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Complete solutions of the simultaneous Pell equations x2−24y2=1 and y2−pz2=1
Journal of Number Theory 147 (2015) 103–108
Contents lists available at ScienceDirect
Journal of Number Theory www.elsevier.com/locate/jnt
Complete solutions of the simultaneous Pell equations x2 − 24y 2 = 1 and y 2 − pz 2 = 1 Xiaochuan Ai a,c , Jianhua Chen a , Silan Zhang b,∗ , Hao Hu a a b c
School of Mathematics and Statistics, Wuhan University, Wuhan, 430068, China College of Science, Huazhong Agricultural University, Wuhan, 430070, China School of Science, Navy University of Engineering, Wuhan, 430033, China
a r t i c l e
i n f o
Article history: Received 13 February 2014 Received in revised form 21 July 2014 Accepted 21 July 2014 Available online 6 September 2014 Communicated by David Goss MSC: 11D61
a b s t r a c t We prove that the simultaneous Pell equations
x2 − 24y 2 = 1 y 2 − pz 2 = 1
,
where p is a prime, have positive integer solutions only in the cases of p = 11 and p = 2. Furthermore, the only solutions are (x, y, z, p) = (49, 10, 3, 11) and (x, y, z, p) = (485, 99, 70, 2). © 2014 Elsevier Inc. All rights reserved.
Keywords: Diophantine equations Pell equation Simultaneous Pell equations
1. Introduction According to M.A. Bennett’s results, i.e., Theorem 7.1 [2], the simultaneous Diophantine equations x2 − ay 2 = 1 , (1) y 2 − bz 2 = 1 * Corresponding author. E-mail address: [email protected] (S. Zhang). http://dx.doi.org/10.1016/j.jnt.2014.07.009 0022-314X/© 2014 Elsevier Inc. All rights reserved.
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X. Ai et al. / Journal of Number Theory 147 (2015) 103–108
where a and b are distinct non-zero integers, possess at most three positive integer solutions (x, y, z). Using Ljunggren’s result and certain results on primitive prime factors of Lucas sequences, P. Yuan [6] showed that (1) has at most one positive integer solution (x, y, z) for a = 4m(m + 1). Likewise, M. Cipu [4] obtained the same result in the case of a = 4m2 − 1. Based on these results, P. Yuan extended a conjecture: Yuan’s conjecture. (See Conjecture 1.1 in [6].) For any nonzero positive integers a and b, (1) possesses at most one positive integer solution (x, y, z). A substantial improvement was provided by M. Cipu and M. Mignotte [4], who furthered Bennett’s result and demonstrated that (1) has at most two positive integer solutions (x, y, z). Moreover, these authors restated Yuan’s conjecture, making it an attractive problem. In this paper, we completely solve the simultaneous Pell equations 2 x − 24y 2 = 1 , (2) y 2 − pz 2 = 1 where p is a prime, hence confirming Yuan’s conjecture for a class of coefficients. The main result of this paper is as follows: Theorem 1.1. When p is a prime, the simultaneous Pell equations (2) have positive integer solutions only in the case of p = 11 and p = 2. Furthermore, (x, y, z, p) = (49, 10, 3, 11) and (x, y, z, p) = (485, 99, 70, 2) are the only solutions. 2. Lemmas The first equation of (2), x2 − 24y 2 = 1,
(3)
√ has a fundamental solution denoted by ε1 = 5 + 24. All positive integer solutions (x, y) of (3) are given by m εm 1 + ε1 , 2 εm − εm = 1√ 1 , 2 24
x = xm = y = ym
for m = 1, 2, 3... . All the positive integer solutions (x, y, z) of the simultaneous Pell equations (2) satisfy x = xm = where m ∈ Z+ .
m εm 1 + ε1 , 2
y = ym =
m εm 1 − ε1 √ , 2 24
X. Ai et al. / Journal of Number Theory 147 (2015) 103–108
105
We present a few lemmas, that will be used in the proof of the theorem. Lemma 2.1. (See J.H.E. Cohn [5].) Let D be a positive integer that is not a square, the √ fundamental solution of the equation v 2 − Du2 = 1 be a + b D. Then the only possible solutions of the equation x4 − Dy 2 = 1 are given by x2 = a and x2 = 2a2 − 1. Both of these solutions exist only in one case, D = 1785. Lemma 2.2. (See J.H. Chen [3].) Let a > 1 and B > 0 be positive integers that are √ square-free. Suppose ε = u + v B > 1 is the fundamental solution of the Pell equation x2 − By 2 = 1. Define √ εn = un + vn B,
n = 1, 2, ... .
