CUBO A Mathematical Journal
Vol.15, No¯ 02, (79–88). June 2013

Squares in Euler triples from Fibonacci and Lucas numbers
Zvonko Čerin

University of Zagreb,

Kopernikova 7, 10010 Zagreb, CROATIA, Europe,

cerin@math.hr

ABSTRACT

In this paper we shall continue to study from [4], for k = −1 and k = 5, the infi-

nite sequences of triples A = (F2n+1, F2n+3, F2n+5), B = (F2n+1, 5F2n+3, F2n+5),

C = (L2n+1, L2n+3, L2n+5), D = (L2n+1, 5L2n+3, L2n+5) with the property that the

product of any two different components of them increased by k are squares. The se-

quences A and B are built from the Fibonacci numbers Fn while the sequences C and

D from the Lucas numbers Ln. We show some interesting properties of these sequences

that give various methods how to get squares from them.

RESUMEN

En este art́ıculo continuaremos el estudio de [4], para k = −1 y k = 5, las secuen-

cias infinitas de tripletas A = (F2n+1, F2n+3, F2n+5), B = (F2n+1, 5F2n+3, F2n+5),

C = (L2n+1, L2n+3, L2n+5), D = (L2n+1, 5L2n+3, L2n+5) con la propiedad que el pro-

ducto de dos componentes diferentes que se aumenta en k son cuadrados. Las secuencias

A y B se construyen con los números de Fibonacci Fn mientras que las secuencias C

y D se construyen con los números de Lucas Ln. Mostramos algunas propiedades in-

teresantes de estas secuencias que entregan muchos métodos de cómo conseguir los

cuadrados de ellos.

Keywords and Phrases: D(k)-triple, Fibonacci numbers, Lucas numbers, square, symmetric

sum, alternating sum, product, component

2010 AMS Mathematics Subject Classification: 11B37, 11B39, 11D09.



80 Zvonko Čerin CUBO
15, 2 (2013)

1 Introduction

For integers a, b and c, let us write a
b
∼ c provided a + b = c2. For the triples A = (a,b,c),

D = (d,e,f) and à = (ã, b̃, c̃) the notation A
D
∼ Ã means that bc

d
∼ ã, ca

e
∼ b̃ and ab

f
∼ c̃. When

D = (k,k,k), let us write A
k
∼ Ã for A

D
∼ Ã. Hence, A is the D(k)-triple (see [1]) if and only if

there is a triple à such that A
k
∼ Ã.

In the paper [4] we constructed infinite sequences α = {α(n)}∞n=0 and β = {β(n)}
∞

n=0 of D(−1)-

triples and γ = {γ(n)}∞n=0 and δ = {δ(n)}
∞

n=0 of D(5)-triples. Here, α(n) = A = (F2n+1, F2n+3,

F2n+5), β(n) = B = (F2n+1, 5F2n+3, F2n+5) and γ(n) = C = (L2n+1, L2n+3, L2n+5), δ(n) =

D = (L2n+1, 5L2n+3, L2n+5), where the Fibonacci and Lucas sequences of natural numbers Fn

and Ln are defined by the recurrence relations F0 = 0, F1 = 1, Fn = Fn−1 + Fn−2 for n > 2 and

L0 = 2, L1 = 1, Ln = Ln−1 + Ln−2 for n > 2.

The numbers Fk make the integer sequence A000045 from [6] while the numbers Lk make

A000032.

The goal of this article is to further explore the properties of the sequences α, β, γ and δ. Each

member of these sequences is an Euler D(−1)- or D(5)-triple (see [2] and [3]) so that many of their

properties follow from the properties of the general (pencils of) Euler triples. It is therefore interest-

ing to look for those properties in which at least two of the sequences appear. This paper presents

several results of this kind giving many squares from the components, various sums and products of

the sequences α, β, γ and δ. Most of our theorems have also versions for the associated sequences

α̃, β̃, γ̃ and δ̃, where α̃(n) = Ã = (F2n+4,F2n+3,F2n+2), β̃(n) = B̃ = (L2n+4,F2n+3,L2n+2),

γ̃(n) = C̃ = (L2n+4,L2n+3,L2n+2), δ̃(n) = D̃ = (5F2n+4,L2n+3, 5F2n+2) satisfy A
−1
∼ Ã, B

−1
∼ B̃,

C
5
∼ C̃ and D

5
∼ D̃.

