research article on the periods of biperiodic fibonacci and biperiodic lucas...

Research ArticleOn the Periods of Biperiodic Fibonacci andBiperiodic Lucas Numbers

Dursun TascJ and Gul Ozkan KJzJlJrmak

Department of Mathematics Gazi University 06500 Ankara Turkey

Correspondence should be addressed to Gul Ozkan Kızılırmak gulozkangaziedutr

Received 1 October 2016 Accepted 22 November 2016

Academic Editor Allan C Peterson

Copyright copy 2016 D Tascı and G Ozkan Kızılırmak This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

This paper is concerned with periods of Biperiodic Fibonacci and Biperiodic Lucas sequences taken as modulo prime and primepower By using Fermatrsquos little theorem quadratic reciprocity many results are obtained

1 Introduction

Fibonacci sequence and Lucas sequence are well-knownsequences among integer sequences The Fibonacci numberssatisfy the recurrence relation 119865119899 = 119865119899minus1 + 119865119899minus2 with theinitial conditions 1198650 = 0 and 1198651 = 1 Binetrsquos Formula for theFibonacci sequence is

119865119899 = 120572119899 minus 120573119899120572 minus 120573 (1)

where 120572 = (1 + radic5)2 and 120573 = (1 minus radic5)2 are roots ofthe characteristic equation 1199092 minus 119909 minus 1 = 0 The positiveroot 120572 is known as ldquogolden ratiordquo Fibonacci numbers andtheir generalizations have many interesting properties andapplications to almost every field of science and art [1ndash3] Inparticular Edson and Yayenie [4] introduced the BiperiodicFibonacci sequence as follows

1199020 = 01199021 = 1119902119899 =

119886119902119899minus1 + 119902119899minus2 if 119899 is even

119887119902119899minus1 + 119902119899minus2 if 119899 is oddfor 119899 ge 2

(2)

where 119886 and 119887 are two nonzero real numbers We take 119886 and119887 as integers If we take 119886 = 119887 = 1 we get the Fibonaccisequence 0 1 1 2 3 5 8

If we take 119886 = 119887 = 2 we get the Pell sequence0 1 2 5 12 29 70 Similarly if we take 119886 = 119887 = 119896 weget the 119896ndashFibonacci sequence [5ndash8] Binetrsquos Formula for theBiperiodic Fibonacci sequence is given as

119902119898 = ( 1198861minus120585(119898)(119886119887)lfloor1198982rfloor)

120572119898 minus 120573119898120572 minus 120573= 1119886lfloor(119898minus1)2rfloor119887lfloor(119898+1)2rfloor

120572119898 minus 120573119898120572 minus 120573 (3)

where 120572 120573 = (119886119887 plusmn radic11988621198872 + 4119886119887)2 are the roots of thecharacteristic equation 1199092 minus 119886119887119909 minus 119886119887 = 0 and 120585(119898) =119898 minus 2lfloor1198982rfloorMoreover

(120572120573) = minus119886119887(120572 + 120573) = 119886119887 (4)

This sequence and its properties can be found in [1 4]Another well-known sequence is the Lucas sequence

which satisfies the same recurrence relation as the Fibonaccisequence 119871119899 = 119871119899minus1 + 119871119899minus2 with the initial conditions 1198710 = 2and 1198711 = 1 Binetrsquos Formula for the Lucas sequence is

119871119899 = 120572119899 + 120573119899 (5)

Hindawi Publishing CorporationDiscrete Dynamics in Nature and SocietyVolume 2016 Article ID 7341729 5 pageshttpdxdoiorg10115520167341729

2 Discrete Dynamics in Nature and Society

where 120572 and 120573 are defined in (1) Bilgici defined generaliza-tion of Lucas sequence similar to the Biperiodic Fibonaccisequence as follows

