On the diophantine equation

(1)

LXXVI.1 (1996)

On the diophantine equation D

1

x

⁴

− D

2

y

²

= 1

by

Maohua Le (Zhanjiang)

1. Introduction. Let Z, N, Q, R be the sets of integers, positive integers, rational numbers and real numbers respectively. Let D

1

, D

2

∈ N with gcd(D

₁

, D

₂

) = 1. There were many papers concerned with the equation (1) D

1

x

⁴

− D

2

y

²

= 1, x, y ∈ N,

written by Ljunggren, Bumby, Cohn, Ke and Sun. Concerning the solvability of (1), Zhu [7] and Le [2] proved independently that if D

₁

= 1, then (1) has solutions (x, y) if and only if the fundamental solution u

1

+ v

1

√ D

2

of Pell’s equation

u

²

− D

2

v

²

= 1, u, v ∈ Z,

satisfies either u

₁

= x

²₁

or u

²₁

+ D

₂

v

₁²

= x

²₁

, where x

₁

∈ N. In addition, Zhu [7] showed that if D

2

= 1, then (1) has solutions (x, y) if and only if the equation

u

⁰²

− D

1

v

⁰²

= −1, u

⁰

, v

⁰

∈ Z,

has solutions (u

⁰

, v

⁰

) and its least positive integer solution (u

⁰₁

, v

⁰₁

) satisfies v

₁⁰

= x

²₁

, where x

₁

∈ N. In this paper we prove a general result as follows.

Theorem 1. If min(D

₁

, D

₂

) > 1, then (1) has solutions (x, y) if and only if the equation

(2) D

1

U

²

− D

2

V

²

= 1, U, V ∈ Z,

has solutions (U, V ) and its least positive integer solution (U

₁

, V

₁

) satisfies U

₁

= x

²₁

, where x

₁

∈ N.

Let N (D

₁

, D

₂

) denote the number of solutions (x, y) of (1). Ljunggren [4] showed that N (1, D

₂

) ≤ 2. In [3], Le proved that if D

₂

> e

⁶⁴

, then N (1, D

2

) ≤ 1. Recently, Wu [6] relaxed the condition D

2

> e

⁶⁴

to D

2

> e

³⁷

. In this paper we prove the following result.

Supported by the National Natural Science Foundation of China and Guangdong Provincial Natural Science Foundation.

[1]

(2)

Theorem 2. If D

1

is a square and max(D

₁

, D

₂

) ≥

9.379 · 10

⁸

if min(D

₁

, D

₂

) = 1, 2.374 · 10

¹⁰

if min(D

1

, D

2

) > 1, then N (D

1

, D

2

) ≤ 1.

2. Preliminaries

Lemma 1. For any odd prime p with p ≡ 1 (mod 4), there exists a

₁

∈ N such that p > a

₁

> 1, 2 - a

₁

and (a

₁

/p) = −1, where (a

₁

/p) is Legendre’s symbol.

P r o o f. It is a well known fact that there exists a ∈ N with p > a > 1 and (a/p) = −1. Since (a/p) = ((p − a)/p) for p ≡ 1 (mod 4), we get

a

1

=

a if 2 - a, p − a if 2 | a.

The lemma is proved.

Lemma 2 ([3, Lemma 3]). Let d ∈ N be square-free. If (u, v) and (u

⁰

, v

⁰

) are solutions of the equation

(3) u

²

− dv

²

= 1, u, v ∈ N,

with u

⁰

≡ 0 (mod u), then there exist fixed d

₁

, d

₂

∈ N such that

(4) d

₁

d

₂

= d, u + 1 = δd

₁

v

₁²

, u − 1 = δd

₂

v

₂²

, u

⁰

+ 1 = δd

1

v

₁⁰²

, u

⁰

− 1 = δd

2

v

₂⁰²

, where δ, v

₁

, v

₂

, v

₁⁰

, v

₂⁰

∈ N satisfy

(5) δv

1

v

2

= v, δv

₁⁰

v

⁰₂

= v

⁰

, δ =

1 if 2 - v, 2 if 2 | v.

