Journal of Number Theory 147 (2015) 287–299

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Journal of Number Theory

www.elsevier.com/locate/jnt

On products of consecutive arithmetic

progressions ✩

Yong Zhang a,b, Tianxin Cai b,∗

a College of Mathematics and Computing Science, Changsha University of Science and Technology, Changsha 410114, People’s Republic of Chinab Department of Mathematics, Zhejiang University, Hangzhou 310027,People’s Republic of China

a r t i c l e i n f o a b s t r a c t

Article history:Received 30 May 2014Received in revised form 1 July 2014Accepted 1 July 2014Available online 6 September 2014Communicated by David Goss

MSC:primary 11D25secondary 11D72, 11G05

Keywords:Diophantine equationConsecutive arithmetic progressionPell’s equationElliptic curve

In this paper, first, we show the Diophantine equation

x(x + b)y(y + b) = z(z + b)

has infinitely many nontrivial positive integer solutions for b ≥ 3. Second, we prove the Diophantine equation

(x− b)x(x + b)(y − b)y(y + b) = (z − b)z(z + b)

has infinitely many nontrivial positive integer solutions for b = 1, and the set of rational solutions of it is dense in the set of real solutions for b ≥ 1. Third, we get infinitely many nontrivial positive integer solutions of the Diophantine equation

(x− b)x(x + b)(y − b)y(y + b) = z2

for even number b ≥ 2. At last, we raise some unsolved questions.

© 2014 Elsevier Inc. All rights reserved.

✩ This research was supported by China National Science Foundation Grant (No. 11351002), Natural Science Foundation of Zhejiang Province (No. LQ13A010012), and Zhejiang Projects for Postdoctoral Research Preferred Funds (No. Bsh1201021).* Corresponding author.

E-mail addresses: zhangyongzju@163.com (Y. Zhang), txcai@zju.edu.cn (T. Cai).

http://dx.doi.org/10.1016/j.jnt.2014.07.0030022-314X/© 2014 Elsevier Inc. All rights reserved.

288 Y. Zhang, T. Cai / Journal of Number Theory 147 (2015) 287–299

1. Introduction

Diophantine equations on products of consecutive integers have been studied by many authors. In 1990, S. Katayama [7] investigated the Diophantine equation on products of 2-term consecutive integers, i.e.,

x(x + 1)y(y + 1) = z(z + 1). (1.1)

He proved that Eq. (1.1) has infinitely many integer solutions and there exists an algo-rithm to obtain all the integer solutions. In 1996, A. Baragar [1] investigated Eq. (1.1)by using the Markoff equation, and his method is interesting.

In 1990, K. Kashihara [6] considered the Diophantine equation on products of 2-term consecutive arithmetic progressions with common difference 2, i.e.,

(x2 − 1

)(y2 − 1

)= z2 − 1. (1.2)

He showed that all the integer solutions of Eq. (1.2) can be derived from the trivial solutions (n, 1, 1) and (1, n, 1). We can refer to S. Katayama and K. Kashihara’s paper [8]for more specific detail on Eq. (1.2). Some related information on Eqs. (1.1) and (1.2)could be found in [5]: D23 Some quartic equations.

In this paper, first, we study the Diophantine equation on products of 2-term consec-utive arithmetic progressions with common difference b, i.e.,

x(x + b)y(y + b) = z(z + b), b ≥ 1. (1.3)

It’s easy to see that Eq. (1.3) has solutions (x, y, z) = (n, n +1, n2 +(b +1)n), where n is an integer parameter. We call them trivial solutions, and the others nontrivial. For b ≥ 3, we can’t prove the same result as in [1,6]. But we have the following theorem by using a similar method of [1,6].

Theorem 1.1. For b ≥ 3, Eq. (1.3) has infinitely many nontrivial positive integer solu-tions.

Second, we consider the Diophantine equation on products of 3-term consecutive arith-metic progressions with common difference b, i.e.,

(x− b)x(x + b)(y − b)y(y + b) = (z − b)z(z + b), b ≥ 1. (1.4)

Eq. (1.4) has trivial solutions (x, y, z) = (x; −b, 0, b; −b, 0, b), (−b, 0, b; y; −b, 0, b). For b = 1, (x, y, z) = (2, 5, 9), (5, 13, 64), (13, 34, 441) are nontrivial solutions. By using a transformation and Pell’s equation, we prove the following theorem, which is another form of Theorem 4 in our previous paper [16].

Y. Zhang, T. Cai / Journal of Number Theory 147 (2015) 287–299 289

Theorem 1.2. For b = 1, Eq. (1.4) has infinitely many nontrivial positive integer solu-tions.

