A Generalization of the Question of Sierpiński on Geometric Progressions

Jiagui Luo^{1} College of Mathematics and Information Science Zhaoqing University Zhaoqing 526061 P. R. China

                     Luojg62@yahoo.com.cn

Pingzhi Yuan^{1} School of Mathematics South China Normal University Guangzhou 510631 P. R. China

                       mcsypz@zsu.edu.cn

Abstract In this paper we prove that there is no geometric progression that contains four

distinct integers of the form Dm^{2} + C, D, m ∈ ℕ, C = ±1, ±2, ±4.

1 Introduction

The integers of the form T_{n} = n(n + 1)/2, n ∈ ℕ, are called triangular numbers. Sierpiński [7, D23] asked whether or not there exist four (distinct) triangular numbers in geometric progression. Szymiczek [10] conjectured that the answer is negative. The problem of finding three such triangular numbers is readily reduced to finding solutions to a Pell equation (by an old result of Gérardin [6]; see also [9, 4]). This implies that there are infinitely many such triples, the smallest of which is (T_{1}, T_{3}, T_{8}). In fact, an easy calculation shows that if T_{n} = m^{2} then

1Supported by NSF of China (No. 10971072) and the Guangdong Provincial Natural Science Foundation (No. 8151027501000114).

(T_{n}, T_{n+2m} = m(2n + 3m + 1), T_{3n+4m+1} = (2n + 3m + 1)^{2})

forms a geometric progression.

Recently M. Bennett [1] proved that there do not exist four distinct triangular numbers in geometric progression with the ratio being a positive integer. Chen and Fang [3] extended Bennett’s result to the rational ratio and proved that there do not exist four distinct triangular numbers in geometric progression. Using the theory of Pell equations and a result of Bilu-Hanrot-Voutier [2] on primitive divisors of Lucas and Lehmer numbers, Yang-He [15] and Yang [14] claimed that there is no geometric progression that contains four distinct triangular numbers. But their proof is under the assumption that the geometric progression has an integral common ratio. Fang [5], using only the Störmer theorem on Pell’s equation, showed that there is no geometric progression which contains four distinct triangular numbers.

Note that if T_{n} = n(n + 1)/2, n ∈ ℕ is a triangular number, then 8T_{n} = m^{2} − 1, where m = 2n + 1. Thus the Sierpiński problem is equivalent to whether or not there exist four distinct integers of the form m^{2} − 1 in geometric progression. In this paper, we consider the more general question whether or not there exists a geometric progression which contains four distinct integers of the form Dm^{2} + C with D, m ∈ ℕ, C = ±1, ±2, ±4. We use Störmer theory on Pell equations to prove the following results:

Theorem 1. Let D be a positive integer. Then there is no geometric progression which contains four distinct integers of the form Dm^{2} + C, m ∈ ℕ, C ∈ {−4, 4}.

By Theorem 1, we have the following two Corollaries immediately.

Corollary 2. Let D be a positive integer. Then there is no geometric progression which contains four distinct integers of the form Dm^{2} + C, m ∈ ℕ, C ∈ {−1, 1}.

Corollary 3. Let D be a positive integer. Then there is no geometric progression which contains four distinct integers of the form Dm^{2} + C, m ∈ ℕ, C ∈ {−2, 2}.

2 Some Lemmas

To prove the above theorem, we need the following lemmas. Throughout this paper, we assume that k, l are coprime positive integers and kl nonsquare; and let 2 ∤ kl when C = 2 or 4. We need some results on the solutions of the diophantine equations

kx^{2} − ly^{2} = C, C = 1, 2, 4.  (1)

We recall that the minimal positive solution of Diophantine equation (1) is the positive integer solution (x, y) of equation (1) such that x\sqrt{k} + y\sqrt{l} is the smallest. One can easily see that this is equivalent to determining a positive integer solution (x, y) of equation (1) such that x and y are the smallest. By abuse of language, we shall also refer to x\sqrt{k} + y\sqrt{l} instead of the pair (x, y) as a solution to (1) and call x\sqrt{k}+y\sqrt{l} the minimal positive solution.

If x_{1}\sqrt{k} + y_{1}\sqrt{l} is the minimal positive solution of (1), then we have the following result.

Lemma 4. ([11]) All positive integer solutions of (1) are given by

\frac{x\sqrt{k} + y\sqrt{l}}{\sqrt{C}} = (\frac{x_{1}\sqrt{k} + y_{1}\sqrt{l}}{\sqrt{C}})^{n} , n ∈ ℕ.

Moreover, we have 2 ∤ n when k > 1 or C = 2.

Störmer (see [4, p. 391]) proved a result on divisibility properties of solutions of Pell equations. More new results extending Störmer theory had been obtained over the years. We will list some known results that will be used in the proofs in this paper.

Lemma 5. (Störmer’s theorem [4, p. 391]) Let D be a positive nonsquare integer. Let (x_{1}, y_{1}) be a positive integer solution of Pell equation

x^{2} − Dy^{2} = C, C ∈ {−1, 1}.

