On the Extension of the Diophantine Pair \{1, 3\} in ℤ[\sqrt{d}]

Zrinka Franušić Department of Mathematics University of Zagreb Bijenička cesta 30 10000 Zagreb Croatia

                         fran@math.hr

Abstract

In this paper, we consider Diophantine triples of the form \{1, 3, c\} in the ring ℤ[\sqrt{d}]. We prove that the Diophantine pair {1, 3} cannot be extended to the Diophantine quintuple in ℤ[\sqrt{d}] with d < 0 and d ≠ −2.

1 Introduction

A Diophantine m-tuple in a commutative ring R with the unit 1 is a set of m distinct nonzero elements with the property that the product of each two distinct elements increased by 1 is a perfect square in R. These sets are studied in many different rings: the ring of integers ℤ, the ring of rationals ℚ, the ring of Gaussian integers ℤ[i] ([2, 8]), the ring of integers of quadratic fields ℤ[\sqrt{d}] and ℤ[(1 + \sqrt{d})/2] ([7, 9]), polynomial rings ([4, 5]). The most famous Diophantine m-tuples and historical examples are quadruples \{\frac{1}{16}, \frac{33}{16}, \frac{17}{4}, \frac{105}{16}\} (found by Diophant) and {1, 3, 8, 120} (found by Fermat). More information about these sets can be found on Dujella’s web page on Diophantine m-tuples [3].

In 1969, Baker and Davenport [1] showed that the Diophantine triple {1, 3, 8} can be extended uniquely to the Diophantine quadruple \{1, 3, 8, 120\} \underline{A030063} . Hence, the triple {1, 3, 8} cannot be extended to a Diophantine quintuple. Jones [10] proved that if the set {1, 3, c} is a Diophantine triple then c must be of the form

c_{k} = \frac{1}{6}((2 + \sqrt{3})(7 + 4\sqrt{3})^{k} + (2 − \sqrt{3})(7 − 4\sqrt{3})^{k} − 4),

for some k ∈ ℕ (\underline{A045899}). It can be easily verified that sets \{1, 3, c_{k}, c_{k−1}\} and \{1, 3, c_{k}, c_{k+1}\} are Diophantine quadruples. Indeed, c_{k}c_{k+1} + 1 = t_{k}^{2}, where (t_{k}) is \underline{A051048}. Moreover, the Diophantine triple {1, 3, c_{k}} can be extended to a quadruple only by the elements c_{k−1} and c_{k+1}. This assertion is proved in 1998. by Dujella and Pethö [6]. A direct consequence of their assertion is the fact that the pair {1, 3} cannot be extended to a Diophantine quintuple.

A natural question that may arise is can we extend the pair {1, 3} to a quintuple in a larger ring. In this paper, we have chosen the ring ℤ[\sqrt{d}] = \{a+b\sqrt{d} : a, b ∈ ℤ\} where d ∈ ℤ and d is not a perfect square. So, we assume that \{1, 3, a + b\sqrt{d}\} is a Diophantine triple in ℤ[\sqrt{d}] and according to a definition it means that there exist integers ξ_{1}, η_{1}, ξ_{2}, η_{2} such that

a + b\sqrt{d} + 1 = (ξ_{1} + η_{1}\sqrt{d})^{2}  (1)
3a + 3b\sqrt{d} + 1 = (ξ_{2} + η_{2}\sqrt{d})^{2}.  (2)

Although, it seams that we have much more ’freedom’ in larger rings, the result for some rings stays the same.

Theorem 1. Let d be a negative integer and d ≠ −2. The Diophantine pair {1, 3} cannot be extended to a Diophantine quintuple in the ring ℤ[\sqrt{d}].

In certain rings ℤ[\sqrt{d}] for positive d, Theorem 1 is not valid. For example \{1, 3, 8, 120, 1680\} is the Diophantine quintuple in ℤ[\sqrt{8 · 1680 + 1}].

In last two sections, we describe the set of all Diophantine triples \{1, 3, c\} in ℤ[\sqrt{−2}] and in ℤ[\sqrt{d}] for some positive integer d, where c is an integer and such that \{1, 3, c\} is not a Diophantine triple in ℤ. The existence of such triples is related to solvability of certain Pellian equations.

