On k-traid sequences

October 16, 2017 | Autor: Hansraj Gupta | Categoría: Applied Mathematics, Mathematical Physics, Pure Mathematics
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Internat. J. Math. & Math. Sci. Vol. 8 No. 4 (1985) 799-804

799

ON k-TRIAD SEQUENCES HANSRAJ GUPTA Panjab University, Chandigarh, India 160014 and

K. SlNGH Department of Mathematics and Statistics University of New Brunswick, Fredericton, Canada E3B 5A3 (Received May 29, 1984 and in revised June 29, 1984)

At the conference of the Indian Mathematical Society held at Allahabad

ABSTRACT.

in December 1981, S. P. Mohanty and A. M. S. Ramasamy pointed out that the three numbers

I, 2, 7, have the following property:

increased by 2 is a perfect square.

the product of any two of them

They then showed that there is no fourth

integer which hsares this property with all of them.

They used Pell’s equation

and the theory of quadratic residues to prove their statement.

In this paper,

we show that their statement holds for a very large set of triads and our proof

of the statement is very simple.

KEY WORDS AND PHPASES.

Pell’s equation, congruence, Fibonacci, sequence. 1980 AMS SUBJECT CLASSIFICATION CODE. 10A10. i.

INTRODUCTION. DEFINITION.

Given any integer k, three numbers a I,

a2,

a

3

are said to form

a k-triad, if the numbers

ala 2 +

ala 3 +

k,

k

and

a2a 3 +

k

(I.I)

are all perfect squares.

An ascending sequence of integers a

I, a 2, a 3,

a

n

is said to be a k-triad sequence if every three consecutive elements of the sequence form a k-triad.

Evidently, if (I.i) is a k-triad sequence, then

a2, a3,

a

4

a

n

is also a k-triad sequence. This elementary statement will prove useful to us in our work.

(1.2)

H. GUPTA AND K. SINGH

800

2.

CONSTRUCTION OF SEQUENCES (I.i).

In what follows, small letters denote integers and c.’s are positive. i Given any integer k, we can choose two integers a

such that

ala 2 +

I

and

a2,

a

2

>

al,

k is a perfect square:

+

ala 2

k

c

2

(2.1)

I

The problem of constructing the sequence (I.I) then reduces to finding a number a

3

such that both

ala 3 + ala 3 +

Let Set

x

k and k

x

2

a2a 3 + k will a2a 3 + k

+ Cl,

a

y

a

2

be squares.

y

2

(2.2)

+ cI

Then from (2.2) we have

(a

(a so that

a

x)(y + x)

(y

al)a 3

2

a

3

+ 2c I +

a2);

+ a 2 + 2c I

I

We assert that this value of a

-al)(a I

2

(2.3)

actually satisfies our requirements.

3

In

fact, we have

ala 3 +

+ 2c I + a 2) + k 2 a + 2alc I + (ala 2 + k)

al(a I

k

a

2

+

(a

+ c 2I

2alc

+

ci)2

+

Cl )2

Similarly

a2a 3 + Writing

c c

2 2 2

(a

k

a2a 3 + a

2

2

(2.4)

k, we have

+ cI

(2.5)

Notice that (2.3) provides a formula for writing the third element of a

k-triad sequence interms of

a

and c

al, 2 I. Applying this procedure to (1.2), we get a

4

a a

2

2

a

+

a

3

+ 2c 2

+ (a I + a 2 + 2c I) + 2(a 2 + + 4a 2 + 4c

c

I) (2.6)

Treating this as a formula for the fourth element of a k-triad sequence and applying it to (1.2), we get a

5

+ 4a 3 + 4c 2 4a + 9a + 12c I 2 I a

2

The process can be repeated until the desired number of been obtained.

(2.7) elements of (I.I)

has

k-TRIAD SEQUENCES

801

Then assuming that

aj

uja I + vja 2 + wjc I

aj

uja 2 + vja 3 + wjc 2

we obtain

uja 2 + +

vja

+ a 2 + 2c I) + wj(a 2 + c I) + vj + wj)a 2 + (2vj + wj)c

vj(a I (uj

This gives us the following recurrence relations: v

uj+ I

v

j,

+

u.

j+

+

v.

Wo,

wj +I

2v. +w.

Identically:

uj+ I

0

vj+

I

wj+

0

0

u.

I

I

v

2

I

wj

(2.9)

or

uj

1

-I

uj+ 1

vj

0

0

jv LWj+l

i

wj

(2.10)

+I

0

Using the same technique, from (2.5), we obtain c

3

a

3

(a a

+

c

2

a 2c I) + (a + I + 2 + 2 + 2a 2 + 3c

c

I)

In general, c. =a.

+c.

(2.11)

j-I

Assuming that

rja I + sja 2 + tjc I

cj

there is no difficulty in obtaining the recurrence relations: r s t

+I

+I

+I

0

I

I

I

0

2

0

Irl s

(2.12)

it_

and

rj

1

sj

tj

2

1

-I

0

0

sj+

0

I

tj+

rj+ 1

While relations (2.9) and (2.12) enable us to find expressions for ao’S and

c.’s

as

takes value in ascending order of magnitude, the relations (2.10)

and (2.13) enable us to extend them in the opposite direction.

(2.13)

802

H. GUPTA AND K. SINGH Our k-triad sequence can thus be defined for all integral values of the

subscript; therefore, the sequence of

c.’s is defined

for all integral

values of the subscript.

IDENTIFICATION OF THE COEFFICIENTS u, v, w AND r, s, t.

3.

We hardly would expect that the Fibonacci sequence has anything to do with the coefficients u, v, w and r, s, t introduced in the preceding section.