If every prime divisor of y_{1} divides D, then x_{1} + y_{1}\sqrt{D} is the minimal positive solution.

Considering the Diophantine equation

kx^{2} − ly^{2} = 1, k > 1,  (2)

D. T. Walker [12] obtained a result similar to Störmer’s theorem. See also Q. Sun and P. Yuan [11].

Lemma 6. ([12, 11]) Let (x, y) be a positive integer solution of (2) .

(i) If every prime divisor of x divides k, then either

x\sqrt{k} + y\sqrt{l} = ε

or

x\sqrt{k} + y\sqrt{l} = ε^{3}, and x = 3^{s}x_{1}, 3 ∤ x_{1}, 3^{s} + 3 = 4kx_{1}^{2},

where in both cases ε = x_{1}\sqrt{k} + y_{1}\sqrt{l} is the minimal positive solution of (2) , s ∈ ℕ. (ii) If every prime divisor of y divides l, then either

x\sqrt{k} + y\sqrt{l} = ε

or

x\sqrt{k} + y\sqrt{l} = ε^{3}, and y = 3^{s}y_{1}, 3 ∤ y_{1}, 3^{s} − 3 = 4ly_{1}^{2}, s ≥ 2.

Using the method in [11], the first author proved the following results.

Lemma 7. ([8]) Let k, l be coprime positive odd integers and kl nonsquare. Suppose that (x, y) is a positive integer solution of the Diophantine equation

kx^{2} − ly^{2} = 2.  (3)

(i) If every prime divisor of x divides k , then either

x\sqrt{k} + y\sqrt{l} = ε

or

\frac{x\sqrt{k} + y\sqrt{l}}{\sqrt{2}} = (\frac{ε}{\sqrt{2}})^{3} , and x = 3^{s}x_{1}, 3^{s} + 3 = 2kx_{1}^{2},

where in both cases ε = x_{1}\sqrt{k} + y_{1}\sqrt{l} is the minimal positive solution of (3) , s ∈ ℕ. (ii) If every prime divisor of y divides l , then either

x\sqrt{k} + y\sqrt{l} = ε

or

\frac{x\sqrt{k} + y\sqrt{l}}{\sqrt{2}} = (\frac{ε}{\sqrt{2}})^{3} , and y = 3^{s}y_{1}, 3^{s} − 3 = 2ly_{1}^{2}, s ≥ 2.

Lemma 8. ([8]) Let k, l be coprime positive odd integers and kl nonsquare. Suppose that (x, y) is a positive integer solution of the Diophantine equation

kx^{2} − ly^{2} = 4.  (4)

(i) If every prime divisor of x divides k , then x\sqrt{k}+y\sqrt{l} = ε is the minimal positive solution of equation (4) except for the case (k, l, x, y) = (5, 1, 5, 11). (ii) If every prime divisor of y divides l , then x\sqrt{k}+y\sqrt{l} = ε is the minimal positive solution of equation (4) .

Lemma 9. Let k, l = a_{0}a^{m} be coprime positive integers and kl nonsquare with m > 1 an integer. If (x, a^{r}) is a positive integer solution of the Diophantine equation

kx^{2} − ly^{2} = C,  C ∈ {−1, 1, −2, 2, −4, 4},  (5)

where r is a non-negative integer. Then x\sqrt{k} + a^{r}\sqrt{l} = ε is the minimal positive solution of equation (5).

Proof. We only consider the case of C = 1 (the proofs of the other cases are similar). If x\sqrt{k} + a^{r}\sqrt{l} is not the minimal positive solution of equation (5) , then by Lemma 6(ii) and since u\sqrt{k} +v\sqrt{l} is the minimal positive solution, we have a^{r} = 3^{s}v and 3 ∤ v. Therefore, 3|a. Since m ≥ 2, 3^{s} − 3 = 4a_{0}a^{m}v^{2} is also divisible by 9. Hence 9|3, which is a contradiction. This completes the proof of Lemma 9.

Remark 10. Lemma 9 is also true for l = a_{0}a^{m}/2^{t}, C = ±1 with t ≤ m is a nonnegative integer and 2|l.

Lemma 11. ([13]) Let x_{1}\sqrt{k} +y_{1}\sqrt{l} be the minimal positive solution of (1) such that 2 ∤ x_{1}y_{1} when C = 2 or 4. If x\sqrt{k} + y\sqrt{l} is a positive integer solution of (1), then y_{1}|y. And if k > 1 or C = 2, then x_{1}|x.

