Extending Theorems of Serret and Pavone

Keith R. Matthews School of Mathematics and Physics

University of Queensland

Brisbane 4072

Australia

keithmatt@gmail.com

John P. Robertson

Actuarial and Economic Services Division National Council on Compensation Insurance

Boca Raton, FL 33487

USA

jpr2718@gmail.com

Anitha Srinivasan Department of Mathematics

Saint Louis University-Madrid Campus

Avenida del Valle 34

28003 Madrid

Spain

rsrinivasan.anitha@gmail.com

Abstract

We extend theorems of Serret and Pavone for solving f(x, y) = ax^{2} +bxy +cy^{2} = μ, where a > 0, gcd(x, y) = 1, y > 0. Here d = b^{2} − 4ac > 0 is not a perfect square and 0 < |μ| < \sqrt{d}/2. If μ > 0, Serret proved that x/y is a convergent to ρ = (−b + \sqrt{d})/2a or σ = (−b − \sqrt{d})/2a. If μ < 0, we are able to modify Pavone’s approach and show that with at most one exception, the solutions are convergents to ρ or σ.

1 Introduction

In 1885, Serret [13] studied the quadratic diophantine equation

f(x, y) = ax^{2} + bxy + cy^{2} = μ,  (1)

where a > 0, gcd(a, b, c) = 1, d = b^{2} − 4ac is positive and nonsquare and 0 < |μ| < \sqrt{d}/2. Serret showed that if μ > 0, then any relatively prime solution (x, y) with y > 0 is a convergent to ρ = (−b + \sqrt{d})/2a or σ = (−b − \sqrt{d})/2a. However he was unable to deal conclusively with the case μ < 0. This was done by Pavone [11] in the special case when |μ| = m(f) is the least of the absolute values of integers represented by f for integers x and y, not both zero. We remark that Lagrange [3, Thm. 86] proved m(f) < \sqrt{d}/2.

We modify Pavone’s proof when −\sqrt{d}/2 < μ < 0, to show that either x/y is a convergent to ρ or σ, or has the form (p_{m} − p_{m−1}, q_{m} − q_{m−1}) or (P_{r} − P_{r−1}, Q_{r} − Q_{r−1}), where

ρ = [a_{0}, …, a_{m}, \overline{b_{1}, …, b_{n}}], σ = [c_{0}, …, c_{r}, \overline{d_{1}, …, d_{n}}],

where a_{m} ≠ b_{n}, c_{r} ≠ d_{n} and p_{h}/q_{h} and P_{k}/Q_{k} denote convergents of ρ and σ, respectively. Barnes [1, Lemma 16] gave a result that overlaps with Serret’s theorem when a > 0 > c. Finally, we remark that there is a continued fraction algorithm [8] for solving (1), irre-

spective of the size of μ.

2 Definitions and Lemmas

Definition 1. We call an indefinite form g(x, y) = Ax^{2} + Bxy + Cy^{2} Hermite reduced if the roots θ_{1} and θ_{2} of g(x, 1) = 0 satisfy θ_{1} > 1 and −1 < θ_{2} < 0. Equivalently, θ_{1} = [\overline{b_{1}, …, b_{n}}] and θ_{2} = −[0, \overline{b_{n}, …, b_{1}}], where the b_{i} are positive integers. See [12, pp. 73–76].

Remark 2. The term Hermite reduced was introduced in [4]. Markov had previously used the concept in his Master’s dissertation [2] and [7].

The following sequences were introduced in [4] and [11].

Definition 3. Let θ_{1} = [a_{0}, a_{1}, …,] and θ_{2} = −[0, a_{−1}, a_{−2}, …,] be as in Definition 1 and let the doubly-infinite sequences (S_{k}), (T_{k}) be defined as follows:

S_{0} = T_{−1} = 1, S_{−1} = T_{0} = 0,
S_{k+1} = a_{k}S_{k} + S_{k−1}, T_{k+1} = a_{k}T_{k} + T_{k−1}, k ≥ 0,  (2)
S_{−k−1} = −a_{−k}S_{−k} + S_{−k+1}, T_{−k−1} = −a_{−k}T_{−k} + T_{−k+1}, k ≥ 1.