Lemma 3 ([5]). For min(D

1

, D

2

) > 1, if (2) has solutions (U, V ), then it has a unique positive integer solution (U

₁

, V

₁

) such that U

₁

√

D

₁

+ V

₁

√ D

₂

≤ U √

D

1

+ V √

D

2

for all positive integer solutions (U, V ) of (2). (U

1

, V

1

) is called the least solution of (2). Moreover , all positive integer solutions (U, V ) of (2) are given by

U p

D

₁

+ V p

D

₂

= (U

₁

p

D

₁

+ V

₁

p

D

₂

)

^t

, t ∈ N, 2 - t.

Lemma 4. For min(D

₁

, D

₂

) > 1, let (U, V ) be a solution of (2), and let

(6) ε = U p

D

₁

+ V p

D

₂

, ε = U p

D

₁

− V p D

₂

. Further , for any m ∈ Z with 2 - m, let

(7) E(m) = ε

^m

+ ε

^m

ε + ε .

Then E(m) ∈ N and:

(3)

(i) E(m) = E(−m).

(ii) E(m) ≡ 1 (mod 4), E(m) ≡ (−1)

^(m−1)/2

m (mod U ).

(iii) For any m, m

⁰

∈ Z with 2 - mm

⁰

, E(m) ≡ −E(m−2m

⁰

) (mod E(m

⁰

)).

(iv) For any m, m

⁰

∈ Z with 2 - mm

⁰

and gcd(m, m

⁰

) = 1, (E(m)/E(m

⁰

))

= 1, where (∗/∗) is the Jacobi symbol.

P r o o f. Since εε = 1 by (2) and (6), we get (i) by (7). Since we have E(m + 4) + E(m) = (ε

²

+ ε

²

)E(m + 2) ≡ −2E(m + 2) (mod 4U ), (ii) follows by induction on m in view of the fact that E(−1) = E(1) = 1.

Notice that

(8) E(m) + E(m − 2m

⁰

) = (ε

^m−m⁰

+ ε

^m−m⁰

)E(m

⁰

),

where ε

^m−m⁰

+ ε

^m−m⁰

∈ Z, since m − m

⁰

is even. So we have (iii). Moreover, using (8) and (ii), we obtain (iv) by induction. The lemma is proved.

By much the same argument as in the proof of Lemma 4, we can prove the following lemma.

Lemma 5. For min(D

₁

, D

₂

) > 1, let U, V, ε, ε be defined as in Lemma 4, and let

F (m) = ε

^m

− ε

^m

ε − ε for any m ∈ Z with 2 - m. Then F (m) ∈ Z and:

(i) F (m) = −F (−m) and F (m) > 0 if m > 0.

(ii) For any m, m

⁰

∈ Z with 2 - mm

⁰

, F (m) ≡ F (m − 2m

⁰

) (mod F (m)).

Lemma 6. If (U, V ) and (U

⁰

, V

⁰

) are positive integer solutions of (2) satisfying U

⁰

≡ 0 (mod U ) or V

⁰

≡ 0 (mod V ), then there exists t

⁰

∈ N such that

(9) U

⁰

p

D

1

+ V

⁰

p

D

2

= (U p

D

1

+ V p

D

2

)

^t⁰

, 2 - t

⁰

. P r o o f. Let ε = U

₁

√

D

₁

+ V

₁

√

D

₂

and ε = U

₁

√

D

₁

− V

₁

√

D

₂

, where (U

₁

, V

₁

) is the least solution of (2). By Lemma 3, there exist m, m

⁰

∈ N such that

(10) U p

D

1

+ V p

D

2

= ε

^m

, U

⁰

p

D

1

+ V

⁰

p

D

2

= ε

^m⁰

, 2 - mm

⁰

. Then we have U = U

₁

E(m), V = V

₁

F (m), U

⁰

= U

₁

E(m

⁰

) and V

⁰

= V

1

F (m

⁰

).