2 Squares from products of components

The relations A
−1
∼ Ã and C

5
∼ C̃ imply that the components of A and C satisfy A2A3

−1
∼ Ã1

and C2C3
5
∼ C̃1. Our first theorem shows that the product A2A3C2C3 is in a similar relation

with respect to 1. Of course, the other products A3A1C3C1, A1A2C1C2 as well as B2B3D2D3,

B3B1D3D1 and B1B2D1D2 exhibit a similar property.

Theorem 1. The following hold for the products of components:

A2 A3 C2 C3
1
∼ F4n+8,

1
5
B2 B3 D2 D3

9
∼ L4n+8,

A3 A1 C3 C1
9
∼ F4n+6, B3 B1 D3 D1

9
∼ F4n+6,

A1 A2 C1 C2
1
∼ F4n+4,

1
5
B1 B2 D1 D2

9
∼ L4n+4.



CUBO
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Squares in Euler triples ... 81

Proof. Let ϕ = 1+
√
5

2
and ψ = 1−

√
5

2
= − 1

ϕ
. Since Fj =

ϕ
j
−ψ

j

ϕ−ψ
and Lj = ϕ

j + ψj, it follows that

A2 =
ϕ

2 n+3
−ψ

2 n+3

ϕ−ψ
, A3 =

ϕ
2 n+5

−ψ
2 n+5

ϕ−ψ
and C2 = ϕ

2n+3 + ψ2n+3, C3 = ϕ
2n+5 + ψ2n+5.

After the substitutions ψ = − 1
ϕ
and M = ϕn, the sum of A2A3C2C3 and 1 becomes

ϕ
16

(M
8
−ψ

16
)
2

20M8
.

However, this is precisely the square of F4n+8. This shows the first relation. The other relations

have similar proofs.

The version of the previous theorem for the sequences α̃, β̃, γ̃ and δ̃ is the following result.

Theorem 2. The products of components of Ã, B̃, C̃ and D̃ satisfy:

Ã2Ã3C̃2C̃3
1
∼ F4n+5, B̃2B̃3D̃2D̃3

1
∼ L4n+5,

Ã3Ã1C̃3C̃1
1
∼ F4n+6,

1
25
B̃3B̃1D̃3D̃1

1
∼ F4n+6,

Ã1Ã2C̃1C̃2
1
∼ F4n+7, B̃1B̃2D̃1D̃2

1
∼ L4n+7.

Proof. Since Ã2 =
ϕ

2 n+3
−ψ

2 n+3

ϕ−ψ
, Ã3 =

ϕ
2 n+2

−ψ
2 n+2

ϕ−ψ
, C̃2 = ϕ

2n+3+ ψ2n+3 and C̃3 = ϕ
2n+2+

ψ2n+2, the sum of Ã2Ã3C̃2C̃3 and 1, after the substitutions ψ = −
1
ϕ

and M = ϕn, becomes
ϕ

10
(M

8
+ψ

10
)
2

5M8
. However, the square of F4n+5 has the same value. This proves the first relation

Ã2Ã3C̃2C̃3
1
∼ F4n+5. The remaining five relations in this theorem have similar proofs.

The same kind of relations hold also for the products of components from all four sequences

α, β, γ and δ.

Theorem 3. The following relations for products of components hold:

A2B3C2D3
1
∼ F4n+8,

1
25
A3B2C3D2

1
∼ F4n+8,

A3B1C3D1
9
∼ F4n+6, A1B3C1D3

9
∼ F4n+6,

1
25
A1B2C1D2

1
∼ F4n+4, A2B1C2D1

1
∼ F4n+4.

Proof. Since B3 = A3 and D3 = C3, the first relation is the consequence of the first relation in

Theorem 1.

In order to prove the second relation, notice that B2 = 5A2 and D2 = 5C2 so that the multi-

plication of the identity behind the first relation in Theorem 1 with 25 we conclude that the second

relation holds. The other relations in this theorem have similar proofs.

There is again the version of the previous theorem for the products of components from all

four sequences α̃, β̃, γ̃ and δ̃.