.

Remark 4. For k ≥ 1, the convergents to θ_{1} are given by

S_{k}/T_{k} = A_{k−1}/B_{k−1} = [a_{0}, …, a_{k−1}].

To determine the convergents to θ_{2}, we note that

S_{−k−1}/T_{−k−1} = −[0, a_{−1}, …, a_{−k}]

and use the following result from [5].

θ_{2} = \begin{cases}[−1, 1, a_{−1} − 1, a_{−2}, …], & if a_{−1} > 1; \\ [−1, a_{−2} + 1, a_{−3}, …], & if a_{−1} = 1. \end{cases}

Then for k ≥ 0, the convergents to θ_{2} are given by

(A_{0}, B_{0}) = (−1, 1) and (A_{k}, B_{k}) = (−1)^{k+1}(S_{−k}, T_{−k}), k ≥ 1 if a_{−1} > 1,
(A_{k}, B_{k}) = (−1)^{k+1}(S_{−k−2}, T_{−k−2}), k ≥ 0, if a_{−1} = 1.  (3)

We note that for k ≥ 1, S_{−k−1} is positive exactly when k is odd, and T_{−k−1} is negative exactly when k is odd.

We now give a simple proof of Serret’s theorem.

Proposition 5. Assume 0 < μ < \sqrt{d}/2. Let (p, q) be a relatively prime solution of f(x, y) = ax^{2} +bxy +cy^{2} = μ, with q > 0, where a > 0. Then p/q is a convergent to ρ = (−b+\sqrt{d})/2a or σ = (−b − \sqrt{d})/2a.

Proof. We have

a(\frac{p}{q} − ρ)(\frac{p}{q} − σ) = \frac{μ}{q^{2}} .

We cannot have σ < p/q < ρ. First assume p/q > ρ. Then

\frac{p}{q} − ρ = \frac{μ}{a(\frac{p}{q} − ρ + ρ − σ)q^{2}}

< \frac{\sqrt{d}}{2a(ρ − σ)q^{2}} = \frac{1}{2q^{2}} .

Hence p/q is a convergent p_{k}/q_{k}, k ≥ 0, to ρ by Lagrange [6, Thm. 184]. There is a similar argument if p/q < σ.

We replace Pavone’s Lemma 2 by a more general result:

Lemma 6. Let g(x, y) be a Hermite reduced form Ax^{2} + Bxy + Cy^{2}, D = B^{2} − 4AC with roots θ_{1} = [\overline{b_{1}, …, b_{n}}] and θ_{2} = −[0, \overline{b_{n}, …, b_{1}}] and let (S_{k}), (T_{k}) be the sequences defined in (2). Suppose g(p, q) = μ, where 0 < |μ| < \sqrt{D}/2, with gcd(p, q) = 1. Then there exists a k such that

(p, q) = ±(S_{k}, T_{k}).

Proof. If (p, q) = (±1, 0), we can take k = 0, whereas if (p, q) = (0, ±1), we can take k = −1. So we assume p and q are nonzero. Then p/q is a convergent to θ_{1} or θ_{2}. For if A and μ have the same sign, the result follows from Serret’s theorem. If A and μ have opposite signs, then as A and C have opposite sign, C and μ have the same sign and we instead consider the equation Cq^{2} + Bqp + Ap^{2} = μ. We know by Serret’s theorem that q/p is a convergent to one of 1/θ_{1} or 1/θ_{2}, so p/q will be a convergent A_{h}/B_{h}, h ≥ 0, to one of θ_{1} or θ_{2}. In the former case, (p, q) = ±(S_{k}, T_{k}) for some k ≥ 1. From Remark 4, the only convergent to θ_{2} that is not S_{k}/T_{k} is A_{0}/B_{0} = −1/1, and this occurs when b_{n} > 1. We show that this is impossible here.