This is an interesting result. It is not only related to the Fibonacci numbers and the Lucas numbers, but also related to the congruent numbers.

A congruent number is a positive integer that is the area of a right triangle with three rational number sides. It’s easy to prove that (x − 1)x(x + 1) = x3 − x is a congruent number for every integer x > 1. From Theorem 1.2, we have

Corollary 1.3. There are infinitely many congruent numbers in the form x3 − x, which are products of two congruent numbers in the same form.

For b > 1, we don’t find a similar structure of the integer solutions of Eq. (1.4). But we get the following theorem, by using the theory of elliptic curves.

Theorem 1.4. For b ≥ 1, the set of rational solutions S(Q) of Eq. (1.4) is dense in the set of real solutions S(R).

Moreover, we investigate the rational solutions of the products of 4-term consecutive integers, and prove

Theorem 1.5. There is a rational parametric solutions of the Diophantine equation

x(x− 1)(x− 2)(x− 3)y(y − 1)(y − 2)(y − 3) = z(z − 1)(z − 2)(z − 3). (1.5)

At last, we study when the products of two disjoint blocks of 3-term consecutive arithmetic progressions with common difference b can be a square, i.e.,

(x− b)x(x + b)(y − b)y(y + b) = z2, b ≥ 1. (1.6)

For b = 1, this is a special case of a problem raised by P. Erdős and R.L. Graham [4]. K.R.S. Sastry [5] showed that Eq. (1.6) has infinitely many positive integer solutions (x, y, z) for b = 1, where x, y satisfying y = 2x − 1 and (x + 1)(2x − 1) is a square. For more information on Eq. (1.6), we can refer to problem D17 in [5] and the references therein. We are looking for the integer solutions which satisfy b � x or b � y. If the solutions (x, y, z) satisfy b | x and b | y, we call them trivial. By using the same method as K.R.S. Sastry, we have

Theorem 1.6. For even number b ≥ 2, Eq. (1.6) has infinitely many nontrivial positive integer solutions.

For odd number b ≥ 3, we don’t get the same result. It seems that there are also infinitely many nontrivial positive integer solutions of Eq. (1.6).

290 Y. Zhang, T. Cai / Journal of Number Theory 147 (2015) 287–299

2. Proofs of the theorems

Proof of Theorem 1.1. Let

X = 2x + b, Y = 2y + b, V = 4z + 2b.

From Eq. (1.3) we have

(X2 − b2

)(Y 2 − b2

)= V 2 − 4b2.

Let X be a constant such that X2 − b2 is not a perfect square. Put α2 = X2 − b2

and L = Q(α). Then

NL/Q(V + αY ) = 4b2 −(X2 − b2

)b2.

Noting that

1 = X + α

b

X − α

b,

thus

X + α

b

could be considered as an element with norm 1 of the field L = Q(α). So V + αY and

(V + αY )(X + α

b

)

have the same norm. If the Diophantine equation (X2 − b2)(Y 2 − b2) = V 2 − 4b2 has a solution (X, Y, V ), then it has another solution

(X ′, Y ′, V ′) =

(X,

XY + V

b,(X2 − b2)Y + XV

b

).

From the above process, we get the conclusion: If (x, y, z) is an integer solution of Eq. (1.3) and 2(z + xy) ≡ 0 (mod b), then (y, x, z) and

(x′, y′, z′

)=

(x, x + y + 1 + 2(z + xy)

b, x2 + 2xy + bx + x + z + 2x(z + xy)

b

)

are integer solutions of Eq. (1.3).Note that (x, y, z) = (n, n + 1, n2 + (b + 1)n) are trivial integer solutions of Eq. (1.3).

We can get two classes of solutions

Y. Zhang, T. Cai / Journal of Number Theory 147 (2015) 287–299 291

(n, 4n + 2 + 4n(n + 1)

b, 6n2 + 4n + 2bn + 4n2(n + 1)

b

)

and(n + 1, 4n + 2 + 4n(n + 1)

b, 6n2 + 8n + 2bn + 2 + b + 4n(n + 1)2

b

).

When 4n(n + 1) ≡ 0 (mod b), these give out two classes of integer solutions. This congruence equation has infinitely many integer solutions n for each b ≥ 3. Then Eq. (1.3)has infinitely many nontrivial positive integer solutions. �Example 2.1. For b = 3, Eq. (1.3) has a trivial integer solution (2, 3, 12) satisfying

2(z + xy) ≡ 0 (mod 3).

Then we obtain two nontrivial positive integer solutions

(2, 18, 60), (3, 18, 81).

