The Diophantine equation x 2 (t 2 + t)y 2 (4t + 2)x + (4t 2 + 4t)y = 0

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1 Rev Mat Complut 00) 3: 5 60 DOI 0.007/s The Diophantine equation x t + t)y 4t + )x + 4t + 4t)y 0 Ahmet Tekcan Arzu Özkoç Receive: 4 April 009 / Accepte: 5 May 009 / Publishe online: 5 November 009 The Authors) 009. This article is publishe with open access at Springerlink.com Abstract Let t be an integer. In this work, we consier the number of integer solutions of Diophantine equation x t + t)y 4t + )x + 4t + 4t)y 0 over Z an also over finite fiels F p for primes p 5. Keywors Diophantine equation Pell equation Mathematics Subject Classification 000) D09 D4 D45 Preliminaries A Diophantine equation is an ineterminate polynomial equation that allows the variables to be integers only. Diophantine problems have fewer equations than unknown variables an involve fining integers that work correctly for all equations. In more technical language, they efine an algebraic curve, algebraic surface or more general object, an ask about the lattice points on it. The wor Diophantine refers to the Hellenistic mathematician of the 3r century, Diophantus of Alexanria, who mae a stuy of such equations an was one of the first mathematicians to introuce symbolism into algebra. The mathematical stuy of Diophantine problems Diophantus initiate is now calle Diophantine analysis. A linear Diophantine equation is an A. Tekcan ) A. Özkoç Faculty of Science, Department of Mathematics, Uluag University, Görükle, Bursa, Turkey tekcan@uluag.eu.tr url: A. Özkoç arzuozkoc@uluag.eu.tr

2 5 A. Tekcan, A. Özkoç equation between two sums of monomials of egree zero or one. While iniviual equations present a kin of puzzle an have been consiere throughout history, the formulation of general theories of Diophantine equations was an achievement of the twentieth century. For example, the equation ax + by is known the linear Diophantine equation. In general, the Diophantine equation is the equation given by ax + bxy + cy + x + ey + f 0..) Also for n, there are infinitely many solutions x,y,z) of the Diophantine equation x n + y n z n. For larger values of n, Fermat s last theorem see [4]) states that no positive integer solutions x,y,z satisfying the equation exist. The Diophantine equation x y.) is known the Pell equation see [3, 6]) which is name after the English mathematician John Pell a mathematician who searche for integer solutions to equations of this type in the seventeenth century. The Pell equation in.) has infinitely many integer solutions x n,y n ) for n. The first non-trivial positive integer solution x,y ) of this equation is calle the funamental solution, because all other solutions can be easily) erive from it. In fact if x,y ) is the funamental solution, then the n-th positive solution of it, say x n,y n ), is efine by the equality x n + y n x + y ) n.3) for integer n. There are several methos for fining the funamental solution of x y. For example, the cyclic metho known in Inia in the th century, an the slightly less efficient but more regular English metho 7th century, prouce all solutions of x y see[4]). But the most efficient metho for fining the funamental solution is base on the simple finite continue fraction expansion of see [, 5, 6, 0 3]). We can escribe it as follows: Let [a 0 ; a,...,a r, a 0 ] be the simple continue fraction of a 0 ). Let p 0 a 0, p + a 0 a, q 0, q a, p n a n p n + p n an q n a n q n + q n for n. If r is o, then the funamental solution is x,y ) p r,q r ), where p r /q r is the rth convergent of an if r is even, then the funamental solution is x,y ) p r+,q r+ ). The Pell equation was first stuie by Brahmagupta ) an Bhaskara 4 85) see []). Its complete theory was worke out by Lagrange ), not Pell. It is often sai that Euler ) mistakenly attribute Brouncker s ) work on this equation to Pell. However the equation appears in a book by Rahn 6 676) which was certainly written with Pell s help: some say entirely written by Pell. Perhaps Euler knew what he was oing in naming the equation. Baltus [], Kaplan an Williams [6], Lenstra [7], Matthews [8], Mollin also Poorten an Williams) [9, 4, 5], Stevenhagen [7], Tekcan [8 3], an the others consiere some specific Pell an Diophantine) equations an their integer solutions.