We can assume A > 0 and that g(−1, 1) = μ, where |μ| < \sqrt{d}/2. Then A(−1 − θ_{1})(−1 − θ_{2}) = μ, so (1 + θ_{1})(1 + θ_{2}) = μ/A. Then as θ_{1} > 1 and −1 < θ_{2} < 0, we have μ > 0. Hence

1 + θ_{2} = \frac{μ}{A(1 + θ_{1})} < \frac{1}{2},

as 1+θ_{1} = 1+θ_{2}+θ_{1}−θ_{2} > θ_{1}−θ_{2} = \sqrt{d}/A. Hence 1/1 is a convergent to −θ_{2} = [0, \overline{b_{n}, …, b_{1}}], which means that b_{n} = 1.

We replace Pavone’s Lemma 4 by the following result:

Lemma 7. Let g(x, y) be a Hermite reduced form with roots θ_{1} = [\overline{b_{1}, …, b_{n}}] and θ_{2} = −[0, \overline{b_{n}, …, b_{1}}]. Suppose b_{n} > 1, b_{n−1} = 1 and that g(S_{−2}, T_{−2}) = μ, where |μ| < \sqrt{d}/2. Then b_{n−2} = 1.

Proof. For 1 ≤ i ≤ n, let

W_{i} = [\overline{b_{i}, b_{i+1}, …, b_{n}, b_{1}, …, b_{i−1}}] + [0, \overline{b_{i−1}, …, b_{1}, b_{n}, …, b_{i}}].

Then W_{i} = \sqrt{d}/|g(S_{i−1}, T_{i−1}) (see [7, p. 385]). Hence, as the W_{i} are periodic with period n, we have

W_{n−1} = [b_{n−1}, b_{n}, b_{1}, …, b_{n−2}] + [0, b_{n−2}, …, b_{1}, b_{n}, b_{n−1}]
= \sqrt{d}/|g(S_{−2}, T_{−2})|
= \sqrt{d}/|μ| > 2.

Hence if b_{n−1} = 1, we have

W_{n−1} = [1, b_{n}, b_{1}, …, b_{n−2}] + [0, b_{n−2}, …, b_{1}, b_{n}, b_{n−1}]
< 1 + 1/b_{n} + 1/b_{n−2}.

So if b_{n} > 1, we have

2 < W_{n−1} < 1 + 1/b_{n} + 1/b_{n−2} ≤ 1 + 1/2 + 1/b_{n−2}.

Hence 1/2 < 1/b_{n−2} and so b_{n−2} < 2, giving b_{n−2} = 1.

The next result is due to Pavone [11].

Lemma 8. Let f(x, y) = ax^{2} + bxy + cy^{2} and let ρ = [a_{0}, …, a_{m}, \overline{b_{1}, …, b_{n}}] and σ = [c_{0}, …, c_{r}, \overline{d_{1}, …, d_{n}}] be the roots of f(x, 1) = 0. Also let p_{h}/q_{h} and P_{h}/Q_{h} denote the convergents of ρ and σ, respectively. We do not require the periods to have minimal lengths, but assume m and r are minimal, i.e., a_{m} ≠ b_{n} and c_{r} ≠ d_{n}. It is also convenient to assume n ≥ 4.