Lemma 12. Let k, l, a, b, r_{2}, r_{3}, r_{4} be positive integers such that gcd(k, lab) = 1, gcd(a, b) = 1, a > b, r_{2} < r_{3} < r_{4}. If 2|ab but 2 ∤ l, then following system of Diophantine equations

kx_{1}^{2} − la^{r_{4}} = C,  (6)

kx_{2}^{2} − la^{r_{4}−r_{2}}b^{r_{2}} = C,  (7)
kx_{3}^{2} − la^{r_{4}−r_{3}}b^{r_{3}} = C,  (8)
kx_{4}^{2} − lb^{r_{4}} = C,  (9)

where C ∈ {−4, 4}, has no positive integer solutions (x_{1}, x_{2}, x_{3}, x_{4}) with x_{1} > x_{2} > x_{3} > x_{4}.

Proof. We now suppose that 2|a but 2 ∤ b.

Case 1: 2|r_{4}.

If 2|r_{2}, then we get k ≡ l ≡ C/4 (mod 4) by considering the equations k (\frac{x_{1}}{2} )^{2} −

l (\frac{a^{r_{4}/2}}{2} )^{2} = C/4 and (9) mod 4. And so by taking mod 4 for k (\frac{x_{2}}{2} )^{2} − l (\frac{a^{(r_{4}−r_{2})/2}}{2} )^{2} = C/4,

we have (\frac{x_{2}}{2} )^{2} − (\frac{a^{(r_{4}−r_{2})/2}}{2} )^{2} ≡ 1 (mod 4). This follows that 2|\frac{a^{r_{4}−r_{2}}}{4} . Thus by (6) and

Remark 10 of Lemma 9, we know that (\frac{x_{1}}{2} , a^{\frac{r_{2}}{2}} ) is the minimal positive solution of the Diophantine equation

kx^{2} − \frac{la^{r_{4}−r_{2}}}{4} y^{2} = C/4.  (10)

By (7), (\frac{x_{2}}{2} , b^{\frac{r_{2}}{2}} ) is a positive integer solution of (10). By Lemma 11, we obtain a^{\frac{r_{2}}{2}} |b^{\frac{r_{2}}{2}} , which contradicts the assumption that a > b.

Similarly we have that 2 ∤ r_{3} from (8).

We now suppose that 2 ∤ r_{2} and 2 ∤ r_{3}. Then, by (7), (8), both (\frac{x_{2}}{2} , a^{\frac{r_{3}−r_{2}}{2}} b^{\frac{r_{2}−1}{2}} ) and (\frac{x_{3}}{2} , b^{\frac{r_{3}−1}{2}} ) are positive integer solutions of the Diophantine equation

kx^{2} − \frac{la^{r_{4}−r_{3}}b}{4}y^{2} = C/4.

Noting that x_{2} > x_{3}, by Lemmas 6 and 7, \frac{x_{3}}{2} \sqrt{k} + b^{\frac{r_{3}−1}{2}} \sqrt{\frac{la^{r_{4}−r_{3}}b}{4}} = ε must be the minimal

positive solution. Therefore again by Lemmas 6 and 7, \frac{x_{3}}{2} \sqrt{k} + a^{\frac{r_{3}−r_{2}}{2}} b^{\frac{r_{2}−1}{2}} \sqrt{\frac{la^{r_{4}−r_{3}}b}{4}} = ε^{3}

and a^{\frac{r_{3}−r_{2}}{2}} b^{\frac{r_{2}−1}{2}} = 3^{s}b^{\frac{r_{3}−1}{2}} , 3^{s} ∓ 3 = la^{r_{4}−r_{3}}b^{r_{3}}. It follows that a^{(r_{3}−r_{2})/2} = 3^{s}b^{(r_{3}−r_{2})/2}, and thus b = 1, a = 3, which contradicts the assumption that 2|a. This concludes the analysis of Case 1.

Case 2: 2 ∤ r_{4}. We can prove that 2 ∤ r_{2}r_{3} is impossible by using the same method of proving Case 1.

If 2|r_{2}, then, since r_{4} − r_{2} > 2, we have 2|\frac{a^{r_{4}−r_{2}}}{4} . Therefor by (6) and Remark 10 of Lemma 9, we know that (\frac{x_{1}}{2} , a^{\frac{r_{2}}{2}} ) is the minimal positive solution of the Diophantine equation

kx^{2} − \frac{la^{r_{4}−r_{2}}}{4} y^{2} = C/4.  (11)

By (7), (\frac{x_{2}}{2} , b^{\frac{r_{2}}{2}} ) is a positive integer solution of (11). So by Lemma 11, we obtain a^{\frac{r_{2}}{2}} |b^{\frac{r_{2}}{2}} , which contradicts the assumption that a > b.