3 The Diophantine equation x t + t)y 4t + )x + 4t + 4t)y 0 53 The Diophantine equation x t + t)y 4t + )x + 4t + 4t)y 0 Let t be an integer. In this section, we will consier the integer solutions of Diophantine equation D : x t + t)y 4t + )x + 4t + 4t)y 0.) over Z. Note that D has always two integer solutions 0, 0) an 0, 4) for every arbitrary t. So we omit these two solutions.) Note that D represents a conic. So before consiering our main problem, we give some preliminaries on conics. Recall that a conic C is given by an equation C : ax + bxy + cy + x + ey + f 0,.) where a,b,c,,e an f are real numbers. The iscriminant of C is C) b 4ac. If C) < 0, then C represents an ellipse; C) > 0, then C represents a hyperbole an if C) 0, then C represents a parabola. If b 0, then we can transform C to a centripetal conic on the uv-plane via the transformation x u + h, T : y v + k.3) for some h an k. Here we call the pair h, k} as the base of T an enote it by T [h; k]h, k}..4) Since a Diophantine equation D represents a conic C, the corresponing conic is hence C : x t + t)y 4t + )x + 4t + 4t)y 0..5) Now we want to carry out C to a centripetal conic on uv-plane. To get this let x u + h an y v + k for some h an k. Then.5) becomes C : u + h) t + t)v + k) 4t + )u + h) + 4t + 4t)v + k) 0..6) In.6), we obtain uh 4t) an v kt kt + 4t + 4t).Soweget h t + an k.7) since the coefficients u an v must be zero. Therefore for x u + t + an y v +,.8) we get C : u t + t)v.9) which is a centripetal conic in fact a hyperbole) on uv-plane with focuses an. So the corresponing Diophantine equation is hence t +t) D : u t + t)v..0) In fact this a Pell equation by.). Now we can consier the integer solutions of D.

4 54 A. Tekcan, A. Özkoç Theorem. Let D be the Diophantine equation in.0). Then ) The funamental solution of D is u,v ) t +, ). ) Define the sequence u n,v n )}, where un v n ) t + t ) n + t t + 0).) for n. Then u n,v n ) is a solution of D. 3) The solutions u n,v n ) satisfy u n t + )u n + t + t)v n an v n u n + t + )v n for n. 4) The solutions u n,v n ) satisfy the recurrence relations u n 4t + )u n + u n ) u n 3 an v n 4t + )v n + v n ) v n 3 for n 4. 5) The continue fraction expansion of t + t is [; ] if t, t + t [t;, t] if t>. 6) The n-th solution u n,v n ) can be given by u [ n v n t; t,t,...,t,t }} n times ],.) for n. Proof. It is easily seen that u,v ) t +, ) is the funamental solution of D since t + ) t + t).. We prove it by inuction on n. Letn. Then by.), we get u,v ) t +, ) which is a solution of D since u,v ) is the funamental solution. Let us assume that the Diophantine equation in.0) issatisfieforn, that is, D : u n t + t)vn..3) We want to show that this equation is also satisfie for n. Applying.), we fin that ) un t + t ) n + t v n t + 0) t + t ) + t t + t ) n + t t + t + 0) t + t ) ) + t un t + v n t + )un + t ) + t)v n..4) u n + t + )v n

5 The Diophantine equation x t + t)y 4t + )x + 4t + 4t)y 0 55 Hence we conclue that u n t + t)v n t + )u n + t + t)v n ) t + t) u n + t + )v n ) t + ) u n + t + )t + t)u n v n + t + t) vn t + t) 4u n + 4t + )u n v n + t + ) vn ) u n t + ) 4t + t) ) + u n v n t + )t + t) t + t)4t + ) ) + vn t + t) t + t)t + ) ) u n t + t)vn since u n t + t)v n by.3). So u n,v n ) is also a solution of D. 3. Applying.4), we fin that u n t + )u n + t + t)v n an v n u n + t + )v n for n. 4. We only prove that u n satisfy the recurrence relation u n 4t + )u n + u n ) u n 3 for n 4. For n 4, we fin from.) that u t +,u 8t + 8t +,u 3 3t t + 8t + an u 4 8t t t + 3t +. Hence u 4 4t + )u 3 + u ) u 4t + )3t t + 6t + ) t + ) 8t t t + 3t +. So u n 4t + )u n + u n ) u n 3 is satisfie for n 4. Let us assume that this relation is satisfie for n, that is, u n 4t + )u n + u n 3 ) u n 4..5) Then applying the previous assertion,.4) an.5), we fin that u n 4t + )u n + u n ) u n 3 for n Let t. Then it is easily seen that [; ]. Nowlett>. Then t + t t + t + t t) t + t + + t t +t t So t + t [t;, t]. t + + t +t t t +t+t t + t +t+t t t + + t + + t+ t +t t) t +t t t.

