Integer Solutions of the Equation y^{2} = Ax^{4} + B

Paraskevas K. Alvanos 1st Model and Experimental High School of Thessaloniki Kleanthous 30 54453 Thessaloniki Greece

                        alvanos@sch.gr

Konstantinos A. Draziotis Department of Informatics Aristotle University of Thessaloniki 54124 Thessaloniki Greece

                     drazioti@csd.auth.gr

Abstract Let A ∈ {k^{2}(k^{2}l^{2} + 1), 4k^{2}(k^{2}(2l − 1)^{2} + 1)}, where k and l are positive integers, and let B be a non-zero square-free integer such that |B| < \sqrt{A}. In this paper we determine all the possible integer solutions of the equation y^{2} = Ax^{4} + B by using terms of Lucas sequences of the form mx^{2}.

1 Introduction and statement of results

Diophantine equations of the form Dy^{2} = Ax^{4}+B have been widely studied with a variety of methods. A number of innovative ideas have been developed in order to study such equations. For instance, Ljunggren [13, 14, 15], by studying units in quadratic and biquadratic fields, showed that the equation x^{2} − dy^{4} = ±1 has at most two solutions. Some equations of the previous form are given in Table 1.

Equation  Ref.  Equation  Ref.
x^{4} ± 1 = (4r^{2} + 1)y^{2} [10]  y^{2} − dx^{4} = 1  [8, 22]
x^{4} − 1 = dy^{2}  [9]  y^{2} − dx^{4} = −1  [6]
x^{4} + 1 = dy^{2}  [7]  (m^{2} + m + 1)x^{4} − (m^{2} + m)y^{2} = 1  [25]
ty^{2} = (t + 2)x^{4} − 2 [1]  x^{4} − dy^{2} = 1, d prime  [17]
b^{2}x^{4} − 1 = dy^{2}  [2] x^{4} − dy^{2} = 1, d prime or twice a prime [18]

Table 1: Quartic equations of the form αX^{2} + βY^{4} = ±1.

Moreover, Tzanakis [23, 24] introduced a method for solving quartic elliptic equations by generalizing the elliptic logarithm method. Another method to treat equations of this type is to apply a reduction to the study of the squares of some binary linear recurrent sequence or to a family of Thue equations. In that case, either the Baker or the Thue-Siegel method is used. Also, there are elementary approaches, such as manipulations with Legendre symbol and reduction mod p. Note that in all the previous equations the constant term is very small (it belongs to the set \{±1, ±2\}). In this paper we allow |B| to be smaller than \sqrt{A}.

We shall determine all possible solutions of a large family of equations of the form

y^{2} = Ax^{4} + B,  (1)

where A ∈ {k^{2}(k^{2}l^{2} + 1), 4k^{2}(k^{2}(2l − 1)^{2} + 1)} (k, l ∈ ℤ_{>0}) and |B| is a non-zero square-free integer smaller than \sqrt{A}. In particular, for the cases where k ∈ \{1, 2, 3, 6\}, we give the set of possible solutions without any hard computation. For the rest of the cases the computation of a minimal unit in the ring of integers of a quadratic extension of ℚ is needed. The idea is that in order to compute the solutions of equation (1) we need to study the squares that occur to the denominators of the continued fraction of \sqrt{A}. Furthermore, A is chosen with period 1 or 2. The case with period 1 (part 1(a) of Theorem 1) is very easy and was already covered by the general result of Togbe et al. [22]. The case with period 2 gives two families of quartic elliptic curves. We provide the exact solutions for all possible square-free B’s with |B| < \sqrt{A}.

One can also apply well known methods (e.g., reducing to finite number of Thue equations) in order to solve the equations y^{2} = Ax^{4} + B for all the possible B’s but one, one by one. In addition, we have the Magma [4] function, SIntegralLjunggrenPoints([1,A,0,B], []), which returns the integer solutions of (1), but this function makes sense if we let A, B run in some (relatively) small intervals (the function uses linear forms of logarithms).

By K we denote the quadratic number field ℚ(\sqrt{A}). Furthermore, let 𝕆_{K} be the ring of integers of K, ε_{d} = y_{1} + x_{1}\sqrt{d} ∈ ℤ[\sqrt{A}] is the minimal unit > 1, with norm 1 and ε_{d}^{t} = y_{t} + x_{t}\sqrt{d}. The main theorem that we prove is the following.

Theorem 1. Let k, l be positive integers, and (x, y) an integer solution of

y^{2} = Ax^{4} + B,

where 0 ≠ |B| < \sqrt{A} and B is square-free.