But

the unexpected happens.

Recall that the Fibonacci sequence is defined by the recurrence relations:

I,

fl

0

f0

f

f

m

m-2

+

m > 2

fm-i

(3.1)

This definiton can be extended to negative integral subscripts by noting that f

(-i)

-m

m+l

f

(3.2)

m

From results (2.3), (2.6) and (2.7), it will be seen that for the values

3, 4 and 5;

f2 2 al

aj

+ f2.j-1

a2 + 2fj -2

f

j-1

c

I

so that for these values of j,

f2.j-2

uj

v

f2.j-I

2fj_2 fj-I

wj

From (2.9), we will now have

2

and

f2.j-2 + f2. -i + 2fj -2 fj -I (fj-2 + fj-i )2 f2.

Uj+l

fj-I

Vj+l

wj+

2f

2.

+ 2f j-2 (fj-i +

j-I

2f.j_l

fj_ i fj-2

2fj_l f’j

We now leave it to the reader to complete the induction and show that

f2j-2

aj

al +

f2.]-I a2 + 2fj -2 fj -i

c

i

(3.3)

for all integral values of j.

It is no more difficult to prove that for all integral values of

fj_2fj_lal

cj 4.

S! OF CONSECUTIVE

+

fj_ifja 2

+

(f2_j fj-2fj-I )el

(3.4)

a’s.

From (2.11), we have

cj

a.

cj_ 1

Hence we have an interesting summation formula n

a.

j=m+l

c

n

c

m

n > m

This provides a good check on the values of a.’s and c.’s. A GENERALIZATION OF THE STATEMENT OF MOHANTY AND RAMASAMY. Our generalization can be stated in the form of the following. THEOREM. If al, a2, a3 is a k-triad and k 2 (mod 4), then there is no integer a for which 5.

(4.1)

k-TRIAD SEQUENCES

ala +

a2a +

k,

k,

a3a

803

+ k

are all perfect squares.

We need two lemmas for the proof of our theorem.

LEMMA I. PROOF.

Only two of the three numbers a I, a 2, a 3 are odd. If a and a 2 are both even, let 2 ala 2 + k c I

This implies that c I is even.

Modulo 4, we have

0 (mod 4)

2

cannot be even.

and a

Hence both a

This is impossible.

2 I and a are of opposite parity, then since 2 a + a + 2c a I 2

If a

3

a

must be odd, and our lemma holds.

3 If a

and a are both odd, then a 3 is even. 2 This completes the proof of our lemma. LEMMA 2. The difference of the two odd elements of the given k-triad is congruent to 2 (mod 4).

First let a I and a 2 be the odd elements of the k-triad. Then 0 or 2 (mod 4). a H 0 (mod 4) and a cannot be congruent to 0 (mod 4). Suppose a 2 I

PROOF. a

a

I 2 But a 2 let

I

a

Since

ala 2

+

As

a

2

2

c

k

I

I + 2

c

a

even.

must be odd.

I

al(a I

a

2

and a

3 I Again since a

Hence

a

Now

a

Hence

a

2

+ 4d)

a

E

I (mod 4)

I (mod 4).

Hence

This is impossible.

Next let a

c

ala 2

we will have

for some integer d.

+ 4d

I

2 (mod 4)

be the two odd elements of the k-triad. Then a is necessarily 2 2 ala 2 + k c I c I must be even. We must, therefore, have

+ 2

0 (mod 4)

2 (mod 4)

2

+ a 2 + 2c

a

3

a

3

The case in which a

2

I

and

is even.

c

2 (mod 4). and a

3

are the odd elements, can be dealt with in the

same manner and the lemma is proven.

PROOF OF THE THEOREM.

Since the existence or non-existence of the number

a does not depend on the order in which the three expressions are

can assume that a

k-triad.

I

is the even and a

2

and a

3

taken, we

the odd elements of the given

804

H. GUPTA AND K. SINGH

For some positive integers x, y, z, let

ala +

k

a2a + a3a +

x

k

y

k

z

2

(i)

2

(ii)

2

(iii)

From (i) it is evident that x is even.

Replace x by 2g, a

(which is even)

by 2h, and k (which is congruent to 2 (mod 4)) by 2q where q is odd. (i) takes the form

(2g) 2

+ q

ha

Then,

(iv)

Since q is odd and the right-hand side is even, h and a must both be odd.

This implies that y and z are both odd.

Now substracting (ii) and (iii),

we have

(a

3

a2)a

z

2

y

2

0 (mod 4).

(v)

a 2 (mod 4), (v) implies that a is even. This contradicts 3 2 the earlier statement that a is odd. Hence a does not exis and we are

Since a

through.

-4

-3

-2

-I

0

i

2

3

4

5

6

-67

-25

-i0

-3

-i

2

5

15

38

I01

263

690

c.

41

16

6

3

2

4

9

24

62

163

426

1116

f.

-3

2

-I

I

0

i

2

3

5

8

13

u v

64

25

9

4

i

I I

0

i

I

4

9

25

25

9

4

i

I

0

i

I

4

9

25

64

w.

-80

-30

-12

-4

-2

0

0

2

4

12

30

80

r.

-40

-15

-6

-2

-I

0

0

i

2

6

15

40

s.

-15

-6

-2

-i

0

0

I

2

6

15

40

104

t.

49

19

7

3

i

i

i

3

7

19

49

129

2,

a

a.

k

ACKNOWLEDGEMENT.

6,

a

I

2

5,

c

I

4

The second author was partially funded by a travel grant from the

University of New Brunswick. REFERENCE

I.

ANDREWS, George E.

Number Theory

W. B. Saunders Company, Philadelphia, 1971.

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