Some New Properties of Balancing Numbers and Square Triangular Numbers

Refik Keskin and Olcay Karaatlı Department of Mathematics Faculty of Arts and Science Sakarya University 54187 Serdivan Sakarya Turkey

                    rkeskin@sakarya.edu.tr
                   okaraatli@sakarya.edu.tr

Abstract A number N is a square if it can be written as N = n^{2} for some natural number n; it is a triangular number if it can be written as N = n(n + 1)/2 for some natural number n; and it is a balancing number if 8N^{2} + 1 is a square. In this paper, we study some properties of balancing numbers and square triangular numbers.

1 Introduction

A triangular number is a number of the form T_{n} = n(n + 1)/2, where n is a natural number. So the first few triangular numbers are 1, 3, 6, 10, 15, 21, 28, 36, 45,… (sequence \underline{A000217} in [20]). A well known fact about the triangular numbers is that x is a triangular number if and only if 8x + 1 is a perfect square. Triangular numbers can be thought of as the numbers of dots needed to make a triangle. In a similar way, square numbers can be thought of as the numbers of dots that can be arranged in the shape of a square. The m-th square number is formed using an outher square whose sides have m dots. Let us denote the expression for m-th square number by S_{m} = m^{2} [15]. Behera and Panda [1] introduced balancing numbers m ∈ ℤ^{+} as solutions of the equation

1 + 2 + ⋯ + (m − 1) = (m + 1) + (m + 2) + ⋯ + (m + r),  (1)

calling r ∈ ℤ^{+}, the balancer corresponding to the balancing number m. For instance 6, 35, and 204 are balancing numbers with balancers 2, 14, and 84, respectively. It is clear from (1) that m is a balancing number with balancer r if and only if

m^{2} = \frac{(m + r)(m + r + 1)}{2},

which when solved for r gives

r = \frac{−(2m + 1) + \sqrt{8m^{2} + 1}}{2}.  (2)

It follows from (2) that m is a balancing number if and only if 8m^{2} + 1 is a perfect square. Since 8 × 1^{2} + 1 = 9 is a perfect square, we accept 1 as a balancing number. In what follows, we introduce cobalancing numbers in a way similar to the balancing numbers. By modifying (1), we call m ∈ ℤ^{+}, a cobalancing number if

1 + 2 + ⋯ + (m − 1) + m = (m + 1) + (m + 2) + ⋯ + (m + r)  (3)

for some r ∈ ℤ^{+}. Here, r ∈ ℤ^{+} is called a cobalancer corresponding to the cobalancing number m. A few of the cobalancing numbers are 2, 14, and 84 with cobalancers 6, 35, and 204, respectively. It is clear from (3) that m is a cobalancing number with cobalancer r if and only if

m(m + 1) = \frac{(m + r)(m + r + 1)}{2},

which when solved for r gives

r = \frac{−(2m + 1) + \sqrt{8m^{2} + 8m + 1}}{2}.  (4)

It follows from (4) that m is a cobalancing number if and only if 8m^{2} + 8m + 1 is a perfect square, that is, m(m + 1) is a triangular number. Since 8 × 0^{2} + 8 × 0 + 1 = 1 is a perfect square, we accept 0 is a cobalancing number [7, 8]. Also since m(m + 1)/2 is known as a triangular number by the very definition of triangular number, the above discussion means that if m is a cobalancing number, then both m(m + 1) and m(m + 1)/2 are triangular numbers. Panda and Ray [7] proved that every cobalancing number is even. And also they showed that every balancer is a cobalancing number and every cobalancer is a balancing number.

Oblong numbers are numbers of the form O_{n} = n(n + 1), where n is a positive integer. The n-th oblong number represents the number of points in a rectangular array having n columns and n+1 rows. The first few oblong numbers are 2, 6, 12, 20, 30, 42, 56, 72, 90, 110,… (sequence \underline{A002378} in [20]). Since 2 + 4 + 6+⋯ +2n = 2(1 + 2 + 3+⋯ +n) = 2n(n + 1)/2 = n(n+1) = O_{n}, the sum of the first n even numbers equals the n-th oblong number. Actually it is clear from the definition of oblong numbers and triangular numbers that an oblong number is twice a triangular number. After the definition of oblong numbers, we can say from (4) that if m is a cobalancing number, then m(m + 1) is both an oblong and triangular number. Well then, what about the square triangular numbers? Since triangular numbers are of the form T_{n} = n(n + 1)/2 and square numbers are of the form S_{m} = m^{2}, square triangular numbers are integer solutions of the equation

m^{2} = \frac{n(n + 1)}{2}.  (5)