1. Let A = k^{2}(k^{2}l^{2} + 1).

(a) If k = 1, then |x| ∈ \{0, 1, 13, \sqrt{2l}\}. Furthermore, |x| = 13 occurs only if A = 2, B = −1.

(b) If k ∈ \{2, 3, 6\}, then |x| ∈ \{0, 1, 6, 68, \sqrt{2l}\}. Furthermore, |x| = 6 occurs only if A = 20, B = 1 and |x| = 68 occurs only if A = 1305, B = 1.

(c) If k ∈ ℕ−\{1, 2, 3, 6\} then |x| ∈ \{0, 1, \sqrt{2l}, \sqrt{x_{1}}, \sqrt{x_{2}}, \sqrt{x_{p}}\} where y_{1}+x_{1}\sqrt{A} > 1 is the minimal unit of 𝕆_{K} of norm 1 and p is prime ≡ 3 (mod 4) such that x_{1} = pu^{2} for some integer u.

2. Let A = 4k^{2}(k^{2}(2l − 1)^{2} + 1).

(a) If k ∈ \{1, 3\}, then |x| ∈ \{0, 1, \sqrt{2l − 1}\}.

(b) If k ∈ ℕ−\{1, 3\} then |x| ∈ \{0, 1, \sqrt{2l − 1}, \sqrt{x_{1}}, \sqrt{x_{2}}, \sqrt{x_{p}}\} where y_{1}+x_{1}\sqrt{A/4} > 1 is the minimal unit of 𝕆_{K} of norm 1 and p is prime ≡ 3 (mod 4) such that x_{1} = pu^{2} for some integer u.

The previous result uses the following proposition.

Proposition 2. Let B be non-zero square-free integer, and let A be a non-square positive integer. Assume that |B| < \sqrt{A}. Let (x, y) ∈ ℤ^{2}, such that y^{2} = Ax^{2m} + B. Then x^{m} = q_{n}, for some positive integer n, where p_{n}/q_{n} is the n-th convergent of \sqrt{A}.

We shall prove the following proposition.

Proposition 3. 1. Let (x, y) be an integer solution to equation

y^{2} = k^{2}(k^{2}l^{2} + 1)x^{4} + B,

with B square-free, k > 1, and 0 ≠ |B| ≤ k^{2}l. Then B = 1.

2. Let (x, y) be an integer solution to equation y^{2} = (l^{2} + 1)x^{4} + B, with B square-free, and 0 ≠ |B| ≤ l. Then |B| = 1.

3. Let A = 4k^{2}(k^{2}(2l − 1)^{2} + 1). Then the equation y^{2} = Ax^{4} + B, with B square-free, k > 3, and 0 ≠ |B| ≤ 2k^{2}(2l − 1) has integers solutions only for B = 1.

Another way to write Proposition 3 is as follows:

1. For A = k^{2}(k^{2}l^{2} + 1), if y^{2}, Ax^{4} are not consecutive integers and their difference is square free, then |y^{2} − Ax^{4}| > \sqrt{A}.

2. For A = 4k^{2}(k^{2}(2l − 1)^{2} + 1) (k > 3), if y^{2}, Ax^{4} are not consecutive integers and the difference is square free, then |y^{2} − Ax^{4}| > \sqrt{A}.

The previous proposition is not true for any value of A. For instance, the equation y^{2} = 6x^{4} − 2 has the solutions |x| ∈ {1, 3}.

The paper is organized as follows: in Section 2 we give some auxiliary results, in Section 3 we obtain the proof of Proposition 2 and in Section 4 the proof of the theorem. In Section 5 we provide the proof of Proposition 3, and finally in Section 6 we provide some examples.

2 Auxiliary results

We denote by u_{n}(r, s) the Lucas sequence

u_{0} = 0, u_{1} = 1, u_{n+2} = ru_{n+1} + su_{n},

and by v_{n}(r, s) the companion Lucas sequence

v_{0} = 2, v_{1} = r, v_{n+2} = rv_{n+1} + sv_{n},

where r, s are non-zero integers and n ≥ 0. A well-known and useful identity containing both of the above sequences is

v_{n}^{2} − (r^{2} + 4s)u_{n}^{2} = 4(−s)^{n}.  (2)

Lemma 4. The even-indexed terms of the form kx^{2} of a Lucas sequence are solutions of the equation

y^{2} − dx^{4} = 1,  (3)

where

1. d = k^{2}(k^{2}l^{2} + 1), if the Lucas sequence is (u_{n}(2kl, 1))_{n≥0}.

