THE EXPONENTIAL DIOPHANTINE EQUATION x2 +

DOI: 10.2478/s12175-014-0265-z Math. Slovaca 64 (2014), No. 5, 1145–1152

THE EXPONENTIAL DIOPHANTINE EQUATION x2 + (3n2 + 1)y = (4n2 + 1)z Wang Jianping* — Wang Tingting** — Zhang Wenpeng** (Communicated by Georges Grekos ) ABSTRACT. Let n be a positive integer. In this paper, using the results on the existence of primitive divisors of Lucas numbers and some properties of quadratic and exponential diophantine equations, we prove that if n ≡ 3 (mod 6), then the equation x2 + (3n2 + 1)y = (4n2 + 1)z has only the positive integer solutions (x, y, z) = (n, 1, 1) and (8n3 + 3n, 1, 3). c 2014 Mathematical Institute Slovak Academy of Sciences

1. Introduction Let Z, N be the sets of all integers and positive integers respectively. Let n be a positive integer. Many papers have as object the solutions (x, y, z) of the equation x2 + (3n2 + 1)y = (4n2 + 1)z , x, y, z ∈ N. (1.1) 2 Thus, P.-Z. Yuan and Y.-Z. Hu [11] proved that if 4n + 1 is an odd prime, then (1.1) has only the solutions (x, y, z) = (n, 1, 1),

(8n3 + 3n, 1, 3).

(1.2)

Afterwards, Z.-W. Liu [6] and S.-C. Yang [10] independently prove the same result. This result completely solves one family of generalized Ramanujan-Nagell equations (see [2], [4]). In addition, Y.-Z. Hu and R.-X. Liu [3] proved that if 3n2 + 1 is an odd prime or the square of an odd prime, then (1.1) has only the solutions (1.2). However, up to now no one has known whether there exist infinitely many n which are able to accord with the above mentioned hypotheses. Thus, it can be seen that the resolution of (1.1) is a difficult problem. 2010 M a t h e m a t i c s S u b j e c t C l a s s i f i c a t i o n: Primary 11M20. K e y w o r d s: exponential Diophantine equation, Pell equation, Lucas number. This work is supported by the National Natural Science Foundation of P. R. China (No. 11071194).

Unauthenticated Download Date | 9/24/15 11:38 PM

WANG JIANPING — WANG TINGTING — ZHANG WENPENG

In this paper, using the results on the existence of primitive divisors of Lucas numbers and some properties of quadratic and exponential diophantine equations, we prove the following general result:

If n ≡ 3 (mod 6), then (1.1) has only the solutions (1.2). Obviously, by our theorem, (1.1) is solved for at least one sixth of n’s. 2. Preliminaries Let D and k be coprime positive integers.

1

([7: Section 8.1]). If D is not a square, then the equation u2 − Dv 2 = 1,

u, v ∈ Z.

(2.1)

has solutions (u, v) with v√= 0, and it√has a unique solution (u1 , v1 ) satisfying u1 > 0, v1 > 0 and u1 + v1 D ≤ u + v D, where (u, v) runs through all positive integer solutions of (2.1). Such (u1 , v1 ) is called the least solution of (2.1). Then, every solution (u, v) of (2.1) can be expressed as √ √ r u + v D = λ u1 + v1 D , λ ∈ {±1} , r ∈ Z.

2

([7: Section 8.1]). If the equation u2 − Dv 2 = −1,

u, v ∈ Z

(2.2)

has solutions (u, v), √ a unique solution (u1 , v1 ) satisfying u1 > 0, √ then it has v1 > 0 and u1 + v1 D ≤ u + v D, where (u, v) runs through all positive integer solutions of (2.2). Such (u1 , v1 ) is called the least solution of (2.2). Then we have: (i) Every solution (u, v) of (2.2) can be expressed as √ √ r u + v D = λ u1 + v1 D , λ ∈ {±1} , r ∈ Z, 2 r. √ √ 2 (ii) The least solution of (2.1) is u1 + v1 D = u1 + v1 D .

