A note on the simultaneous Pell equations $$x^2-(a^2-1)y^2=1$$ x 2 - ( a 2 - 1 ) y 2 = 1 and $$y^2-bz^2=1$$
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A note on the simultaneous Pell equations x 2 − (a2 − 1)y 2 = 1 and y 2 − bz 2 = 1 Xing-Wang Jiang1
© Akadémiai Kiadó, Budapest, Hungary 2020
Abstract Let a > 1, b be two positive integers where the square-free part of b is 2 pq with p, q two distinct odd primes. Recently, Cipu (Proc Am Math Soc 146:983–992, 2018) proved that if one of the following conditions holds: (i) 2a 2 − 1 is not a perfect square, (ii) { p (mod 8), q (mod 8)} = {1, 3}, then the equations x 2 − (a 2 − 1)y 2 = 1 and y 2 − bz 2 = 1 2 − 1)/b is a perfect square. When it have solutions in positive integers if and only if 8a 2 (2a 3 2 exists, the solution is (x, y, z) = (4a − 3a, 4a − 1, 8a 2 (2a 2 − 1)/b). In this paper, we completely solve these equations when the square-free part of b is 2 pq.
Keywords Simultaneous Pell equations · Diophantine equations · Pell equations Mathematics Subject Classification 11D09 · 11D25 · 11D45
1 Introduction Let D1 and D2 be two positive integers. In this paper, we study positive integer solutions (x, y, z) of the simultaneous Pell equations x 2 − D1 y 2 = 1,
y 2 − D2 z 2 = 1.
(1.1)
In 1998, Bennett [2] proved that (1.1) has at most three solutions. In 2004, Yuan [10] conjectured that (1.1) possesses at most one solution for any positive integers D1 and D2 , and he proved that it is true for D1 = 4m(m + 1). Afterward, Cipu and Mignotte [5] and He [7] proved that (1.1) has at most two solutions, independently. In 2007, Cipu [3] proved that the conjecture of Yuan [10] holds for D1 = 4m 2 − 1.
B 1
Xing-Wang Jiang [email protected] School of Mathematical Sciences and Institute of Mathematics, Nanjing Normal University, Nanjing 210023, People’s Republic of China
123
X.-W. Jiang
Clearly, the systems of equations considered by Cipu [3] and Yuan [10] are of the following form x 2 − (a 2 − 1)y 2 = 1,
y 2 − bz 2 = 1,
(1.2)
where a > 1, b are two positive integers. In 2018, Cipu [4] gave the explicit formula for the solution of (1.2) when the square-free part of b, denoted by sqf(b), has at most two prime divisors. Cipu [4] also dealt with the case sqf(b) = 2 pq, where p, q are two distinct odd primes. He proved that Theorem 1.1 [4, Theorem 4.1 (b)] Let sqf(b) = 2 pq with p, q two distinct odd primes. If one of the following conditions holds: (i) there is no integer c such that 2a 2 − 1 = c2 , (ii) { p (mod 8), q (mod 8)} = {1, 3}, then (1.2) is solvable in positive integers if and only if 8a 2 (2a 2 − 1)/b is a perfect square. When it exists, this solution is ⎛ ⎞ 2 2 8a (2a − 1) ⎠ (x, y, z) = ⎝4a 3 − 3a, 4a 2 − 1, . b For more related results, see [1,8,9]. In this paper, we completely solve the system of equations (1.2) for sqf(b) = 2 pq. Our result reads as follows. Theorem 1.2 Let sqf(b) = 2 pq with p, q two distinct odd primes. Then (1.2) has solutions in positive integers if and only if either 8a 2 (2a 2 − 1)/b or 8a 2 (2a 2 − 1)(4a 2 − 3)(4a 2 − 1)/b is a perfect square. Furthermore, the unique solution is ⎛ ⎞ 2 (2a 2 − 1) 8a 3 2 ⎠ (x, y, z) = ⎝4a − 3a, 4a − 1, b or
⎛
(x, y, z) = ⎝16a 5
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