If 2 ∤ r_{2}, then 2|r_{3}, noting that b^{\frac{r_{4}−1}{2}} ≠ 5, by (9) and Lemma 2.5, (x_{4}, b^{\frac{r_{4}−1}{2}} ) is the minimal positive solution of the Diophantine equation kx^{2} − lby^{2} = C. By (7) , (x_{2}, a^{\frac{r_{4}−r_{2}}{2}} b^{\frac{r_{2}−1}{2}} ) is a positive integer solution of kx^{2} − lby^{2} = C. Thus by Lemma 11, we obtain b^{\frac{r_{4}−1}{2}} |a^{\frac{r_{4}−r_{2}}{2}} b^{\frac{r_{2}−1}{2}} . This follows that b^{\frac{r_{4}−r_{2}}{2}} |a^{\frac{r_{4}−r_{2}}{2}} , and so, since gcd(a, b) = 1, b = 1. We have 2|\frac{a^{r_{4}−r_{3}}}{4} as shown at the beginning of Case 1. Proceeding as before, we can prove 2 ∤ r_{3}, which is a contradiction. This concludes the analysis of Case 2. The proof of 2 ∤ a and 2|b is similar. This completes the proof of Lemma 12.

3 Proof of Theorem 1

Proof. Suppose that there is a geometric progression {a_{n}} which contains four distinct integers Dm_{1}^{2} + C = a_{1}q^{t_{1}}, Dm_{2}^{2} + C = a_{1}q^{t_{2}}, Dm_{3}^{2} + C = a_{1}q^{t_{3}}, Dm_{4}^{2} + C = a_{1}q^{t_{4}} with 0 ≤ t_{1} < t_{2} < t_{3} < t_{4}, where q = b/a is the common ratio such that a ≥ 1 and gcd(a, b) = 1. It is easy to see that both a_{1} and q are not zero and that |q| ≠ 1. Without loss of generality, we may assume that 0 < |q| < 1, so a > |b| > 0. Let Dm_{1}^{2} + C = A, t_{2} − t_{1} = r_{2}, t_{3} − t_{1} = r_{3}, t_{4} − t_{1} = r_{4}, then A ≠ 0 and 0 < r_{2} < r_{3} < r_{4} satisfying

Dm_{1}^{2} + C = A, Dm_{2}^{2} + C = Aq^{r_{2}}, Dm_{3}^{2} + C = Aq^{r_{3}}, Dm_{4}^{2} + C = Aq^{r_{4}}.  (12)

Since Aq^{r_{4}} is an integer, then a^{r_{4}}|Ab^{r_{4}}, and so a^{r_{4}}|A since gcd(a, b) = 1. Let A = a_{0}a^{r_{4}}. We can derive that all the numbers Dm_{i} + C, i ∈ {1, 2, 3, 4} are positive integers. If not, then since a > 1, r_{4} ≥ 3, r_{4} − r_{2} ≥ 2, r_{4} − r_{3} ≥ 1, we must have either

Dm_{3}^{2} + C = a_{0}a^{r_{4}−r_{3}}b^{r_{3}} = −2, −3

or

Dm_{4}^{2} + C = a_{0}b^{r_{4}} = −1, −2 or − 3.

This follows either

(D, m_{3}, C, a_{0}, a, b) = (2, 1, −4, 1, 2, −1), (1, 1, −4, 1, 3, −1)

such that r_{4} is even since Dm_{4}^{2} + C = a_{0}b^{r_{4}} is a positive integer, or

(D, m_{4}, C, a_{0}, b) = (3, 1, −4, 1, −1), (2, 1, −4, 2, −1), or (1, 1, −4, 3, −1)

such that r_{4} is odd integer and such that both r_{2} and r_{3} are even integers since both Dm_{2}^{2} + C = a_{0}a^{r_{4}−r_{2}}b^{r_{2}} and Dm_{3}^{2} + C = a_{0}a^{r_{4}−r_{3}}b^{r_{3}} are positive integers.

If (D, m_{3}, C, a_{0}, a, b) = (2, 1, −4, 1, 2, −1), we will get m_{1}^{2} ≡ 2 (mod 4) by considering equation m_{1}^{2} − 2 = 2^{r_{4}−1} mod 4, which is impossible.

If (D, m_{3}, C, a_{0}, a, b) = (1, 1, −4, 1, 3, −1), we will get m_{1}^{2} − (3^{r_{4}/2})^{2} = 4, which is impossible.

If (D, m_{4}, C, a_{0}, b) = (3, 1, −4, 1, −1), we have that both (m_{2}, a^{(r_{4}−r_{2}−1)/2}) and (m_{3}, a^{(r_{4}−r_{3}−1)/2}) are positive integer solutions of Diophantine equation

3x^{2} − ay^{2} = 4.

Thus by Lemma 8, (m_{2}, a^{\frac{r_{4}−r_{2}−1}{2}} ) = (m_{3}, a^{\frac{r_{4}−r_{3}−1}{2}} ) is the minimal positive solution of 3x^{2} − ay^{2} = 4, which contradicts the assumption that m_{2} ≠ m_{3}.

If (D, m_{4}, C, a_{0}, b) = (2, 1, −4, 1, −1), we have that both (m_{1}, a^{(r_{4}−3)/2}) and (m_{2}, a^{(r_{4}−r_{2}−3)/2}) are positive integer solutions of Diophantine equation

x^{2} − a^{3}y^{2} = 2.