Eq.(5) says something about the relation between balancing and square triangular numbers. Behera and Panda [1] proved that a positive integer m is a balancing number if and only if m^{2} is a triangular number, that is, 8m^{2} + 1 is a perfect square. Here, we will get Eq.(5) again using an amusing problem and we will see the interesting relation between balancing and square triangular numbers by means of this problem. In equation (1), if we make the substitution m + r = n, then we get 1 + 2+⋯ +(m − 1) = (m + 1) + (m + 2)+⋯ +n. Thus this equation gives us a problem as follows.

I live on a street whose houses are numbered in order 1, 2, 3,…, n − 1, n; so the houses at the ends of the street are numbered 1 and n. My own house number is m and of course 0 < m < n. One day, I add up the house numbers of all the houses to the left of my house; then I do the same for all the houses to the right of my house. I find that the sums are the same. So how can we find m and n [14]? Since 1+2+3+⋯ +m−1 = (m+1)+⋯ +(n−1)+n, it follows that

\frac{(m − 1)m}{2} = \frac{n(n + 1)}{2} − \frac{m(m + 1)}{2}.

Thus we get m^{2} = n(n + 1)/2. Here, m^{2} is both a triangular number and a square number. That is, m^{2} is a square triangular number. In Eq.(1), since m is a balancing number, it is easy to see that a balancing number is the square root of a square triangular number. For more information about triangular, square triangular and balancing numbers, one can consult [1, 13, 16, 17, 18].

2 Preliminaries

In this section, we introduce two kinds of sequences named generalized Fibonacci and Lucas sequences (U_{n}) and (V_{n}), respectively. Let k and t be two nonzero integers. The generalized Fibonacci sequence is defined by U_{0} = 0, U_{1} = 1 and U_{n+1} = kU_{n} + tU_{n−1} for n ≥ 1 and generalized Lucas sequence is defined by V_{0} = 2, V_{1} = k and V_{n+1} = kV_{n} + tV_{n−1} for n ≥ 1, respectively. Also generalized Fibonacci and Lucas numbers for negative subscript are defined as

U_{−n} = \frac{−U_{n}}{(−t)^{n}} and V_{−n} = \frac{V_{n}}{(−t)^{n}}  (6)

for n ⩾ 1. For k = t = 1, the sequences (U_{n}) and (V_{n}) are called classic Fibonacci and Lucas sequences and they are denoted as (F_{n}) and (L_{n}) , respectively. The first Fibonacci numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34,… (sequence \underline{A000045} in [20]) and the first Lucas numbers are 2, 1, 3, 4, 7, 11, 18, 29, 47, 76,… (sequence \underline{A000032} in [20]). For k = 2 and t = 1, the sequences (U_{n}) and (V_{n}) are called Pell and Pell-Lucas sequences and they are denoted as (P_{n}) and (Q_{n}) , respectively. Thus P_{0} = 0, P_{1} = 1 and P_{n+1} = 2P_{n} + P_{n−1} for n ⩾ 1 and Q_{0} = 2, Q_{1} = 2 and Q_{n+1} = 2Q_{n} + Q_{n−1} for n ⩾ 1. The first few terms of Pell sequence are 0, 1, 2, 5, 12, 29, 70, 169, 408, 985,… (sequence \underline{A000129} in [20]) and the first few terms of Pell-Lucas sequence are 2, 2, 6, 14, 34, 82, 198, 478, 1154, 2786,… (sequence \underline{A002203} in [20]). Moreover, for k = 6 and t = −1, we represent (U_{n}) and (V_{n}) by (u_{n}) and (v_{n}), respectively. Thus u_{0} = 0, u_{1} = 1 and u_{n+1} = 6u_{n} − u_{n−1} and v_{0} = 2, v_{1} = 6 and v_{n+1} = 6v_{n} − v_{n−1} for all n ⩾ 1. The first few terms of the sequence (u_{n}) are 0, 1, 6, 35, 204,… (sequence \underline{A001109} in [20]) and the first few terms of the sequence (v_{n}) are 2, 6, 34, 198, 1154,… (sequence \underline{A003499} in [20]). Furthermore, from the equation (6), it clearly follows that

u_{−n} = −u_{n} and v_{−n} = v_{n}

for all n ⩾ 1. For more information about generalized Fibonacci and Lucas sequences, one can consult [4, 5, 6, 10, 11, 19]. Now we present some well known theorems and identities regarding the sequences (P_{n}), (Q_{n}), (u_{n}), and (v_{n}), which will be useful during the proofs of the main theorems and the new properties of the sequence (y_{n}), where y_{n} = (v_{n} − 2)/4.