2. d = k^{2}(k^{2}(2l − 1)^{2} + 1), if the Lucas sequence is (u_{n}(2k(2l − 1), 1))_{n≥0}.

Proof. Starting with equation (2) and by substituting r with 2kl and s with 1 we get

v_{n}^{2} − (4k^{2}l^{2} + 4)u_{n}^{2} = 4.

Obviously v_{n} is even and therefore we can substitute v_{n} by 2y. Moreover, we are interested in terms u_{n} of the form kx^{2}, so by substituting u_{n} by kx^{2} we get y^{2} − k^{2}(k^{2}l^{2} + 1)x^{4} = 1. The proof of part (2) is similar and it is omitted.

Lemma 5. For d ∈ {k^{2}(k^{2}l^{2} + 1), k^{2}(k^{2}(2l − 1)^{2} + 1)} the equation y^{2} − dx^{4} = 1 has at most two solutions such that y + x^{2}\sqrt{d} ∈ \{ε_{d}, ε_{d}^{2}, ε_{d}^{p}\}. Where, ε_{d} = y_{1} + x_{1}\sqrt{d} is the minimal unit greater than 1 of norm 1 in ℤ[\sqrt{d}] and p is a prime ≡ 3 (mod 4) such that x_{1} = pu^{2}, for some u ∈ ℤ.

Proof. If we have exactly two solutions of y^{2} − dx^{4} = 1 then we get y + x^{2}\sqrt{d} ∈ \{ε_{d}, ε_{d}^{2}\} except if d = 1785 or 16 · 1785 [22, Thm. 1.1]. In that case we have y + x^{2}\sqrt{d} ∈ \{ε_{d}, ε_{d}^{4}\}. If we have exactly one solution, we get y + x^{2}\sqrt{d} ∈ \{ε_{d}, ε_{d}^{2}, ε_{d}^{p}\}, where p is prime such that x_{1} = pu^{2} for some u ∈ ℤ, and p ≡ 3 (mod 4). Since d cannot be equal to 1785 or 16 · 1785, the result follows.

Lemma 6. Let r ≥ 1, n ≥ 4, and let m ∈ {1, 2, 3, 6}. Assume that for some integer x we get u_{n}(r, 1) = mx^{2}. Then (n, r, m) = (4, 1, 3), (4, 2, 3), (4, 4, 2), (4, 24, 3), (6, 1, 2), (7, 2, 1) or (12, 1, 1).

Proof. [16, Thm. 1].

Lemma 7. The doubling rate of the recurrence sequence u_{n+2} = ru_{n+1} + su_{n} is

u_{n+4} = (r^{2} + 2s)u_{n+2} − s^{2}u_{n}.

Proof. By definition, we have that u_{n+4} = ru_{n+3} +su_{n+2}. By substituting u_{n+3} with ru_{n+2} + su_{n+1} we get

u_{n+4} = r^{2}u_{n+2} + sru_{n+1} + su_{n+2}.

Notice that ru_{n+1} = u_{n+2} − su_{n}. So,

u_{n+4} = r^{2}u_{n+2} + s(u_{n+2} − su_{n}) + su_{n+2}.

The result follows.

Lemma 8. Let k, l ∈ ℤ_{>0}. The continued fraction of \sqrt{A} is equal to:

1. [k^{2}l; \overline{2l, 2k^{2}l}] for A = k^{2}(k^{2}l^{2} + 1) and k ≠ 1.

2. [4k^{2}l − 2k^{2}; \overline{2l − 1, 8k^{2}l − 4k^{2}}] for A = 4k^{2}(k^{2}(2l − 1)^{2} + 1).

3. [l; \overline{2l}] for A = l^{2} + 1.

Proof. Let [k^{2}l; \overline{2l, 2k^{2}l}] be the continued fraction of \sqrt{A}. Then

\sqrt{A} = k^{2}l + \frac{1}{2l + \frac{1}{2k^{2}l+ \frac{1}{2l+···}}} .

So the quantity \sqrt{A} − k^{2}l is a periodic fraction and by substituting it, in the initial fraction we get

\sqrt{A} − k^{2}l = \frac{1}{2l + \frac{1}{2k^{2}l+(\sqrt{A}−k^{2}l)}} .

The result follows from the above equation.

The proof of part (2) is similar and it is omitted. The continued fraction in part (3) is a specific case of the first continued fraction for k = 1. The only difference is that in this case the period of the continued fraction is 1. Following the same procedure as in the first proof it turns out that the result follows from the equation

\sqrt{A} − l = \frac{1}{2l + (\sqrt{A} − l)}.