2 Main result

Proposition 2. Let d be a negative integer. If {1, 3, c} is a Diophantine triple in the ring ℤ[\sqrt{d}], then c is an integer.

Proof. Let c = a + b\sqrt{d}. There exist ξ_{1}, η_{1}, ξ_{2}, η_{2} ∈ ℤ such that

a + 1 = ξ_{1}^{2} + η_{1}^{2}d, b = 2ξ_{1}η_{1}, 3a + 1 = ξ_{2}^{2} + η_{2}^{2}d, 3b = 2ξ_{2}η_{2}.  (3)

Evidently, b must be an even number. By squaring and subtracting corresponding relations in (3), we get

(a + 1)^{2} − db^{2} = x^{2}, (3a + 1)^{2} − d(3b)^{2} = y^{2},  (4)

where x = ξ_{1}^{2} − η_{1}^{2}d and y = ξ_{2}^{2} − η_{2}^{2}d. By eliminating the quantities b and d from (4), we obtain

(3x)^{2} − y^{2} = (3a + 3)^{2} − (3a + 1)^{2} = 4(3a + 2),  (5)

It is clear, from (5), that 3x − y and 3x + y must be even. So, we have

3x − y = 2^{k}c_{1}, 3x + y = 2^{l}c_{2},

where k, l are positive integers and c_{1}, c_{2} are odd. Thus, we have

(3x − y)(3x + y) = 2^{k+l}n,

where and c_{1}c_{2} = n. Furthermore, it follows that

y = 2^{l−1}c_{2} − 2^{k−1}c_{1}.

The expression (3a + 1)^{2} − y^{2} can be given in terms of c_{1} and c_{2}:

(3a + 1)^{2} − y^{2} = (2^{k+l−2}n − 1)^{2} − (2^{l−1}c_{2} − 2^{k−1}c_{1})^{2}
= 4^{k+l−2}n^{2} − 2^{k+l−1}n + 1 − 4^{l−1}c_{2}^{2} + 2^{k+l−1}c_{1}c_{2} − 4^{k−1}c_{1}^{2}
= 4^{k+l−2}n^{2} + 1 − 4^{l−1}c_{2}^{2} − 4^{k−1}c_{1}^{2}
= (4^{k−1}c_{1}^{2} − 1)(4^{l−1}c_{2}^{2} − 1).

On the other hand, (3a + 1)^{2} − y^{2} = d(3b)^{2} (by (4)). Since |c_{i}| ≥ 1 and k, l ≥ 1, we obtain that d < 0 implies b = 0.

According to the previous proposition, it may have sense to study Diophantine triples of the form \{1, 3, c\} in ℤ[\sqrt{d}] such that c is an integer.

The triple \{1, 3, c\}, c ∈ ℤ, will be called a proper triple in ℤ[\sqrt{d}] if it is a Diophantine triple ℤ[\sqrt{d}], but not a Diophantine triple in ℤ. For instance, \{1, 3, 161\} is a proper triple in ℤ[\sqrt{2}], \{1, 3, −3\} is a proper triple in ℤ[\sqrt{−2}], but \{1, 3, 120\} is not a proper triple in ℤ[\sqrt{d}] (d ≠ 1).

Let us, now, assume that \{1, 3, c\} is proper triple in ℤ[\sqrt{d}]. So, one of the following cases may occur

\begin{aligned}c + 1 & = & ξ^{2} \\ 3c + 1 & = & dη^{2}, \end{aligned}  (6)

\begin{aligned}c + 1 & = & dη^{2} \\ 3c + 1 & = & ξ^{2}, \end{aligned}  (7)

\begin{aligned}c + 1 & = & dη_{1}^{2} \\ 3c + 1 & = & dη_{2}^{2}. \end{aligned}  (8)

Following proposition gives a connection between the extensibility of the Diophantine pair {1, 3} and solvability of certain Pellian equations.

Proposition 3. The Diophantine pair {1, 3} can be extended to a proper Diophantine triple \{1, 3, c\} ⊂ ℤ in ℤ[\sqrt{d}] if and only if one of the following equations

3x^{2} − dy^{2} = 2  (9)

and

x^{2} − 3dy^{2} = −2  (10)

is solvable in ℤ.