Theorem 1. Let γ and δ be the roots of the characteristic equation x^{2} − 2x − 1 = 0. Then

we have P_{n} = \frac{γ^{n} − δ^{n}}{2\sqrt{2}} and Q_{n} = γ^{n} + δ^{n} for all n ≥ 0.

Theorem 2. Let α and β be the roots of the characteristic equation x^{2} − 6x + 1 = 0. Then

u_{n} = \frac{α^{n} − β^{n}}{4\sqrt{2}}  (7)

and

v_{n} = α^{n} + β^{n}

for all n ≥ 0.

The formulas given in the above theorems are known as Binet’s formulas. Let B_{n} denote the n−th balancing number. From [8], we know that

B_{n} = \frac{(3 + \sqrt{8})^{n} − (3 − \sqrt{8})^{n}}{2\sqrt{8}} .  (8)

From Theorems 1 and 2, it is easily seen that u_{n} = B_{n} = P_{2n}/2 and v_{n} = Q_{2n} for n ≥ 0. Moreover, from identities (7) and (8), it is easily seen that B_{n} = u_{n} for negative integer n. Then well known identities for (P_{n}) , (Q_{n}) , (B_{n}) and (v_{n}) are

Q_{n}^{2} − 8P_{n}^{2} = 4(−1)^{n},  (9)
v_{n}^{2} − 32B_{n}^{2} = 4,  (10)
B_{n}^{2} − 6B_{n}B_{n−1} + B_{n−1}^{2} = 1,  (11)

Q_{n}^{2} = Q_{2n} + 2(−1)^{n},  (12)
B_{2n} = B_{n}v_{n},  (13)
P_{2n} = P_{n}Q_{n},  (14)

and

v_{n}^{2} = v_{2n} + 2.  (15)

In order to see close relations between balancing numbers and square triangular numbers, we can give the following well known theorem which characterizes all square triangular numbers. We omit the proof of this theorem due to Karaatlı and Keskin [5].

Theorem 3. A natural number x is a square triangular number if and only if x = B_{n}^{2} for some natural number n.

Since y_{n} = (v_{n} − 2)/4, it follows that

B_{n}^{2} = \frac{v_{n}^{2} − 4}{32} = \frac{1}{2} \frac{(v_{n} − 2)}{4} (\frac{(v_{n} − 2)}{4} + 1) = \frac{y_{n}(y_{n} + 1)}{2}.

Then, it is seen that x^{2} = \frac{y(y + 1)}{2} for some positive integers x and y if and only if x = B_{n} and y = y_{n} for some natural number n. Now we prove the following lemma given in [9].

Lemma 4. The sequence (y_{n}) satisfies the recurrence relation y_{n+1} = 6y_{n} − y_{n−1} + 2 for n ⩾ 1 where y_{0} = 0 and y_{1} = 1.

Proof. Using the fact that y_{n} = \frac{v_{n} − 2}{4}, we get

6y_{n} − y_{n−1} + 2 = 6(v_{n} − 2)/4 − (v_{n−1} − 2)/4 + 2
= (6v_{n} − v_{n−1} − 2)/4 = (v_{n+1} − 2)/4
= y_{n+1}.

The first few terms of the sequence (y_{n}) are 0, 1, 8, 49, 288,… (sequence \underline{A001108} in [20]). For n = 1, 2,…, let b_{n} be n−th cobalancing number and so let (b_{n}) denote the cobalancing number sequence. Then, the cobalancing numbers satisfy the similar recurrence relation given in Lemma 4. That is, b_{n+1} = 6b_{n} − b_{n−1} + 2 for n ≥ 1 where b_{0} = 0 and b_{1} = 2 (see [7, p. 1191]). The first few terms of the cobalancing number sequence are 0, 2, 14, 84, 492,… (sequence \underline{A053141} in [20]). Moreover, there is a close relation between cobalancing numbers, balancing numbers and the sequence (y_{n}). In order to see this relation, we can give the following lemma without proof.

Lemma 5. For every n ⩾ 1, b_{n} = y_{n} + B_{n} and b_{n} = y_{n+1} − B_{n+1}.

