ISSN 2319-8133 (Online)
(An International Research Journal), www.compmath-journal.org
Square Numbers in − Jacobsthal Sequence
G. Upender Reddy
Assistant Professor, Department of Mathematics, Nizam College (A), Osmania University Basheerbagh, Hyderabad – 500001, Telangana, INDIA.
email:[email protected].
(Received on: November 14, 2018) ABSTRACT
Two new - Jacobsthal Sequence and - Jacobsthal Lucas Sequence are defined. Based on the properties of the Jacobsthal and Jacobsthal Lucas Sequence the possible square numbers are studied.
2010 Mathematics Subject Classification: 11B37.
Keywords: Jacobsthal Sequence and Jacobsthal Lucas Sequence, Square Number.
1. INTRODUCTION
We know that Jacobsthal Sequence { } and the Jacobsthal Lucas Sequences { } which are defined by
= 0, = 1 and = + 2 , ≥ 0 (1.1) and
= 0, = 1 and = + 2 , ≥ 0 (1.2)
Although the two sequences of natural numbers were introduced by E. Jacobsthal
2in 1919, they have not drawn much attention until their applications to curve was studied by Horadam
3,4,5in 1988. B. Srinivasa Rao
6has proved the squares in the two pairs of sequences given in (1.1) and
(1.2) that { } is a square only if = 1 or 2 while { } is a square only for n =1.
For a fixed integer > 0, we define two new sequences
called − Jacobsthal Sequence
( )defined by
( )= 0,
( )= 1 and
( )
=
( )+ ( + 1)
( ), ≥ 0 (1.3) And the − Jacobsthal Lucas Sequence
( )is defined by
( )= 2,
( )= 1 and
( )
=
( )+ ( + 1)
( ), ≥ 0 (1.4)
2. For the first few values of the terms of
( )and
( )are as follows.
( ) ( )
0 1 2 3 4 5
0 1
1 + +
+ 2 + 2
+ 3 + 4 + 2 + 1
2 1 3 + 2 3 + 3 + 1
3 + 6 + 6 + 2
3 + 9 + 12 + 6 + 1
3. The following properties of the sequences { } and { } are well known, which are useful in proving the Lemmas and the Theorems of this paper. These properties are defined for all integers and .
= ( − ) and = + (3.1)
Where = 2 and = −1
2 = + (3.2)
= 3 + 2(−1) (3.3)
The following are the congruent properties of { } and { } . 3.4 Lemma: (i) ≡ 5 ( 8) for ≥ 2
(ii) ≡ 1 ( 8) for ≥ 2 (iii) ≡ 3 ( 8) for ≥ 1 (iv) ≡ 7 ( 8) for ≥ 2
Proof: We prove (i) by induction on , clearly (i) holds for = 2, since = 5. Assume that it is true for = , where ≥ 2. Now by (3.2) and (3.3) , the induction hypothesis we get
( )
= 1
2 ( + )
= (5 + ) = (5 + 3 + 2) = (8 + 2)
= 4 + 1 ≡ 4(5) + 1 ( 8) ≡ 5 ( 8) Proving (i) for = + 1.
Hence, (i) holds for all ≥ 2.
(ii) By (3.3) and part (i) of the lemma, for ≥ 2, = 3 + 2 ≡ 3.5 + 2 ( 8) ≡ 1 ( 8) Which proves(ii).
(iii) It is trivial if = 1, since = 3 For ≥ 2, we have by (3.2) ,
= 1
2 ( + ) = 1
2 ( + ) ≡ 1
2 (5 + 17) ( 8) ≡ 3 ( 8) Which proves (iii).
(iv) Let = 3 + 2 (−1) ≡ 3 − 2 ≡ 3.3 − 2 ≡ 7 ( 8) Which proves (iv).
Which completes the proof of the lemma.
3.5 Note: For any integer , we have
(i) ≡ 0,1 4 ( 8)
(ii) ≡ 0, 1, 4, 9, 16, 17 25 ( 32)
3.6 Theorem:
( )is a square if and only if = 1 2.
Proof: Note that
( )= , we can observe that and are squares.
Conversely, suppose = for some integer . By (i) and (ii) of Lemma 3.4, which cannot be true for above Note 3.5.
Hence the theorem is proved.
3.7 Theorem:
( )is a square if and only if = 1.
Proof: Note that
( )= , clearly is a square.
Conversely, suppose = for some integer . Let neither is even nor an odd integer greater than 1. In fact, if = 2 , where ≥ 0. By (3.1), = 2 + 1,
Which is obviously not a square.
If = 2 + 1 and ≥ 1 then (iv) of lemma 3.4, gives = 7 ( 8), Proving that cannot be square.
Which completes the proof of the theorem for = 1.
3.8 Theorem:
( )is a square if = 0 1.
Proof: We prove the theorem by the Principle of Mathematical Induction on . We have
( )= 0 and
( )= 1.
Consider = 1, from Theorem 3.6 and we observe that
( )and
( )are squares.
Therefore the result is true for = 1.
Assume that it is true for = .
We have to prove that it is true for = + 1.
