If we insert the expressions for X 1, X, X 3 an X 4 into the first equation of (13) we obtain 4(a + 3k) ( (a + 5k)y 4(a + k)z ) ( xy(x y)(a + 5k) + 4xz(z x)(a + k) + 3yz(y z)(a + 3k) ) = 0. (16) The first factor a + 3k of (16) cannot be zero because we assume a 3 k. Also the secon factor (a + 5k)y 4(a + k)z is not zero, since otherwise = 4(y x)((a + 5k)y 4(a + k)z) k(a + k) ( (a + 5k)xy 4(a + k)(x + y)z + 4(a + k)z ) = 0. Now let us insert the special values for a an k. Then (16) turns into 3x y 3xy + 5xz 5x z + 6y z 6yz = 0 (a = 1, k= ), (17) 5x y 5xy + 8xz 8x z + 9y z 9yz = 0 (a = 0, k= 1). (18) Using Hoeij s algorithm (see [8]) we transform these elliptic curves into Weierstrass normal form, with the substitutions in the case a = 1, k =, respectively in the case a = 0, k = 1, into the elliptic curves x z := ξ + 6ξζ + 6ηζ 108ζ ξ 81ζ, y z := 6ξζ + 36ζ ηζ 3ξζ + 7ζ, x z := 81ξ + 594ξ ζ + 34η ζ 19647ζ 45ξ + 150ξ ζ 14455ζ, y z := 7ξ ζ 3η ζ + 333ζ 15ξ ζ + 45ζ,

19 A. Pethő, V. Ziegler / Journal of Number Theory 18 (008) η ζ = ξ 3 63ξζ + 16ζ 3 (a = 1, k= ), (19) η ζ = ξ ξ ζ ζ 3 (a = 0, k= 1). (0) We use the implementation of Hoeij s algorithm in Maple 8 to compute these transformations. A computation in PARI [1] shows that the elliptic curve (19) is a minimal integral moel of an elliptic curve (see [9, Section X.1]). On the other han the elliptic curve (0) can be transforme into its minimal integral moel by the transformation So we have to consier ξ = 4ξ + ζ 3, η = 8η + 4ξ + 4ζ, ζ = ζ. η ζ = ξ 3 63ξζ + 16ζ 3 (a = 1, k= ), (1) η ζ + ξηζ + ηζ = ξ 3 19ξζ + 6ζ 3 (a = 0, k= 1), () the minimal integral moels of the elliptic curves (19) an (0). Further computations in PARI show that the torsion group of (1) is isomorphic to Z/Z Z/Z with generators (ξ,η,ζ)= (3, 0, 1) an (6, 0, 1). The torsion group of () is isomorphic to Z/6Z Z/Z with generators (, 8, 1) an (3,, 1). A look on Cremona s tables [4] shows that the curve () has rank 0 an the curve (1) has rank 1 an its free group is generate by ( 3, 18, 1). With this information we are able to compute all rational points of () an sufficiently many rational points on (1). If we go back all substitutions for all rational points on () we get all solutions for (13). But no solution yiels a pair (, m) that satisfies the conitions of Theorem 4. Therefore we have prove this theorem. At the en of this section we want to show some examples of pairs (, m) such that 1, 3, 5 an7arey-components of solutions to the Pell equation X Y = m. LetT 1 = (3, 0, 1), T = (6, 0, 1) an P = ( 3, 18, 1), then P = (7, 8, 1) yiels the pair (, m) = ( 105, 5434). Although this is not an acceptable example, since it results to a negative, it furnishes us with the interesting Diophantine equation X + 105Y = 5434 having solutions with Y = 1, 3, 5, 7. The smallest acceptable example we have foun is T 3P = ( 71 9, 350 7, 1), which yiels = an m = The next acceptable example is T 1 4P = ( , , 1) which yiels = an m = Dual theorem Theorems 3, 4 an 5 only take care on the case for which the Y -component forms an arithmetic progression. In this section we consier the ual case, where the X-component forms an arithmetic progression. It turns out that in this case things are much more simple. In particular, we prove the following theorem. Theorem 7. Let X 1 <X <X 3 be in arithmetic progression such that X i X j for any i j. Then there are at most finitely many,m Z such that X 1, X, X 3 are the X-components of solutions to X Y = m.