Lemma 6. For every n ⩾ 1, y_{2n} = 8B_{n}^{2} and y_{2n+1} = 8B_{n}B_{n+1} + 1.

Proof. By identities (10) and (15), we get

B_{n}^{2} = (v_{n}^{2} − 4)/32 = (v_{2n} − 2)/32 = y_{2n}/8.

Thus it follows that y_{2n} = 8B_{n}^{2}. Also since y_{n+1} = 6y_{n} − y_{n−1} + 2, it is easy to see that y_{n} = (y_{n+1} + y_{n−1} − 2)/6. By using y_{2n} = 8B_{n}^{2}, we find that

y_{2n+1} = (y_{2n+2} + y_{2n} − 2)/6 = (8B_{n+1}^{2} + 8B_{n}^{2} − 2)/6 = (8(B_{n+1}^{2} + B_{n}^{2}) − 2)/6

Since B_{n+1}^{2} + B_{n}^{2} = 6B_{n}B_{n+1} + 1 by identity (11), it follows that

y_{2n+1} = (8(B_{n+1}^{2} + B_{n}^{2}) − 2)/6 = (8(6B_{n}B_{n+1} + 1) − 2) /6 = 8B_{n}B_{n+1} + 1.

This completes the proof.

Now we can give the following theorem. Since its proof is easy, we omit it.

Theorem 7. If n is an odd natural number, then y_{n} = Q_{n}^{2}/4 and if n is an even natural number, then y_{n} = Q_{n}^{2}/4 − 1.

Since y_{2n+1} = 8B_{n}B_{n+1} + 1 and y_{2n+1} = Q_{2n+1}^{2}/4, it follows that B_{n}B_{n+1} is a triangular number. Moreover, it follows from Lemma 6 that y_{n} is odd if and only if n is odd and y_{n} is even if and only if n is even.

3 Main Theorems

In the previous sections, we mentioned the well known elementary properties about triangular, square triangular, balancing and cobalancing numbers. In this chapter, by using the previous theorems, lemmas and identities we prove some new properties concerning balancing numbers and square triangular numbers. The principal question of our interest is whether the product of two balancing numbers greater than 1 is another balancing number. We will show that the answer to this question is negative. Similarly, we will show that the product of two square triangular numbers greater than 1 is not a triangular number. The product of two oblong numbers may be another oblong number and similarly, the product of two triangular numbers may be another one. For a simple example, 2 and 6 are two oblong numbers. The product of them is 2 × 6 = 12 and 12 = 3(3 + 1) is another oblong number. Similarly, 3 and 15 are two triangular numbers. The product of 3 and 15 is 3 × 15 = 45

and 45 = \frac{9(9 + 1)}{2} is another triangular number. Also it is obvious that the product of two consecutive oblong numbers is another oblong:

[(x − 1)x] [x(x + 1)] = (x^{2} − 1)x^{2}.

For solving the general problem, we need to solve the Diophantine equation

x(x + 1)y(y + 1) = z(z + 1).  (16)

In [2], Breiteig gave recursion formulae for the solutions x, y, and z satisfying the equation (16). For more information about the product of two oblong numbers, one can consult [2].

The question of when the product of two oblong numbers is another one suggests an analogous question for balancing numbers. When is the product of two balancing numbers another balancing number? Now before giving these properties concerning balancing numbers and square triangular numbers, we present some theorems which will be needed in the proof of the main theorems. Since the following two theorems are given in [19], we omit their proofs.

Theorem 8. Let n ∈ ℕ ∪ {0} and m, r ∈ ℤ. Then

P_{2mn+r} ≡ (−1)^{(m+1)n}P_{r} (mod Q_{m}),  (17)
Q_{2mn+r} ≡ (−1)^{(m+1)n}Q_{r} (mod Q_{m}),  (18)
P_{2mn+r} ≡ (−1)^{mn}P_{r} (mod P_{m}),  (19)

and

Q_{2mn+r} ≡ (−1)^{mn}Q_{r} (mod P_{m}).  (20)

Theorem 9. Let n ∈ ℕ ∪ {0} and m, r ∈ ℤ. Then

B_{2mn+r} ≡ B_{r} (mod B_{m}),  (21)
v_{2mn+r} ≡ v_{r} (mod u_{m}),  (22)
B_{2mn+r} ≡ (−1)^{n}B_{r} (mod v_{m}),  (23)

and

v_{2mn+r} ≡ (−1)^{n}v_{r} (mod v_{m}).  (24)

The proofs of the following theorems can be given by using the above two theorems. Also, we can find some of their proofs in [3]. Moreover, some of them are given in [12] without proof.