Proof. ⇒ If {1, 3, c} is a proper triple and (6) is fulfilled, then by eliminating c, we get 3ξ^{2} − dη^{2} = 2. So, the equation (9) is solvable in ℤ.

Analogously, by eliminating c from (7), we obtain that the equation (10) is solvable in ℤ. Finally, from (8) we have

d(η_{2}^{2} − 3η_{1}^{2}) = −2,

and the only possibilities are:

(a) d = −1 and η_{2}^{2} − 3η_{1}^{2} = 2, which is impossible (because the x^{2} − 3y^{2} = 2 is not solvable in ℤ) ,

(b) d = 2 and η_{2}^{2} − 3η_{1}^{2} = −1, which is also impossible for the same reason,

(c) d = −2 and η_{2}^{2} − 3η_{1}^{2} = 1, which is possible because the Pell’s equation x^{2} − 3y^{2} = 1 has infinitely many solution. Note that the equation (9) is also solvable if d = −2, it has a unique solution (0, 1).

⇐ Assume that (ξ, η) ∈ ℤ^{2} is a solution of the equation (9). Then the set {1, 3, ξ^{2} − 1} represents a proper triple in ℤ[\sqrt{d}]. Indeed, 3(ξ^{2} − 1) + 1 = dη^{2} = (n\sqrt{d})^{2}.

Similarly, if (ξ, η) ∈ ℤ^{2} is a solution of (10), then we get a proper triple {1, 3, dη^{2}−1}.

In order to prove Theorem 1, we need a special case of Theorem 8 from [10].

Lemma 4 (Theorem 8, [10]). If {1, 3, c} is a Diophantine triple in ℤ, then c = c_{k} for some integer k ≥ 2 where

c_{k} = 14c_{k−1} − c_{k−2} + 6, c_{1} = 8, c_{0} = 0.

By solving the recursion from Lemma 4 we get

c_{k} = \frac{1}{6}((2 + \sqrt{3})(7 + 4\sqrt{3})^{k} + (2 − \sqrt{3})(7 − 4\sqrt{3})^{k} − 4), k ≥ 0  (11)

Proposition 5. Let d be a negative integer and d ≠ −2. If {1, 3, c} is a Diophantine triple in ℤ[\sqrt{d}], then c = c_{k} for some positive integer k, where c_{k} is given by (11).

Proof. According to Proposition 2, c is an integer. Hence, {1, 3, c} is a triple in ℤ or a proper triple in ℤ[\sqrt{d}]. Since the equations (9) and (10) are not solvable in ℤ for d < 0 and d ≠ −2, Proposition 3 implies that \{1, 3, c\} is not a proper triple in ℤ[\sqrt{d}]. Hence, it is a Diophantine triple in ℤ. Finally, from Lemma 4 we get that c = c_{k} for some k.

Lemma 6 (Theorem 1, [6]). If {1, 3, c_{k}, d} is a Diophantine quadruple in ℤ where c_{k} is given by (11) and k is a positive integer, then d ∈ {c_{k−1}, c_{k+1}}.

At this point we can conclude that Theorem 1 is proved. (It follows directly from Proposition 5 and Lemma 6.)

Since the extension of the pair {1, 3} is closely related to the solvability of the Pellian equation x^{2} − 3y^{2} = −2, our result can be understood as the following statement.

Corollary 7. Let d be a negative integer and d ≠ −2. The equation

x^{2} − 3y^{2} = −2  (12)

in ℤ[\sqrt{d}] has only real solutions.

Proof. Suppose that (ξ_{1} + η_{1}\sqrt{d}, ξ_{2} + η_{2}\sqrt{d}) is a solution of the equation (12). For

a + b\sqrt{d} = (ξ_{2} + η_{2}\sqrt{d})^{2} − 1,

we obtain that \{1, 3, a + b\sqrt{d}\} is a Diophantine triple u ℤ[\sqrt{d}]. Indeed,

3(a + b\sqrt{d}) + 1 = 3(ξ_{2} + η_{2}\sqrt{d})^{2} − 2 = (ξ_{1} + η_{1}\sqrt{d})^{2}.

According to Proposition 5, there exists some k such that a + b\sqrt{d} = c_{k}. Immediately, we conclude that η_{1} = η_{2} = 0.