1198970 = 21198971 = 119886119897119899 =



(6)

where 119886 and 119887 are two nonzero real numbers We take 119886and 119887 as integers This sequence and other generalizations ofLucas sequence with their properties can be found in [9 10]If we take 119886 = 119887 = 1 we get the Lucas sequence 2 1 3 47 11 18 Also if we take 119886 = 119887 = 119896 we get the 119896ndashLucassequence [11]

119897119899 = ( 119886120585(119899)(119886119887)lfloor(119899+1)2rfloor)120572119899 + 120573119899

= 1119886lfloor1198992rfloor119887lfloor(119899+1)2rfloor (120572119899 + 120573119899) (7)

where 120572 120573 and 120585 are defined in (3)On the other hand several researchers have made signif-

icant studies about the period of the recurrence sequences[2]Wall [12] defined the period-length of the recurring seriesobtained by reducing a Fibonacci series by a modulus 119898 Asan example the Fibonacci sequence mod3 is

0 1 1 2 0 2 2 1 0 1 1 (8)

and has period 8 Vinson [3] and Robinson [13] bothextended Wallrsquos study Moreover they studied the Fibonaccisequence for prime moduli and showed that for primes119901 equiv 1 4 (mod5) the period-length of the Fibonacci sequencemod119901 divides 119901 minus 1 while for primes 119901 equiv 2 3 (mod5) theperiod-length of the Fibonacci sequence mod 119901divides 2(119901+1)

Gupta et al [14] give alternative proofs of this resultsthat also use the Fibonacci matrix They place the roots ofits characteristic polynomial in an appropriate splitting fieldsRenault [15] examined the behaviour of the (119886 119887)-Fibonaccisequence under a modulus

Lucas studied the (119886 119887)-Fibonacci sequence extensivelyHe assigned Δ = 1198862 + 4119887 and deduced that if Δ isquadratic residue (that is a nonzero perfect square) mod119901 then 120572(119901) | 119901 minus 1 If Δ is quadratic nonresidue then120572(119901) | 119901 + 1 Also Rogers and Campbell studied the periodof the Fibonacci sequence mod119895 [16] They investigated theFibonacci sequence modulo 119901 prime and then generalized toprime powers

2 Period of Biperiodic Fibonacci Sequence

In this section we investigate the Biperiodic Fibonaccisequence modulo 119901 prime and then generalize to primepowers

Definition 1 Theperiod of the Biperiodic Fibonacci sequencemodulo a positive integer 119895 is the smallest positive integer 119898such that

119902119898 equiv 0 (mod119895) 119902119898+1 equiv 1 (mod119895) (9)

By the definition above the only members that can possiblycome back to the starting point are multiples of 119898This canbe summed up in the statement that if 119898 is the period of119902119899 (mod119895) then

119902119896 equiv 0 (mod119895)119902119896+1 equiv 1 (mod119895)

119898 | 119896

for any 119896 isin 119885

(10)

Theorem 2 Let 119901 be a prime and let 119899 be a positive integer If119886 equiv 1 (mod119901) (11)

then

119886119901119899 equiv 1 (mod119901119899+1) (12)

We remark that the proof of the Theorem 2 can be seen in[16]

Theorem 3 Let 119901 be a prime let 119896 be a positive integer and let120572 and 120573 be the fundamental roots of the Biperiodic Fibonaccisequence If119898 is the period of 119902119899 (mod119901)

120572119898119901119896minus1 equiv 120573119898119901119896minus1 equiv 1 (mod119901119896) (13)

Proof For 119886 = 0 and 119901 is a prime integer we have

119902119898 = ((1198861minus120585(119898))

(119886119887)lfloor1198982rfloor)120572119898 minus 120573119898120572 minus 120573 equiv 0 (mod119901) (14)

and then

120572119898 equiv 120573119898 (mod119901) (15)

Also we obtain119902119898 equiv 119902119898+1 minus 1199021 (mod119901)equiv ( 1198861minus120585(119898+1)

(119886119887)lfloor(119898+1)2rfloor)120572119898+1 minus 120573119898+1120572 minus 120573 minus 120572 minus 120573120572 minus 120573