Let θ_{1} = [\overline{b_{1}, …, b_{n}}], θ_{2} = −[0, \overline{b_{n}, …, b_{1}}] and let S_{k}, T_{k} be the sequences (2) for θ_{1} and θ_{2}. Then

( \begin{matrix}p_{m} & p_{m−1} \\ q_{m} & q_{m−1} \end{matrix} )( \begin{matrix}S_{k} \\ T_{k} \end{matrix} ) = ( \begin{matrix}p_{m+k} \\ q_{m+k} \end{matrix} ), k ≥ −1.  (4)

Moreover, there exists i, 1 ≤ i ≤ 3, such that

σ = [c_{0}, …, c_{r}, \overline{b_{n−i}, …, b_{1}, b_{n}, b_{n−1}, …, b_{n−i+1}}],  (5)

and

( \begin{matrix}p_{m} & p_{m−1} \\ q_{m} & q_{m−1} \end{matrix} )( \begin{matrix}S_{−k} \\ T_{−k} \end{matrix} ) = ±( \begin{matrix}P_{r+k−(i+1)} \\ Q_{r+k−(i+1)} \end{matrix} ), k ≥ i.  (6)

Also i = 3 implies b_{n−1} = 1, while b_{n} = b_{n−1} = 1 implies i = 3. Remark 9. A list of the possible continued fraction expansions (5) for σ is given at [9].

3 Extending Pavone’s theorem

Theorem 10. Let f(x, y) = ax^{2} + bxy + cy^{2}, a > 0, d = b^{2} − 4ac > 0 and not square. Let p and q > 0 be relatively prime integers, such that f(p, q) = μ. Let the roots of f(x, 1) = 0 be ρ = [a_{0}, …, a_{m}, \overline{b_{1}, …, b_{n}}] and σ = [c_{0}, …, c_{r}, \overline{d_{1}, …, d_{n}}], where a_{m} ≠ b_{n} and c_{r} ≠ d_{n}. Let the convergents of ρ and σ be denoted by p_{h}/q_{h} and P_{h}/Q_{h}, respectively.

(i) If 0 < μ < \sqrt{d}/2, then p/q is a convergent to ρ or σ.

(ii) If −\sqrt{d}/2 < μ < 0, then p/q is a convergent to ρ or σ, or

(p, q) = (p_{m} − p_{m−1}, q_{m} − q_{m−1}) or (P_{r} − P_{r−1}, Q_{r} − Q_{r−1}).  (7)

Proof. We assume f(p, q) = ap^{2} + bpq + cq^{2} = μ, where 0 < |μ| < \sqrt{d}/2 and gcd(p, q) = 1. We follow Pavone’s argument closely and define g(x, y) by

g(x, y) = f(p_{m}x + p_{m−1}y, q_{m}x + q_{m−1}y).

Then g is Hermite reduced with roots θ_{1} = [\overline{b_{1}, …, b_{n}}] and θ_{2} = −[0, \overline{b_{n}, …, b_{1}}] and with sequences S_{k}, T_{k} for θ_{1} and θ_{2} defined in (2). Define integers α and β by

p_{m}α + p_{m−1}β = p, q_{m}α + q_{m−1}β = q.

Then g(α, β) = μ and by Lemma 6, there exists an integer k such that (α, β) = ±(S_{k}, T_{k}). Hence

( \begin{matrix}p \\ q \end{matrix} ) = ±( \begin{matrix}p_{m} & p_{m−1} \\ q_{m} & q_{m−1} \end{matrix} )( \begin{matrix}S_{k} \\ T_{k} \end{matrix} ).

Let i be the integer satisfying equations (5) and (6). If i = 1 or 2, then by (4) and (6)

( \begin{matrix}p \\ q \end{matrix} ) = ±( \begin{matrix}p_{h} \\ q_{h} \end{matrix} ) or ± ( \begin{matrix}P_{h} \\ Q_{h} \end{matrix} )

for some h, and p/q is a convergent to ρ or σ. If i = 3, then the pair ( \begin{matrix}S_{k} \\ T_{k} \end{matrix} ) occurs in (4)

or (6) for all k ≠ −2. Hence either p/q is a convergent to ρ or σ, or

( \begin{matrix}p \\ q \end{matrix} ) = ±( \begin{matrix}p_{m} & p_{m−1} \\ q_{m} & q_{m−1} \end{matrix} )( \begin{matrix}S_{−2} \\ T_{−2} \end{matrix} ) = ±( \begin{matrix}p_{m} & p_{m−1} \\ q_{m} & q_{m−1} \end{matrix} )( \begin{matrix}1 \\ −b_{n} \end{matrix} )

and

(p, q) = ±(p_{m} − b_{n}p_{m−1}, q_{m} − b_{n}q_{m−1}).