These two solutions also satisfy the above congruence relation, so we can get four integer solutions. In Fig. 1, we display some integer solutions which are derived from (2, 3, 12).

Fig. 1. The construction of solutions of Eq. (1.3).

If there are solutions satisfying 2(z + xy) ≡ 0 (mod 3), then the above process could be repeated, which is like the Markoff tree in [1]. This may give another proof of Theo-rem 1.1, but it seems difficult. As the referee pointed out that (2, 85, 272) does not satisfy the congruence relation, so it cannot produce other solutions.

Remark 2.2. In fact, let y be a fixed positive integer. Eq. (1.3) can be considered as Pell’sequation. Then we can use the theory of Pell’s equation to get infinitely many integer solutions.

292 Y. Zhang, T. Cai / Journal of Number Theory 147 (2015) 287–299

Before giving the proof of Theorem 1.2, we introduce the initial idea to deal with Eq. (1.4) for b = 1. Note that z − 1, z, and z + 1 are the divisors of

(x− 1)x(x + 1)(y − 1)y(y + 1).

By some calculations, we find that the transformation

z + 1 = xy

is suitable to solve Eq. (1.4).

Proof of Theorem 1.2. Let

z = xy − 1.

Eq. (1.4) equals

xy(x2 + y2 − 3xy + 1

)= 0.

Suppose that xy �= 0, we have x2 + y2 − 3xy + 1 = 0. Hence,

(2x− 3y)2 − 5y2 = −4.

Taking 2x − 3y = X, y = Y , we get Pell’s equation

X2 − 5Y 2 = −4. (2.1)

From Eq. (2.1), it is natural to take

Xn + Yn

√5

2 =(

1 +√

52

)2n−1

, n ≥ 1.

Then y = F2n−1 and Xn = 2x − 3y = L2n−1 = F2n + F2n−2 lead to

2x = F2n + 3F2n−1 + F2n−2 = (F2n + F2n−1) + (F2n−1 + F2n−2) + F2n−1 = 2F2n+1.

Hence, x = F2n+1, y = F2n−1. Moreover, Binet’s formula implies

z = xy − 1 = F2n+1F2n−1 − 1 = F 22n.

Thus one can conclude the integer solution (x, y, z) of Eq. (1.4) for the case b = 1 and z = xy − 1 is

(x, y, z) =(F2n+1, F2n−1, F

22n).

Y. Zhang, T. Cai / Journal of Number Theory 147 (2015) 287–299 293

Consequently, they have shown an interesting result that the integer solution of Eq. (1.4)with b = 1, z = xy − 1 can be parameterized by the Fibonacci numbers. �Remark 2.3. In the proof of Theorem 1.2, we get an identity about the Fibonacci numbers

(F2n+1 − 1)F2n+1(F2n+1 + 1)(F2n−1 − 1)F2n−1(F2n−1 + 1) =(F 2

2n − 1)F 2

2n(F 2

2n + 1),

where n ≥ 1. It’s worth to note that M. Skałba [13] may know this identity.

Next, we give the proof of Theorem 1.4.

Proof of Theorem 1.4. It is enough to consider the case b = 1. Let y = u, z = tx, where u and t are rational parameters. From Eq. (1.4) we have

x((−u + u3 − t3

)x2 + t + u− u3) = 0.

Considering it as a cubic equation of x, we get

x = 0,±√

(t3 + u− u3)(t + u− u3)t3 + u− u3 .

To make x be a nonzero rational number, let(t3 + u− u3)(t + u− u3) = s2.

Hence, we need to study the following curve

Cu : s2 = t4 +(u− u3)t3 +

(u− u3)t + u2 − 2u4 + u6.

The rational point P = (t, s) = (u −u3, 0) lies on Cu. If we treat P as a point at infinity on Cu and use the method of [10] (see p. 77), then Cu is birationally equivalent to the elliptic curve

Eu : Y 2 = X3 − 3(u− u3)2X +

(1 + u2 − 2u4 + u6)(u− u3)2. (2.2)

The mapping is⎧⎪⎪⎪⎨⎪⎪⎪⎩

t = (−u + u3)X + 2Y − 2u + 2u3

4X − u2 + 2u4 − u6 ,

s =(4Y 2 −

(u− u3)(8 + u2 − 2u4 + u6)Y − 8X3 + 3

(u− u3)2X2

+(4 + u2 − 2u4 + u6)(u− u3)2)/(4X − u2 + 2u4 − u6)2,

and its inverse transformation is{X = 2t2 +

(u− u3)t− 2s,

3 ( 3) 2 ( 3) 3

Y = 4t + 3 u− u t − 4st− u− u s + u− u .