If a2 x4 − By 2 = 1 is solvable, then it has at most one solution (x, y) in positive integers, √ and then ax2 + v B = εt , where t is the least positive integer such that ut ≡ 0 (mod a). Lemma 2.3. The equations x2m − 1 ≡ 0 (mod 2), x2m + 2 ≡ 0 (mod 3) hold for all positive integers m. 2 Proof. As x2m − 24ym = 1, modulo 2, we have x2m ≡ 1 (mod 2). Similarly, modulo 3, we 2 obtain xm ≡ 1 (mod 3). Thus, Lemma 2.3 is proven. 2
Lemma 2.4. If (xm , ym , z) is a positive integer solution of (2), where p is prime and m is even, then m = 2, (xm , ym , z) = (49, 10, 3) and p = 11. Proof. Let m = 2k. As
x22k −1 24
2 2 = y2k and y2k − 1 = pz 2 , we have
x22k − 25 = pz 2 , 24 where x22k − 25 = (x2k − 5)(x2k + 5) and x2k = 2x2k − 1. Direct substitution leads to (x2k − 3)(x2k + 2) = pz 2 . 6
(4)
By Lemma 2.3, we rewrite (4) as x2k − 3 x2k + 2 · = pz 2 , 2 3 where
x2k −3 2
and
x2k +2 3
(5)
are integers. x2 −3 x2 +2
The greatest common divisor, d = gcd( k2 , k3 ) is 1 or 5. If d = 5, then x2k ≡ 3 (mod 5), and this equation does not hold, as 3 is a quadratic non-residue modulo 5.
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X. Ai et al. / Journal of Number Theory 147 (2015) 103–108
When d = 1, factoring (5), we have ⎧ 2 x −3 ⎪ ⎪ ⎨ k = z12 2 2 ⎪ ⎪ ⎩ xk + 2 = pz22 3
(6)
⎧ 2 x −3 ⎪ ⎪ ⎨ k = pz12 2 , 2 ⎪ x + 2 ⎪ 2 k ⎩ = z2 3
(7)
or
where z = z1 z2 . From (6), we have x2k − 3 = 2z12 . This equation does not hold, as 2 is a quadratic non-residue modulo 3. Hence, (6) does not hold. When p is even, the first equation of (7) x2k −3 = 4z12 does not hold, as 3 is a quadratic non-residue modulo 4. When p is odd, from x2k − 24yk2 = 1 and the second equation x2k + 2 = 3z22 of (7), we have z22 − 8yk2 = 1.
(8)
From Anglin’s work, i.e., Theorem 4 [1], the simultaneous Pell equations
x2k − 24yk2 = 1
(9)
z22 − 8yk2 = 1
√ √ (5+ 24)4 −(5− 24)4 √ = 980 > 120, 2 24 √ √ (5+ 24)3 −(5− 24)3 √ 10, y3 = = 99 2 24
have no positive integer solutions with yk > 120. As y4 = √
√
24) substituting y0 = 0, y1 = 1, y2 = (5+ 24)2√−(5− = 24 into (9), we know that the positive integer solutions of (9) are 2
2
(xk , yk , z2 ) = (1, 0, 1), (5, 1, 3). Substituting (xk , z2 ) = (1, 1), (5, 3) into (7), we find that (xk , z2 ) = (1, 1) leads to no solution and that (xk , z2 ) = (5, 3) leads to z1 = 1, p = 11 and k = 1. Thus, the solution of (2) is (x, y, z) = (x2 , y2 , z) = (49, 10, 3) for p = 11. Thus, Lemma 2.4 is proven. 2 Lemma 2.5. If (2) has a positive integer solution (xm , ym , z), where p is prime and m is odd, then m = 3, p = 2 and (xm , ym , z) = (485, 99, 2).
