#A47 INTEGERS 15 (015) QUADRATIC DIOPHANTINE EQUATIONS WITH INFINITELY MANY SOLUTIONS IN POSITIVE INTEGERS Mihai Ciu Simion Stoilow Institute of Mathematics of the Romanian Academy, Research Unit No. 5, Bucharest, Romania Mihai.Ciu@imar.ro Received: 1/8/15, Revised: 7/6/15, Acceted: 10/17/15, Published: 11/13/15 Abstract For an arbitrary ositive integers k, any integer a for which the equation x axy + y +kx = 0 has infinitely many solutions in ositive integers satisfies either a = k+ or 3 ale a ale b k+5 c. Moreover, each of 3, b k+5 c, and k + has this roerty. 1. Introduction For fixed integers a and k, we shall call the equation x axy + y + kx = 0 (1) admissible if it has infinitely many solutions in ositive integers. For a ositive integer k, let S(k) denote the set of integers a for which the above equation is admissible. To the best of our knowledge, the earliest study of such a set is due to Marlewski and Zarzycki []. In the resent terminology, their main result can be hrased as follows. Theorem 1. ([]) S(1) = {3}. In an equivalent form, this has already been found by Owings [5] in his solution to a comletely di erent roblem. More sets of the kind are determined by Yuan and Hu [7]. Theorem. ([7]) S() = {3, 4} and S(4) = {3, 4, 6}. Imortant qualitative information has been rovided by Feng, Yuan, and Hu [1], who have shown that S(k) is a finite set for any ositive integer k. Moreover, the arguments emloyed to justify this statement rovide the inequality a ale 4k + for any a S(k). The same bound is exlicitly stated in Corollary 1.1 from [3].
INTEGERS: 15 (015) The aim in this aer is to give the best ossible result in this vein. Additionally, we show that the extreme values always belong to the set in question. Theorem 3. Let k be a ositive integer. Then, for any a S(k) one has either a = k + or 3 ale a ale b k+5 k+5 c. Moreover, 3, b c, and k + belong to S(k). Note that Theorems 1 and are obtained by secialization from our result, which also gives {3, 4, 5} S(3) {3, 4, 5}, {3, 5, 7} S(5) {3, 4, 5, 7}, and {3, 6, 9} S(7) {3, 4, 5, 6, 9}. From what we shall rove in Section 4 it will result S(5) = {3, 5, 7}, S(7) = {3, 6, 9}, in accordance with [1].. Prearations for the Proof of the Main Result Throughout the aer k will denote a ositive integer. A key remark of Marlewski and Zarzycki (see [, Theorem 1]) in their study of the equation x kxy + y + x = 0 has been extended by Feng et al. so as to aly to the more general equation (1). Lemma 1. ([1, Lemma 3]) Suose ositive integers a, k x, y satisfy (1). If gcd(x, y, k) = 1 then there exist ositive integers c, e such that x = c, y = ce, and gcd(c, e) = 1. Let a S(k). Since k has finitely many divisors, there exists a divisor d of k such that gcd(x, y, k) = k/d for infinitely many solutions (x, y) to the equation (1). Hence, both relations u auv + v + du = 0 and gcd(u, v, d) = 1 are satisfied by infinitely many airs (u, v) of ositive integers. Having in view Lemma 1, we obtain the following result. Lemma. For any a S(k) there exists a ositive divisor d of k such that the equation c ace + e + d = 0, gcd(c, e) = 1, has infinitely many solutions in ositive integers. Another remark valid for all sets of interests is the following. Lemma 3. If k, K are ositive integers with K multile of k, then S(k) S(K). Proof. If K = uk and (, ) is a solution in ositive integers to x axy+y +kx = 0, then (u, u ) is a solution in ositive integers to x axy + y + Kx = 0. We record a direct consequence of this result for future reference.
