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GAZETA MATEMATICA˘ - SSMR

GAZETA MATEMATIC ASERIA AANUL XXXIV (CXIII)Nr. 1 2/ 2016 ARTICOLEOn a two parameter class of quadratic diophantine equations Arp ad B enyi1),2),Ioan Cas u3)Abstract. We characterize the solution set of a two parameter class ofquadratic diophantine equations. Our proof relies on the solvability of thepositive Pell : diophantine equation, quadratic equation, Pell equationMSC:Primary 11D09; Secondary this note we will be interested in exploring the solutionsof quadraticDiophantine equations of the formax2+by2+cx+dy+f= 0,wherea, b, c, d, f Z. Naturally, there is no hope in solving such a problemfor arbitrary coefficients, and simple examples show that there are such equa-tions which have zero, finitely many or infinitely many solutions (x, y) Z , we can see this to be the case by inspecting thelattice pointssitting on the algebraic curve defined by the quadratic equation above, butthis geometric intuition is not the one w

2 Articole We wish to show that the intuition behind these problems lends itself nicely to completely solving a two parameter class of quadratic Diophantine

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Transcription of GAZETA MATEMATICA˘ - SSMR

1 GAZETA MATEMATIC ASERIA AANUL XXXIV (CXIII)Nr. 1 2/ 2016 ARTICOLEOn a two parameter class of quadratic diophantine equations Arp ad B enyi1),2),Ioan Cas u3)Abstract. We characterize the solution set of a two parameter class ofquadratic diophantine equations. Our proof relies on the solvability of thepositive Pell : diophantine equation, quadratic equation, Pell equationMSC:Primary 11D09; Secondary this note we will be interested in exploring the solutionsof quadraticDiophantine equations of the formax2+by2+cx+dy+f= 0,wherea, b, c, d, f Z. Naturally, there is no hope in solving such a problemfor arbitrary coefficients, and simple examples show that there are such equa-tions which have zero, finitely many or infinitely many solutions (x, y) Z , we can see this to be the case by inspecting thelattice pointssitting on the algebraic curve defined by the quadratic equation above, butthis geometric intuition is not the one we will be using motiva-tion for this note stems from the following two problems proposed at the 7 8grade level:[3].

2 Leta, b Nbe such that4a2+ 9a+ 1 = 3b2+ 7b. Show that3a+ 3b+ 7and4a+ 4b+ 9are perfect squares.[4].Letx, y Nsuch thatx > yandx+ 4x2=y+ 5y2. Show thatx yis a perfect )Department of Mathematics, Western Washington University, 516 High Street, Belling-ham, Washington 98225, work is partially supported by a grant from the Simons Foundation (No. 246024to Arp ad B enyi)3)Department of Mathematics, West University of Timi wish to show that the intuition behind these problems lends itselfnicely to completely solving a two parameter class of quadratic Diophantineequations. Specifically, let , N {0}and consider the following equationin variables (x, y) Z Z: x2 ( + 1)y2+ (2 + 1)x (2 + 2 + 1)y 2= 0.

3 (1)Clearly, for = 3 and = 1 we obtain the equation in [3], while for = 4, = 0 we get the equation in [4].Let us remark right away that when = 0, the equation (1) reducestox=y2+ (2 + 1)y+ 2, which already produces the parametrizationfor all the infinitely many integer solutions of the equation. In fact, slightlyre-writing, we have in this casex= (y+ )2+y,y Z, which suggeststhat understanding the solutions of the two parameter diophantine equation(1) could potentially be reduced to the solutions of just a one parameterDiophantine equation by appropriately substituting forxandy. We showfirst that this is indeed the to a one parameter diophantine us makethe change of variablesX=x+ andY=y+.

4 Thus, replacingx=X andy=Y in (1) we obtain (X )2 ( +1)(Y )2+(2 +1)(X ) (2 +2 +1)(Y ) = 2,which after some straightforward algebra simplifies to X2 ( + 1)Y2+X Y= simply means that in the two parameter diophantine equation (1) wecan assume without loss of generality = 0. Moreover, if a solution of (1)exists for = 0, the general solution for 6= 0 is obtained from that one bytranslating by in both also that if we let = 0 and = 0 in (1), we already have theparametrization of all the integer solutions:x=y2 y, withy Z. Withthese considerations, we will assume in the remainder of this note that Nand = 0 in (1), that is we will investigate the integer solutions of x2 ( + 1)y2+x y= 0.