82 Zvonko Čerin CUBO
15, 2 (2013)

Theorem 4. The products of components of Ã, B̃, C̃ and D̃ satisfy:

Ã2B̃3C̃2D̃3
1
∼ L4n+5, Ã3B̃2C̃3D̃2

1
∼ F4n+5,

Ã3B̃1C̃3D̃1
9
∼ L4n+6, Ã1B̃3C̃1D̃3

9
∼ L4n+6,

Ã1B̃2C̃1D̃2
1
∼ F4n+7, Ã2B̃1C̃2D̃1

1
∼ L4n+7.

Proof. Since Ã2 = B̃2 and C̃2 = D̃2, the first, the second, the fifth and the sixth relations are the

consequence of the second, the first, the fifth and the sixth relations in Theorem 2.

In order to prove the third relation, note that the components Ã3, B̃1 and C̃3, D̃1 are
ϕ

2 n+2
−ψ

2 n+2
)

ϕ−ψ
, ϕ2n+4 + ψ2n+4, ϕ2n+2 + ψ2n+2 and

5(ϕ
2 n+4

−ψ
2 n+4

)

ϕ−ψ
. It is now clear from the

proof of Theorem 1 that the sum of Ã3B̃1C̃3D̃1 and 9 is precisely the square of L4n+6. This shows

the third relation. The fourth relation has a similar proof.

Nice relationships of the same kind hold also for the products of components with other choices

of indices.

Theorem 5.

1
5
A2 B2 C3 D3

0
∼ F2n+3L2n+5,

1
5
A3 B3 C2 D2

0
∼ F2n+5L2n+3,

A3 B3 C1 D1
0
∼ F2n+5L2n+1, A1 B1 C3 D3

0
∼ F2n+1L2n+5,

1
5
A1 B1 C2 D2

0
∼ F2n+1L2n+3,

1
5
A2 B2 C1 D1

0
∼ F2n+3L2n+1.

Proof. Since B2 = 5A2, A2 = F2n+3 and C3 = D3 = L2n+5, the product
1
5
A2 B2 C3 D3 is the

square of F2n+3L2n+5. The other claims in this theorem have similar proofs.

The version of the previous theorem for the products of components from all four associated

sequences is the following result.

Theorem 6. The products of components of Ã, B̃, C̃ and D̃ satisfy:

Ã2B̃2C̃3D̃3

5F4n+4

0
∼ F2n+3,

Ã3B̃3C̃2D̃2

F4n+4

0
∼ L2n+3,

1
5
Ã3B̃3C̃1D̃1

1
∼ F4n+6,

1
5
Ã1B̃1C̃3D̃3

1
∼ F4n+6,

Ã1B̃1C̃2D̃2

F4n+8

0
∼ L2n+3,

Ã2B̃2C̃1D̃1

5F4n+8

0
∼ F2n+3.

Proof. Since Ã2 = B̃2 = F2n+3, C̃3 = L2n+2, D̃3 = 5F2n+2, we see that the first relation clearly

holds. The others in this theorem are proved similarly.



CUBO
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Squares in Euler triples ... 83

This time the pairs (A,D) and (B,C) have equal indices.

Theorem 7. The following hold for the products of components:

1
5
A2 B3 C3 D2

1
∼ F4n+8,

1
5
A3 B2 C2 D3

1
∼ F4n+8,

A3 B1 C1 D3
9
∼ F4n+6, A1 B3 C3 D1

9
∼ F4n+6,

1
5
A1 B2 C2 D1

1
∼ F4n+4,

1
5
A2 B1 C1 D2

1
∼ F4n+4.

Proof. Since A2 = F2n+3, B3 = F2n+5, C3 = L2n+5 and D2 = 5L2n+3, the sum of
1
5
A2 B3 C3 D2

and 1 is F4n+6F4n+10 + 1 = F
2
4n+8. The other claims in this theorem have similar proofs.

Once again the version of the previous theorem for the associated sequences includes interesting

relations.

Theorem 8. The products of components of Ã, B̃, C̃ and D̃ satisfy:

Ã2B̃3C̃3D̃2

F4n+6

0
∼ L2n+2,

Ã3B̃2C̃2D̃3

5F4n+6

0
∼ F2n+2,

1
5
Ã3B̃1C̃1D̃3

0
∼ F2n+2L2n+4,

1
5
Ã1B̃3C̃3D̃1

0
∼ F2n+4L2n+2,

Ã1B̃2C̃2D̃1

5F4n+6

0
∼ F2n+4,

Ã2B̃1C̃1D̃2

F4n+6

0
∼ L2n+4.