3

For any positive integer n, the equation u2 − (4n2 + 1)v 2 = 1,

u, v ∈ Z

(2.3) 2

has solutions (u, v) with v = 0, and its least solution is (u1 , v1 ) = (8n + 1, 4n). P r o o f. Since n = 0, 4n2 +1 is not a square and, by Lemma 1, (2.3) has solutions (u, v) with v = 0. On the other hand, since (2n)2 − (4n2 + 1) · 12 = −1, by (i) of Lemma 2, the equation U 2 − (4n2 + 1)V 2 = −1,

U, V ∈ Z

1146


THE EXPONENTIAL DIOPHANTINE EQUATION x2 + (3n2 + 1)y = (4n2 + 1)z

has solutions (U, V ) and its least solution is (U1 , V1 ) = (2n, 1) since u ≥ 2n. √ √ 2 Therefore, by (ii) of Lemma 2, we get u1 + v1 4n2 + 1 = 2n + 4n2 + 1 = √ (8n2 + 1) + 4n 4n2 + 1. The Lemma is proved.

4

([8]) If D is not a square and the equation X 2 − DY 2 = −k,

X, Y ∈ Z, k ∈ N gcd(X, Y ) = 1

(2.4)

has solutions (X, Y ), then every solution (X, Y ) of (2.4) can be expressed as √ √ √ X + Y D = X0 + Y0 D u + v D , where (u, v) is a solution of (2.1), (X0 , Y0 ) is a solution of (2.4) satisfying k(u1 − 1) k 0 ≤ |X0 | ≤ , 0 < Y0 ≤ v1 , 2 2(u1 − 1) where (u1 , v1 ) is the least solution of (2.1).

5

For any positive integer n, the equation

X 2 − (4n2 + 1)Y 2 = −(3n2 + 1),

X, Y ∈ Z, gcd(X, Y ) = 1

(2.5)

has solutions (X, Y ) and every solution (X, Y ) of (2.5) can be expressed as (2.6) X + Y 4n2 + 1 = X0 + Y0 4n2 + 1 u + v 4n2 + 1 , where (u, v) is a solution of (2.3) and (X0 , Y0 ) is a solution of (2.5) satisfying 0 < Y0 ≤ 3n2 + 1. (2.7) 0 ≤ |X0 | < 2n 3n2 + 1, P r o o f. Since n2 − (4n2 + 1) · 12 = −(3n2 + 1), (2.5) has (X, Y ) as solution. Further, since 4n2 + 1 is not a square, by Lemma 4, every solution (X, Y ) of (2.5) can be expressed as in (2.6), where (X0 , Y0 ) is a solution of (2.5) satisfying (3n2 + 1)(u1 − 1) 3n2 + 1 0 ≤ |X0 | ≤ , 0 < Y0 ≤ v1 , (2.8) 2 2(u1 − 1) with (u1 , v1 ) the least solution of (2.3). Furthermore, by Lemma 3, we have (u1 , v1 ) = (8n2 + 1, 4n). Substituting it into (2.8), we obtain (2.7) immediately. The lemma is proved.

6

([5]) If D > 3, 2 k and the equation X 2 + DY 2 = k Z ,

X, Y, Z ∈ Z, gcd(X, Y ) = 1, Z > 0

(2.9)