5 (2)Our main result is the N. The non-trivial solutions(xn, yn)n 1of the qua-dratic diophantine equation(2)are given byxn= ( + 1)v2n unvn, yn= v2n unvn,whereun=( + 1 + )2n+ ( + 1 )2n2, A. B enyi, I. Cas u, Quadratic diophantine equations3vn=( + 1 + )2n ( + 1 )2n2p ( + 1).Moreover, the solution set of(1)is given by{(xn , yn ) :n N} {( , )},withxn, ynas word trivial above refers to the pair (0,0) which obviously satis-fies the given equation. We point out also that the signs in the expressionsofxnandyn, respectively, coincide. We have the following immediate conse-quence of our main (x, y) Zbe a solution of(2).

6 Thenx yis a perfectsquare and the values ofx ybelong precisely to the set{0} {v2n:n N},withvndefined in Theorem that, in particular, we recover the statement in [4]. Infact, the ideaof proving Theorem 1 is guided by the intuition contained in the elementaryproblems [3] and [4], combined with the well-understood theory of positivePell equations. As we shall see, the statement in [3] is also aby-product ofour arguments showing the main result. Before proceeding with the proof ofTheorem 1, we take a brief excursion into the theory of Pell general positive Pell a positive integer that isnot a perfect square. It is a known fact that thepositive Pell equationu2 Dv2= 1(3)has infinitely many solutions inZ Z.

7 Clearly, if (u, v) N Nis a solutionto the Pell equation, then (u, v),( u, v) and ( u, v) are also , without loss of generality, it suffices to consider onlyits solutions in thenatural lattice. Any equation of the form (3) has thetrivialsolution (1,0).Besides the trivial solution, the other solutions in the natural lattice can beobtained from thefundamental solution(u0, v0), which is the least positiveinteger solution to (3) different from (1,0) the so-calledfundamental so-lutionof (3) for which the expressionu+v Dis minimal, via the followingformulas:un=(u0+v0 D)n+ (u0 v0 D)n2vn=(u0+v0 D)n (u0 v0 D)n2 D, n a brief introduction to Pell equations, the interested reader can consult,for example, [1] and the references these prerequisites we are ready to proceed with the proof of ourmain of Theorem what follow, we assume that both unknowns,xandy, are non-zero.

8 In particular, this means thatx6= first claim, following the statement in [4], is that ifx, ysatisfy (2),thenx ymust be a perfect square. Let us start by re-writing the equation(2) in two ways:(x y)[1 + (x+y)] =y2,(x y)[1 + ( + 1)(x+y) = 1 + (x+y) andB= 1 + ( + 1)(x+y) and multiplying thesetwo equations we obtain (x y)2AB= (xy)2, thus showing thatABmustbe a perfect square. In particular, we find thatAandBmust be either bothpositive or both us assume for the moment thatbothAandBare positive. Notenow that ( +1)A B= 1. Therefore,AandBare relatively prime naturalnumbers. Combining this with the fact thatABis a perfect square showsthatAandBhave to be perfect squares as well; this is a simple consequenceof the Fundamental Theorem of Arithmetic.]

9 Incidentally, the statement thatAandBmust be perfect squares is precisely the content of the problem [3]if we take into account also the natural change of variables reducing (1) to(2). Now, sinceAis a perfect square and (x y)Ais a perfect square, weconclude thatx y >0 must also be a perfect square. This proves our this information in hand, let us write thenx y=v2for somev N. Our next claim is thatvmust be of the formvn,n 1, withvnas inthe statement of our Theorem 1. Substitutingx=y+v2in (2) we find that (y+v2)2 ( + 1)y2+v2= 0 y2 2 v2y ( v4+v2) = are interested in the integer solutions of the quadratic equation iny,which are given byy= v2 vp ( + 1)v2+ 1.

10 (4)Clearly, foryto be an integer, we need now ( + 1)v2+ 1 =u2 u2 ( + 1)v2= 1,(5)for someu Z. But this is exactly a positive Pell equation of the form (3),withD= ( + 1) obviously not being a perfect square. It is equally easyto see that the fundamental solution of (5) is given byu0= 2 + 1, v0= , the general solution of (5) inN Ncan be expressed asun=(2 + 1 + 2p ( + 1))n+ (2 + 1 2p ( + 1))n2, A. B enyi, I. Cas u, Quadratic diophantine equations5vn=(2 + 1 + 2p ( + 1))n (2 + 1 2p ( + 1))n2p ( + 1).It is straightforward to see that these expressions are precisely the ones statedin Theorem 1, since2 + 1 2p ( + 1) = ( + 1 ) now to the formula (4), we thus find that the integersolutions areprecisely those of the formyn= v2n vnun,and consequentlyxn=y2n+v2n= ( + 1)v2n the argument above, we assumed that bothAandBare strictlypositive.


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