20 1408 A. Pethő, V. Ziegler / Journal of Number Theory 18 (008) Proof. The proof of this theorem starts in the same way as the proof of the ual Theorems 3, 4 an 5. As in Section 6 we have which implies the system a Y 1 = m, (a + k) Y = m, (a + k) Y 3 = m, Y Y 1 = k(a + k)/, Y 3 Y = k(a + 3k)/. We may assume Y 1,Y,Y 3 > 0. If k a then k(a + k) 0. As Y 1,Y Z the integer ivies the fixe integer k(a + k), i.e. there are finitely many possibilities for. Keeping fixe, Y 1 + Y is a positive integer ivisor of k(a + k)/, i.e. it is boune. Thus there are only finitely many possibilities for Y 1 an Y. Hence there are only finitely many possibilities for m. If k = a, we apply the same consieration to the secon equation. 8. Open questions There are a lot of questions that arise reaing this paper. In this section we want to iscuss some of them. Is there an absolute constant C such that there is no Pell equation with arithmetic progression of length C? Although Theorem 1 an Corollary 1 suggest that C epens on an m, Theorem 5 inicates that only exceptional curves will have a certain arithmetic progression of length 5. So we guess that such a constant C exists. Can one prove or isprove that there are an m with >0 an not a perfect square such that Y = 1, 3, 5, 7, 9 is in arithmetic progression on the curve X Y = m? We have foun an arithmetic progression of length 4 that lies on some curve X Y = m an we have foun an arithmetic progression such that there oes not exist such a curve. Do both cases arise with the same probability or is one of them an exception? Are there simple criteria for a an k such that the arithmetic progression a,a + k,a + k,a + 3k lies on a hyperbola? Or is there a criterion for which the associate elliptic curve has rank 1? Acknowlegment We are very inebte the anonymous referee, who rea the earlier versions of this paper carefully, correcte several mistakes an gave etaile hints to make our arguments, especially the proof of Corollary 1, clearer. References [1] A. Bremner, On arithmetic progressions on elliptic curves, Experiment. Math. 8 (4) (1999) [] A. Bremner, J. Silverman, N. Tzanakis, Integral points in arithmetic progressions on y = x(x n ),J.Number Theory 80 (000) [3] G. Campbell, A note on arithmetic progressions on elliptic curves, J. Integer Seq. 6 (003), Article , 5 pp. (electronic). [4] J.E. Cremona, Elliptic curve ata, available at [5] A.Dujella,A.Pethő, P. Taić, On arithmetic progressions on Pellian equations, Acta Math. Hungar., in press.

21 A. Pethő, V. Ziegler / Journal of Number Theory 18 (008) [6] G. Faltings, Enlichkeitssätze für abelsche Varietäten über Zahlkörpern, Invent. Math. 73 (3) (1983) [7] R. Hartshorne, Algebraic Geometry, Gra. Texts in Math., vol. 5, Springer-Verlag, New York, [8] M. van Hoeij, An algorithm for computing the Weierstrass normal form, in: Proceeings of the 1995 International Symposium on Symbolic an Algebraic Computation, ISSAC 95, Montreal, Canaa, July 10 1, 1995, ACM, New York, 1995, pp [9] A.W. Knapp, Elliptic Curves, Math. Notes, vol. 40, Princeton Univ. Press, Princeton, NJ, 199. [10] R.A. Mollin, Funamental Number Theory with Applications, CRC Press, New York, [11] T. Nagell, Introuction to Number Theory, Almqvist & Wiksell Boktryckerei, Stockholm, [1] The PARI Group, Boreaux, PARI/GP, version.1.5, available from [13] M. Ulas, A note on arithmetic progressions on quartic elliptic curves, J. Integer Seq. 8 (005), Article , 5 pp. (electronic).