has solutions (X, Y, Z), then we have: 1147



(i) All solutions (X, Y, Z) of (2.9) can be put into finitely many classes. For all solutions (X, Y, Z) belonging to the same class, say S, there is a unique positive integer l satisfying k l2 ≡ −D (mod k), X ≡ ±lY (mod k), 0 0, Y1 > 0 and Z1 ≤ Z, where Z runs through all solutions (X, Y, Z) of S. Such (X1 , Y1 , Z1 ) is called the least solution of S. Then, every solution (X, Y, Z) of S can be expressed as √ √ t Z = Z1 t, X + Y −D = λ1 X1 + λ2 Y1 −D , t ∈ N, λ1 , λ2 ∈ {±1} . Let α, β be algebraic integers. If α + β and αβ are nonzero coprime integers and α/β is not a root of unity, then (α, β) is called a Lucas pair. Further let a = α + β and c = αβ. Then we have √ √ 1 1 α= a+λ b , β = a−λ b , λ ∈ {±1} , 2 2 where b = a2 − 4c. We call (a, b) the parameters of the Lucas pair (α, β). Two Lucas pairs (α1 , β1 ) and (α2 , β2 ) are equivalent if α1 /α2 = β1 /β2 = ±1. Given a Lucas pair (α, β), one defines the corresponding sequence of Lucas numbers by αm − β m Lm (α, β) = , m = 0, 1, 2, . . . . (2.10) α−β For equivalent Lucas pairs (α1 , β1 ) and (α2 , β2 ), we have Lm (α1 , β1 ) = ±Lm (α2 , β2 ) for any m. A prime p is called a primitive divisor of Lm (α, β) (m > 1) if p | Lm (α, β) and p bL1 (α, β) . . . Lm−1 (α, β). A Lucas pair (α, β) such that Lm (α, β) has no primitive divisor will be called a m-defective Lucas pair. Further, a positive integer m is called totally non-defective if no Lucas pair is m-defective.

7

([9]) Let m satisfy 4 < m ≤ 30 and m = 6. Then, up to equivalence, all parameters of m−defective Lucas pairs are given as follows: (i) m = 5, (a, b) = (1, 5), (1, −7), (2, −40), (1, −11), (1, −15), (12, −76), (12, −1364). (ii) m = 7, (iii) m = 8,

(a, b) = (1, −7), (1, −19). (a, b) = (2, −24), (1, −7).

(iv) m = 10,

(a, b) = (2, −8), (5, −3), (5, −47).

(v) m = 12,

(a, b) = (1, 5), (1, −7), (1, −11), (2, −56), (1, −15), (1, −19).

(vi) m ∈ {13, 18, 30},

(a, b) = (1, −7).

1148



8 9

([1]) If m > 30, then m is totally non-defective.

The equation z x+ −(3n2 + 1) = λ1 n+λ2 −(3n2 + 1) ,

x, z ∈ N, 2 z, λ1 , λ2 ∈ {±1} (2.11) has only the solutions (x, z) = (n, 1) and (8n3 + 3n, 3). P r o o f. Let α =n+

−(3n2 + 1),

β =n−

−(3n2 + 1).

(2.12)

2

Then, α + β = 2n and αβ = 4n + 1 are coprime positive integers, and α/β satisfies (4n2 + 1)(α/β)2 − 2(2n2 − 1)(α/β) + (4n2 + 1) = 0, hence it is not a root of unity. Therefore, (α, β) is a Lucas pair with parameters (2n, −4(3n2 + 1)). Let (x, z) be a solution of (2.11). Since z (2.13) x − −(3n2 + 1) = λ1 n − λ2 −(3n2 + 1) , by the definition of Lucas numbers, we get from (2.10)–(2.13) that |Lz (α, β)| = 1.

(2.14)

It implies that the Lucas number Lz (α, β) has no primitive divisor. Since 2 z, applying Lemmas 7 and 8 to (2.14), we obtain z ∈ {1, 3}. Thus, (2.11) has only the solutions (x, z) = (n, 1) and (8n3 + 3n, 3). The lemma is proved.

3. Proof of the theorems Since n ≡ 3 (mod 6), we have 2 n and 3n2 + 1 ≡ 4

(mod 8),

4n2 + 1 ≡ 5

(mod 8).