6 56 A. Tekcan, A. Özkoç 6. It is easily seen that u,v ) t +, ) is a solution since u v [t; ]t + t+. Let us assume that u n,v n ) is a solution of D, that is, u n t + t)vn. Then by.), we erive u n+ v n+ t + + t+ + t+ +t+ t + + t+t+ + t+ +t+ t + + t+ un vn t + )u n + t + t)v n u n + t + )v n..6) So u n+,v n+ ) is also a solution of D since u n+ t +t)v n+ t +)u n +t +t)v n ) t +t) u n +t +)v n ). This completes the proof. Example. Let t 5. Then u,v ), ) is the funamental solution of D : u 30v an some other solutions are u,v ) 4, 44), u 3,v 3 ) 59, 966), u 3,v 3 ) 59, 966), u 4,v 4 ) 66, 08), u 5,v 5 ) 5505, 46560), u 6,v 6 ) , 0). Also 30 [5;, 0]. The solutions u n,v n ) satisfy the relations u n u n + u n ) u n 3 an v n v n + v n ) v n 3 for n 4. Also u n u n + 60v n an v n u n + v n for n. Further u [5; ] v, u [5;, 0, ] 4 v 44, u 3 [5;, 0,, 0, ] 59 v 3 966, u 4 [5;, 0,, 0,, 0, ] 66 v 4 08, u 5 [5;, 0,, 0,, 0,, 0, ] 5505 v , u 6 [5;, 0,, 0,, 0,, 0,, 0, ] v 6 0. From above theorem we can give the following result.

7 The Diophantine equation x t + t)y 4t + )x + 4t + 4t)y 0 57 Corollary. The base of the transformation T in.3) is the funamental solution of D, that is T [h; k]h, k}u,v }. Proof We prove in above theorem that u,v ) t +, ) is the funamental solution of D.Wealsoshowein.7) that h t + an k. So the base of T is T [h; k]h, k}t +, } as we claime. We see as above that the Diophantine equation D : x t + t)y 4t + )x + 4t + 4t)y 0 can be transforme into the Diophantine equation D : u t + t)v via the transformation T efine in.3). Also we show in.8) that x u+t + an y v +. Further we prove in Theorem. that the funamental solution of D is u,v ) t +, ) an the other solutions are u n,v n ) which is efine in.). So we can retransfer all results from D to D. Thus we can give the following main theorem. Theorem.3 Let D be the Diophantine equation in.). Then ) The funamental minimal) solution of D is x,y ) 4t +, 4). ) Define the sequence x n,y n )} n u n + t +,v n + )}, where u n,v n )} efine in.). Then x n,y n ) is a solution of D. So it has infinitely many many integer solutions x n,y n ) Z Z. 3) The solutions x n,y n ) satisfy x n t + )x n + t + t)y n 8t 6t, y n x n + t + )y n 8t for n. 4) The solutions x n,y n ) satisfy the recurrence relations for n 4. x n 4t + )x n + x n ) x n 3 6t 8t, y n 4t + )y n + y n ) y n 3 6t Example. Some integer solutions of D are x,y ) 4t +, 4), x,y ) 8t + 0t +, 8t + 6), x 3,y 3 ) 3t t + 0t +, 3t + 3t + 8), x 4,y 4 ) 8t t t + 34t +, 8t 3 + 9t + 80t + 0), x 5,y 5 ) 5t t 4 + 0t t + 5t +, 5t t t + 60t + ).