Let θ be a quadratic irrational number and [a_{0}, a_{1}, a_{2}, …] its simple continued fraction. We denote by (p_{n}) the sequence of the numerators, and by (q_{n}) the sequence of the denominators of the convergents to θ. By the theory of the continued fractions we have that

\frac{p_{n}}{q_{n}} = [a_{0}, a_{1}, …, a_{n}],

while gcd(p_{n}, q_{n}) = 1. The convergents p_{n} and q_{n} can be computed from the recurrence relations

p_{−2} = 0, p_{−1} = 1, p_{n} = a_{n}p_{n−1} + p_{n−2}, n ≥ 0,
q_{−2} = 1, q_{−1} = 0, q_{n} = a_{n}q_{n−1} + q_{n−2}, n ≥ 0  (4)

The continued fraction of a quadratic irrational θ has the form

[a_{0}; \overline{a_{1}, a_{2}, …, a_{2}, a_{1}, 2a_{0}}],

where the central term might appear either once or twice, see Burger [5, Lemma 8.6, p. 51].

Proposition 9. Let θ be a quadratic irrational number, and let [a_{0}; \overline{a_{1}, a_{2}, …, a_{s}}] be its simple continued fraction with period s. Then the terms of the sequence (q_{n}) satisfy the linear recurrence

q_{n+2s} − tq_{n+s} + (−1)^{s}q_{n} = 0, n ≥ 1

where t is the trace of the matrix

M = ∏_{1≤j≤s} [ \begin{matrix}a_{j} & 1 \\ 1 & 0 \end{matrix} ] .

Proof. [12, Thm. 1].

Lemma 10. Let (q_{n}) be the sequence of the denominators of the convergents to \sqrt{A}.

1. Let \sqrt{A} = [k^{2}l; \overline{2l, 2k^{2}l}]. Then

q_{n} = \begin{cases}u_{n+1}, & n even; \\ \frac{u_{n+1}}{k} , & n odd. \end{cases}

Where u_{n} are the terms of the Lucas sequence u_{n}(2kl, 1).

2. Let \sqrt{A} = [4k^{2}l − 2k^{2}; \overline{2l − 1, 8k^{2}l − 4k^{2}}]. Then

q_{n} = \begin{cases}u_{n+1}, & n even; \\ \frac{u_{n+1}}{2k} , & n odd. \end{cases}

Where u_{n} are the terms of the Lucas sequence u_{n}(2k(2l − 1), 1).

3. Let \sqrt{A} = [l; \overline{2l}]. Then q_{n} = u_{n+1}, where u_{n} are the terms of the Lucas sequence u_{n}(2l, 1).

Proof. From relation (4) we get q_{0} = 1, q_{1} = 2l, q_{2} = 4k^{2}l^{2} + 1. According to Proposition 9 the denominators (q_{n})_{n≥0} of the convergents to \sqrt{A} satisfy the recurrence sequence

q_{n+2s} − tq_{n+s} + (−1)^{s}q_{n} = 0, n ≥ 1,

where s = 2, and

t = trace ([ \begin{matrix}2l & 1 \\ 1 & 0 \end{matrix} ] · [ \begin{matrix}2k^{2}l & 1 \\ 1 & 0 \end{matrix} ]) = 4k^{2}l^{2} + 2.

Thus, we get

q_{n+4} = (4k^{2}l^{2} + 2)q_{n+2} − q_{n} = 0, n ≥ 1.  (5)

On the other hand, the first terms of the Lucas sequence u_{n}(2kl, 1) are u_{0} = 0, u_{1} = 1, u_{2} = 2klu_{1} + u_{0} = 2kl, u_{3} = 2klu_{2} + u_{1} = 4k^{2}l^{2} + 1. From Lemma 7 we have

u_{n+4} = (4k^{2}l^{2} + 2)u_{n+2} − u_{n} = 0, n ≥ 1.  (6)

Since u_{1} = q_{0} and u_{3} = q_{2}, inductively using ((5) and (6)), we can easily conclude that q_{2n} = u_{2n+1}, for all n ≥ 0.

Next, we will show that q_{2n−1} = u_{2n}/k. Since \sqrt{A} = [k^{2}l; \overline{2l, 2k^{2}l}], we have that a_{n} = 2l for n odd and a_{n} = 2k^{2}l for n even, n > 1. Now, let r be a positive integer. From (4) we have that q_{2r+2} = a_{2r+2}q_{2r+1} + q_{2r} which implies that

q_{2r+2} = 2k^{2}lq_{2r+1} + q_{2r}

and by substituting q_{2r+1} with

a_{2r+1}q_{2r} + q_{2r−1} = 2lq_{2r+1} + q_{2r},

we get

q_{2r+2} = (4k^{2}l^{2} + 1)q_{2r} + 2k^{2}lq_{2r−1}.