Thus by Lemma 9, (m_{1}, a^{\frac{r_{4}−3}{2}} ) = (m_{2}, a^{\frac{r_{4}−r_{2}−3}{2}} ) is the minimal positive solution of x^{2}−a^{3}y^{2} = 2, which contradicts the assumption that m_{1} ≠ m_{2}.

If (D, m_{4}, C, a_{0}, b) = (1, 1, −4, 3, −1) and 2 ∤ a, then, both (m_{1}, a^{(r_{4}−3)/2}) and (m_{2}, a^{(r_{4}−r_{2}−3)/2}) are positive integer solutions of Diophantine equation

x^{2} − 3a^{3}y^{2} = 4.

Thus by Lemma 9, (m_{1}, a^{\frac{r_{4}−3}{2}} ) = (m_{2}, a^{\frac{r_{4}−r_{2}−3}{2}} ) is the minimal positive solution, which contradicts the assumption that m_{1} ≠ m_{2}.

If (D, m_{4}, C, a_{0}, b) = (1, 1, −4, 3, −1) and 2|a, then, both (m_{1}/2, a^{(r_{4}−3)/2}) and (m_{2}/2, a^{(r_{4}−r_{2}−3)/2}) are positive integer solutions of Diophantine equation

x^{2} − \frac{3a^{3}}{4} y^{2} = 1.

Thus by Remark 4 of Lemma 9, (m_{1}/2, a^{\frac{r_{4}−3}{2}} ) = (m_{2}/2, a^{\frac{r_{4}−r_{2}−3}{2}} ) is the minimal positive solution, which contradicts the assumption that m_{1} ≠ m_{2}.

Therefore we can assume that 0 < q < 1, which follows that 0 < b < a and m_{1} > m_{2} > m_{3} > m_{4} > 0. It follows from (12) that

Dm_{1}^{2} − a_{0}a^{r_{4}} = −C,  (13)

Dm_{2}^{2} − a_{0}a^{r_{4}−r_{2}}b^{r_{2}} = −C,  (14)
Dm_{3}^{2} − a_{0}a^{r_{4}−r_{3}}b^{r_{3}} = −C,  (15)
Dm_{4}^{2} − a_{0}b^{r_{4}} = −C.  (16)

It is easy to see that gcd(a_{0}a, a_{0}b) = a_{0} since gcd(a, b) = 1. We will consider three cases according to the divisibility of a_{0} by 2.

If 2 ∤ a_{0}, we must have either 2 ∤ a_{0}a or 2 ∤ a_{0}b. Suppose now that 2 ∤ a_{0}a, then D is odd, by (13). Thus by Lemma 12, we have that (D, a_{0}ab) = 1 and 2 ∤ Da_{0}ab. The case of 2 ∤ a_{0}b is similar, using (16).

If 2||a_{0}, we can derive that 2 ∤ a and 2 ∤ b. Assume to the contrary, we let 2|a (the case that 2|b is similar), then 2 ∤ b. We get Dm_{4}^{2} ≡ 2 (mod 4) by considering equation (16) mod 4, which implies that 2||D since 2|m_{4} would imply Dm_{4}^{2} ≢ 2 (mod 4). Therefore we obtain from (13) that either 2m_{1}^{2} ≡ ±4 (mod 8) or 6m_{1}^{2} ≡ ±4 (mod 8) which is impossible. Hence 2 ∤ a and 2 ∤ b, and so 2||D. Let a_{0} = 2l_{1}, D = 2D_{1}, where l_{1} and D_{1} are odd positive integers. Thus we have from (13), (14), (15), (16) that

D_{1}m_{1}^{2} − l_{1}a^{r_{4}} = −C/2,  (17)
D_{1}m_{2}^{2} − l_{1}a^{r_{4}−r_{2}}b^{r_{2}} = −C/2,  (18)
D_{1}m_{3}^{2} − l_{1}a^{r_{4}−r_{3}}b^{r_{3}} = −C/2,  (19)
D_{1}m_{4}^{2} − l_{1}b^{r_{4}} = −C/2,  (20)

where l_{1}, a, b and D_{1} are odd positive integers.

If 4|a_{0}, we must have either 4|D or 2| gcd(m_{1}, m_{2}, m_{3}, m_{4}). Let D = 4D_{2}, a_{0} = 4l_{2} and (n_{1}, n_{2}, n_{3}, n_{4}) = (m_{1}, m_{2}, m_{3}, m_{4}) when 4|D, and let D_{2} = D, a_{0} = 4l_{2} and (n_{1}, n_{2}, n_{3}, n_{4}) = (m_{1}/2, m_{2}/2, m_{3}/2, m_{4}/2) when 2| gcd(m_{1}, m_{2}, m_{3}, m_{4}). Thus we have from (13), (14), (15), (16) that

D_{2}n_{1}^{2} − l_{2}a^{r_{4}} = −C/4,  (21)
D_{2}n_{2}^{2} − l_{2}a^{r_{4}−r_{2}}b^{r_{2}} = −C/4,  (22)
D_{2}n_{3}^{2} − l_{2}a^{r_{4}−r_{3}}b^{r_{3}} = −C/4,  (23)