8 58 A. Tekcan, A. Özkoç 3 The Diophantine equation x t + t)y 4t + )x + 4t + 4t)y 0 over F p In this section, we will consier the integer solutions of D efine in.) over finite fiels F p for primes p 5. We show that D becomes D which is efine in.0). Now let t F p an let t + t mo p). Then D becomes D p : u v mo p). 3.) Set D p F p) u, v) F p F p : u v mo p)}. Then we can give the following theorem. Theorem 3. Let D p be the Diophantine equation in 3.).. If p, 7mo 8), then. If p 3, 5mo 8), then p + for / # D p Qp F, p) p for Q p. p + for # D p Qp F, p) p for / Q p, where Q p enote the set of quaratic resiues. Proof. Let p, 7mo 8). Then we have two cases: Case ) Let p mo 8) an let let Q p.ifv 0, then u mo p) an hence u an u p. So the Diophantine equation D p has two integer solutions, 0) an p, 0). Ifu 0, then v mo p) has two solutions v,v.so D p has two integer solutions 0,v ) an 0,v ).NowletS p F p,p }. Then there are p 5 points u in S p such that u is a square. Set u c for some c 0. Then v c mo p) v ±cmo p). So the Diophantine equation D p has two solutions u, c) an u, c), that is, for each u in S p, D p has two solutions. So it has p 5 ) p 5 solutions. We see as above that it has also four solutions, 0), p, 0), 0,v ) an 0,v ). Therefore D p has p p integer solutions. Now let / Q p.ifv 0, then u mo p) an hence u an u p. So the Diophantine equation D p has two integer solutions, 0) an p, 0). If u 0, then v mo p) has no solution. So D p has no integer solution 0,v).LetL p F p p,p }. Then there are points u in L p such that u is a square. Set u j for some j 0. Then v j mo p) v ±jmo p). So the Diophantine equation D p has two solutions u, j) an u, j), that is, for each u in L p, D p p has two solutions. So it has ) p solutions. We see as

9 The Diophantine equation x t + t)y 4t + )x + 4t + 4t)y 0 59 above that it has also two solutions, 0) an p, 0).So D p has p + p + integer solutions. Case ) Let p 7mo 8) an let Q p.ifv 0, then u mo p) an hence u an u p. So the Diophantine equation D p has two integer solutions, 0) an p, 0). Ifu 0, then v mo p) has no solution v. So D p has no integer solution 0,v).NowletH p F p 3 p,p }. Then there are points u in H p such that u is a square. Set u k for some k 0. Then v k mo p) v ±kmo p). So the Diophantine equation D p has two solutions u, k) an u, k), that is, for each u in H p, D p p 3 has two solutions. So it has ) p 3 solutions. We see as above that it has also two solutions, 0) an p, 0). Therefore D p has p 3 + p integer solutions. Let / Q p.ifv 0, then u mo p) an hence u an u p. So the Diophantine equation D p has two integer solutions, 0) an p, 0). Ifu 0, then v mo p) has two solutions v,v.so D p has two integer solutions 0,v ) an 0,v ).Nowlet K p F p 3 p,p }. Then there are points u in K p such that u is a square. Set u m for some m 0. Then v m mo p) v ±mmo p). So the has two solutions u, k) an u, k), that is, for each u Diophantine equation D p in K p, D p that it has also four solutions, 0), p, 0), 0,v ) an 0,v ). Therefore D p has p p + integer solutions.. It can be prove as in same way that the previous assertion was prove. p 3 has two solutions. So it has ) p 3 solutions. We see as above Example 3. Let t 5. Then 30 8mo ), 4mo 3), 3mo 7), 7mo 3).So } D 8 0, ), 0, 9),, 0), 3, ), 3, 0), 5, 5), 5, 6), 6, 5), 6, 6), F ), 8, ), 8, 0), 0, 0) } D 3 4 0, 4), 0, 9),, 0),, ),, ), 6, 5), 6, 8), 7, 5), 7, 8), F 3),, ),, ),, 0) } D 7 3 0, 8), 0, 9),, 0), 3, 7), 3, 0), 4, 3), 4, 4), 6, ), 6, 5), F 7),, ),, 5), 3, 3), 3, 4), 4, 7), 4, 0), 6, 0) 0, 6), 0, 7),, 0), 4, 9), 4, 4), 8, 3), 8, 0), 9, 8), 9, 5), D 3 7 F 3) 0, ), 0, ),, ),, ),, ),, ), 3, ), 3, ), 4, 8), 4, 5), 5, 3), 5, 0), 9, 9), 9, 4),, 0) an hence # D 8 F ), # D 4 3 F 3), # D 3 7 F 7) 6 an # D 7 3 F 3) 4. For the Diophantine equation D,weset DF p ) x, y) F p F p : x t + t)y 4t + )x + 4t + 4t)y 0mo p)}. Then we can give the following theorem.