(3.1)

Let (x, y, z) be a solution of (1.1). Since x2 · (4n2 + 1)z − (3n2 + 1)y ≡ 1 − 1 ≡ 0 (mod n2 ), we get n | x. (3.2) If y > 1, then from (1.1) and (3.1) we get 5z ≡ (4n2 +1)z ≡ x2 +(3n2 +1)y ≡ x2 (mod 8). It implies that 2 x and 2|z. Hence, by (1.1), we get z

(4n2 + 1) 2 + λx = 2ay ,

z

(4n2 + 1) 2 − λx = 22y−1 by ,

(3.3)

2

3n + 1 = 4ab, a, b ∈ N, gcd(a, b) = 1, 2 ab, λ ∈ {±1} . (3.4) gcd(a, b) = 1 since the first members of 2 in (3.3) have just p = 2 as prime common factor; the exponents of 2 in (3.3) come from the fact that one and z only one of the two even numbers (4n2 + 1) 2 ± x is not divisible by 4. By (3.3), we have z ay + 22y−2 by = (4n2 + 1) 2 . (3.5) 1149



On the other hand, we see from (3.4) that (−3/a) = 1, where (∗/∗) is the Jacobi a−1 symbol. So we have 1 = (−1) 2 (3/a) = (a/3), from which we get a ≡ 1 (mod 3). Hence, by (3.4), we get b ≡ 1 (mod 3). Therefore, since 3|n, we get from (3.5) that z

1 ≡ (4n2 + 1) 2 ≡ ay + 22y−2 by ≡ 1 + 22y−2 ≡ 2 (mod 3),

(3.6)

a contradiction. Thus, we obtain y = 1.

(3.7)

x2 + (3n2 + 1) = (4n2 + 1)z .

(3.8)

Substitute (3.7) into (1.1); we have Further, by (3.1) and (3.8), we get 5z ≡ (4n2 + 1)z · x2 + (3n2 + 1) ≡ 1 + 4 ≡ 5 (mod 8) and 2 z. (3.9) From (3.8) and (3.9), we obtain z−1 2 x2 − (4n2 + 1) (4n2 + 1) 2 = −(3n2 + 1).

(3.10)

It implies that (2.5) has the solution z−1 (X, Y ) = x, (4n2 + 1) 2 .

(3.11)

Applying Lemma 5 to (3.11), we get z−1 x + (4n2 + 1) 2 4n2 + 1 = X0 + Y0 4n2 + 1 u + v 4n2 + 1 ,

(3.12)

where (u, v) is a solution of (2.3), X0 and Y0 are integers satisfying (2.7) and X02 − (4n2 + 1)Y02 = −(3n2 + 1),

gcd(X0 , Y0 ) = 1.

(3.13)

By (3.12), we have x = X0 u + (4n2 + 1)Y0 v and (4n2 + 1)

z−1 2

(3.14)

= X0 v + Y0 u.

(3.15)

Since (u, v) is a solution of (2.3), by Lemma 3, we get r u + v 4n2 + 1 = λ (8n2 + 1) + 4n 4n2 + 1 |r| = λ (8n2 + 1) + 4nλ 4n2 + 1 , r r ∈ Z, λ ∈ {±1} , λ = . |r|

(3.16)

By (3.16), we have n|v. Further, since gcd(u, v) = 1, we see from (3.2) and (3.14) that n|X0 . Hence, by (3.15), we get Y0 u · (4n2 + 1)

z−1 2

− X0 v ≡ 1 − 0 ≡ 1

(mod n2 ).

1150


(3.17)


Furthermore, by (3.16), we have 2 ] |r| (8n2 + 1)|r|−2i (16n2 (4n2 + 1))i , u=λ 2i i=0 [

|r|

where [ |r| 2 ] is the integral part of

|r| 2 .

(3.18)

It implies that

u ≡ λ(8n2 + 1)|r| ≡ λ

(mod n2 ).

(3.19)

λ ∈ {±1} .

(3.20)

Therefore, by (3.17) and (3.19), we obtain Y0 ≡ λ (mod n2 ),

If Y0 = λ, then from (3.20) we get Y0 + 1 ≥ Y0 − λ ≥ n2 . But, since n ≥ 3, this is impossible by (2.7). So we have Y0 = λ and Y0 = 1.

(3.21)

By substituting (3.21) into (3.13), we get X0 = λ n,

λ ∈ {±1} .