Theorem 10. Let m, n ∈ ℕ and m ⩾ 2. Then P_{m} | P_{n} if and only if m | n.

Theorem 11. Let m, n ∈ ℕ and m ⩾ 2. Then Q_{m} | Q_{n} if and only if m | n and \frac{n}{m} is an odd integer.

Theorem 12. Let m, n ∈ ℕ and m ⩾ 2. Then Q_{m} | P_{n} if and only if m | n and \frac{n}{m} is an even integer.

Since B_{n} = P_{2n}/2 and v_{n} = Q_{2n}, the proofs of the following theorems can be given by using the above theorems and identity (23).

Theorem 13. Let m, n ∈ ℕ and m ⩾ 2. Then B_{m} | B_{n} if and only if m | n.

Theorem 14. Let m, n ∈ ℕ and m ⩾ 1. Then v_{m} | v_{n} if and only if m | n and \frac{n}{m} is an odd integer.

Theorem 15. Let m, n ∈ ℕ and m ⩾ 1. Then v_{m} | u_{n} if and only if m | n and \frac{n}{m} is an even integer.

The following theorem is a well known theorem (see [8, 12]).

Theorem 16. Let m ⩾ 1 and n ⩾ 1. Then (B_{m}, B_{n}) = B_{(m,n)}.

Corollary 17. Let m ⩾ 1 and n ⩾ 1. Then (B_{m}^{2}, B_{n}^{2}) = B_{(m,n)}^{2}.

Theorem 16 says that the greatest common divisor of any two balancing numbers is again a balancing number. As a conclusion of this theorem, Corollary 17 says that the greatest common divisor of any two square triangular numbers is again a square triangular number. Now we will discuss the least common multiple of any two balancing numbers. The least common multiple of any two triangular numbers may be a triangular number. For instance, 15 and 21 are two triangular numbers and [15, 21] = 105 is again a triangular number. Note that 15 ∤ 21. Similarly, the least common multiple of any two oblong numbers may be an oblong number. For a simple example, 6 and 15 are two oblong numbers and [6, 15] = 30 is again an oblong number. But this is not true in general for any two balancing numbers. This can be seen from the following theorem.

Theorem 18. Let B_{n} > 1, B_{m} > 1 and B_{n} < B_{m}. Then [B_{n}, B_{m}] is a balancing number if and only if B_{n} | B_{m}.

Proof. Assume that B_{n} | B_{m}. Then [B_{n}, B_{m}] = B_{m} is again a balancing number. Conversely, assume that B_{n} > 1, B_{m} > 1 and B_{n} ∤ B_{m}. Then by Theorem 13, n ∤ m. Let d = (m, n). Then by Theorem 16, we get (B_{n}, B_{m}) = B_{d}. Therefore

[B_{n}, B_{m}] = \frac{B_{n}B_{m}}{(B_{n}, B_{m})} = \frac{B_{n}B_{m}}{B_{d}} .  (25)

Assume that [B_{n}, B_{m}] is a balancing number. Thus [B_{n}, B_{m}] = B_{r} for some natural number

r. Then by (25), we have B_{n}B_{m}/B_{d} = B_{r}. That is, B_{n}B_{m} = B_{d}B_{r}. Thus \frac{B_{n}}{B_{d}} B_{m} = B_{r}

and therefore B_{m} | B_{r}. This implies that r = mt for some natural number t by Theorem 13. Assume that t is an odd integer. Then t = 4q ∓ 1 for some q ⩾ 1. Thus B_{r} = B_{mt} = B_{4qm∓m} = B_{2(2qm)∓m} ≡ B_{∓m} (mod B_{2m}) by (21). This shows that B_{r} ≡ ∓B_{m} (mod B_{2m}).

Since B_{2m} = B_{m}v_{m} by (13), we see that \frac{B_{n}}{B_{d}} B_{m} = B_{r} ≡ ∓B_{m} (mod B_{m}v_{m}). Then it follows that \frac{B_{n}}{B_{d}} = ∓1 (modv_{m}). We assert that B_{n} ≠ B_{d}. On the contrary, assume that B_{n} = B_{d}.