On the other hand, since (u_{n})_{n≥0} is a Lucas sequence, we have

u_{2r+3} = 2klu_{2r+2} + u_{2r+1}

which implies that

u_{2r+3} = (4k^{2}l^{2} + 1)u_{2r+1} + 2klu_{2r}.

We have already proved that the q_{2r+2} = u_{2r+3} and q_{2r} = u_{2r+1}. Hence, q_{2r−1} = u_{2r}/k. This proves part (1). The proof of part (2) is similar and we omit it.

Now we shall prove part (3). By Proposition 9 the denominators (q_{n})_{n≥0} of the convergents to \sqrt{A} satisfy the recurrence sequence

q_{n+2s} − tq_{n+s} + (−1)^{s}q_{n} = 0, n ≥ 1,

where s = 1 and t = 2l. Since q_{−1} = 0 and q_{−2} = 1 we get that q_{0} = 1, q_{1} = 2l, and so on. On the other hand the terms of the Lucas sequence are u_{0} = 0, u_{1} = 1, u_{2} = 2l etc. The result follows.

3 The proof of Proposition 2

Let A, B be non-zero integers. Assume that A > 0 non-square and B square-free with |B| < \sqrt{A}. From Lagrange [19, p. 377-535] or from Matthews [11], we get that the fraction y/x, where (x, y) is a positive solution of the Pell equation y^{2} − Ax^{2} = B, is equal to a convergent p_{n}/q_{n} for some n, of the continued fraction of \sqrt{A}. So, in our case it is \frac{y}{x^{m}} = \frac{p_{n}}{q_{n}},

where \frac{p_{n}}{q_{n}} is the n-th convergent of \sqrt{A}. Let gcd(y, x^{m}) = d > 1. Then

d^{2}y^{′2} = d^{4}Ax^{′2m} + B,

so d^{2}|B. But B is square-free, which leads us to a contradiction. We conclude that q_{n} = x^{m}. The proposition follows.

Remark 11. A similar result holds for Ay^{2} = x^{2m} + B, under the same constraints for B. So we can have a similar result, as in Theorem 1, for equations of the form Ay^{2} = x^{4} + B under the same constraints for A, B.

4 The proof of Theorem 1

By p_{n}/q_{n} we denote the convergents of the continued fraction of \sqrt{A}, where A ∈ \{k^{2}(k^{2}l^{2} + 1), 4k^{2}(k^{2}(2l − 1)^{2} + 1)\} (k, l ∈ ℤ_{>0}). Since 0 ≠ |B| < \sqrt{A}, by Proposition 2 for m = 2 we have that x^{2} = q_{s} for some positive integer s. That is, in order to compute the solutions (x, y) of the equation y^{2} = Ax^{4} + B it is enough to find the denominators of the convergents of \sqrt{A} that are perfect squares.

Proof. Part 1. According to Lemma 10 (1), the even-indexed terms q_{s} are equal to the odd-indexed terms u_{n} of the Lucas sequence u_{n}(2kl, 1), while the odd-indexed terms q_{s} are equal to the even-indexed terms u_{n} of the Lucas sequence u_{n}(2kl, 1) divided by k. So, in order to compute q_{s} that are perfect squares, we need to compute the odd-indexed terms of u_{n}(2kl, 1) that are of the form x^{2} and the even-indexed terms of u_{n}(2kl, 1) that are of the form kx^{2}.

In order to prove the case (a) of part (1), we consider the case where k = 1, which implies that A = l^{2} + 1. From Lemma 6 we conclude that the only possible case for n ≥ 4 is (n, 2l, 1) = (7, 2, 1). Thus l = 1, A = 2 and u_{7} = 169, which implies |x| = 13. For l = k = 1 we get the sequence u_{n}(2, 1) which is \underline{A000129} in Sloane’s Online Encyclopedia of Integer Sequences [21]. Now let n < 4. For n = 3 we have u_{3} = 4l^{2} + 1 which is not a perfect square unless l = 0 which contradicts the fact that l is positive. For n = 2 we have u_{2} = 2l and for n = 1 we have u_{1} = 1. Including the trivial solution x = 0, we conclude that |x| ∈ \{0, 1, 13, \sqrt{2l}\}, while |x| = 13 occurs only if A = 2, B = 1.