D_{2}n_{4}^{2} − l_{2}b^{r_{4}} = −C/4.  (24)

From consideration of these three cases, let (k, l, u_{i}, C_{1}) = (D, a_{0}, m_{i}, −C) or (D_{1}, l_{1}, m_{i}, −C/2) or (D_{2}, l_{2}, n_{i}, −C/4), then one can easily see that the problem is equivalent to proving that the following questions

ku_{1}^{2} − la^{r_{4}} = C_{1},  (25)
ku_{2}^{2} − la^{r_{4}−r_{2}}b^{r_{2}} = C_{1},  (26)
ku_{3}^{2} − la^{r_{4}−r_{3}}b^{r_{3}} = C_{1},  (27)
ku_{4}^{2} − lb^{r_{4}} = C_{1},  (28)

where C_{1} ∈ {−1, 1, −2, 2, −4, 4}, u_{1} > u_{2} > u_{3} > u_{4} > 0, gcd(k, lab) = 1 and 2 ∤ klab if 2|C_{1}, cannot be simultaneously satisfied.

Case 1: 2|r_{4}. It is easy to see that kl is not a square. Otherwise both k and l are squares. And so (\sqrt{k}u_{1}, \sqrt{l}a^{r_{4}/2}) is a positive integer solution of equation X^{2} − Y^{2} = C_{1} by (25), which is impossible.

If 2|r_{2}, then, by (25) and Lemma 9, we know that (u_{1}, a^{r_{2}/2}) is the minimal positive solution of the Diophantine equation

kx^{2} − la^{r_{4}−r_{2}}y^{2} = C_{1}.  (29)

By (26), (u_{2}, b^{r_{2}/2}) is a positive integer solution of (29). So by Lemma 11, we obtain a^{r_{2}/2}|b^{r_{2}/2}, which contradicts the assumption that a > b.

Similarly we have 2 ∤ r_{3} from (27).

We now suppose that 2 ∤ r_{2} and 2 ∤ r_{3}.

If C_{1} = 4, then, by (26), (27) and Lemma 8, (u_{2}, a^{\frac{r_{4}−r_{2}−1}{2}} b^{\frac{r_{2}−1}{2}} ) = (u_{3}, a^{\frac{r_{4}−r_{3}−1}{2}} b^{\frac{r_{3}−1}{2}} ) is the minimal positive solution of the Diophantine equation

kx^{2} − laby^{2} = 4,

which contradicts the assumption that u_{2} > u_{3}.

If C_{1} = −4, then, by (26), (27) and Lemma 8, we have that (a^{\frac{r_{4}−r_{3}−1}{2}} b^{\frac{r_{3}−1}{2}} , u_{3}) is the minimal positive solution of the Diophantine equation labx^{2}−ky^{2} = 4, and (lab, k, a^{\frac{r_{4}−r_{2}−1}{2}} b^{\frac{r_{2}−1}{2}} , u_{2}) = (5, 1, 5, 11), and so l = b = k = 1, a = 5. Thus (1, 1) = (a^{\frac{r_{4}−r_{3}−1}{2}} b^{\frac{r_{3}−1}{2}} , u_{3}) is the minimal positive solution of 5x^{2} − y^{2} = 4, which follows a = 1, which is a contradiction.

If C_{1} = ±2 or C_{1} = ±1, then, by (26), (27), both (u_{2}, a^{\frac{r_{4}−r_{2}−1}{2}} b^{\frac{r_{2}−1}{2}} ) and (u_{3}, a^{\frac{r_{4}−r_{3}−1}{2}} b^{\frac{r_{3}−1}{2}} ) are positive integer solutions of the Diophantine equation

kx^{2} − laby^{2} = C_{1}.

Noting that u_{2} > u_{3}, by Lemmas 6 and 7,

u_{3}\sqrt{k} + a^{(r_{4}−r_{3}−1)/2}b^{(r_{3}−1)/2}\sqrt{lab} = ε

must be the minimal positive solution. Therefore again by Lemmas 6 and 7, u_{2}\sqrt{k} + a^{(r_{4}−r_{2}−1)/2}b^{(r_{2}−1)/2}\sqrt{lab} = ε^{3} and

a^{(r_{4}−r_{2}−1)/2}b^{(r_{2}−1)/2} = 3^{s}a^{(r_{4}−r_{3}−1)/2}b^{(r_{3}−1)/2}, 3^{s} ∓ 3 = \frac{4}{|C_{1}|}la^{r_{4}−r_{3}}b^{r_{3}}.

It follows that a^{(r_{3}−r_{2})/2} = 3^{s}b^{(r_{3}−r_{2})/2}, and thus

b = 1, a = 3, r_{3} = 2s + r_{2}, l = (3^{s−1} ∓ 1)|C_{1}|/4,

since gcd(a, b) = 1. By (28), we get 4ku_{4}^{2}/|C_{1}| = 3^{s−1} ± 3, and so 3|k and 3| gcd(k, a), which is impossible since gcd(k, a) = 1. This concludes the analysis of Case 1.