INTEGERS: 15 (015) 3 Corollary 1. For any ositive integers k one has [ S(d) S(k). d k d<k A key ingredient in our roofs is a well-known bound on the fundamental solutions to generalized Pell equations, due to Nagell, and its analogue roved by Stolt. Lemma 4. ([4, Theorem 108a]) Let N, D be ositive integers with D not a square. Suose that x 0 +y 0 D is the fundamental solution of the Pell equation X DY = 1 and the equation U DV = N () is solvable in corime integers. Then () has a solution u 0 +v 0 D with the following roerty: r y 0 N 1 0 < v 0 ale (x0 1), 0 ale u 0 ale (x 0 1)N. Lemma 5. ([6, Theorem ]) Let N, D be odd ositive integers with D not a square. Suose that the equation X DY = 4 has solutions in corime integers and let x 0 + y 0 D be the least solution in ositive integers. If the Diohantine equation U DV = 4N, gcd(u, V ), (3) is solvable, then (3) has a solution u 0 + v 0 D with the following roerty: 0 < v 0 ale y 0 N x0, 0 ale u 0 ale (x 0 )N. 3. Proof of the Main Result First of all, note that if a ale then the left side of (1) is a sum of non-negative integers for ositive x, y, so that a 6 S(k). Now we roceed to establish a tight uer bound on elements from S(k). Since the conclusion of Theorem 3 is true for k = 1 by Theorem 1, we suose below k. According to Lemma, there exists a ositive divisor d of k with the roerty that the equation c ace + e + d = 0 (4) has infinitely many solutions in corime ositive integers. Take an arbitrary a S(k).
INTEGERS: 15 (015) 4 If a is even, say, a = b, then Equation (4) is equivalent to (c be) (b 1)e = d, whose associated Pell equation X (b 1)Y = 1 has the fundamental solution (x 0, y 0 ) = (b, 1). Lemma 4 ensures the existence of a solution (c 0, e 0 ) to (4) with 1 ale e 0 ale d (b 1) = d a. This inequality imlies a ale d + ale k +. If d < k then a ale k k+5 + ale b c, while if e 0 then a ale k 4 + < b k+5 c. In the remaining case e 0 = 1 and d = k, we have to show that for a < k + one actually has a ale b k+5 c. From Equation (4), which becomes c 0 ac 0 + 1 + k = 0 in this case, we obtain a = c 0 + k + 1 c 0, so that c 0 is a (ositive) divisor of k + 1. On noting that a = k + is tantamount to c 0 {1, k + 1}, it results that a = c 0 + k + 1 ale + k + 1, c 0 whence a ale b k+5 c. If a is odd then the equation (4) is rewritten as (c ae) (a 4)e = 4d. (5) Since either none or two of the first three terms in (4) can be even, d must be odd. The least ositive solution to the equation X (a 4)Y = 4, gcd(x, Y ) = 1, is (x 0, y 0 ) = (a, 1). According to Lemma 5, the equation (5) has a solution (c 0, e 0 ) with 1 ale e 0 ale d a. As above, we conclude that in this case one also has either a = k + or a ale b k+5 c. It remains to show that the three extremal elements indicated in the statement of Theorem 3 always belong to S(k). By Corollary 1 and Theorem 1, one has 3 S(1) S(k). To rove k+ S(k), note that from the seed solution (x 0, y 0 ) = (k+1), k+1 one obtains infinitely many solutions in ositive integers to the equation x (k + )xy + y + kx = 0 by alying the transformations (x, y) 7! x, (k + )x y and (x, y) 7! (k+)y x k, y. Exlicitly, they are given by the recursion x n+1 = x n, y n+1 = (k + )x n y n, x n+ = (k + )y n+1 x n+1 k, y n+ = y n+1. From Corollary 1 and what we have just roved, it results that k + 5 = k + S(k) S(k),
INTEGERS: 15 (015) 5 so we are left to show that the equation x (t + )xy + y + (t 1)x = 0 is admissible. Indeed, it has a ositive solution (x 0, y 0 ) = (t, t) to which one can aly the transformations (x, y) 7! x, (t + )x y and (x, y) 7! (t + )y x t + 1, y. 4. Other Results Hu and Le characterize in [3] the elements from S(k) in terms of solvability of certain generalized Pell equations. Since deciding this solvability can occasionally be very costly, we give several criteria for establishing whether a given integer can or can not aear in a secific set. As a direct alication of our main theorem we have the next result. Proosition 1. For each ositive integer k one has [ d + 5, d + S(k). d k In [1] one can find a list with S(k) for k ale 33. Comarison with the sets roduced with the hel of Proosition 1 reveals that sometimes all elements of S(k) are thus roduced. For instance, one has S(4) = {3, 4, 5, 6, 8, 10, 14, 6} = [ d 4 {d + }, S(5) = {3, 5, 7, 15, 7} = [ d 5 d + 5 S(30) = {3, 4, 5, 7, 8, 10, 1, 17, 3} = [ d 30, d +, d + 5, d +. However, there are examles where additional elements o u: S(0) = {3, 4, 5, 6, 7, 10, 1, } = [ d + 5, d + [ {10}, d 0 S(3) = {3, 4, 6, 10, 14, 18, 34} = S(16) [ {14, 34} = [ d + 5, d + [ {14}, d 3 S(33) = {3, 4, 5, 7, 8, 13, 19, 35} = S(3) [ S(11) [ {19, 35} = [ d + 5, d + [ {7}. d 33 Our next results hel to decide in certain situations whether an integer is or is not in a set of interest.