In order to prove the case (b) we consider k ∈ {2, 3, 6}. We assume first that n is odd ≥ 4. Thus we have to solve u_{n} = x^{2} for some integer x and u_{n} = u_{n}(2kl, 1). According to Lemma 6 (applying it for m = 1) we get (n, 2kl, 1) = (7, 2, 1), which implies that k = l = 1, contradicting the fact that k > 1. Thus there are no possible terms u_{n} of the form x^{2}.

For n even we need to compute the terms of the Lucas sequence u_{n}(2kl, 1) such that u_{n} = kx^{2} for some integer x. For n even and ≥ 4 by Lemma 6 we get the triplets (n, 2kl, 2) = (4, 4, 2) and (n, 2kl, 3) = (4, 2, 3), (4, 24, 3). If we have the triplet (4, 4, 2), we get u_{4}(4, 1) = 72 = 2x^{2}, which implies |x| = 6. The sequence u_{4}(4, 1) appears in Sloane’s database [21] as \underline{A001076}. In this case we have to solve the equation y^{2} = A · 6^{4} + B for A = 20 and |B| ≤ 4. For this equation there is a unique solution, which is (x, y, B) = (6, 161, 1). For the triplet (4, 2, 3) we have kl = 1, which is a contradiction. For the triplet (4, 24, 3) we have u_{4}(24, 1) = 13872 = 3x^{2} which implies that |x| = 68.

For x = 68, k = 3, l = 4, we get A = k^{2}(k^{2}l^{2} + 1) = 1305. For |B| ≤ \sqrt{1305} < 37 the equation y^{2} = Ax^{4} + B has a unique solution (|x| = 68, |y| = 167041).

Now let n < 4. For n = 3 we have u_{3} = 4k^{2}l^{2} + 1, which is not a perfect square, since k, l are positive integers. For n = 2, u_{2} = 2kl must be of the form kx^{2}, which implies that |x| = \sqrt{2l}. For n = 1 we have that |x| = 1. Including the trivial solution x = 0, we conclude that |x| ∈ \{0, 1, 6, \sqrt{2l}\}, where the solution |x| = 6 occurs only when A = 20, B = 1.

Finally, to prove the case (c) of part (1), we consider the case where k ∉ {1, 2, 3, 6}. As in the previous case, we need to compute the square terms of the Lucas sequence u_{n}(2kl, 1) with an odd index. By Lemma 6 we conclude that there are no possible solutions for n ≥ 4, and the cases where n = 3 or n = 1 are the same with case (a).

For n even we need to compute the terms of the form kx^{2} of the Lucas sequence u_{n}(2kl, 1). The integer x is a solution of the equation y^{2} − dx^{4} = 1, where d = k^{2}(k^{2}l^{2} + 1) (Lemma 4). By Lemma 5 we get |x| ∈ \{\sqrt{x_{1}}, \sqrt{x_{2}}, \sqrt{x_{p}}\}, where y_{1} + x_{1}\sqrt{d} = ε_{d} is the minimal unit (> 1) of the ring ℤ[\sqrt{d}] and ε_{d}^{t} = y_{t} + x_{t}\sqrt{d}. The solutions for n < 4 and m ≠ 1, 2, 3, 6 are analogous to the previous cases, so |x| ∈ \{0, 1, \sqrt{2l}, \sqrt{x_{1}}, \sqrt{x_{2}}, \sqrt{x_{p}}\}.

Part 2. According to Lemma 10 (2), the even-indexed terms q_{s} are equal to the oddindexed terms u_{n} of the Lucas sequence u_{n}(2k(2l −1), 1), while the odd-indexed terms q_{s} are equal to the even-indexed terms u_{n} of the Lucas sequence u_{n}(2k(2l − 1), 1) divided by 2k.

So, in order to compute q_{s} that are perfect squares, we need to compute the odd-indexed terms of u_{n}(2kl, 1) that are of the form x^{2}, and the even-indexed terms of u_{n}(2k(2l − 1), 1) that are of the form 2kx^{2}.

Now consider the case (a) of part (2). Since k ∈ {1, 3}, by Lemma 6 for m = 1 we get that the only possible case for n odd and n ≥ 4 is (n, 2k(2l − 1), 1) = (7, 2, 1). So l = k = 1, A = 8 and u_{7} = 169 = x^{2}. We easily conclude that there are no integral solutions of y^{2} = 8x^{4} + B for x = 13 and |B| < \sqrt{A}. For n = 3 we have u_{3} = 4k^{2}(2l − 1)^{2} + 1, which is clearly not a perfect square since k ≠ 0, and for n = 1, we have u_{1} = 1, which is obviously a perfect square.