By Lemma 4(iii), if U

⁰

≡ 0 (mod U ), then we have

(11) 0 ≡ U

⁰

= U

1

E(m

⁰

) ≡ −U

1

E(m

⁰

− 2m) ≡ . . . ≡ ±U

1

E(s) (mod U ), where s ∈ Z satisfies 2 - s and −m < s ≤ m. Since 0 < E(s) < E(m) if

|s| < m, we obtain s = m by (11). Therefore, m | m

⁰

and (9) holds by (10).

(4)

The proof in the case V

⁰

≡ 0 (mod V ) is analogous. By Lemma 5(ii), we then have

0 ≡ V

⁰

= V

1

F (m

⁰

) ≡ V

1

F (m

⁰

− 2m) ≡ . . . ≡ V

1

F (s) (mod V ), where s ∈ Z satisfies 2 - s and −m < s ≤ m. Since 0 < |F (s)| < F (m) if

|s| < m, we get s = m, and hence, m | m

⁰

. Thus (9) holds in this case. The lemma is proved.

Let α be an algebraic number with the minimal polynomial a

0

z

^d

+ . . . + a

d

= a

0

Y

d i=1

(z − σ

i

α), a

0

> 0, where σ

₁

α, . . . , σ

_d

α are all conjugates of α. Then

h(α) = 1 d

log a

0

+ X

d i=1

log max(1, |σ

i

α|)

is called the logarithmic absolute height of α.

Lemma 7 ([1, Corollary 2]). Let α

₁

, α

₂

be real algebraic numbers with α

1

> 1 and α

2

> 1 which are multiplicatively independent, and let log A

_j

≥ max(h(α

_j

), |log α

_j

|/r, 1/r) for j = 1, 2, where r = [Q(α

₁

, α

₂

) : Q]/[R(α

₁

, α

₂

) : R]. If Λ = b

₁

log α

₁

− b

₂

log α

₂

6= 0 for some b

₁

, b

₂

∈ N, then

|Λ| ≥ exp(−24.34r

⁴

(log A

₁

)(log A

₂

)(max(log b

⁰

+ 0.69, 21/r, 1/2))

²

), where b

⁰

= b

1

/(r log A

2

) + b

2

/(r log A

1

).

3. Proof of Theorem 1. The sufficiency of the theorem is clear; it suffices to prove the necessity. Now we assume that (1) has solutions (x, y).

Then (1) has a unique solution (x

1

, y

1

) such that

(12) x

²₁

p

D

₁

+ y

₁

p

D

₂

≤ x

²

p

D

₁

+ y p D

₂

for all solutions (x, y) of (1). Clearly, (x

²₁

, y

1

) is a positive integer solution of (2). Let (U

₁

, V

₁

) be the least solution of (2). By Lemma 3, we have (13) x

²₁

p

D

₁

+ y

₁

p

D

₂

= (U

₁

p

D

₁

+ V

₁

p

D

₂

)

^t

, t ∈ N, 2 - t.

If t = 1, then the theorem is proved. Otherwise, t has an odd prime fac- tor p. By Lemma 3, (2) has a positive integer solution (U, V ) which satisfies

(14) U p

D

₁

+ V p

D

₂

= (U

₁

p

D

₁

+ V

₁

p

D

₂

)

^t/p

. From (13) and (14), we get

(15) x

²₁

p

D

₁

+ y

₁

p

D

₂

= (U p

D

₁

+ V p

D

₂

)

^p

.

(5)

For any m ∈ Z with 2 - m, let ε, ε and E(m) be defined as in (6) and (7), respectively. From (15) we get

(16) x

²₁

= ε

^p

+ ε

^p

2 √

D

₁

= U E(p).