3 The case d = −2

In this section, we will take a closer look at the Diophantine triples \{1, 3, c\} in ℤ[\sqrt{−2}]. In fact, we are able to describe all Diophantine triples in ℤ[\sqrt{−2}].

Proposition 8. If \{1, 3, c\} is a Diophantine triple in ℤ[\sqrt{−2}], then c ∈ \{−1, c_{k}, d_{k}\} for some k ∈ ℕ, where c_{k} is given by (11) and

d_{k} = −\frac{1}{6} ((7 + 4\sqrt{3})^{k} + (7 − 4\sqrt{3})^{k} + 4)  (13)

Proof. By Proposition 2, we have that c is an integer. Since the equation (9) is solvable in ℤ, Proposition 3 implies that \{1, 3, c\} can be a triple in ℤ or a proper triple in ℤ[\sqrt{−2}]. If \{1, 3, c\} is a triple in ℤ, then c = c_{k} from Lemma 4. If \{1, 3, c\} is a proper triple in ℤ[\sqrt{−2}], then one of the cases (6) and (8) may occur. The case (6) implies that c = −1, because (0, 1) is the unique solution of (9).

In the case of (8), c = (−2x^{2} − 1)/3 where x is a solution of the Pell’s equation

x^{2} − 3y^{2} = 1.  (14)

All solutions of (14) are

x_{k+2} = 4x_{k+1} − x_{k}, x_{1} = 2, x_{0} = 1,

for k ≥ 0. By solving the recursion we obtain that

x_{k} = \frac{1}{2}((2 − \sqrt{3})^{k} + (2 + \sqrt{3})^{k}),

and d_{k} = (−2x_{k}^{2} − 1)/3.

Note that for k = 0, we have d_{0} = −1. Also, it is interesting that (−d_{k}) is \underline{A011922}.

Remark 9. It is known that {1, 3, c_{k}, c_{k+1}} is a Diophantine quadruple in ℤ for k ≥ 1 (Lemma 6). The same can be proven for the set \{1, 3, d_{k}, d_{k+1}\} in ℤ[\sqrt{−2}] for k ≥ 0. We have

36(d_{k}d_{k+1} + 1) = 66 + (7 − 4\sqrt{3})^{2k+1} + (7 + 4\sqrt{3})^{2k+1} +
16((7 − 4\sqrt{3})^{k}(2 − \sqrt{3}) + (2 + \sqrt{3})(7 + 4\sqrt{3})^{k})
= 64 + 16((2 + \sqrt{3})^{2k+1} + (2 − \sqrt{3})^{2k+1}) +
((2 + \sqrt{3})^{2k+1} + (2 − \sqrt{3})^{2k+1})^{2}
= 4(x_{2k+1}^{2} + 8x_{2k+1} + 16) = 4(x_{2k+1} + 4)^{2},

where x_{2k+1} is a solution of x^{2} − 3y^{2} = 1. It can be easily shown that x_{2k+1} ≡ 2 (mod 3) (because x_{n+2} = 4x_{n+1} − x_{n}, x_{0} = 1, x_{1} = 2), so

\sqrt{d_{k}d_{k+1} + 1} = \frac{1}{3}(x_{2k+1} + 4)

is an integer.

Further, the set {1, 3, c_{k}, d_{l}} is not a Diophantine quadruple for k ≥ 1 and l ≥ 0. Indeed, it can be easily seen that c_{k}d_{l} + 1 is a negative odd number (from Lemma 4 it follows that c_{k} is even, and from (13) that d_{l} < 0). Hence, \sqrt{c_{k}d_{l} + 1} ∉ ℤ[\sqrt{−2}].

4 Proper triples in ℤ[\sqrt{d}] for some d > 0

According to Proposition 3, the set \{1, 3, c\} ⊂ ℤ is a proper Diophantine triple in ℤ[\sqrt{d}] if and only if one of the equations (9), (10) is solvable in ℤ.

Example 10. Determine all proper Diophantine triples \{1, 3, c\} in ℤ[\sqrt{10}].