10 60 A. Tekcan, A. Özkoç Theorem 3. Let D be the Diophantine equation in.). ) If p, 7mo 8), then p + for t + t/ Q p, #DF p ) p for t + t Q p. ) If p 3, 5mo 8), then p + for t + t Q p, #DF p ) p for t + t/ Q p. Open Access This article is istribute uner the terms of the Creative Commons Attribution Noncommercial License which permits any noncommercial use, istribution, an reprouction in any meium, provie the original authors) an source are creite. References. Arya, S.P.: On the Brahmagupta-Bhaskara equation. Math. Euc. 8), ). Baltus, C.: Continue fractions an the Pell equations: the work of Euler an Lagrange. Comm. Anal. Theory Contin. Fractions 3, ) 3. Barbeau, E.: Pell s Equation. Springer, Berlin 003) 4. Ewars, H.M.: Fermat s Last Theorem. A Genetic Introuction to Algebraic Number Theory. Grauate Texts in Mathematics, vol. 50. Springer, New York 996). Correcte reprint of the 977 original 5. Hensley, D.: Continue Fractions. Worl Scientific, Singapore 006) 6. Kaplan, P., Williams, K.S.: Pell s equations x my, 4 an continue fractions. J. Number Theory 3, ) 7. Lenstra, H.W.: Solving the Pell equation. Not. AMS 49), ) 8. Matthews, K.: The Diophantine equation x Dy N,D >0. Expo. Math. 8, ) 9. Mollin, R.A., Poorten, A.J., Williams, H.C.: Halfway to a solution of x Dy 3. J. Theor. Nr. Borx. 6, ) 0. Mollin, R.A.: Simple continue fraction solutions for Diophantine equations. Expo. Math. 9), ). Mollin, R.A., Cheng, K., Goar, B.: The Diophantine equation AX BY C solve via continue fractions. Acta Math. Univ. Comen. 7), 38 00). Mollin, R.A.: The Diophantine equation ax by c an simple continue fractions. Int. Math. J. ), 6 00) 3. Mollin, R.A.: A continue fraction approach to the Diophantine equation ax by ±. JP J. Algebra Number Theory Appl. 4), ) 4. Mollin, R.A.: A note on the Diophantine equation D x + D ak n. Acta Math. Aca. Paeagog. Nyireg., 4 005) 5. Mollin, R.A.: Quaratic Diophantine equations x Dy c n. Ir. Math. Soc. Bull. 58, ) 6. Niven, I., Zuckerman, H.S., Montgomery, H.L.: An Introuction to the Theory of Numbers, 5th en. Wiley, New York 99) 7. Stevenhagen, P.: A ensity conjecture for the negative Pell equation. Comput. Algebra Number Theory Math. Appl. 35, ) 8. Tekcan, A.: Pell equation x Dy, II. Bull. Ir. Math. Soc. 54, ) 9. Tekcan, A., Bizim, O., Bayraktar, M.: Solving the Pell equation using the funamental element of the fiel Q ). South East Asian Bull. Math. 30, ) 0. Tekcan, A.: The Pell equation x Dy ±4. Appl. Math. Sci. 8), ). Tekcan, A., Gezer, B., Bizim, O.: On the integer solutions of the Pell equation x y t.int.j. Comput. Math. Sci. 3), ). Tekcan, A.: The Pell equation x k k)y t. Int. J. Comput. Math. Sci. ), ) 3. Tekcan, A.: The Diophantine equation 4y 4yx 0, curves an conics over finite fiels. Math. Rep. accepte for publication)