However we can interchange ρ and σ and similarly deduce that

(p, q) = ±(P_{r} − b_{n−2}P_{r−1}, Q_{r} − b_{n−2}Q_{r−1}).

If b_{n} = 1, we have q_{m} − b_{n}q_{m−1} = q_{m} − q_{m−1} > 0 and

(p, q) = (p_{m} − p_{m−1}, q_{m} − q_{m−1}).

If b_{n} > 1, then b_{n−1} = 1 by Lemma 8, and as g(S_{−2}, T_{−2}) = μ, it follows from Lemma 7 that b_{n−2} = 1. Hence Q_{r} − b_{n−2}Q_{r−1} = Q_{r} − Q_{r−1} > 0 and

(p, q) = (P_{r} − P_{r−1}, Q_{r} − Q_{r−1}).

Definition 11. A solution (p, q) of (1) with gcd(p, q) = 1 and q > 0 that is not a convergent to ρ or σ is called an exceptional solution.

Remark 12. From the above proof, we see that if a solution is given by a convergent p_{k}/q_{k} to ρ, then k ≥ m − 1. Similarly, if a solution is given by a convergent P_{j}/Q_{j} to σ, then j ≥ r − 1.

4 Fundamental solutions

We need some definitions and lemmas associated with the diophantine equation

ax^{2} + bxy + cy^{2} = μ.  (8)

The integer solutions of (8) divide into equivalence classes under the relation (x_{1}, y_{1}) and (x_{2}, y_{2}) are equivalent if and only if

x_{2} = \frac{x_{1}(u − bv)}{2} − cvy_{1}, y_{2} = \frac{y_{1}(u + bv)}{2} + avx_{1},

where u and v are integers satisfying u^{2} − dv^{2} = 4.

Definition 13. A fundamental solution (u, v) of a class of solutions K of (8), is one where v has least non-negative value when (u, v) belongs to K. Let u′ = −(au + bv)/a be the conjugate solution to u. If u′ is not integral, or if (u′, v) is not equivalent to (u, v), this determines (u, v). If u′ is integral and (u′, v) is equivalent to (u, v), where u ≠ u′, we choose u > u′. There are finitely many equivalence classes, each indexed by a fundamental solution.

Reference [10] contains more information about the fundamental solutions.

Lemma 14. ([14, p. 383]). Solutions (x_{1}, y_{1}) and (x_{2}, y_{2}) of (8) are equivalent if and only if x_{1}y_{2} − x_{2}y_{1} ≡ 0 (mod |μ|).

We find the fundamental solutions when |μ| < \sqrt{d}/2, by examining the continued fractions for ρ and σ. It can be proved that the different classes will be represented in the first period (and second period if the period is odd). We first check to see if an exceptional solution exists. Then we examine the the convergents p_{m}/q_{m}, … and P_{r}/Q_{r}, … of the first period of ρ and σ, and additionally the second period, if the period-length is odd, using the equations

aA_{n−1}^{2} + bA_{n−1}B_{n−1} + cB_{n−1}^{2} = \begin{cases}(−1)^{n}Q_{n}/2, & with ρ = (−b + \sqrt{d})/2a; \\ (−1)^{n+1}Q_{n}/2, & with σ = (−b − \sqrt{d})/2a, \end{cases}

to check if for ρ, we have μ = (−1)^{n}Q_{n}/2, or for σ, we have μ = (−1)^{n+1}Q_{n}/2.

We then use Lemma 14 to test for equivalence of solutions using (p_{m−1}, q_{m−1}), (P_{r−1}, Q_{r−1}) and any exceptional solution.

