Pure Mathematical Sciences, Vol. 5, 2016, no. 1, 27 - 32
HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/pms.2016.51117
A Note on the Simultaneous Pell Equations
x2 − 24y 2 = 1 and y 2 − pz 2 = 1
Liqun Tao
Mathematical Sciences, Luoyang Normal University
Luoyang, Henan, 471022, P. R. China
c 2015 Liqun Tao. This article is distributed under the Creative Commons
Copyright Attribution License, which permits unrestricted use, distribution, and reproduction in any
medium, provided the original work is properly cited.
Abstract
We give an alternative solution to the simultaneous Pell equations
x2 − 24y 2 = 1
,
y 2 − pz 2 = 1
where p is a prime, which has been solved by Ai et al. in [1].
Mathematics Subject Classification: 11D61
Keywords: Diophantine equations, Simultaneous Pell equations
1
Introduction
Many authors have considered the solution of simultaneous Pell equations.
Among the constant appearance of vast amount of literature on this topic, the
following conjecture proposed by Yuan is particularly interesting:
Yuan’s Conjecture [9]. For any distinct nonzero integers a, b, the system of
simultaneous Diophantine equations
x2 − ay 2 = 1
,
y 2 − bz 2 = 1
28
Liqun Tao
has at most one positive integer solution (x, y, z).
This conjecture is based on Bennett’s result [3] that the above system of simultaneous equations has at most three positive integer solutions (x, y, z), and
Yuan’s own result [9] that this system of equations has at most one positive
integer solution for a = 4m(m + 1). Subsequently Cipu and Mignotte’s contribution [5] in the case a = 4m2 − 1 makes this conjecture more credible.
Towards this direction, recently Ai et al. proved in [1] the following theorem:
Theorem 1.1 The system of simultaneous equations
2
x − 24y 2 = 1
,
y 2 − pz 2 = 1
where p is a prime, has at most one integer solution. In fact the only solutions
are (x, y, z) = (485, 99, 70) for p = 2 and (x, y, z) = (49, 10, 3) for p = 11.
Thus they confirmed Yuan’s conjecture for a family of simultaneous equations.
The aim of this note is to give an elementary proof independent of Anglin’s
work [2], which is needed in the proof of the above theorem in [1]. Actually
this method has been used in [7] to simplify the proof in [8].
2
Proof of the theorem
During the proof of the above theorem, we need the following lemmas:
Lemma 2.1 (Cohn [6]) Let D be a positive integer that √
is not a square, the
2
2
fundamental solution of the equation v − Du = 1 be a + b D. Then the only
possible integer 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 (Bumby [4]) The only positive integers of 3x4 − 2y 2 = 1 are
given by x = 1, 3.
Proof of Theorem 1.1. Since x2 − 24y 2 = 1 = y 2 − pz 2 , we have the
equivalent system of simultaneous equations
2
x − 24y 2 = 1
,
x2 + pz 2 = 25y 2
From 25y 2 − x2 = pz 2 , we have (5y + x)(5y − x) = pz 2 . Let d = (5y +
x, 5y − x). Thus d|2(5y, x). Since (x, y) = 1, we see that if 5 - x, d = 1, 2 and
5y−x
if 5|x, d = 5, 10. From 5y+x
= p( dz )2 , we have
d
d
A note on the simultaneous Pell equations ...
5y+x
d
5y−x
d
= pu2
,
= v2
5y+x
d
5y−x
d
= v2
,
= pu2
29
or
where
z
d
= uv. So in both cases we have
(
2
2
x = ±d pu 2−v
.
2
2
y = d pu 10+v
Substituting x, y into x2 − 24y 2 = 1 and simplifying, we obtain
)2 ,
(pu2 − 49v 2 )2 − 2400v 4 = ( 10
d
where d = 1, 2, 5, 10.
2
2
−49v )
Let w = | d(pu 10
|. It suffices to consider w2 − 24d2 v 4 = 1 in four cases.
(i) Case d = 1:
From w2 − 24v 4 = 1, we have
w+1 w−1
2
2
= 6v 4 . Thus
w+1
2
w−1
2
= 6s4
,
= t4
w+1
2
w−1
2
= t4
,
= 6s4
w+1
2
w−1
2
= 3s4
,
= 2t4
w+1
2
w−1
2
= 2t4
,
= 3s4
or
or
or
where v = st. Thus we have 6s4 − t4 = ±1, or 3s4 − 2t4 = ±1. By modulo 3
argument, we can only have 6s4 − t4 = −1, or 3s4 − 2t4 = 1. By Lemma 2.1,
the solution of 6s4 −t4 = −1 is given by t = 7, s2 = 20, which is impossible. By
Lemma 2.2, 3s4 − 2t4 = 1 has only one positive integer solution (s, t) = (1, 1).
Thus v = st = 1, w = 5 and hence from |pu2 − 49| = 50, we have pu2 = 99, so
p = 11, u = 3. Now z = uv = 3, so from y 2 − pz 2 = 1, we obtain y = 10 and
from x2 − 24y 2 = 1, we obtain x = 49. Therefore we get one positive integer
30
Liqun Tao
solution (x, y, z) = (49, 10, 3), and in this case p = 11.
