EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 6, No. 4, 2013, 400-404 ISSN 1307-5543 – www.ejpam.com On the Number of Pairs of Points in a Quadratic Equation with Rational Distance Juan Sebastian Beleño Diaz Escuela de Ingeniería de Sistemas e Informática, Facultad de Ingenierías Físico-Mecánicas, Industrial University of Santander, Bucaramanga, Colombia Abstract. In this paper is shown a solution to the number of pair of points in a quadratic equation with rational distance, this result have an important impact to solve the open problem “Points on a parabola” [3] proposed in The Center for Discrete Mathematics and Theoretical Computer Science (DIMACS), because it’s an approach to set down basis in the problem. 2010 Mathematics Subject Classifications: 97F40,11D09,41A20 Key Words and Phrases: rational numbers, quadratic equations, Approximation by rational functions 1. Introduction Define f : ℜ −→ ℜ by the function f (x) = ax2 + bx + c; where x > 0 and a, b, c ∈ ℜ with a 6= 0, then the question is: how many pairs of points, so that the distance between them is a rational number?, although exist some references about quadratic equations and distances[1–3, 6, 7, 9, 10] , there is no information about this specifically question and the solution of this problem allows to start to solve the still open problem “Points on a parabola” [3], that it’s about to find the maximum number of points that satisfies the condition to have a rational distance between any of them. 2. Main Result Theorem 1. Let f : ℜ −→ ℜ by f (x) = ax2 + bx + c; where x ∈ Z+ and a, b, c ∈ ℜ with a 6= 0 ⇒exist infinite pairs of points within the polynomial, where the distance between them is a rational number. By reductio ad absurdum, we suppose that the quadratic equation with form f (x) = ax2 + bx + c have finite pairs of points that satisfies the condition that its distance is a rational number. Email address: juan.beleno@correo.uis.edu.co http://www.ejpam.com 400 c© 2013 EJPAM All rights reserved. J. Diaz / Eur. J. Pure Appl. Math, 6 (2013), 400-404 401 Select two points in the polynomial: (r, ar2+ br+ c) and (s, as2+ bs+ c); where r, s ∈ Z+ Define the distance function for this case [5] d = p (s− r)2+ (as2+ bs+ c− ar2− br − c)2 Cancel the constants: c− c = 0 d = p (s− r)2+ (as2+ bs− ar2− br)2 Factorize by common factor: d = p (s− r)2+ (a(s2− r2) + b(s− r))2 Factorize by difference of squares: d = p (s− r)2+ (a(s− r)(s+ r) + b(s− r))2 Again, factorize by common factor: d = p (s− r)2+ ((s− r)(a(s+ r) + b))2 (1) Now define d = p q ; where p, q ∈ Z and q 6= 0, to find all points where the distance between them is a rational number. Replace d in the equation by (1). p q = p (s− r)2+ ((s− r)(a(s+ r) + b))2 Squaring both sides: p2 q2 = (s− r)2+ ((s− r)(a(s+ r) + b))2 Reorganizing the equation: � p q �2 = (s− r)2+ ((s− r)(a(s+ r) + b))2 (2) Without loss of generality, we will use the equation [5] (5n)2 = (−3n)2+ (4n)2 (3) Where n ∈ Q, to represent that this family of pairs of points is infinite even if it’s a subset of all points that satisfies the condition to have a rational distance between them inside the quadratic equation. Matching the equations (2) and (3) p q = 5n (4) J. Diaz / Eur. J. Pure Appl. Math, 6 (2013), 400-404 402 s− r =−3n (5) (s− r)(a(s+ r) + b) = 4n (6) Replace (5) in (6): −3n(a(s+ r) + b) = 4n Divide by 3n in both sides: a(s+ r) + b =− 4 3 Deduct b in both sides: a(s+ r) =− 4 3 − b Divide by a in both sides: s+ r =− 4 3 + b a Reorganizing the equation: s+ r =− 4 3 + 3b 3 a =− 4−3b 3 a =− 4− 3b 3a (7) Define j =−4−3b 3a ; where j ∈ ℜ and replace in the equation (7). s+ r = j (8) Do (5)+(8) 2s = j− 3n Divide by 2 in both sides: s = j− 3n 2 (9) Now, do (8)-(5) 2r = j+ 3n Divide by 2 in both sides: r = j+ 3n 2 (10) The equations (9) and (10) present some restriction: j > 0 (11) r > 0 (12) J. Diaz / Eur. J. Pure Appl. Math, 6 (2013), 400-404 403 s > 0 (13) Replace (10) in (12) j+ 3n 2 > 0 Multiply 2 in both sides: j+ 3n> 0 Deduct j in both sides: 3n>− j Divide by 3 in both sides n> − j 3 (14) Now, replace (9) in (13) j− 3n 2 > 0 Multiply 2 in both sides: j− 3n> 0 Add 3n in both sides: j > 3n Divide by 3 in both sides: j 3 > n (15) From (14) and (15) − j 3 < n< j 3 (16) Lemma 1. The set Q ∩ � − j 3 , j 3 � is countably infinite. Proof. Let s ∈ Q ∩ � − j 3 , j 3 � , then each s will be written in the (unique) form p q , where p, q ∈ Z+ and have no common divisor other than 1 [8, 11]. Now, define f : Q ∩ � − j 3 , j 3 � → Z+ × Z+ by f � p q � = (p, q), and let K = range f . For p q , u v ∈ Q ∩ � − j 3 , j 3 � , we find that f � p q � = f � u v � ⇒ (p, q) = (u, v) ⇒ p = u and q = v ⇒ � p q � = � u v � , so f is a one-to-one function. Therefore |Q∩ � − j 3 , j 3 � |= |K |, a subset of the countable set Z+× Z+ (by Theorem A3.5 in [4] we know that Z+ × Z+ is countable). 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