Note that d = 10 is the smallest positive integer, d ≠ 1, such that the equation (9) is solvable in ℤ. If (x_{n}, y_{n}) is a solution of (9), then \{1, 3, x_{n}^{2} − 1\} is a proper triple in ℤ[\sqrt{10}]. All solutions of (9) are

x_{n+2} = 22x_{n+1} − x_{n}, x_{1} = 42, x_{0} = 2,

for n ≥ 0. Hence,

x_{n} = \frac{1}{6}((6 + \sqrt{30})(11 + 2\sqrt{30})^{n} + (6 − \sqrt{30})(11 − 2\sqrt{30})^{n}), n ≥ 0.

So, all proper triples \{1, 3, c\} in ℤ[\sqrt{10}] are \{1, 3, d_{n}\} where

d_{n} = x_{n}^{2} − 1 = \frac{1}{6}((11 + 2\sqrt{30})^{2n+1} + (11 − 2\sqrt{30})^{2n+1} − 4), n ≥ 0.  (15)

Besides these triples, sets \{1, 3, 25 ± 8\sqrt{10}\}, \{1, 3, 160 ± 44\sqrt{10}\}, \{1, 3, 355 ± 112\sqrt{10}\} are also Diophantine triples in ℤ[\sqrt{10}]. The third element of these triples is related to the solution in ℤ[\sqrt{10}] of the equation x^{2} − 3y^{2} = −2. For instance, (6 + 2\sqrt{10}, 4 + \sqrt{10}) is solution of x^{2} −3y^{2} = −2 and 25+8\sqrt{10} = (4+\sqrt{10})^{2} −1. More generally, if X^{2} −3Y^{2} = −2 and X, Y ∈ ℤ[\sqrt{10}] then \{1, 3, Y^{2} − 1\} is a Diophantine triple in ℤ[\sqrt{10}].

Example 11. Determine all proper Diophantine triples \{1, 3, c\} in ℤ[\sqrt{2}].

Since the equation (10) is solvable in ℤ, \{1, 3, 2y_{n}^{2} − 1\} is a proper triple in ℤ[\sqrt{2}] where (x_{n}, y_{n}) is a solution of (10). All solutions of (10) are

y_{n+2} = 10y_{n+1} − y_{n}, y_{1} = 9, y_{0} = 1,

for n ≥ 0, i.e.,

y_{n} = \frac{1}{6}((5 − 2\sqrt{6})^{n}(3 − \sqrt{6}) + (3 + \sqrt{6})(5 + 2\sqrt{6})^{n}).

So, all proper triples \{1, 3, c\} in ℤ[\sqrt{2}] are \{1, 3, d_{n}\} where

d_{n} = 2y_{n}^{2} − 1 = \frac{1}{6}((5 + 2\sqrt{6})^{2n+1} + (5 − 2\sqrt{6})^{2n+1} − 4).  (16)

Also, because the equation (10) is solvable in ℤ[\sqrt{2}], there exist triples like \{1, 3, 11 ± 8\sqrt{2}\}, \{1, 3, 32 ± 20\sqrt{2}\}, \{1, 3, 161 ± 112\sqrt{2}\}. Like in Example 10, the third element of these triples is obtained from the solution (X, Y) ∈ ℤ[\sqrt{2}]^{2} of the equation x^{2} − 3y^{2} = −2, i.e., \{1, 3, Y^{2} − 1\} is a triple in ℤ[\sqrt{2}].

It is interesting to note that the expression (15) contains 11 + 2\sqrt{30} (i.e. (11, 2)) which is a fundamental solution of the Pell’s equation x^{2} − 30y^{2} = 1. Similarly, in (16), 5 + 2\sqrt{6} represents a fundamental solution of x^{2} − 6y^{2} = 1.

In what follows, we will give the complete set of proper Diophantine triples {1, 3, c} in ℤ[\sqrt{d}] for positive d which satisfies certain conditions. This problem is related to the solvability of the equation x^{2} − 3dy^{2} = 6. For this equation, we have the following Nagell’s result.