X. Ai et al. / Journal of Number Theory 147 (2015) 103–108
107
Proof. Denote m = 2k + 1 and suppose (xm , ym , z) is a positive integer solution of (2). 2 We rewrite y2k+1 − 1 = pz 2 as (y2k+1 + 1)(y2k+1 − 1) = pz 2 . Noting that ⎧ k+1 k ε2k+1 − ε2k+1 ε1 − εk+1 ε1 + εk1 ⎪ 1 1 1 ⎪ ⎪ y2k+1 + 1 = = 2xk yk+1 +1=2 ⎨ ε1 − ε1 2 ε1 − ε1 , k k+1 ⎪ ε2k+1 − ε2k+1 ε1 − εk1 ε1 + εk+1 ⎪ 1 1 1 ⎪ = 2xk+1 yk −1=2 ⎩ y2k+1 − 1 = ε1 − ε1 2 ε1 − ε1
(10)
+1 y2k+1 −1 and as gcd( y2k+1 , ) = 1, we have 2 2
xk yk+1 = pz12
(11)
xk+1 yk = z22 or
xk yk+1 = z12 xk+1 yk = pz22
,
(12)
where z = 2z1 z2 . √ √ From εk+1 = εk1 ε1 = (xk + yk 24)(5 + 24), we know 1 xk+1 = 5xk + 24yk ,
yk+1 = 5yk + xk .
As gcd(xk , yk ) = 1, we have gcd(xk+1 , yk ) = 1 or 5 and gcd(xk , yk+1 ) = 1 or 5. If gcd(xk+1 , yk ) = 1 and (11) holds, then xk+1 = u2 and hence, by Lemma 2.1, k = 1, p = 2. If gcd(xk+1 , yk ) = 5 and (11) holds, then xk+1 = 5u2 and hence, by Lemma 2.2, k = −1 and no corresponding positive integer solutions of (2) exist. If gcd(xk , yk+1 ) = 1 and (12) holds, then xk = u2 and again by Lemma 2.1, k = 2 and x2 y3 = 49 ×99 which is not a square. If gcd(xk , yk+1 ) = 5 and (12) holds, then xk = 5u2 and again by Lemma 2.2, k = 0, and no corresponding integer positive solutions of (2) exist. Consequently, in this case the only positive integer solution of (2) is (x, y, z) = (x3 , y3 , z) = (485, 99, 2) for p = 2. Thus, Lemma 2.5 is proven. 2 3. Proof of Theorem 1.1 All the positive integer solutions (x, y, z) of the simultaneous Pell equations (2) satisfy x = xm =
m εm 1 + ε1 , 2
y = ym =
m εm 1 − ε1 √ , 2 24
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X. Ai et al. / Journal of Number Theory 147 (2015) 103–108
where m ∈ Z+ . When m is even, we have m = 2, (xm , ym , z) = (49, 10, 3) and p = 11 by Lemma 2.4; when m is odd, we have m = 3, (xm , ym , z) = (485, 99, 2) and p = 2 by Lemma 2.5. Hence the theorem follows. Acknowledgments The authors express their gratitude to the anonymous reviewer for the instructive suggestions. They also thank H.L. Zhu, P. Abigail and M. Laura for a careful reading of the manuscript and help in improving the English. This research is partly supported by the Fundamental Research Funds for the Central Universities (No. 2662014QC010) and NSF grants to Navy University of Engineering (No. HGDQNEQJJ13001). References [1] W.S. Anglin, Simultaneous Pell equations, Math. Comp. 65 (1996) 355–359. [2] M.A. Bennett, On the number of solutions of simultaneous Pell equations, J. Reine Angew. Math. 498 (1998) 173–199. [3] J.H. Chen, A note on the Diophantine equation a2 x4 − By 2 = 1, Acta Arith. 3 (2001) 205–212. [4] M. Cipu, M. Mignotte, On the number of solutions to systems of Pell equations, J. Number Theory 125 (2007) 356–392. [5] J.H.E. Cohn, The Diophantine equation x4 − Dy 2 = 1, II, Acta Arith. 78 (4) (1997) 401–403. [6] P. Yuan, On the number of solutions of x2 − 4m(m + 1)y 2 = y 2 − byz 2 = 1, Proc. Amer. Math. Soc. 132 (2004) 1561–1566.