Then n = d and this implies that n | m, which is impossible since n ∤ m. Since \frac{B_{n}}{B_{d}} ≠ 1 and

\frac{B_{n}}{B_{d}} ≡ ∓1 (mod v_{m}), it follows that v_{m} ⩽ \frac{B_{n}}{B_{d}} ∓ 1 ⩽ \frac{B_{n}}{B_{d}} + 1 ⩽ B_{n} + 1. Since B_{n} < B_{m},

we get n < m. This shows that v_{n} < v_{m} ⩽ B_{n} + 1. On the other hand, by identity (10), we get v_{n} > 2B_{n}. Therefore 2B_{n} < v_{n} < B_{n} + 1, which implies that B_{n} < 1. But this is a contradiction since B_{n} > 1. Now assume that t is an even integer. Then t = 2k and

thus r = mt = 2mk. Therefore \frac{B_{n}}{B_{d}} B_{m} = B_{r} = B_{2km} = B_{km}v_{km} ⩾ B_{m}v_{m}. This shows that

\frac{B_{n}}{B_{d}} ⩾ v_{m} and thus v_{m} ⩽ \frac{B_{n}}{B_{d}} ⩽ B_{n}. Since n < m, we get v_{n} < v_{m} ⩽ B_{n}. That is, v_{n} < B_{n},

which is impossible by identity (10). This completes the proof.

Now as a result of the above theorem, we can give the following corollary which says something about the least common multiple of any two square triangular numbers. The proof of the following corollary is straightforward, using the fact that

[a^{2}, b^{2}] = \frac{a^{2}b^{2}}{(a^{2}, b^{2})} = \frac{a^{2}b^{2}}{(a, b)^{2}} = (\frac{ab}{(a, b)})^{2} = [a, b]^{2}

where a and b are positive integers.

Corollary 19. Let B_{n} > 1, B_{m} > 1 and B_{n} < B_{m}. Then [B_{n}^{2}, B_{m}^{2}] is a triangular number if and only if B_{n}^{2} | B_{m}^{2}.

In order to answer the main question which is about the product of two balancing numbers, we give the following theorem. This theorem says something more than the above theorem.

Theorem 20. Let n > 1, m > 1 and m ⩾ n. Then there is no integer r such that B_{n}B_{m} = B_{r}.

Proof. Assume that m > 1, n > 1 and B_{n}B_{m} = B_{r} for some r > 1. Then B_{m} | B_{r} and therefore m | r by Theorem 13. Thus r = mt for some positive integer t. Assume that t is an even integer. Then t = 2k and therefore r = mt = 2mk. Thus

B_{n}B_{m} = B_{r} = B_{2km} = B_{km}v_{km}

by identity (13). This shows that B_{n} = \frac{B_{km}}{B_{m}} v_{km} and therefore v_{km} | B_{n}. By Theorem 15,

we get km | n and n/km = 2s for some integer s. Then n = 2kms. Since n = 2kms and r = 2km, we get n = rs. Thus r | n. On the other hand, since B_{n}B_{m} = B_{r}, it follows that B_{n} | B_{r} and therefore n | r by Theorem 13. This implies that n = r and B_{n} = B_{r}. Since B_{n}B_{m} = B_{r}, we get B_{m} = 1, which is a contradiction. Now assume that t is an odd integer. Then t = 4q ∓ 1 for some positive integer q. Thus r = mt = 4qm ∓ m and therefore

B_{r} = B_{4qm∓m} = B_{2(2qm)∓m} ≡ B_{∓m} (mod B_{2m})

by (21). This shows that B_{m}B_{n} ≡ ∓B_{m} (mod B_{2m}). Since B_{2m} = B_{m}v_{m}, we get B_{m}B_{n} ≡ ∓B_{m} (mod B_{m}v_{m}), which implies that B_{n} ≡ ∓1 (mod v_{m}). Therefore v_{m} | B_{n} ∓ 1 and thus v_{m} ⩽ B_{n} ∓ 1. Since v_{n} > 2B_{n} and m ≥ n, we get B_{n} + 1 ⩾ B_{n} ∓ 1 ⩾ v_{m} ⩾ v_{n} > 2B_{n}. This implies that B_{n} + 1 > 2B_{n}. Then B_{n} < 1, which is a contradiction. This completes the proof.

Since square triangular numbers are square of the balancing numbers, the above theorem says that the product of two square triangular numbers greater than one is not a triangular number. Now we can give the following corollary easily.

Corollary 21. The only positive integer solution of the system of Diophantine equations 2u^{2} = x(x + 1), 2v^{2} = y(y + 1) and 2u^{2}v^{2} = z(z + 1) is given by (x, y, u, v, z) = (1, 1, 1, 1, 1).