Case 2: 2 ∤ r_{4}.

Subcase 2.1: C_{1} = ±4. Similarly, by (25) and Lemma 8, we can derive that

u_{1}\sqrt{k} + a^{(r_{4}−1)/2}\sqrt{la} = ε  (30)

is the minimal positive solution of the Diophantine equation

kx^{2} − lay^{2} = C_{1}.  (31)

If not, we must have (la, k, a^{(r_{4}−1)/2}, u_{1}, C_{1}) = (5, 1, 5, 11, −4). This follows that

l = k = 1, a = 5, r_{4} = 3,

and thus

r_{2} = 1, r_{3} = 2, b = 1 or 3.

Hence by (26), we get either u_{2}^{2} = 21 or u_{2}^{2} = 71, which is impossible.

If 2|r_{2}, then, by (26), (u_{2}, a^{(r_{4}−r_{2}−1)/2}b^{r_{2}/2}) is a positive integer solutions of (31). We have by Lemma 11 that a^{(r_{4}−1)/2}|a^{(r_{4}−r_{2}−1)/2}b^{r_{2}/2}. Therefore a^{r_{2}/2}|b^{r_{2}/2}, contradicting with a > b.

Similarly we have 2 ∤ r_{3} from (27).

Now we assume that 2 ∤ r_{2} and 2 ∤ r_{3}, then since la^{2}b ≠ 5, by (26), (27) and Lemma 8, (u_{2}, a^{\frac{r_{4}−r_{2}−2}{2}} b^{\frac{r_{2}−1}{2}} ) = (u_{3}, a^{\frac{r_{4}−r_{3}−2}{2}} b^{\frac{r_{3}−1}{2}} ) is the minimal positive solution of the Diophantine equation

kx^{2} − la^{2}by^{2} = C_{1},

which contradicts the assumption that u_{2} > u_{3}.

Subcase 2.2: C_{1} = ±1 or C_{1} = ±2.

If 2|r_{2}, then, by (26), (u_{2}, a^{\frac{r_{4}−r_{2}−3}{2}} b^{\frac{r_{2}}{2}} ) is a positive integer solution of Diophantine equation

kx^{2} − la^{3}y^{2} = C_{1}.  (32)

By (25) and Lemma 9, u_{1}\sqrt{k} + a^{(r_{4}−3)/2}\sqrt{la^{3}} must be the minimal positive solution of (32). Therefore we have by Lemma 11 that a^{(r_{4}−3)/2}|a^{\frac{r_{4}−r_{2}−3}{2}} b^{\frac{r_{2}}{2}} . So a^{r_{2}/2}|b^{r_{2}/2}, which contradicts the assumption that a > b.

If 2 ∤ r_{2} and 2 ∤ r_{3}, then, by (26) and (27), both (u_{2}, a^{\frac{r_{4}−r_{2}−2}{2}} b^{\frac{r_{2}−1}{2}} ) and (u_{3}, a^{\frac{r_{4}−r_{3}−2}{2}} b^{\frac{r_{3}−1}{2}} ) are positive integer solutions of Diophantine equation

kx^{2} − la^{2}by^{2} = C_{1}.  (33)

Noting that u_{2} > u_{3}, by Lemmas 6 and 7,

u_{3}\sqrt{k} + a^{(r_{4}−r_{3}−2)/2}b^{(r_{3}−1)/2}\sqrt{la^{2}b} = ε

must be the minimal positive solution. Therefore again by Lemmas 6 and 7, u_{2}\sqrt{k} + a^{(r_{4}−r_{2}−2)/2}b^{(r_{2}−1)/2}\sqrt{la^{2}b} = ε^{3} and that

a^{\frac{r_{4}−r_{2}−2}{2}} b^{\frac{r_{2}−1}{2}} = 3^{s}a^{\frac{r_{4}−r_{3}−2}{2}} b^{\frac{r_{3}−1}{2}} , 3^{s} ∓ 3 = \frac{4}{|C_{1}|}la^{r_{4}−r_{3}}b^{r_{3}}.

This follows that a^{\frac{r_{3}−r_{2}}{2}} = 3^{s}b^{\frac{r_{3}−r_{2}}{2}} , and so 3|a. Since r_{4} − r_{3} is even, 3^{s} ∓ 3 = \frac{4}{|C_{1}|}la^{r_{4}−r_{3}}b^{r_{3}} is also divisible by 9. Hence 9|3, which is a contradiction.