Proposition 15. If (p, q) is an exceptional solution of (1), then it is a fundamental solution if |μ| > 1.

Proof. Suppose (p, q) is an exceptional solution of (1). Then from the proof of Theorem 10, we have

(p, q) = (p_{m} − b_{n}p_{m−1}, q_{m} − b_{n}q_{m−1}) = (P_{r} − b_{n−2}P_{r−1}, Q_{r} − b_{n−2}Q_{r−1}).

We have to compare (p, q) with the convergents p_{h}/q_{h} and P_{k}/Q_{k} of ρ and σ that give solutions of (1). By Remark 12, we know that h ≥ m − 1 and k ≥ r − 1.

Now q = q_{m} − b_{n}q_{m−1} < q_{m} ≤ q_{k} if k ≥ m and q = Q_{r} − b_{n−2}Q_{r−1} < Q_{r} ≤ Q_{j} if j ≥ r. Also (p, q) is not equivalent to (p_{m−1}, q_{m−1}) or (P_{r−1}, Q_{r−1}) if |μ| > 1. For

(p, q) ∼ (p_{m−1}, q_{m−1}) ⟺ p_{m−1}(q_{m} − b_{n}q_{m−1}) − q_{m−1}(p_{m} − b_{n}p_{m−1}) ≡ 0 (mod |μ|)
⟺ p_{m−1}q_{m} − q_{m−1}p_{m} ≡ 0 (mod |μ|)
⟺ (−1)^{m} ≡ 0 (mod |μ|).

Similarly with (P_{r−1}, Q_{r−1}).

Remark 16. If μ = −1, the situation is more complicated. For example, the equation x^{2} +xy −y^{2} = −1 has one solution class, with fundamental solution (0, 1) and an exceptional solution (−1, 1).

5 Examples

Example 17. Consider the equation x^{2}+xy−100y^{2} = −10. Here d = 401, |μ| = 10 < \sqrt{d}/2. Also

ρ = (−1 + \sqrt{401})/2 = [9, \overline{1, 1, 19}], σ = (−1 − \sqrt{401})/2 = [−11, 2, \overline{19, 1, 1}].

The double periods for ρ and σ give the solutions

\begin{matrix}(p_{0}, q_{0}) = (9, 1) & (P_{3}, Q_{3}) = (−431, 41) \\ (p_{4}, q_{4}) = (390, 41) & (P_{5}, Q_{5}) = (−16410, 1561) \end{matrix}

Also (−10, 1) = (P_{1} − P_{0}, Q_{1} − Q_{0}) is an exceptional solution, where P_{1}/Q_{1} = −21/2 and P_{0}/Q_{0} = −11/1; also neither (p_{m−1}, q_{m−1}) = (p_{−1}, q_{−1}) = (1, 0) nor (P_{r−1}, Q_{r−1}) = (P_{0}, Q_{0}) = (−11, 1) is a solution of x^{2} + xy − 100y^{2} = −10. We also have

(9, 1) ∼ (−431, 41) ≁ (−10, 1),
(390, 41) ∼ (−16410, 1561) ∼ (−10, 1).

Hence the fundamental solutions are (9, 1) and (−10, 1), and the complete solution is

x = (9u + 191v)/2; x = (−10u + 210v)/2,  (9)
y = (u + 19v)/2;  y = (u − 19v)/2,  (10)

where u^{2} − 401v^{2} = 4.

Example 18. Consider the equation 69x^{2} + 71xy + 15y^{2} = −13. Here d = 901, |μ| = 13 < \sqrt{d}/2. Then

ρ = (−71 + \sqrt{901})/138 = [−1, 1, 2, \overline{2, 1, 2, 1, 1, 1, 1}],
σ = (−71 − \sqrt{901})/138 = [−1, 3, \overline{1, 2, 1, 2, 1, 1, 1}],

and we have the exceptional solution

(−1, 2) = (P_{1} − P_{0}, Q_{1} − Q_{0}) = (p_{2} − p_{1}, q_{2} − q_{1}),

where P_{1}/Q_{1} = −2/3, P_{0}/Q_{0} = −1/1 and p_{2}/q_{2} = −1/3, p_{1}/q_{1} = 0/1.