Contents lists available at ScienceDirect
Journal of Number Theory www.elsevier.com/locate/jnt
Complete solutions of the simultaneous Pell equations x2 − 24y 2 = 1 and y 2 − pz 2 = 1 Xiaochuan Ai a,c , Jianhua Chen a , Silan Zhang b,∗ , Hao Hu a a b c
School of Mathematics and Statistics, Wuhan University, Wuhan, 430068, China College of Science, Huazhong Agricultural University, Wuhan, 430070, China School of Science, Navy University of Engineering, Wuhan, 430033, China
a r t i c l e
i n f o
Article history: Received 13 February 2014 Received in revised form 21 July 2014 Accepted 21 July 2014 Available online 6 September 2014 Communicated by David Goss MSC: 11D61
a b s t r a c t We prove that the simultaneous Pell equations
x2 − 24y 2 = 1 y 2 − pz 2 = 1
,
where p is a prime, have positive integer solutions only in the cases of p = 11 and p = 2. Furthermore, the only solutions are (x, y, z, p) = (49, 10, 3, 11) and (x, y, z, p) = (485, 99, 70, 2). © 2014 Elsevier Inc. All rights reserved.
Keywords: Diophantine equations Pell equation Simultaneous Pell equations
1. Introduction According to M.A. Bennett’s results, i.e., Theorem 7.1 [2], the simultaneous Diophantine equations x2 − ay 2 = 1 , (1) y 2 − bz 2 = 1 * Corresponding author. E-mail address: [email protected] (S. Zhang). http://dx.doi.org/10.1016/j.jnt.2014.07.009 0022-314X/© 2014 Elsevier Inc. All rights reserved.
104
X. Ai et al. / Journal of Number Theory 147 (2015) 103–108
where a and b are distinct non-zero integers, possess at most three positive integer solutions (x, y, z). Using Ljunggren’s result and certain results on primitive prime factors of Lucas sequences, P. Yuan [6] showed that (1) has at most one positive integer solution (x, y, z) for a = 4m(m + 1). Likewise, M. Cipu [4] obtained the same result in the case of a = 4m2 − 1. Based on these results, P. Yuan extended a conjecture: Yuan’s conjecture. (See Conjecture 1.1 in [6].) For any nonzero positive integers a and b, (1) possesses at most one positive integer solution (x, y, z). A substantial improvement was provided by M. Cipu and M. Mignotte [4], who furthered Bennett’s result and demonstrated that (1) has at most two positive integer solutions (x, y, z). Moreover, these authors restated Yuan’s conjecture, making it an attractive problem. In this paper, we completely solve the simultaneous Pell equations 2 x − 24y 2 = 1 , (2) y 2 − pz 2 = 1 where p is a prime, hence confirming Yuan’s conjecture for a class of coefficients. The main result of this paper is as follows: Theorem 1.1. When p is a prime, the simultaneous Pell equations (2) have positive integer solutions only in the case of p = 11 and p = 2. Furthermore, (x, y, z, p) = (49, 10, 3, 11) and (x, y, z, p) = (485, 99, 70, 2) are the only solutions. 2. Lemmas The first equation of (2), x2 − 24y 2 = 1,
(3)
√ has a fundamental solution denoted by ε1 = 5 + 24. All positive integer solutions (x, y) of (3) are given by m εm 1 + ε1 , 2 εm − εm = 1√ 1 , 2 24
x = xm = y = ym
for m = 1, 2, 3... . All the positive integer solutions (x, y, z) of the simultaneous Pell equations (2) satisfy x = xm = where m ∈ Z+ .
m εm 1 + ε1 , 2
y = ym =
m εm 1 − ε1 √ , 2 24
X. Ai et al. / Journal of Number Theory 147 (2015) 103–108
105
We present a few lemmas, that will be used in the proof of the theorem. Lemma 2.1. (See J.H.E. Cohn [5].) Let D be a positive integer that is not a square, the √ fundamental solution of the equation v 2 − Du2 = 1 be a + b D. Then the only possible solutions of the equation x4 − Dy 2 = 1 are given by x2 = a and x2 = 2a2 − 1. Both of these solutions exist only in one case, D = 1785. Lemma 2.2. (See J.H. Chen [3].) Let a > 1 and B > 0 be positive integers that are √ square-free. Suppose ε = u + v B > 1 is the fundamental solution of the Pell equation x2 − By 2 = 1. Define √ εn = un + vn B,
n = 1, 2, ... .