Now let n be even. We need to compute the terms of the Lucas sequence u_{n}(2k(2l−1), 1), such that u_{n} = 2kx^{2} for some integer x. For n ≥ 4 according to Lemma 6 for m = 2k (since k ∈ {1, 3} we get m ∈ {2, 6}) we can easily deduce that there is no possible triplet of the form (n, 2k(2l − 1), 2k). For n = 2, we have u_{2} = 2k(2l − 1), which is of the form 2kx^{2} for |x| = \sqrt{2l − 1}. Including the trivial solution x = 0, we conclude that |x| ∈ \{0, 1, \sqrt{2l − 1}\}.

Finally, consider the case where k ∉ {1, 3} (case (b)). As in the previous case (a), according to Lemma 6 there are no possible solutions for n odd, n ≥ 4, and the cases where n = 3 or n = 1 are also the same with the previous case. For n even we work as in case (c) of part (1). Including the trivial solution x = 0, we conclude that |x| ∈ \{0, 1, \sqrt{2l − 1}, \sqrt{x_{1}}, \sqrt{x_{2}}, \sqrt{x_{p}}\}.

5 The proof of Proposition 3

Initially, we recall Euler’s theorem about continued fractions as it is provided by Burger [5, p. 52].

Theorem 12. (Euler’s theorem). Let α be a quadratic irrational, and let α_{0} = α, a_{0} = ⌊α_{0}⌋. Then α_{0} = (r_{0}+\sqrt{d})/s_{0}, where d is not a perfect square and s_{0}|d−r_{0}^{2}. Let α_{n+1} = 1/(α_{n}−a_{n}), a_{n+1} = ⌊α_{n+1}⌋. Further, we consider two sequences s_{n}, r_{n} such that

r_{n+1} = a_{n}s_{n} − r_{n}, s_{n+1} = \frac{d − r_{n+1}^{2}}{s_{n}} .

Then for all n ≥ 0 it follows that:

1. α = [a_{0}, a_{1}, a_{2}, …].

2. r_{n}, s_{n} ∈ ℤ with s_{n} ≠ 0 and s_{n}|d − r_{n}^{2}.

3. α_{n} = (r_{n} + \sqrt{d})/s_{n} and a_{n} = ⌊(r_{n} + \sqrt{d})/s_{n}⌋.

4. If r_{0} = 0, s_{0} = 1 we set p_{n}/q_{n} to be the n-th convergent of \sqrt{d}. Then for all n ≥ 0 we

get

p_{n}^{2} − dq_{n}^{2} = (−1)^{n−1}s_{n+1}.  (7)

Further, if the second part of equation (7) is ±1, then n = ts − 1 for some t ≥ 0, where s is the period of the continued fraction of \sqrt{d}.

5. s_{n} = 1 ⇔ s|n. Proof. For part (1),(2) and (3) see [5, Euler’s theorem, Module 8]. For part (4) see [5, Thm. 8.11 and Cor. 8.12] and for part (5) see [5, Lemma 8.9 (iv)].

Let x, y be integers that satisfy equation (1), and let |B| < \sqrt{A}. Then y = p_{n} and x^{2} = q_{n}. Thus p_{n}^{2} − Aq_{n}^{2} = y^{2} − Ax^{4} = B. Part 4 of the previous theorem, give us

p_{n}^{2} − Aq_{n}^{2} = (−1)^{n−1}s_{n+1},

so

B = (−1)^{n−1}s_{n+1}.  (8)

Therefore B is allowed to take specific discrete values. We shall prove that for the A’s we have considered, B is finally equal to 1. Lemma 13. Let A = k^{2}(k^{2}l^{2} + 1), and let n be a positive integer. Then we get α_{n} = k^{2}α_{1} for n even and α_{n} = α_{1} for n odd. Proof. From Lemma 8 (1) we get \sqrt{A} = [k^{2}l; \overline{2l, 2k^{2}l}]. According to the notation of Theorem 12, α = \sqrt{A} = α_{0}. So

a_{0} = k^{2}l, a_{2n+1} = 2l, a_{2n} = 2k^{2}l.

It is easy to see that

α_{1} = \frac{1}{α_{0} − a_{0}} = \frac{\sqrt{A} + k^{2}l}{k^{2}}, α_{2} = \frac{1}{α_{1} − a_{1}} = k^{2}α_{1} and α_{3} = α_{1}.