Proof. Since Ã2 = F2n+3 B̃3 = C̃3 = L2n+2, D̃2 = L2n+3 and F2n+3L2n+3 = F4n+6 we see that the

first relation clearly holds. The other relations in this theorem are proved similarly.

It is interesting that in some cases we can even mix components of the triples A, B,

C, D and Ã, B̃, C̃, D̃ as the relations Ã2 B3 C3 D̃2
1
∼ F4n+8, Ã2 B̃2 C3 D3

0
∼ F2n+3L2n+5 and

Ã2 B̃2 C1 D1
0
∼ F2n+3L2n+1 show, but we do not see a general pattern here.

3 Squares from symmetric sums

Let σ1, σ2, σ3 : Z
3
→ Z be the basic symmetric functions defined for x=(a, b, c) by

xσ1 = a + b + c, xσ2 = bc + ca + ab, xσ3 = abc. Let σ
∗
2, σ

∗
1 : Z

3
→ Z be defined by

xσ∗
2
= bc − ca + ab and xσ∗

1
= a − b + c. Note that xσ∗

1
is the determinant of the 1 × 3 matrix

[a,b,c] (see [5]).

For the sums σ2 and σ
∗
2 of the components the following relations are true.



84 Zvonko Čerin CUBO
15, 2 (2013)

Theorem 9. The following is true for the sums σ2 of the components:

Aσ2Cσ2
21
∼ 4F4n+6, Bσ2Dσ2

69
∼ 16F4n+6,

Ãσ2C̃σ2
5
∼ 2F4n+7, B̃σ2D̃σ2

45
∼ 10F4n+6.

Proof. Since Aσ2 =
1
5
(4L4n+6 + 13) and Cσ2 = 4L4n+6 − 13, the sum Aσ2Cσ2 + 21 is

1
5
[(4L4n+6)

2 − 64] that we recognize as the square of 4F4n+6. This proves the first relation

Aσ2Cσ2
21
∼ 4F4n+6. The other relations in this theorem have similar proofs.

The sums B̃σ∗
2
and D̃σ∗

2
have constant values −1 and 5. On the other hand, we have the

following result.

Theorem 10. The following is true for the sums σ∗2 of the components:

Aσ∗
2
Cσ∗

2

−3
∼ 2F4n+6, Bσ∗

2
Dσ∗

2

−51
∼ 14F4n+6, Ãσ∗

2
C̃σ∗

2

5
∼ 2F4n+6.

Proof. Since Bσ∗
2
= 1
5
(14L4n+6 + 23) and Dσ∗

2
= 14L4n+6 − 23, the sum Bσ∗

2
Dσ∗

2
− 51 is the quo-

tient
196(L

2
4n+6−4)

5
. It is now easy to check that this is the square of 14F4n+6. This proves the

second relation Bσ∗
2
Cσ2

−51
∼ 14F4n+6. The other relations in this theorem have similar proofs.

Some similar relations where all four letters A, B, C and D appear make the following result.

Theorem 11. The following is true for the sums σ∗2 of the components:

Ãσ∗
2
D̃σ∗

2
+ B̃σ∗

2
C̃σ∗

2
= 6, Ãσ∗

2
C̃σ∗

2
+ B̃σ∗

2
D̃σ∗

2

10
∼ 2F4n+5,

−Ãσ∗
2
B̃σ∗

2
C̃σ∗

2
D̃σ∗

2

9
∼ 2L4n+5.

Proof. Since B̃σ∗
2
= −1, Ãσ∗

2
= 1
5
(2L4n+5 + 3), C̃σ∗

2
= 2L4n+5 − 3 and D̃σ∗

2
= 5, it follows that

Ãσ∗
2
D̃σ∗

2
+ B̃σ∗

2
C̃σ∗

2
= 6. The second and the third relations in this theorem have similar proofs.

Here are two relations which contains both sums σ2 and σ
∗
2.