By Lemma 4(ii), E(p) ∈ N with E(p) ≡ (−1)

^(p−1)/2

p (mod U ). This implies that gcd(U, E(p)) = 1 or p.

If gcd(U, E(p)) = 1, then from (16) we get U = x

²₁₁

and E(p) = x

²₁₂

, where x

11

, x

12

∈ N with x

11

x

12

= x

1

. It follows that (x

11

, V ) is a solution of (1) satisfying

x

²₁₁

p

D

₁

+ V p

D

₂

= U p

D

₁

+ V p

D

₂

= ε < ε

^p

= x

²₁

p

D

₁

+ y

₁

p D

₂

, which contradicts (12).

If gcd(U, E(p)) = p, then we have

(17) U = px

²₁₁

, E(p) = px

²₁₂

,

where x

₁₁

, x

₁₂

∈ N with px

₁₁

x

₁₂

= x

₁

. Since E(p) ≡ 1 (mod 4) by Lemma 4(ii), we see from (17) that p ≡ 1 (mod 4). Therefore, by Lemma 1, there exists a

₁

∈ N such that 2 - a

₁

, p > a

₁

> 1 and (a

₁

/p) = −1. Further, since p | U , by Lemma 4(ii), we get E(a

₁

) ≡ (−1)

^(a¹^−1)/2

a

₁

(mod p). So we have

E(p) E(a

₁

)

=

px

²₁₂

E(a

₁

)

=

p

E(a

₁

)

=

E(a

₁

) p

(18)

=

(−1)

^(a¹^−1)/2

a

1

p

=

a

1

p

= −1,

by (17). However, by Lemma 4(iv), (18) is impossible. The theorem is proved.

4. Proof of Theorem 2. First we consider the case where min(D

₁

, D

₂

)

> 1. By Theorem 1, if (1) has solutions (x, y), then (x

1

, y

1

) = ( √ U

1

, V

1

) is a solution of (1), where (U

₁

, V

₁

) is the least solution of (2). Further, by Lemma 3, if N (D

1

, D

2

) > 1, then (1) has another solution (x

2

, y

2

) which satisfies x

₂

> x

₁

and

(19) x

₂

≡ 0 (mod x

₁

).

Since D

₁

is a square, D

₂

cannot be such, therefore we may also assume, without loss of generality, that D

2

is square-free. Let D

1

= a

²

, where a ∈ N with a > 1. Then (ax

²₁

, y

₁

) and (ax

²₂

, y

₂

) are solutions of the equation

u

²

− D

₂

v

²

= 1, u, v ∈ N.

Notice that ax

²₂

≡ 0 (mod ax

²₁

) by (19). We have

ax

²₁

+ 1 = δD

₂₁

y

²₁₁

, ax

²₁

− 1 = δD

₂₂

y

₁₂²

, (20)

ax

²₂

+ 1 = δD

₂₁

y

²₂₁

, ax

²₂

− 1 = δD

₂₂

y

₂₂²

,

(21)

(6)

by Lemma 2, where δ, D

21

, D

22

, y

11

, y

12

, y

21

, y

22

∈ N satisfy (22) D

₂₁

D

₂₂

= D

₂

, gcd(D

₂₁

, D

₂₂

) = 1, (23) δy

11

y

12

= y

1

, δy

21

y

22

= y

2

, δ =

1 if 2 - y

1

, 2 if 2 | y

₁

.

We see from (20) and (21) that (y

11

, x

1

) and (y

21

, x

2

) are solutions of the equation

δD

21

X

²

− aY

²

= 1, X, Y ∈ N, while (x

₁

, y

₁₂

) and (x

₂

, y

₂₂

) are solutions of the equation

aX

⁰²

− δD

22

Y

⁰²

= 1, X

⁰

, Y

⁰

∈ N.