If a2 x4 − By 2 = 1 is solvable, then it has at most one solution (x, y) in positive integers, √ and then ax2 + v B = εt , where t is the least positive integer such that ut ≡ 0 (mod a). Lemma 2.3. The equations x2m − 1 ≡ 0 (mod 2), x2m + 2 ≡ 0 (mod 3) hold for all positive integers m. 2 Proof. As x2m − 24ym = 1, modulo 2, we have x2m ≡ 1 (mod 2). Similarly, modulo 3, we 2 obtain xm ≡ 1 (mod 3). Thus, Lemma 2.3 is proven. 2
Lemma 2.4. If (xm , ym , z) is a positive integer solution of (2), where p is prime and m is even, then m = 2, (xm , ym , z) = (49, 10, 3) and p = 11. Proof. Let m = 2k. As
x22k −1 24
2 2 = y2k and y2k − 1 = pz 2 , we have
x22k − 25 = pz 2 , 24 where x22k − 25 = (x2k − 5)(x2k + 5) and x2k = 2x2k − 1. Direct substitution leads to (x2k − 3)(x2k + 2) = pz 2 . 6
(4)
By Lemma 2.3, we rewrite (4) as x2k − 3 x2k + 2 · = pz 2 , 2 3 where
x2k −3 2
and
x2k +2 3
(5)
are integers. x2 −3 x2 +2
The greatest common divisor, d = gcd( k2 , k3 ) is 1 or 5. If d = 5, then x2k ≡ 3 (mod 5), and this equation does not hold, as 3 is a quadratic non-residue modulo 5.
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X. Ai et al. / Journal of Number Theory 147 (2015) 103–108
When d = 1, factoring (5), we have ⎧ 2 x −3 ⎪ ⎪ ⎨ k = z12 2 2 ⎪ ⎪ ⎩ xk + 2 = pz22 3
(6)
⎧ 2 x −3 ⎪ ⎪ ⎨ k = pz12 2 , 2 ⎪ x + 2 ⎪ 2 k ⎩ = z2 3
(7)
or
where z = z1 z2 . From (6), we have x2k − 3 = 2z12 . This equation does not hold, as 2 is a quadratic non-residue modulo 3. Hence, (6) does not hold. When p is even, the first equation of (7) x2k −3 = 4z12 does not hold, as 3 is a quadratic non-residue modulo 4. When p is odd, from x2k − 24yk2 = 1 and the second equation x2k + 2 = 3z22 of (7), we have z22 − 8yk2 = 1.
(8)
From Anglin’s work, i.e., Theorem 4 [1], the simultaneous Pell equations
x2k − 24yk2 = 1
(9)
z22 − 8yk2 = 1
√ √ (5+ 24)4 −(5− 24)4 √ = 980 > 120, 2 24 √ √ (5+ 24)3 −(5− 24)3 √ 10, y3 = = 99 2 24
have no positive integer solutions with yk > 120. As y4 = √
√
24) substituting y0 = 0, y1 = 1, y2 = (5+ 24)2√−(5− = 24 into (9), we know that the positive integer solutions of (9) are 2
2
(xk , yk , z2 ) = (1, 0, 1), (5, 1, 3). Substituting (xk , z2 ) = (1, 1), (5, 3) into (7), we find that (xk , z2 ) = (1, 1) leads to no solution and that (xk , z2 ) = (5, 3) leads to z1 = 1, p = 11 and k = 1. Thus, the solution of (2) is (x, y, z) = (x2 , y2 , z) = (49, 10, 3) for p = 11. Thus, Lemma 2.4 is proven. 2 Lemma 2.5. If (2) has a positive integer solution (xm , ym , z), where p is prime and m is odd, then m = 3, p = 2 and (xm , ym , z) = (485, 99, 2).