294 Y. Zhang, T. Cai / Journal of Number Theory 147 (2015) 287–299

The discriminant of Eu is

Δ(u) = −27(u− u3)4(−1 + u− u3)2(1 + u− u3)2.

When u = 0, ±1, we have Δ(u) = 0. So Eu is nonsingular for u ∈ Q − {0, ±1}.The point P can be transformed into

Q =(2(u− u3), (u− u3)(1 + u− u3))

on Eu. By the group law, we have

[2]Q =(

(−u + u3)(16 − 49u + 16u2 + 49u3 − 32u4 + 16u6)4(−1 − u + u3)2 ,

(u− u3)(8 − 184u + 345u2 − 682u4 + 552u5 + 313u6 − 552u7 + 48u8

+ 184u9 − 32u10 + 8u12)/(2(−1 − u + u3))3).Let E2 be the specialization of Eu at u = 2, and the specialization of point [2]Q

at u = 2 is

[2]Q2 =(

132925 ,−47733

125

).

According to the Nagell–Lutz theorem (see p. 56 of [11]), [2]Q2 is a point of infinite order on E2. Then the point Q is a point of infinite order and Eu has positive rank in the field Q(u). By the Specialization Theorem of Silverman (see p. 457, Theorem 20.3 of [12]), for all but finite many u0 ∈ Q, Eu0 has positive rank. Hence, Eu0 has infinitely many rational points.

Let U denote the set of all u such that Eu has positive rank. To get this set, we only need to find all u such that the point Q is a point of finite order. From Mazur’s Theorem (see p. 58 of [11]), the finite order point on elliptic curve has maximal order 12. Let us compute the expression [m]Q = (X(m), Y (m)) with 1 ≤ m ≤ 12, and determine those u ∈ Q such that denominator of X(m) has a zero at u. The u ∈ Q with such a property are u = 0, ±1. Then U = Q − {0, ±1}.

When u = 0, ±1, we have

Eu : Y 2 = X3.

This is a singular curve, which can be parameterized by

X = v2, Y = v3,

where v is a rational parameter. Then the set of rational points on Eu is dense in the set of real points for u = 0, ±1.

Y. Zhang, T. Cai / Journal of Number Theory 147 (2015) 287–299 295

When u ∈ U , Eu has positive rank. In view of a theorem of Poincaré and Hurwitz (see p. 78 of [14]), the set of rational points on Eu is dense in the set of real points for u ∈ U .

Therefore, for any u ∈ Q, the set of rational solutions S(Q) of Eq. (1.4) for b = 1 is dense in the set of real solutions S(R).

For b > 1, the proof is similar. �Proof of Theorem 1.5. Let us consider the solutions of the Diophantine equation

(X2 − 1

)(Y 2 − 1

)= Z2 − 1.

It has a parametric solution

X = n, Y = n + 1, Z = n2 + n− 1,

where n is a rational parameter.Noting that x(x − 1)(x − 2)(x − 3) = (x(x − 3) + 1)2 − 1, let

x(x− 3) + 1 = n, y(y − 3) + 1 = n + 1, z(z − 3) + 1 = n2 + n− 1.

Solving the Diophantine system

x(x− 3) + 1 = n, y(y − 3) + 1 = n + 1,

we have

x(x− 3) + 1 = y(y − 3).

Because (x, y) = (32 ,

52 ) is a solution of the above equation, we can get its rational

parametric solutions

x = 3m2 − 4m− 32(m2 − 1) , y = m2 − 5

2(m2 − 1) ,

where m is a rational parameter. Then

n = − (5m2 − 1)(m2 − 5)4(m2 − 1)2 .

From z(z − 3) + 1 = n2 + n − 1, we have

z = (3m2 + 1)(m2 + 3)4(m2 − 1)2 .

Therefore,

296 Y. Zhang, T. Cai / Journal of Number Theory 147 (2015) 287–299

(x, y, z) =(

3m2 − 4m− 32(m2 − 1) ,

m2 − 52(m2 − 1) ,

(3m2 + 1)(m2 + 3)4(m2 − 1)2

)

satisfies Eq. (1.5). �Proof of Theorem 1.6. For even number b ≥ 2, let y = 2x − b. From Eq. (1.6) we have

(x− b)x(x + b)(2x− 2b)(2x− b)(2x) = z2.

Then we just need to consider

(x + b)(2x− b) = w2,

where z = 2x(x − b)w. This equation is equivalent to Pell’s equation

X2 − 2Y 2 = 9b2,

where X = 4x + b, Y = 2w. An infinity of positive integer solutions are given by

Xn + Yn

√2 = (9b + 6b

√2)(3 + 2

√2)n, n ≥ 0.