Let

ε

1

= x

1

√ a + y

11

p δD

21

, ε

1

= x

1

√ a − y

11

p δD

21

, (24)

ε

2

= x

1

√ a + y

12

p δD

22

, ε

2

= x

1

√ a − y

12

p δD

22

. (25)

Recall that x

₂

≡ 0 (mod x

₁

) by (19). Using Lemma 6, we have (26) x

₂

√

a + y

₂₁

p

δD

₂₁

= ε

^t₁¹

, x

₂

√

a − y

₂₁

p

δD

₂₁

= ε

^t₁¹

, (27) x

2

√ a + y

22

p δD

22

= ε

^t₂²

, x

2

√ a − y

22

p δD

22

= ε

^t₂²

,

where t

₁

, t

₂

∈ N satisfy t

₁

> 1, t

₂

> 1 and 2 - t

₁

t

₂

. From (24)–(27), we obtain

(28) ε

₁

+ ε

₁

= ε

₂

+ ε

₂

,

(29) ε

^t₁¹

+ ε

^t₁¹

= ε

^t₂²

+ ε

^t₂²

.

Let ∆ = ε

2

− ε

1

and ∆

⁰

= ε

^t₂²

− ε

^t₁¹

. Since ε

1

ε

1

= −1 and ε

2

ε

2

= 1, from (28) and (29) we get

log ε

₁

= log ε

₂

+ 2∆

ε

1

+ ε

2

X

∞ i=0

1 2i + 1

∆

ε

1

+ ε

2

_2i

(30)

= log ε

₂

+ 2 ε

₁

ε

₂

X

∞ i=0

1 2i + 1

1 ε

₁

ε

₂

_2i

= log ε

2

+ 2 ε

²₂

ε

2

ε

₁

X

∞ i=0

1 2i + 1

1 ε

₁

ε

₂

_2i

= log ε

2

+ 2 ε

²₂

1 1 + 1/(ε

1

ε

2

) + 1/ε

²₂

X

∞ i=0

1 2i + 1

1 ε

1

ε

2

_2i

< log ε

2

+ 2 ε

²₂

, t

₁

log ε

₁

= t

₂

log ε

₂

+ 2∆

⁰

ε

^t₁¹

+ ε

^t₂²

X

∞ i=0

1 2i + 1

∆

⁰

ε

^t₁¹

+ ε

^t₂²

_2i

(31)

(7)

= t

2

log ε

2

+ 2 ε

^t₁¹

ε

^t₂²

X

∞ i=0

1 2i + 1

1 ε

^t₁¹

ε

^t₂²

_2i

= t

₂

log ε

₂

+ 2 ε

^2t₂²

ε

^t₂²

ε

^t₁¹

X

∞ i=0

1 2i + 1

1 ε

^t₁¹

ε

^t₂²

_2i

= t

₂

log ε

₂

+ 2 ε

^2t₂²

1 1 + 1/(ε

^t₁¹

ε

^t₂²

) + 1/ε

^2t₂²

X

∞ i=0

1 2i + 1

1 ε

^t₁¹

ε

^t₂²

_2i

< t

₂

log ε

₂

+ 2 ε

^2t₂²

,

respectively. By (30) and (31), we get log ε

1

−log ε

2

> t

1

log ε

1

−t

2

log ε

2

> 0.

This implies that (t

₂

− 1) log ε

₂

> (t

₁

− 1) log ε

₁

. Since ε

₁

> ε

₂

> 1 by (28), we obtain t

₂

> t

₁

. Since 2 - t

₁

t

₂

, we get

(32) t

₂

≥ t

₁

+ 2.