By induction our result follows. Lemma 14. For A = k^{2}(k^{2}l^{2} + 1), we get

s_{n} = \begin{cases}1, & n even; \\ k^{2}, & n odd. \end{cases}

Further, r_{n} = k^{2}l for every n ≥ 0. Proof. From the previous lemma

α_{n} = \begin{cases}k^{2}α_{1}, & n even; \\ α_{1}, & n odd. \end{cases}

Also the equality α_{n} = \frac{r_{n}+\sqrt{A}}{s_{n}} holds. Since α_{1} = \frac{k^{2}l+\sqrt{A}}{k^{2}}, our result follows.

In order to prove part (1) of Proposition 3, first we remark that

|B| ≤ \sqrt{A} = \sqrt{k^{2}(k^{2}l^{2} + 1)} = k\sqrt{k^{2}l^{2} + 1}.

Since |B| is an integer, the closest integer to the boundary of |B| is k\sqrt{k^{2}l^{2}}, so |B| ≤ k^{2}l. From Lemma 14 and relation (8) we get B ∈ {1, −k^{2}} and since B is square-free, Proposition 3 part (1) follows. The second part is similarly proved.

For the third part, we repeat some arguments of the proof of part (2b) of Theorem 1. First we consider the equation

p_{n}^{2} − Aq_{n}^{2} = (−1)^{n−1}s_{n+1}  (9)

with n even. So according to Lemma 10 part (2) we have to find the odd-indexed terms of the Lucas sequence u_{n}(2k(2l − 1), 1) which are perfect squares. But, such terms do not exist except when n = 1 (see the proof of part (2b) of Theorem 1). In the specific case where n = 1, a straightforward computation gives s_{2} = 1, thus B = 1. For n odd, the index in the sequence in the right part of relation (9) is even. In that case, since the period of the continued fraction of \sqrt{A} is 2, Theorem 12 part (5) gives s_{n+1} = 1. So in any case, from relation (8), we get B = 1.

6 Examples

Example 15. For k = 1, l = 301, and A = 4k^{2}(k^{2}(2l − 1)^{2} + 1) = 1444808. We consider the family E_{B} : y^{2} = Ax^{4} + B with B square-free, and |B| ≤ 1202. In this family there are elliptic curves with large rank. For instance if B = 1 we get an elliptic curve of rank 5. According to part 2 (a) of Theorem 1, we get |x| ∈ \{0, 1\} (since \sqrt{2l − 1} is not an integer). Solutions occur only for the trivial case where x = 0. So, the solutions (x, y, B) are (0, 1, 1) and (0, −1, 1).

Example 16. Now let k = 12, l = 1, and A = k^{2}(k^{2}l^{2} + 1) = 20880. We consider the family of the curves E_{B} : y^{2} = Ax^{4} + B, where B is as usual square-free and |B| ≤ 144. For B = −5 we get a curve of rank 3. According to part 1 (c) of Theorem 1, we get |x| ∈ \{0, 1, \sqrt{x_{1}}, \sqrt{x_{2}}, \sqrt{x_{p}}\}, and B = 1 from the first part of Proposition 3. The fundamental solution of y^{2} − Ax^{2} = 1 is 289 + 2\sqrt{20880} > 1. So y_{1} = 289 and x_{1} = 2. Since x_{1} is not a perfect square and is not of the form pu^{2} where p ≡ 3 (mod 4), we only need to check x_{2}. Now, (289 + 2\sqrt{20880})^{2} equals to 167041 + 1156\sqrt{20880}. Thus x_{2} = 1156 which is the square of 34. Therefore, the solutions (x, y, B) of y^{2} = 20880x^{4} + B are (0, ±1, 1) and (±34, ±167041, 1).

Example 17. Now we consider the equation y^{2} = (2^{4r+2}+1)x^{4}+n, where −2^{2r+1} ≤ n ≤ 2^{2r+1} and n ≠ 0 square-free. Then our main result for k = 1 and l = 2^{2r+1} gives |x| ∈ {0, 1, 2^{r+1}}. Further, from the second part of Proposition 3 we get |n| = 1. Thus, after some simple calculations we get (|x|, |y|, n) = (0, 1, 1), (1, 2^{2r+1}, −1), (2^{r+1}, 2^{4r+3} + 1, 1).

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2010 Mathematics Subject Classification: Primary 11D25; Secondary 11D41, 11A55, 11B39. Keywords: Lucas sequence, continued fraction, Diophantine equation, quartic elliptic curve.

(Concerned with sequences \underline{A000129} and \underline{A001076}.)

Received November 4 2014; revised versions received January 31 2015; February 23 2015. Published in Journal of Integer Sequences, May 12 2015.

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