Lemma 12 (Theorem 11, [11]). Let D be a positive integer which is not a perfect square, and let C ≠ 1, −D be a square-free integer which divides 2D. Then if (x_{0}, y_{0}) is the least positive (fundamental) solution of the equation

x^{2} − Dy^{2} = C,  (17)

then

\frac{1}{|C|}(x_{0} + y_{0}\sqrt{D})^{2} = \frac{x_{0}^{2} + Dy_{0}^{2}}{|C|} + \frac{2x_{0}y_{0}}{|C|} \sqrt{D}

is a fundamental solution of the related Pell’s equation x^{2} − Dy^{2} = 1. All solutions of (17) are

\frac{x_{n} + y_{n}\sqrt{D}}{\sqrt{|C|}} = (\frac{x_{0} + y_{0}\sqrt{D}}{\sqrt{|C|}})^{n} ,

where n is a positive odd integer.

Theorem 13. Let d be a positive integer such that neither d nor 3d are perfect squares and such that one of the equations (9), (10) is solvable in ℤ. If {1, 3, c} is a proper Diophantine triple in ℤ[\sqrt{d}], d ≠ 2, 10, then there exists nonnegative integer n such that

c = d_{n} = \frac{1}{6}((ξ + η\sqrt{3d})^{2n+1} + (ξ − η\sqrt{3d})^{2n+1} − 4),  (18)

where (ξ, η) is a fundamental solution of the Pell’s equation

x^{2} − 3dy^{2} = 1.  (19)

If d ∈ {2, 10}, then c = d_{n} for some n ≥ 1.

Proof. First, assume that the equation (9) is solvable in ℤ. Hence, c = x^{′2} − 1, where (x′, y′) is a solution of (9). By Lemma 12, we get that all solutions of (9) are

x_{n} = \frac{1}{6}((α + β\sqrt{3d})(ξ + η\sqrt{3d})^{n} + (α − β\sqrt{3d})(ξ − η\sqrt{3d})^{n}), n ≥ 0,

where (α, β) is a fundamental solution of the equation

x^{2} − 3dy^{2} = 6  (20)

and (ξ, η) is a fundamental solution of the related Pell’s equation (19). (Note that (20) has exactly one class of solutions, since 6|x′x′′ − 3dy′y′′ i 6|x′y′′ − y′x′′ for each two solutions (x′, y′) and (x′′, y′′) of (20)). Hence,

d_{n} = \frac{1}{36}((α + β\sqrt{3d})^{2}(ξ + η\sqrt{3d})^{2n} + (α − β\sqrt{3d})^{2}(ξ − η\sqrt{3d})^{2n} + 12) − 1.

According to Lemma 12, \frac{1}{6}(α + β\sqrt{3d})^{2} (i.e., (\frac{1}{6}(α^{2} + 3dβ^{2}), \frac{1}{3}αβ) is a fundamental solution of x^{2} − 3dy^{2} = 1, and so, we get (18).

If (10) is solvable in ℤ, we obtain (18) in similar manner. Indeed, if (x′, y′) is a solution of (10), then c = dy^{′2} − 1. All solutions of (10) are

y_{n} = \frac{1}{2\sqrt{3d}}((x_{0} + y_{0}\sqrt{3d})(ξ + η\sqrt{3d})^{n} + (−x_{0} + y_{0}\sqrt{3d})(ξ − η\sqrt{3d})^{n}), n ≥ 0,

where (x_{0}, y_{0}) and (ξ, η) are fundamental solutions of (10) and (19), respectively. Further, we have

d_{n} = \frac{1}{6} (\frac{1}{2}(x_{0} + y_{0}\sqrt{3d})^{2}(ξ + η\sqrt{3d})^{2n} + \frac{1}{2}(x_{0} − y_{0}\sqrt{3d})^{2}(ξ − η\sqrt{3d})^{2n} − 4) .

From Lemma 12, we get that \frac{1}{2}(x_{0} ± y_{0}\sqrt{3d})^{2} = ξ ± η\sqrt{3d} and again obtain (18).

Finally, we must determine in which cases d_{n} ∈ {0, 1, 3}, since a Diophantine triple consists of three nonzero distinct elements. Obviously, d_{n} > 3 for n ≥ 1, so we have to see if d_{0} ∈ {0, 1, 3}. Because d_{0} = (ξ − 2)/3, it follows that ξ ∈ {2, 5, 11} where ξ is a fundamental solution of (19). If • ξ = 2, then 5 = 3dη^{2} which is not possible,

• ξ = 5, then 8 = dη^{2} which implies that η = 1, d = 8 or η = 2, d = 2,

• ξ = 11, then 40 = dη^{2} which implies that η = 1, d = 40 or η = 2, d = 10. For d = 8, 40 equations (9) and (10) have no solution in ℤ, but (9) is solvable for d = 10 and (10) is solvable for d = 2. That is the reason why the element d_{0} in ℤ[\sqrt{2}] and ℤ[\sqrt{10}] is omitted.