Therefore, we find from (30)–(32) that (33) t

₂

> (t

₂

− t

₁

) log ε

₁

log ε

1

− log ε

2

> ε

²₂

log ε

₁

> ε

²₁

(log ε

₁

)e

^−4/ε²²

. Let K

₁

= Q( √

δD

₂₁

a) and K

₂

= Q( √

δD

₂₂

a). Since D

₂

is not a square, we see from (22) that K

₁

\ Q ∩ K

₂

\ Q = ∅. If there exist k

₁

, k

₂

∈ Q such that ε

^k₁¹

ε

^k₂²

= 1, then ε

^m₁¹

ε

^m₂²

= 1 for some m

1

, m

2

∈ Z with 2 | m

1

and 2 | m

2

. Notice that ε

^m₁

∈ K

₁

\ Q and ε

^m₂

∈ K

₂

\ Q for any m ∈ Z \ {0} with 2 | m.

We get m

₁

= m

₂

= 0 and k

₁

= k

₂

= 0. This implies that ε

₁

and ε

₂

are multiplicatively independent.

Let h(ε

₁

), h(ε

₂

) denote the logarithmic absolute heights of ε

₁

, ε

₂

respectively, and let r denote the degree of Q(ε

₁

, ε

₂

). Then

(34) 4 ≤ r ≤ 8, h(ε

1

) = log ε

1

r/2 , h(ε

2

) = log ε

2

r/2 . Further, let Λ = t

1

log ε

1

− t

2

log ε

2

and

(35) t = t

₁

2 log ε

₂

+ t

₂

2 log ε

₁

. Then we have

(36) t = t

₂

log ε

1

+ Λ

2(log ε

1

)(log ε

2

) < t

₂

log ε

1

1 + 1

t

₂

ε

^2t₂²

log ε

₂

,

by (31). Using Lemma 7, from (34) we get

(37) Λ ≥ exp(−6232(log ε

₁

)(log ε

₂

)(max(log t + 0.69, 21/4))

²

).

(8)

We now suppose that ε

1

≥ 785. By (28), ε

²₁

− (ε

2

+ ε

2

)ε

1

− 1 = 0. Since ε

₂

≥ 1+ √

2, we get ε

₁

< 1.745ε

₂

, whence ε

₂

> 449.856. If log t+0.69 ≤ 21/4, then t < 96 and

(38) t

2

< 96 log ε

1

,

by (35). The combination of (33) and (38) yields 96 > ε

²₁

e

^−4/ε²²

> 6 · 10

⁵

, a contradiction. Hence, log t + 0.69 > 21/4, and

(39) Λ ≥ exp(−6232(log ε

1

)(log ε

2

)(log t + 0.69)

²

), by (37). The combination of (31) and (39) yields

(40) log 2 + 6232(log ε

₁

)(log ε

₂

)(log t + 0.69)

²

> 2t

₂

log ε

₂

. Further, by (31), (36) and (40), we get

1 + 3116

log

t

2

log ε

₁

+ 0.7

₂

> log 2

log ε

₂

+ 3116

log

t

2

log ε

₁

+ Λ 2(log ε

₁

)(log ε

₂

)

+ 0.69

₂

> t

2

log ε

₁

, whence we conclude that

(41) t

₂

log ε

₁

< 615000.

Therefore, from (33) and (41) we get 616000 < ε

²₁

e

^−4/ε²²

< 615000, a contradiction. So we have

(42) ε

1

< 785.

From (22), (24), (25) and (28), we get (43) ε

₁

= ε

₂

− ε

₁

+ ε

₂

> ε

₂

+ ε

₂

= 2x

₁

√

a ≥ 2 √

a = 2D

₁^1/4

and

ε

²₁

> ε

1

ε

2

− ε

1

/ε

2

+ ε

2

/ε

1

− 1/(ε

1

ε

2

) = (ε

1

− ε

1

)(ε

2

− ε

2

) (44)

= (2y

11

p δD

21

)(2y

12

p δD

22

) ≥ 4 p

D

21

D

22

= 4 p D

2

.