If 2 ∤ r_{2} and 2|r_{3}, then, by (25), (u_{1}, a^{(r_{4}−1)/2}) is a positive integer solution of Diophantine equation

kx^{2} − lay^{2} = C_{1},  (34)

and by (28), (u_{4}, b^{(r_{4}−1)/2}) is a positive integer solution of Diophantine equation

kx^{2} − lby^{2} = C_{1}.  (35)

We have by Lemmas 6 and 7 that either

u_{1}\sqrt{k} + a^{(r_{4}−1)/2}\sqrt{la} = ε,

or

\frac{u_{1}\sqrt{k} + a^{(r_{4}−1)/2}\sqrt{la}}{\sqrt{|C_{1}|}} = (\frac{ε}{\sqrt{|C_{1}|}})^{3} , a^{(r_{4}−1)/2} = 3^{s}y_{1}, 3^{s} ∓ 3 = \frac{4}{|C_{1}|}lay_{1}^{2}, s ≥ 2,

where ε = x_{1}\sqrt{k} + y_{1}\sqrt{la} is the minimal positive solution of (34), and that either

u_{4}\sqrt{k} + b^{(r_{4}−1)/2}\sqrt{lb} = δ,

or

\frac{u_{4}\sqrt{k} + b^{(r_{4}−1)/2}\sqrt{lb}}{\sqrt{|C_{1}|}} = (\frac{δ}{\sqrt{|C_{1}|}})^{3} , b^{(r_{4}−1)/2} = 3^{s_{1}}v_{1}, 3^{s_{1}} ∓ 3 = \frac{4}{|C_{1}|}lbv_{1}^{2}, s_{1} ≥ 2,

where δ = d_{1}\sqrt{k} + v_{1}\sqrt{lb} is the minimal positive solution of (35).

If \frac{u_{1}\sqrt{k}+a^{(r_{4}−1)/2}\sqrt{la}}{\sqrt{|C_{1}|}} = (\frac{ε}{\sqrt{|C_{1}|}})^{3} and \frac{u_{4}\sqrt{k}+b^{(r_{4}−1)/2}\sqrt{lb}}{\sqrt{C_{1}}} = (\frac{δ}{\sqrt{|C_{1}|}})^{3} , then a^{(r_{4}−1)/2} = 3^{s}y_{1}^{2}, b^{(r_{4}−1)/2} =

3^{s_{1}}v_{1}^{2}. Thus 3|a and 3|b, which contradicts the assumption that gcd(a, b) = 1.

If u_{1}\sqrt{k} + a^{(r_{4}−1)/2}\sqrt{la} = ε, then, by (27), (u_{3}, a^{\frac{r_{4}−r_{3}−1}{2}} b^{\frac{r_{3}}{2}} ) is a positive integer solution of (34). By Lemma 4, we obtain

\frac{u_{3}\sqrt{k} + a^{(r_{4}−r_{3}−1)/2}b^{r_{3}/2}\sqrt{la}}{\sqrt{|C_{1}|}} = (\frac{u_{1}\sqrt{k} + a^{(r_{4}−1)/2}\sqrt{la}}{\sqrt{|C_{1}|}})^{n}

for some positive integer n. This implies that u_{3} ≥ u_{1}, which is a contradiction.

If u_{4}\sqrt{k} + b^{(r_{4}−1)/2}\sqrt{lb} = δ, then, by (26), (u_{2}, a^{\frac{r_{4}−r_{2}}{2}} b^{\frac{r_{2}−1}{2}} ) is a positive integer solution of (35). We have by Lemma 11 that b^{(r_{4}−1)/2}|a^{(r_{4}−r_{2})/2}b^{(r_{2}−1)/2}. This implies b^{(r_{4}−r_{2})/2}|a^{(r_{4}−r_{2})/2}, and so, since gcd(a, b) = 1, b = 1. By (25) and (27), both (u_{1}, a^{\frac{r_{4}−1}{2}} ) and (u_{3}, a^{\frac{r_{4}−r_{3}−1}{2}} ) are positive integer solutions of Diophantine equation

kx^{2} − lay^{2} = C_{1}.

Noting that u_{1} > u_{3}, by Lemmas 6 and 7,

u_{3}\sqrt{k} + a^{\frac{r_{4}−r_{3}−1}{2}} \sqrt{la} = ε

must be the minimal positive solution. Therefore again by Lemmas 6 and 7, u_{1}\sqrt{k} + a^{(r_{4}−1)/2}\sqrt{la} = ε^{3} and that a^{\frac{r_{4}−1}{2}} = 3^{s}a^{\frac{r_{4}−r_{3}−1}{2}} , 3^{s} ∓ 3 = \frac{4}{|C_{1}|}la^{r_{4}−r_{3}}. It follows that

a = 3, l = (3^{s−1} ∓ 1)|C_{1}|/4,

which is impossible as shown at the end of Case 1. This concludes the analysis of Case 2.

This completes the proof of Theorem 1.

4 Acknowledgement

The authors would like to thank the referee for his valuable suggestions.

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2000 Mathematics Subject Classification: Primary 11D09; Secondary 11B83. Keywords: quadratic diophantine equation, minimal positive solution, triangular numbers, geometric progression.

Received October 11 2009; revised version received February 25 2010. Published in Journal of Integer Sequences, February 26 2010.

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