Theorem 12. The following is true for the sums σ2 and σ
∗
2:

1
36
(Aσ2Dσ2 − Bσ∗2Cσ

∗

2
)
2
∼ F4n+6, 3Bσ∗

2
Cσ∗

2
− Aσ2Dσ2

−74
∼ 2L4n+6 + 6.

Proof. Since the sums Bσ∗
2
, Aσ2, Cσ∗2 and Dσ2 are equal

1
5
(14L4n+6 + 23),

1
5
(4L4n+6 + 13),

2L4n+6 + 1 and 16L4n+6 − 37, we infer that the sum
1
36
(Aσ2Dσ2 − Bσ∗2Cσ

∗

2
) + 2 is the square

of F4n+6. The second relation in this theorem has analogous proof.

In the next result we consider the products of the same components of the triples A, B, C and

D and the product of their components.



CUBO
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Squares in Euler triples ... 85

Theorem 13. The following relations hold:

A1B1C1D1
0
∼ F4n+2, A2B2C2D2

0
∼ 5F4n+6,

A3B3C3D3
0
∼ F4n+10, Aσ3Bσ3Cσ3Dσ3

0
∼ 5F4n+2 F4n+6 F4n+10.

Proof. Since the product F2n+5L2n+5 is F4n+10, A3 = B3 = F2n+5 and C3 = D3 = L2n+5, it fol-

lows that A3B3C3D3 = (F2n+5 L2n+5)
2 = F24n+10. This proves the third relation. The other rela-

tions have similar proofs.

The products of the same components of the triples Ã, B̃, C̃ and D̃ and the product of their

components appear in the following result.

Theorem 14. The following relations are true:

1
5
Ã1B̃1C̃1D̃1

0
∼ F4n+8,

1
5
Ã3B̃3C̃3D̃3

0
∼ F4n+4,

Ã2B̃2C̃2D̃2
0
∼ F4n+6, Ãσ3B̃σ3C̃σ3D̃σ3

0
∼ 5F4n+8F4n+6F4n+4.

Proof. Since Ã3 = F2n+2, B̃3 = L2n+2, C̃3 = L2n+2, D̃3 = 5F2n+2 and F2n+2L2n+2 = F4n+4, it

follows that the product 1
5
Ã3B̃3C̃3D̃3 is the square of F4n+4. This proves the second relation. The

other relations in this theorem have similar proofs.

The products of the sums σ1 and σ
∗ of the components of the triples A, B, C and D show the

same kind of relations. This is also true for the associated triples Ã, B̃, C̃ and D̃.

Theorem 15. The following relations hold for the sums σ1 and σ
∗
1:

Aσ1Bσ1Cσ1Dσ1
0
∼ 32F4n+6,

1
144
Ãσ1B̃σ1C̃σ1D̃σ1

1
∼ F4n+7,

Aσ∗
1
Bσ∗

1
Cσ∗

1
Dσ∗

1

0
∼ 4F4n+6,

1
64
Ãσ∗

1
B̃σ∗

1
C̃σ∗

1
D̃σ∗

1

1
∼ F4n+5.

Proof. The sums of the components Aσ1, Bσ1, Cσ1 and Dσ1 are equal 4F2n+3, 8F2n+3, 4L2n+3 and

8L2n+3. Hence, the product Aσ1Bσ1Cσ1Dσ1 is the square of 32F4n+6 since F2n+3L2n+3 = F4n+6.

This proves the above first relation. The other relations in this theorem have similar proofs.

In the next result we combine the sums σ1 and σ
∗
1 in each product.

Theorem 16. The following relations hold for the sums σ1 and σ
∗
1:

Aσ1Bσ∗1Cσ1Dσ
∗

1

0
∼ 8F4n+6, Aσ∗

1
Bσ1Cσ∗1Dσ1

0
∼ 16F4n+6,

1
24
Ãσ1B̃σ∗1C̃σ

∗

1
D̃σ1

1
∼ 2F4n+6 + 1,

1
24
Ãσ∗

1
B̃σ1C̃σ1D̃σ∗1

1
∼ 2F4n+6 − 1,

1
64
Ãσ1B̃σ∗1C̃σ1D̃σ

∗

1

1
∼ F4n+7,

1
144
Ãσ∗

1
B̃σ1C̃σ∗1D̃σ1

1
∼ F4n+5.