Therefore, by (42)–(44), we obtain max(D

₁

, D

₂

) < ε

⁴₁

/16 < 2.374 · 10

¹⁰

. Thus, if min(D

₁

, D

₂

) > 1 and max(D

₁

, D

₂

) ≥ 2.374 · 10

¹⁰

, then N (D

₁

, D

₂

)

≤ 1.

Next we consider the case where min(D

1

, D

2

) = 1. Since D

1

is a square, we have D

₁

= 1 and D

₂

> 1. By much the same argument as in the proof of the case min(D

₁

, D

₂

) > 1, we can find from the proof of [3, Theorem 1]

that if D

2

6= 1785 and N (1, D

2

) > 1, then there exist t

1

, t

2

∈ N and real

quadratic algebraic numbers %

₁

, %

₂

satisfying 1 < t

₁

< t

₂

, 1 < %

₂

< %

₁

,

D

2

< %

⁴₁

/16, h(%

1

) = (log %

1

)/2, h(%

2

) = (log %

2

)/2, [Q(%

1

, %

2

) : Q] = 4,

(45) 0 < log %

₁

− log %

₂

< 2/%

²₂

, 0 < t

₁

log %

₁

− t

₂

log %

₂

< 2/%

^2t₂²

,

(9)

(46) t

₂

> %

²₁

(log %

₁

)e

^−4/%²²

. Using Lemma 7, we get either %

²₁

e

^−4/%²²

< 96 or

(47) t

₁

log %

₁

− t

₂

log %

₂

≥ exp(−1558(log %

₁

)(log %

₂

)(log t + 0.69)

²

), where

t = t

1

2 log %

2

+ t

2

2 log %

1

= t

2

log %

1

+ t

1

log %

1

− t

2

log %

2

2(log %

1

)(log %

2

) (48)

< t

₂

log %

₁

+ 1

2%

^2t₂²

(log %

₁

)(log %

₂

) .

We now suppose that %

1

≥ 350. Then from (45), (47) and (48) we get (49) t

₂

/ log %

₁

< 122000.

The combination of (46) and (49) yields 122400 < %

²₁

e

^−4/%²²

< 122000, a contradiction. So we have %

₁

< 350 and D

₂

< 9.379 · 10

⁸

. This implies that if min(D

1

, D

2

) = 1 and max(D

1

, D

2

) ≥ 9.379 · 10

⁸

, then N (D

1

, D

2

) ≤ 1.

The proof is complete.

Acknowledgements. The author is grateful to the referees for their valuable suggestions.

References

[1] M. L a u r e n t, M. M i g n o t t e et Y. N e s t e r e n k o, Formes lin´eaires en deux loga- rithmes et d´eterminants d’interpolation, J. Number Theory 55 (1995), 285–321.

[2] M.-H. L e, A necessary and sufficient condition for the equation x

⁴

− Dy

²

= 1 to having positive integer solutions, Chinese Sci. Bull. 30 (1985), 1698.

[3] —, A note on the diophantine equation x

^2p

− Dy

²

= 1, Proc. Amer. Math. Soc. 107 (1989), 27–34.

[4] W. L j u n g g r e n, ¨ Uber die Gleichung x

⁴

− Dy

²

= 1, Arch. Math. Naturv. 45 (5) (1942), 61–70.

[5] K. P e t r, Sur l’´equation de Pell, ˇ Casopis Pest. Mat. Fys. 56 (1927), 57–66 (in Czech).

[6] H.-M. W u, On the number of solutions of the diophantine equation x

⁴

− Dy

²

= 1, J. Zhanjiang Teachers College Nat. Sci. 1995 (1), 12–15 (in Chinese).

[7] W.-S. Z h u, The solvability of equation x

⁴

− Dy

²

= 1, Acta Math. Sinica 28 (1985), 681–683 (in Chinese).

Department of Mathematics Zhanjiang Teachers College P.O. Box 524048

Zhanjiang, Guangdong, P.R. China

Received on 12.1.1994

and in revised form on 19.9.1995 (2558)