The following theorem gives a new property of the sequence (y_{n}). It is about the product of any two elements of the sequence (y_{n}) greater than 1.

Theorem 22. Let n > 1 and m > 1. Then there is no integer r such that y_{n}y_{m} = y_{r}.

Proof. Assume that y_{n}y_{m} = y_{r}. Since y_{k} is odd if and only if k is odd and y_{k} is even if and only if k is even, we see that m, n, and r are odd or r and at least one of the numbers n and m are even. Assume that n and r are even. Then n = 2k and r = 2t for some positive integers k and t. By Lemma 6, we have y_{n} = y_{2k} = 8B_{k}^{2} and y_{r} = y_{2t} = 8B_{t}^{2}. Therefore

y_{m} = \frac{y_{r}}{y_{n}} = \frac{8B_{t}^{2}}{8B_{k}^{2}} = (\frac{B_{t}}{B_{k}} )^{2} .

If m is even, then m = 2l for some positive integer l. This implies that y_{m} = y_{2l} = 8B_{l}^{2}

and therefore (\frac{B_{t}}{B_{u}} )^{2} = 8B_{l}^{2}, which is impossible. So m is odd. Then, by Theorem 7, it

follows that y_{m} = \frac{Q_{m}^{2}}{4} . Thus, we get \frac{Q_{m}}{2} = \frac{B_{t}}{B_{k}} = \frac{P_{2t}/2}{P_{2k}/2} = \frac{P_{2t}}{P_{2k}} = \frac{P_{r}}{P_{n}}. This shows that

2P_{r} = P_{n}Q_{m}. Since n is even, P_{n} is even. Also, since P_{r} = P_{n} \frac{Q_{m}}{2} , we see that P_{n} | P_{r} and Q_{m} | P_{r}. By Theorem 10 and Theorem 12, we get r = nu and r = 2ms for some natural numbers u and s. Since 2P_{r} = P_{n}Q_{m}, we have

P_{n}Q_{m} = 2P_{r} = 2P_{2ms} = 2P_{ms}Q_{ms} > P_{ms}Q_{ms} > P_{ms}Q_{m}.

Therefore P_{n} > P_{ms} and this implies that n > ms. Then 2n > 2ms and thus 2n > r = nu. This shows that u < 2. That is, u = 1. Since u = 1, we get r = nu = n, which is impossible.

Assume that m, n and r are odd integers. Since y_{n}y_{m} = y_{r}, we get \frac{Q_{n}^{2}}{4} \frac{Q_{m}^{2}}{4} = \frac{Q_{r}^{2}}{4} by Theorem 7. Therefore Q_{n}Q_{m} = 2Q_{r}. This shows that Q_{n} | Q_{r} and Q_{m} | Q_{r}. Then r = nt and r = mk for some odd natural numbers t and k by Theorem 11. Since t is an odd integer, t = 4q ∓ 1 for some q ⩾ 1. Thus

Q_{r} = Q_{nt} = Q_{4qn∓n} = Q_{2(2qn)∓n} ≡ ∓Q_{n} (mod Q_{2n})

by the congruence (18). That is, Q_{r} ≡ ∓Q_{n} (mod Q_{2n}). This implies that 2Q_{r} ≡ ∓2Q_{n} (mod Q_{2n}) and thus Q_{n}Q_{m} ≡ ∓2Q_{n} (mod Q_{2n}). Since Q_{2n} = Q_{n}^{2} + 2, when n is odd, by identity (12), we clearly have that (Q_{n}, Q_{2n}) = 2. Then the congruence

Q_{n}Q_{m} ≡ ∓2Q_{n} (mod Q_{2n})

implies that

\frac{Q_{n}}{2} Q_{m} ≡ ∓2\frac{Q_{n}}{2} (mod \frac{Q_{2n}}{2} ) .

This shows that Q_{m} ≡ ∓2 (mod Q_{2n}/2). Since m ≥ 2, we get Q_{m} > 2 and therefore Q_{2n}/2 ⩽ Q_{m} ∓ 2. Therefore Q_{2n} ⩽ 2Q_{m} ∓ 4 < 2Q_{m} + 4. Similarly, by using r = mk, it is seen that Q_{2m} < 2Q_{n} + 4. Then it follows that Q_{2n} + Q_{2m} < 2Q_{m} + 2Q_{n} + 8. Since n and m are odd integers, Q_{2n} = Q_{n}^{2} + 2 and Q_{2m} = Q_{m}^{2} + 2 by identity (12). Thus, we get

Q_{n}^{2} + 2 + Q_{m}^{2} + 2 < 2Q_{m} + 2Q_{n} + 8.