(ii) Case d = 2:
Similarly as Case (i), from w2 − 96v 4 = 1, we have 24s4 − t4 = ±1, or
8s4 − 3t4 = ±1. By modulo 8 argument, 8s4 − 3t4 = ±1 is impossible; by
modulo 3 argument, we can only have 24s4 − t4 = −1. But by Lemma 2.1, the
solution of this equation is given by t = 7, 2s2 = 20, which is impossible.
(iii) Case d = 5:
Similarly as above, from w2 − 600v 4 = 1, we have 150s4 − t4 = ±1, 75s4 −
4
2t = ±1, 50s4 − 3t4 = ±1, or 25s4 − 6t4 = ±1. By modulo 3 argument, we can
only have 150s4 − t4 = −1, 75s4 − 2t4 = 1, 50s4 − 3t4 = −1, or 25s4 − 6t4 = 1.
By Lemma 2.1, the solution of 150s4 − t4 = −1 is given by t = 7, 5s2 = 20,
hence v = st = 14 and w = 4801, thus p = 2, u = 1, z = 70, y = 99, x = 485.
So we obtain one positive integer solution (x, y, z) = (485, 99, 70), and in this
) = ( 35 ) = −1, 75s4 − 2t4 = 1 and
case p = 2. As the Legendre symbol ( −2
5
50s4 −3t4 = −1 are impossible. For the equation 25s4 −6t4 = 1, letting c = 5s2 ,
we may turn to consider c2 − 6t4 = 1. Since c is odd, 6t4 ≡ c2 − 1 ≡ 0 mod 8,
hence d is even. So let d = 2e, it suffices to solve the equation c2 − 96e4 = 1,
which has been treated in Case (ii). This gives no positive integer solutions in
(iii).
(iv) Case d = 10:
Similarly again as above, from w2 − 2400v 4 = 1, we have 600s4 − t4 =
±1, 200s4 − 3t4 = ±1, 75s4 − 8t4 = ±1, or 25s4 − 24t4 = ±1, where st = v.
By modulo 8 argument, 200s4 − 3t4 = ±1, 75s4 − 8t4 = 1 are impossible. By
modulo 3 argument, we can only have 600s4 − t4 = −1, 25s4 − 24t4 = 1. By
Lemma 2.1, the solution of 600s4 − t4 = −1 is given by t = 7, 10s2 = 20,
which is impossible. By modulo 3 argument, we can only have 25s4 − 24t4 = 1.
This equation (5s2 )2 − 24t4 = 1 has been treated in Case (i). We know the
solution is given by 5s2 = 5, t = 1. Hence s = 1, v = st = 1 and therefore from
w2 − 2400v 4 = 1, we have w = 49. Finally from |pu2 − 49| = 49, we obtain
p = 2, u = 7. Now z = duv = 70, so from y 2 − pz 2 = 1 and x2 − 24y 2 = 1,
we have one positive integer solution (x, y, z) = (485, 99, 70), and in this case
p = 2.
This completes the proof of the theorem.
Remark: Using the same method, we have also considered the system of
simultaneous Pell equations
2
x − 24y 2 = 1
,
y 2 − 2pz 2 = 1
A note on the simultaneous Pell equations ...
31
where p is an odd prime and the simultaneous equations
2
x − 24y 2 = 1
,
y 2 − 3pz 2 = 1
where p > 3 is a prime, and we find that in both cases there is no positive
integer solution. Next step we naturally turn to consider the following general
system of simultaneous Pell equations
2
x − 24y 2 = 1
,
y 2 − pqz 2 = 1
where p, q > 3 are distinct primes. However we prefer to discuss them in another place.
Acknowledgement. This work was partially supported by the Natural Science Research Project of the Education Department of Henan Province (Grant:
2011B110022).
References
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simultaneous Pell equations x2 − 24y 2 = 1 and y 2 − pz 2 = 1, J. Number
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[2] W.S. Anglin, Simultaneous Pell equations, Math. Comp., 65 (1996), 355359. http://dx.doi.org/10.1090/s0025-5718-96-00687-4
[3] M.A. Bennett, On the number of solutions of simultaneous Pell equations,
J. Reine. Angew. Math., 1998 (1998), no. 498, 173-199.
http://dx.doi.org/10.1515/crll.1998.049
[4] R.T. Bumby, The diophantine equation 3x4 − 2y 2 = 1, Math. Scan., 21
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[5] M. Cipu and M. Mignotte, On the number of solutions to systems of Pell
equations, J. Number Theory, 125 (2007), 356-392.
http://dx.doi.org/10.1016/j.jnt.2006.09.016
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(1984), 356-359. http://dx.doi.org/10.1016/0022-314x(84)90068-4
32
Liqun Tao
[9] P. Yuan, On the number of solutions of x2 − 4m(m + 1)y 2 = y 2 − bz 2 = 1,
Proc. Amer. Math. Soc., 132 (2004), 1561-1566.
http://dx.doi.org/10.1090/s0002-9939-04-07418-0
Received: December 12, 2015; Published: May 30, 2016