(3.22)

(mod 4n2 + 1).

(3.23)

Hence, by (3.14) and (3.22), we have x ≡ X0 u ≡ λ nu Further, by (3.18), we get u ≡ λ(8n2 + 1)|r| ≡ λ(−1)|r| ≡ λ

(mod 4n2 + 1),

λ ∈ {±1} .

(3.24)

Therefore, by (3.23) and (3.24), we obtain x ≡ θn

(mod 4n2 + 1),

θ ∈ {±1} .

(3.25)

On the other hand, we find from (3.8) that the equation X 2 +(3n2 +1)Y 2 = (4n2 +1)Z ,

X, Y, Z ∈ Z, gcd(X, Y ) = 1, Z > 0 (3.26)

has the solution (X, Y, Z) = (x, 1, z). (3.27) By (i) of Lemma 6, we may assume that the solution (3.27) belongs to a certain class S of solutions (X, Y, Z) of (3.26). Moreover, by (3.25) and (3.27), the characteristic number l of S satisfies l ≡ ± x1 ≡ ±n (mod 4n2 + 1) and 0 < l 2 < 4n 2+1 . So we have l = n. (3.28) 2 2 2 2 Since n + (3n + 1) · 1 = 4n + 1, by (3.28), the class S contains the solution (X, Y, Z) = (n, 1, 1).

(3.29)

Further, by the definitions given in Lemma 6, √ (3.29) is the solution least √ z of S. Therefore, by (ii) of Lemma 6, we see that x + −D = λ1 n + λ2 −D ; from 1151



(3.9) 2 z, hence by Lemma 9 we conclude that (1.1) has only the solutions (1.2). The theorem is proved. Acknowledgement The authors would thank referees for giving the helpful comments.

REFERENCES [1] BILU, Y.—HANROT, G.—VOUTIER, P. M. (with an appendix by MIGNOTTE, M.): Existence of primitive divisors of Lucas and Lehmer numbers, J. Reine Angew. Math. 539 (2001), 75–122. [2] BUGEAUD, Y.: On some exponential diophantine equations, Monatsh. Math. 132 (2001), 93–97. [3] HU, Y.-Z.—LIU, R.-X.: On the solutions of the exponential diophantine equation x2 + (3a2 + 1)m = (4a2 + 1)n , J. Sichuan Univ. Nat. Sci. Ed. 43 (2006), 41–46. [4] HU, Y.-Z.—LIU, R.-X.—LE, M.-H.: The diophantine equation x2 + D m = pn , Acta Arith. 52 (1989), 255–265. [5] LE, M.-H.: some exponential diophantine equations I: The equation D1 x2 − D2 y 2 = λk z , J. Number Theory 55 (1995), 209–221. [6] Z.-W. LIU: On the generalized Ramanujan-Nagell equation x2 + D m = pn , Acta Math. Sinica (Chin. Ser.) 51 (2008), 809–814 (Chinese). [7] MORDELL, L. J.: Diophantine Equations, Academic Press, London, 1969. ¨ [8] NAGELL, T.: Uber die Darstellung ganzer Zahlen durch eine indefinite bin¨ are quadratische Form, Arch. Math. (Basel) 2 (1950), 161–165. [9] VOUTIER, P. M.: Primitive divisors of Lucas and Lehmer sequences, Math. Comp. 64 (1995), 869–888. [10] YANG, S.-C.: A note on the solutions of the generalized Ramanujan-Nagell equation x2 + D m = pn , Acta Math. Sinica (Chin. Ser.) 50 (2007), 943–948 (Chinese). [11] YUAN, P.-Z.—HU, Y.-Z.: On the diophantine equation x2 +D m = pn , J. Number Theory 111 (2005), 144–153. Received 11. 4. 2012 Accepted 19. 7. 2012

* College of Science Chang’an University Xi’an, Shaanxi P.R. CHINA E-mail : [email protected] ** Department of Mathematics Northwest University Xi’an, Shaanxi P.R. CHINA E-mail : [email protected] [email protected]

1152