Since Q_{n}^{2} + 2 + Q_{m}^{2} + 2 < 2Q_{m} + 2Q_{n} + 8, it follows that Q_{n}^{2} + Q_{m}^{2} < 2Q_{m} + 2Q_{n} + 4. This implies that Q_{n}^{2} − 2Q_{n} + Q_{m}^{2} − 2Q_{m} < 4. Then we get

Q_{n}(Q_{n} − 2) + Q_{m}(Q_{m} − 2) < 4,

which implies that Q_{n} + Q_{m} < 4. This is a contradiction since m > 1 and n > 1. This completes the proof.

We easily obtain the following corollary.

Corollary 23. The only positive integer solution of the system of Diophantine equations x(x + 1) = 2u^{2}, y(y + 1) = 2v^{2}, and xy(xy + 1) = 2z^{2} is given by (x, y, u, v, z) = (1, 1, 1, 1, 1).

Balancing numbers and cobalancing numbers are related to the solutions of some Diophantine equations. Solutions of some of the Diophantine equations are given in [5]. Now we give four of them from [5].

Theorem 24. All positive integer solutions of the equation x^{2} = y(y + 1)/2 are given by (x, y) = (B_{n}, y_{n}) with n ≥ 1.

Theorem 25. All positive integer solutions of the equation (x + y − 1)^{2} = 8xy are given by (x, y) = (y_{n}, y_{n+1}) with n ≥ 1.

Theorem 26. All positive integer solutions of the equation x^{2} − 6xy + y^{2} − 1 = 0 are given by (x, y) = (B_{n}, B_{n+1}) with n ≥ 1.

Theorem 27. All positive integer solutions of the equation (x + y − 1)^{2} = 8xy + 1 are given by (x, y) = (b_{n}, b_{n+1}) with n ≥ 1.

From Theorem 27, it follows that b_{n}b_{n+1} is a triangular number for every natural number n.

Moreover, we can easily state the following theorems.

Theorem 28. All positive integer solutions of the equation x^{2} − y^{2} + 2xy + x − y = 0 are given by (x, y) = (B_{n}, b_{n}) with n ≥ 1.

Theorem 29. All positive integer solutions of the equation x^{2} + 2y^{2} − 4xy − x = 0 are given by (x, y) = (y_{n}, b_{n}) with n ≥ 1 or (x, y) = (y_{n}, b_{n−1}) with n ≥ 2.

4 Concluding Remarks

The sum of two triangular numbers may be a triangular number. For instance, 6 and 15 are triangular numbers and 6 + 15 = 21 is again a triangular number. Similarly, the sum of two oblong numbers may be another oblong number. For instance, 12 and 30 are oblong numbers and 12 + 30 = 42 is again an oblong number. But we think that the sum of two square triangular numbers is not a square triangular number. That is, B_{n}^{2} +B_{m}^{2} = B_{r}^{2} has no solution if n ⩾ 1 and m ⩾ 1. We also think that there is no integer r such that B_{n}+B_{m} = B_{r} and b_{n} + b_{m} = b_{r} for n ⩾ 1 and m ⩾ 1. On the other hand, we think that the product of any two cobalancing numbers greater than 1 is not a cobalancing number. That is, there is no integer r such that b_{n}b_{m} = b_{r} for n ⩾ 1 and m ⩾ 1. Moreover, we think that there is no solution of the equation y_{n} + y_{m} = y_{r} if n ⩾ 1 and m ⩾ 1.

5 Acknowledgements

The authors would like to thank the anonymous referee for careful reading and suggestions that improved the clarity of this manuscript.

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2010 Mathematics Subject Classification: Primary 11D25; Secondary 11D41, 11D72. Keywords: triangular number, square triangular number, balancing number, cobalancing number.

(Concerned with sequences \underline{A000032}, \underline{A000045}, \underline{A000129}, \underline{A000217}, \underline{A001108}, \underline{A001109}, \underline{A002203}, \underline{A002378}, \underline{A003499}, and \underline{A053141}.)

Received June 13 2011; revised version received December 2 2011. Published in Journal of Integer Sequences, December 27 2011. Small revision, October 2 2012.

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