In Table 1 we give a list of 1 < d < 100 such that the equation (9) is solvable. In the last column of Table 1, the smallest c of a proper triple \{1, 3, c\} in ℤ[\sqrt{d}] is given.

d fund. solution of (9) fund. solution of (19) c_{min}
10  (2,1)  (11,2)  1763
46  (4,1)  (47,4)  15
58  (22,5)  (1451,110)  483
73  (5,1)  (74,5)  24
94  (28,5)  (2351,140)  783

Table 1: A list of 1 < d < 100 such that (9) is solvable

If the equation (10) is solvable, we obtain Table 2.

d fund. solution of (9) fund. solution of (19) c_{min}
2  (2,1)  (5,2)  161
6  (4,1)  (17,4)  5
17  (7,1)  (50,7)  16
18  (22,3)  (485,66)  161
22  (8,1)  (65,8)  21
34  (10,1)  (101,10)  33
38  (32,3)  (1025,96)  341
41  (11,1)  (122,11)  40
54  (140,11)  (19601,1540)  6533
57  (13,1)  (170,13)  56
66  (14,1)  (197,14)  65
82  (298,19)  (88805,5662)  29601
86  (16,1)  (257,16)  85
89  (49,3)  (2402,147)  800
97  (17,1)  (290,17)  96

Table 2: A list of 1 < d < 100 such that (10) is solvable

5 Acknowledgements

The author would like to thank A. Dujella for his helpful comments.

References

[1] A. Baker and D. Davenport, The equations 3x^{2} − 2 = y^{2} and 8x^{2} − 7 = z^{2} , Quart. J. Math. Oxford Ser. (2) 20 (1969), 129–137.

[2] A. Dujella, The problem of Diophantus and Davenport for Gaussian integers, Glas. Mat. Ser. III 32 (1997), 1–10.

[3] A. Dujella, Diophantine m-tuples, http://www.math.hr/^{∼}duje/dtuples.html.

[4] A. Dujella and C. Fuchs, Complete solution of the polynomial version of a problem of Diophantus, J. Number Theory 106 (2004), 326–344.

[5] A. Dujella and A. Jurasić, On the size of sets in a poynomial variant of a problem of Diophantus, Int. J. Number Theory, to appear.

[6] A. Dujella and A. Pethő, A generalization of a theorem of Baker and Davenport, Quart. J. Math. Oxford Ser. (2) 49 (1998), 291–306.

[7] Z. Franušić, Diophantine quadruples in ℤ[\sqrt{4k + 3}], Ramanujan J. 17 (2008), 77–88.

[8] Z. Franušić, On the extensibility of Diophantine triples {k − 1, k + 1, 4k} for Gaussian integers, Glas. Mat. Ser. III 43 (2008), 265–291.

[9] Z. Franušić, A Diophantine problem in ℤ[(1 + \sqrt{d})/2], Studia Sci. Math. Hungar. 46 (2009), 103–112.

[10] B. W. Jones, A second variation on a problem of Diophantus and Davenport, Fibonacci Quart. 16(1978), 155–165.

[11] T. Nagell, Contributions to the theory of a category of Diophantine equations of the second degree with two unknowns, Nova Acta Reg. Soc. Sc. Upsaliensis (4) 16 (1955), No.2, 1–38.

[12] N. J. A. Sloane, The on-line encyclopedia of integer sequences,

   http://www.research.att.com/^{∼}njas/sequences/.

2000 Mathematics Subject Classification: 11D09, 11R11. Keywords: Diophantine sets, quadratic fields, Pellian equations.

(Concerned with sequences \underline{A011922}, \underline{A030063}, \underline{A045899}, and \underline{A051048}.)

Received August 19 2010; revised version received August 31 2010; November 17 2010. Published in Journal of Integer Sequences, December 7 2010.

Return to Journal of Integer Sequences home page.