86 Zvonko Čerin CUBO
15, 2 (2013)

Proof. The sums of the components Aσ1, Bσ∗1, Cσ1 and Dσ
∗

1
are equal 4F2n+3, −2F2n+3, 4L2n+3

and −2L2n+3. The product Aσ1Bσ∗1Cσ1Dσ
∗

1
is therefore the square of 8F4n+6 since F2n+3L2n+3

= F4n+6. This proves the first relation. The other relations in this theorem have analogous

proofs.

4 Squares from the sums of squares

For a natural number k > 1, let the sums νk, ν
∗
k : Z

3
→ Z of powers be defined for x = (a,b,c) by

xνk = a
k + bk + ck and xν∗

k
= ak − bk + ck.

We proceed with the version of the Theorem 9 for the sums ν2 of the squares of components.

Theorem 17. The following relations are true for the sums ν2:

1
4
Aν2Cν2

−11
∼ 4F4n+6,

1
4
Bν2Dν2

−59
∼ 16F4n+6,

1
4
Ãν2C̃ν2

−3
∼ 2F4n+6,

1
4
B̃ν2D̃ν2

−27
∼ 8F4n+6.

Proof. Since Aν2 and Cν2 are
2
5
(4L4n+6 + 3) and 2(4L4n+6 − 3), the difference of

1
4
Aν2Cν2 and

11 is equal
16(L

2
4n+6−4)

5
. But, one can easily check that

L
2
4n+6−4

5
= F24n+6 so that the above quotient

is the square of 4F4n+6. This concludes the proof of the first relation. The other relations in this

theorem have similar proofs.

The next is the version of the Theorem 10 for the alternating sums ν∗2 of the squares of

components.

Theorem 18. The following relations are true for the sums ν∗2:

1
4
Aν∗

2
Cν∗

2

−7
∼ 3F4n+6,

1
4
Bν∗

2
Dν∗

2

41
∼ 9F4n+6,

1
4
Ãν∗

2
C̃ν∗

2

1
∼ F4n+6,

1
4
B̃ν∗

2
D̃ν∗

2

−23
∼ 7F4n+6.

Proof. Notice that the alternating sums of squares of components Aν∗
2
and Cν∗

2
are 2

5
(3L4n+6 + 1)

and 2(3L4n+6 + 1). Hence, the sum of
1
4
Aν∗

2
Cν∗

2
and −7 is equal to the following quotient

9(L
2
4n+6−4)

5
. This quotient is in fact the square of 3F4n+6. This proves the first relation. The

remaining three relations in this theorem have similar proofs.

Certain sums of products of the sums ν∗2 of components show the same behavior.

Theorem 19. The following relations are true for the sums ν∗2:

1
8
(Aν∗

2
Dν∗

2
+ Bν∗

2
Cν∗

2
)
17
∼
√
−27F4n+6,

1
8
(Ãν∗

2
D̃ν∗

2
+ B̃ν∗

2
C̃ν∗

2
)
−11
∼

√
7F4n+6.



CUBO
15, 2 (2013)

Squares in Euler triples ... 87

Proof. Notice that the alternating sums of squares of components Ãν∗
2
, B̃ν∗

2
, C̃ν∗

2
and D̃ν∗

2
are

2F2n+2F2n+4,
2
5
(7L4n+6 + 9), 2L2n+2L2n+4 and 2(7L4n+6 − 9). Hence, the sum of

1
8
(Ãν∗

2
D̃ν∗

2
+ B̃ν∗

2
C̃ν∗

2
) and −11 is equal to the square of

√
7F4n+6. This proves the second relation.

The first relation has a similar proof.

5 Squares from the products ⊙, ⊲ and ⊳

Let us introduce three binary operations ⊙, ⊲ and ⊳ on the set Z3 of triples of integers by the rules
(a, b, c) ⊙ (u, v, w) = (au, bv, cw), (a, b, c) ⊲ (u, v, w) = (av, bw, cu), and

(a, b, c) ⊳ (u, v, w) = (aw, bu, cv).

This section contains four theorems which show that the operations ⊙, ⊲ and ⊳ are also the
source of squares from components of the eight sequences.