INTEGERS: 15 (015) 6 Proosition. If all rime divisors of k are congruent to 1 modulo 3, then each element of S(k) is divisible by 3. Proof. Consider a S(k). From Lemma we deduce that there exists a divisor d of k such that the equation c ace + e + d = 0 is solvable in ositive integers. Since d inherits the roerties of k, any solution (c, e) consists of integers corime with 3. Reduction modulo 3 yields 1 ± a + 1 + 1 0 (mod 3), hence the desired conclusion. Perusal of the list given in [1] reveals that the hyothesis in the revious roosition cannot be weakened (see, for instance, k = 5, 6, 10, 15, 5). Proosition 3. If gcd(k, 6) > 1 then 4 S(k). Proof. If gcd(k, 6) is even, then from Theorem and Corollary 1 we get that {3, 4} = S() S(gcd(k, 6)) S(k). If gcd(k, 6) is multile of 3, then {3, 4, 5} = S(3) S(k). Proosition 4. If all rime divisors of k are congruent to 1 modulo 4, then no element of S(k) is even. Proof. We argue by contradiction. Suose that one can find a ositive integer k satisfying the hyothesis of our roosition and for which at least one element a of S(k) is even. As argued before, there exists a divisor d of k such that the equation c ace + e + d = 0 is solvable in ositive integers. Since d 1 (mod 4), by reducing this relation modulo one sees that recisely one comonent of each solution (c, e) is even. Reducing modulo 4, one obtains the contradiction 1 + 1 0 (mod 4). Proositions and 4 have the next consequence, which allows one to determine S(5) and S(7) without further exlicit comutations. Corollary. If all rime divisors of k are congruent to either 1 modulo 6 or to 5 modulo 1 then 4 6 S(k). Acknowledgement. The author is grateful to an anonymous referee whose comments have led to an imroved resentation. References [1] L. Feng, P. Yuan, Y. Hu, On the Diohantine equation x kxy + y + lx = 0, Integers 13 (013), #A8, 8.
INTEGERS: 15 (015) 7 [] A. Marlewski, P. Zarzycki, Infinitely many solutions of the Diohantine equation x kxy + y + x = 0, Comut. Math. Al. 47 (004), 115 11. [3] Y. Hu, M. Le, On the Diohantine equation x kxy + y + lx = 0, Chin. Ann. Math. Ser. B 34 (013), 715 718. [4] T. Nagell, Introduction to Number Theory, Chelsea, 1981. [5] J. C. Owings, Jr., Solution of the system a + 1 0 (mod b) and b + 1 0 (mod a), Fibonacci Quart. 5 (1987), 45 49. [6] B. Stolt, On the Diohantine equation u Dv = ±4N, Ark. Mat. (195), 1 3. [7] P. Yuan, Y. Hu, On the Diohantine equation x kxy + y + lx = 0, l {1,, 4}, Comut. Math. Al. 61 (011), 573 577.