The double periods for ρ and σ give the solutions

\begin{matrix}(p_{6}, q_{6}) = (−11, 37) & (P_{7}, Q_{7}) = (−71, 97) \\ (p_{14}, q_{14}) = (−1141, 3842) & (P_{13}, Q_{13}) = (−2461, 3362) \end{matrix}

Also neither (p_{m−1}, q_{m−1}) = (p_{1}, q_{1}) = (0, 1) nor (P_{r−1}, Q_{r−1}) = (P_{0}, Q_{0}) = (−1, 1) is a solution of 69x^{2} + 71xy + 15y^{2} = −13 and

(−11, 37) ∼ (−71, 97) ≁ (−1, 2),
(−1141, 3842) ∼ (−2461, 3362) ∼ (−1, 2).

Hence the fundamental solutions are (−1, 2) and (−11, 37), and the complete solution is given by

x = (−u + 11v)/2; x = (−11u − 329v)/2,  (11)
y = (2u + 4v)/2;  y = (37u + 1109v)/2,  (12)

where u^{2} − 901v^{2} = 4.

Example 19. Consider the equation 2x^{2} + 5xy + y^{2} = −2. Here d = 17, |μ| = 2 < \sqrt{d}/2. Then

ρ = (−5 + \sqrt{17})/4 = [−1, \overline{1, 3, 1}], σ = (−5 − \sqrt{17})/4 = [−3, 1, 2, \overline{1, 1, 3}].

There is no exceptional solution, as i = 2 here.

The double periods for ρ and σ give the solutions

\begin{matrix}(p_{0}, q_{0}) = (−1, 1) & (P_{3}, Q_{3}) = (−9, 4) \\ (p_{2}, q_{2}) = (−1, 4) & (P_{5}, Q_{5}) = (−57, 25) \end{matrix}

Also (p_{m−1}, q_{m−1}) = (p_{−1}, q_{−1}) = (1, 0) and (P_{r−1}, Q_{r−1}) = (P_{1}, Q_{1}) = (−2, 1) are not solutions of 2x^{2} + 5xy + y^{2} = −2. We also have

(−1, 1) ∼ (−57, 25) ≁ (−1, 4) ∼ (−9, 4).

Hence the fundamental solutions are (−1, 1) and (−1, 4) and the complete solution is given by

x = (−u + 3v)/2; x = (−u − 3v)/2,  (13)
y = (u + v)/2;  y = (4u + 16v)/2,  (14)

where u^{2} − 17v^{2} = 4.

If instead we consider the equation 2x^{2} + 5xy + y^{2} = −1, again there are no exceptional solutions. Also the double periods for ρ and σ now give the solutions

(p_{4}, q_{4}) = (−2, 9) (P_{1}, Q_{1}) = (−130, 57)

Here (P_{r−1}, Q_{r−1}) = (P_{1}, Q_{1}) = (−2, 1) is a solution of 2x^{2} + 5xy + y^{2} = −1, whereas (p_{m−1}, q_{m−1}) = (p_{−1}, q_{−1}) = (1, 0) is not a solution. Also

(−2, 9) ∼ (−130, 57) ∼ (−2, 1).

Hence the fundamental solution is (−2, 1) and the complete solution is given by

x = −u + 4v, y = (u − 3v)/2,

where u^{2} − 17v^{2} = 4.

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2000 Mathematics Subject Classification: Primary 11A55; Secondary 11A70, 11D09. Keywords: quadratic diophantine equation, indefinite binary quadratic form, continued fraction, convergent, fundamental solution.

Received July 18 2017; revised version received September 16 2017. Published in Journal of Integer Sequences, October 29 2017.

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