Thus

{Xn = 6Xn−1 −Xn−2, X0 = 9b, X1 = 51b;Yn = 6Yn−1 − Yn−2, Y0 = 6b, Y1 = 36b.

It is easy to prove that

X2n+1 − b ≡ 0 (mod 4), Y2n+1 ≡ 0 (mod 2), n ≥ 0.

Then x2n+1, w2n+1 are integers and

b � x2n+1.

Hence,

z2n+1 = 2x2n+1(x2n+1 − b)w2n+1, y2n+1 = 2x2n+1 − b

are integers. Therefore, Eq. (1.6) has infinitely many nontrivial integer solutions (x2n+1, y2n+1, z2n+1). �

Y. Zhang, T. Cai / Journal of Number Theory 147 (2015) 287–299 297

3. Some related questions

For Eq. (1.3), i.e.,

x(x + b)y(y + b) = z(z + b), b ≥ 1,

we have the following question, which is motivated by the result of [6–8].

Question 3.1. Could all the nontrivial integer solutions of Eq. (1.3) for b ≥ 3 be derived from some trivial integer solutions?

In the range 1 < x ≤ y ≤ 103, 1 < z ≤ 105, we find that Eq. (1.4) has no other nontrivial positive integer solutions for b = 1 except the solutions given in Theorem 1.2. So we raise the following question

Question 3.2. Are all the nontrivial positive integer solutions of Eq. (1.4) for b = 1 with x ≤ y given by (F2n−1, F2n+1, F 2

2n), n ≥ 1?

For b ≥ 2, we have

Question 3.3. Are there infinitely many nontrivial positive integer solutions of Eq. (1.4)for b ≥ 2?

Here we list the nontrivial positive integer solutions of Eq. (1.4) for 2 ≤ b ≤ 10 in Table 1, where 1 < x ≤ y ≤ 103, 1 < z ≤ 104.

Table 1Some solutions of Eq. (1.4).

b (x, y, z)2 (5, 42, 198)3 (3, 15, 45)4 (6, 8, 36), (6, 52, 256), (7, 36, 220)5 (7, 50, 275)6 (7, 18, 78), (9, 28, 202), (9, 32, 234), (9, 442, 3270)7 (8, 8, 25), (8, 14, 63), (8, 91, 448), (11, 15, 128), (11, 42, 385),

(12, 14, 133), (12, 29, 297), (14, 80, 1015), (15, 30, 407)8 (9, 140, 748), (16, 85, 1232), (18, 168, 4008)9 (13, 180, 1881), (19, 102, 1776), (19, 108, 1881)

10 (11, 18, 98), (11, 90, 550), (12, 15, 100), (14, 48, 522)

By the transformation

z = xy

a− 1,

we can prove the Diophantine equation(x3 − x

)(y3 − y

)= a3(z3 − z

), a ≥ 2 (3.1)

298 Y. Zhang, T. Cai / Journal of Number Theory 147 (2015) 287–299

has infinitely many nontrivial positive integer solutions. For example, (x, y, z) =(18a3 − 5a, 6a2 − 1, (18a2 − 5)(6a2 − 1) − 1) is a solution of Eq. (3.1).

Moreover, we can study the Diophantine equation on products of k-term consecutive arithmetic progressions with common difference b for k ≥ 4. We raise the following question

Question 3.4. Does there exist nontrivial positive integer solutions satisfying

x(x + b) · · ·(x + (k − 1)b

)y(y + b) · · ·

(y + (k − 1)b

)= z(z + b) · · ·

(z + (k − 1)b

)(3.2)

for k ≥ 4, b ≥ 1? If exists, are there infinitely many of them?

For k = 4, we get some examples in Table 2 in the range 1 < x ≤ y ≤ 104.

Table 2Some solutions of Eq. (3.2) for k = 4.

b (x, y, z)1 None2 (1, 28, 96)3 (3, 11, 96)4 (1, 16, 100), (8, 14, 252)5 None

For k = 5, 6, 7, and b = 1, Eq. (3.2) does not have nontrivial positive integer solutions with 1 < x ≤ y ≤ 103, 1 < z ≤ 104. It seems difficult to deal with Question 3.4.

Motivated by the question posed by P. Erdős and R.L. Graham (see p. 67 of [4]), we have a similar question

Question 3.5. Can the products of two or more disjoint blocks of k-term consecutive arithmetic progressions with common difference b be a power?

For b = 1, many authors studied this question, and we can refer to [2,3,9,13,15]. For b ≥ 2, we think there exist some interesting results like b = 1.

Acknowledgment

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

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