Put = + 1, then
( )= 0 and
( )= 1 = 1 ( ) Therefore
( )is a square if = 0 1.
Which completes the proof of the theorem.
3.9 Theorem:
( )is a square if = 1.
Proof: We prove the theorem by the Principle of Mathematical Induction on . We have
( )= 1.
Consider = 1, from Theorem 3.6 and we observe that
( )are squares.
Therefore the result is true for = 1.
Assume that it is true for = .
We have to prove that it is true for = + 1.
Put = + 1, then
( )= 1 = 1 ( ) Therefore
( )is a square if = 1.
Which completes the proof of the theorem.
In this section first we proved a lemma then we have to prove square in
( )± and
( )
± .
4.1 Lemma: (i) ≡ 21 ( 32) for ≥ 3 (ii) ≡ 11 ( 32) for ≥ 2
Proof: We prove (i) by induction on . Clearly (i) holds for = 3.
i.e., = 21 ⇒ ≡ 21 ( 32) By (3.2) and (3.3), we get,
( )
= ( + )
= (5 + )
= (8 + 2) = (4 + 1) ≡ 4(21) + 1 ≡ 21 ( 32) Proving (i) for = + 1. Hence (i) holds for all ≥ 3.
(ii) is trivial if = 2, since = 11, for ≥ 3. we have by (3.2), (3.3) and the part (i) of
the lemma that
( )
= 1
2 ( + ) = 1
2 ( + )
= 1
2 ( + 3 + 2(−1) )
= 1
2 (4 + 2) = 2 + 1
≡ 2(21) + 1 ( 32) ≡ 11( 32) Hence, the lemma proved.
4.2 Theorem:
( )− 1 is a square if and only if = 1 or is even.
Proof: Note that
( )= , if = 1 or is even then − 1 is a square, since − 1 = 0 and
− 1 = (2 ) .
Conversely, suppose − 1 is a square, then is odd and greater than 1 cannot hold.
In fact, if = 2 + 1 for some integer ≥ 1, then (iv) of lemma (3.4) shows − 1 =
− 1 ≡ 6 ( 8) and hence − 1 in not a square, by Note 3.5(i).
4.3 Theorem:
( )+ 1 is never a square for ≤ 5.
Proof: Note that
( )=
Let + 1 = 2 + 2 and + 1 = 2 for ≥ 1, by (3.1) it shows that + 1 is never a square.
4.4 Theorem:
( )− 1 is a square onl if = 1, 2 4.
Proof: Note that
( )= , clearly − 1 = − 1 = 0 and − 1 = 2 . Conversely, if = 2 + 1, where ≥ 1. Then by (iii) of lemma 3.4, we have
− 1 = − 1 ≡ 2 ( 8), showing − 1 is not a square for odd ≥ 3, in view of
Note 3.5 (i). Again if = 2 for ≥ 3.
Then lemma 4.1(i), gives − 1 = − 1 ≡ 20 ( 32)
Showing − 1 is not a square for even ≥ 6 by note 3.5(ii).
Thus
( )− 1 is a square if and only if = 1, 2 4.
Which completes the proof of the theorem.
4.5 Theorem:
( )+ 1 is a square if and only if = 0 3.
Proof: Note that
( )= , clearly + 1 = 1 and + 1 = 2 , part of the theorem holds.
Conversely, suppose + 1 is a square. Then cannot be even and > 0, since + 1=2 and for ≥ 2. We have + 1 = 4 ( 8), by lemma 3.4(i) . Again if
= 2 + 1 with ≥ 2, then lemma 4.1(ii) gives + 1 ≡ 12 ( 32).
Therefore + 1 is not a square by Note 3.5(ii).
Which completes the proof of the theorem.
Observations:
(i)
( )+ 1 is a square if = 0.
(ii)
( )+ 1 is a square if = 0 3.
(iii)
( )+ 1 is a square if = 0 2.
(iv)
( )− 1 is a square if = 1 2.
(v)
( )− 1 is a square if = 1, 2 4.
(vi)
( )− 1 is a square if = 0, 1,2 4.
(vii)
( )is a perfect square if = 3 (viii)
( )is a perfect square if = 2
4.7 Theorem:
( )− 1 is asquare if = 0 1.
Proof: We prove the theorem by the principle of Mathematical induction on . We have
( )
= 2 and
( )= 1.
Put = 1, by the Theorem 4.2 ,
( )− 1 is a square if = 0 1.
Therefore the result is true for = 1. Assume that the result is true for = . We have to prove that the result is also true for = + 1
( )
− 1 = 2 − 1 = 1 = 1 and
( )− 1 = 0 (for all ).
Therefore the result is true for = + 1.
By the PMI
( )− 1 is a square if = 0 1.
Which completes the proof of the theorem.
4.8 Theorem:
( )− 1 is asquare if = 1.
Proof: We prove the theorem by the principle of Mathematical induction on . We have
( )
= 1.
Put = 1, by the Theorem 4.4 ,
( )− 1 is a square.
Therefore the result is true for = 1. Assume that the result is true for = . We have to prove that the result is also true for = + 1
( )