X. Ai et al. / Journal of Number Theory 147 (2015) 103–108
107
Proof. Denote m = 2k + 1 and suppose (xm , ym , z) is a positive integer solution of (2). 2 We rewrite y2k+1 − 1 = pz 2 as (y2k+1 + 1)(y2k+1 − 1) = pz 2 . Noting that ⎧ k+1 k ε2k+1 − ε2k+1 ε1 − εk+1 ε1 + εk1 ⎪ 1 1 1 ⎪ ⎪ y2k+1 + 1 = = 2xk yk+1 +1=2 ⎨ ε1 − ε1 2 ε1 − ε1 , k k+1 ⎪ ε2k+1 − ε2k+1 ε1 − εk1 ε1 + εk+1 ⎪ 1 1 1 ⎪ = 2xk+1 yk −1=2 ⎩ y2k+1 − 1 = ε1 − ε1 2 ε1 − ε1
(10)
+1 y2k+1 −1 and as gcd( y2k+1 , ) = 1, we have 2 2
xk yk+1 = pz12
(11)
xk+1 yk = z22 or
xk yk+1 = z12 xk+1 yk = pz22
,
(12)
where z = 2z1 z2 . √ √ From εk+1 = εk1 ε1 = (xk + yk 24)(5 + 24), we know 1 xk+1 = 5xk + 24yk ,
yk+1 = 5yk + xk .
As gcd(xk , yk ) = 1, we have gcd(xk+1 , yk ) = 1 or 5 and gcd(xk , yk+1 ) = 1 or 5. If gcd(xk+1 , yk ) = 1 and (11) holds, then xk+1 = u2 and hence, by Lemma 2.1, k = 1, p = 2. If gcd(xk+1 , yk ) = 5 and (11) holds, then xk+1 = 5u2 and hence, by Lemma 2.2, k = −1 and no corresponding positive integer solutions of (2) exist. If gcd(xk , yk+1 ) = 1 and (12) holds, then xk = u2 and again by Lemma 2.1, k = 2 and x2 y3 = 49 ×99 which is not a square. If gcd(xk , yk+1 ) = 5 and (12) holds, then xk = 5u2 and again by Lemma 2.2, k = 0, and no corresponding integer positive solutions of (2) exist. Consequently, in this case the only positive integer solution of (2) is (x, y, z) = (x3 , y3 , z) = (485, 99, 2) for p = 2. Thus, Lemma 2.5 is proven. 2 3. Proof of Theorem 1.1 All the positive integer solutions (x, y, z) of the simultaneous Pell equations (2) satisfy x = xm =
m εm 1 + ε1 , 2
y = ym =
m εm 1 − ε1 √ , 2 24
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where m ∈ Z+ . When m is even, we have m = 2, (xm , ym , z) = (49, 10, 3) and p = 11 by Lemma 2.4; when m is odd, we have m = 3, (xm , ym , z) = (485, 99, 2) and p = 2 by Lemma 2.5. Hence the theorem follows. Acknowledgments The authors express their gratitude to the anonymous reviewer for the instructive suggestions. They also thank H.L. Zhu, P. Abigail and M. Laura for a careful reading of the manuscript and help in improving the English. This research is partly supported by the Fundamental Research Funds for the Central Universities (No. 2662014QC010) and NSF grants to Navy University of Engineering (No. HGDQNEQJJ13001). References [1] W.S. Anglin, Simultaneous Pell equations, Math. Comp. 65 (1996) 355–359. [2] M.A. Bennett, On the number of solutions of simultaneous Pell equations, J. Reine Angew. Math. 498 (1998) 173–199. [3] J.H. Chen, A note on the Diophantine equation a2 x4 − By 2 = 1, Acta Arith. 3 (2001) 205–212. [4] M. Cipu, M. Mignotte, On the number of solutions to systems of Pell equations, J. Number Theory 125 (2007) 356–392. [5] J.H.E. Cohn, The Diophantine equation x4 − Dy 2 = 1, II, Acta Arith. 78 (4) (1997) 401–403. [6] P. Yuan, On the number of solutions of x2 − 4m(m + 1)y 2 = y 2 − byz 2 = 1, Proc. Amer. Math. Soc. 132 (2004) 1561–1566.