Theorem 20. The following relations for the sequences A, B, C and D hold:

(A ⊙ B)σ1(C ⊙ D)σ1
−76
∼ 12F4n+6, (A ⊲ B)σ1(C ⊲ D)σ1

61
∼ 4L4n+5 and

(A ⊳ B)σ1(C ⊳ D)σ1
61
∼ 4L4n+7.

Proof. Since (A ⊲ B)σ1 = 4F4n+5 + 5 and (C ⊲ D)σ1 = 5(4F4n+5 − 5), it follows that the sum of

(A ⊲ B)σ1(C ⊲ D)σ1 and 61 is the product 16(5F
2
4n+5 − 4), i. e., the square of 4L4n+5. This proves

the second relation. The first and the third could be established similarly.

Theorem 21. The following relations for the sequences A, B, C and D hold:
1
4
(A ⊙ B)σ∗

1
(C ⊙ D)σ∗

1

1
∼ F4n+6, (A ⊲ B)σ∗

1
(C ⊲ D)σ∗

1

69
∼ 2F4n+2 and

(A ⊳ B)σ∗
1
(C ⊳ D)σ∗

1

69
∼ 2F4n+10.

Proof. Since the sums (A ⊲ B)σ∗
1
and (C ⊲ D)σ∗

1
are 1

5
(2L4n+2 + 19) and 2L4n+2 − 19, it follows

that the sum of (A ⊲ B)σ∗
1
(C ⊲ D)σ∗

1
and 69 is the square of 2F4n+2. This is the outline of the

proof of the second relation. The similar proofs of the first and the third relation are left to the

reader.

Theorem 22. The following relations for the triples Ã, B̃, C̃ and D̃ hold:

(Ã ⊙ B̃)σ1(C̃ ⊙ D̃)σ1
36
∼ 31F4n+3 + 7F4n, (Ã ⊲ B̃)σ1(C̃ ⊲ D̃)σ1

−3
∼ 2F4n+8, and

(Ã ⊳ B̃)σ1(C̃ ⊳ D̃)σ1
29
∼ 4F4n+7.

Proof. Since the sums (Ã ⊳ B̃)σ1 and (C̃ ⊳ D̃)σ1 are
1
5
(2L4n+8 + 1) and 2L4n+8 − 1, it follows that

the sum of (Ã ⊳ B̃)σ1(C̃ ⊳ D̃)σ1 and −3 is the square of 2F4n+8. This is the outline of the proof of

the third relation. The similar proofs of the first and the second relation are left to the reader.



88 Zvonko Čerin CUBO
15, 2 (2013)

Theorem 23. The following relations hold for the triples Ã, B̃, C̃ and D̃:

(Ã ⊙ B̃)σ∗
1
(C̃ ⊙ D̃)σ∗

1

36
∼ 23F4n+3 + 5F4n, (Ã ⊲ B̃)σ∗

1
(C̃ ⊲ D̃)σ∗

1

29
∼ 4F4n+5 and

(Ã ⊳ B̃)σ∗
1
(C̃ ⊳ D̃)σ∗

1

−3
∼ 2F4n+4.

Proof. Since the sums (Ã ⊳ B̃)σ∗
1
and (C̃ ⊳ D̃)σ∗

1
are −1

5
(2L4n+4 + 1) and 1 − 2L4n+4, it follows

that the difference of (Ã ⊳ B̃)σ∗
1
(C̃ ⊳ D̃)σ∗

1
and 3 is the square of 2F4n+4. This is the outline of the

proof of the third relation. The similar proofs of the first and the second relation are left to the

reader.

Received: March 2010. Accepted: September 2012.

References

[1] E. Brown, Sets in which xy + k is always a square, Mathematics of Computation, 45 (1985),

613-620.

[2] Z. Čerin, On pencils of Euler triples, I, Sarajevo Journal of Mathematics, 8 (1) (2012), 15–31.

[3] Z. Čerin, On pencils of Euler triples, II, Sarajevo Journal of Mathematics, 8 (2) (2012),

179-192.

[4] Z. Čerin, On Diophantine triples from Fibonacci and Lucas numbers, (preprint).

[5] M. Radić, A definition of determinant of rectangular matrix, Glasnik Mat. 1 (21) (1966),

17-22.

[6] N. Sloane, On-Line Encyclopedia of Integer Sequences,

http://www.research. att.com/∼njas/sequences/.