(mod119901)equiv 0 (mod119901)

(16)

If119898 is even then

119902119898 equiv ( 1(119886119887)1198982)

120572 (120572119898 minus 1) minus 120573 (120573119898 minus 1)120572 minus 120573

(mod119901) (17)

Discrete Dynamics in Nature and Society 3

From 120572119898 equiv 120573119898 (mod119901) we get119902119898 equiv ( 1

(119886119887)1198982) (120572119898 minus 1) equiv 0 (mod119901) (18)

Thus 120572119898 equiv 120573119898 equiv 1 (mod119901) FromTheorem 2


If119898 is odd

119902119898 equiv ( 119886(119886119887)(119898+1)2)(

120572119898+1 minus 120573119898+1120572 minus 120573 minus 120572 minus 120573120572 minus 120573)(mod119901)

(20)

For 120572119898 equiv 120573119898 (mod119901) and 119886(119886119887)(119898+1)2 = 0119902119898 equiv ( 119886

(119886119887)(119898+1)2) (120572119898 minus 1) equiv 0 (mod119901) (21)

Therefore 120572119898 equiv 120573119898 equiv 1 (mod119901) FromTheorem 2


Theorem 4 Let 119901 be an odd prime let 119898 denote the periodof 119902119899 (mod119901) and let 998779 = 11988621198872 + 4119886119887 be a nonzero quadraticresidue mod119901 then119898 | 119901 minus 1Proof As is known from Fermatrsquos little theorem

120572119901minus1 equiv 1 (mod119901) 120573119901minus1 equiv 1 (mod119901) (23)

Thus we have

119902119901minus1 equiv ( 1119886lfloor(119901minus2)2rfloor119887lfloor(119901minus1)2rfloor )120572119901minus1 minus 120573119901minus1120572 minus 120573 equiv 0

(mod119901) 119902119901 = ( 1119886lfloor(119901minus1)2rfloor119887lfloor1199012rfloor )

120572119901 minus 120573119901120572 minus 120573equiv ( 1

(119886119887)(119901minus1)2)120572 minus 120573120572 minus 120573 equiv 1

(119886119887)(119901minus1)2(mod119901)

(24)

From 119886119887 = minus120572120573 we have(119886119887)(119901minus1)2 = ((minus120572120573)119901minus1)12 equiv 1 (mod119901) (25)

and then

119902119901 equiv 1 (mod119901) (26)

So that119902119901minus1 equiv 0 (mod119901) 119902119901 equiv 1 (mod119901) (27)

Therefore (10) implies that119898 | 119901 minus 1

Lemma5 If998779 is a quadratic nonresiduemod119901 then (radic998779)119901 =minusradic998779Lemma 6 Let 120572 and 120573 be the two roots of 1199092minus119886119887119909minus119886119887 = 0 in119865 = 1198651199012 998779 is a quadratic nonresiduemod119901 then 120573119901+1 = 120572119901+1Theorem 7 Let 119901 be an odd prime let119898 denote the period of119902119899 (mod119901) and let 998779 be a quadratic nonresidue mod119901 then119898 | 2119901 + 2Proof From the Binomial theorem we get (120572 minus 120573)119901 equiv 120572119901 minus120573119901 (mod119901) It follows that

119902119901 = ( 1119886lfloor(119901minus1)2rfloor119887lfloor1199012rfloor )120572119901 minus 120573119901120572 minus 120573

= ( 1(119886119887)(119901minus1)2)


(119886119887)(119901minus1)2)(120572 minus 120573)119901120572 minus 120573 (mod119901)

equiv ( 1(119886119887)(119901minus1)2)

(radic998779)119901radic998779 (mod119901)

equiv ( 1(119886119887)(119901minus1)2)

minusradic998779radic998779 (mod119901)equiv minus1 (mod119901)

119902119901+1 = ( 1119886lfloor1199012rfloor119887lfloor(119901+1)2rfloor )120572119901+1 minus 120573119901+1120572 minus 120573

= ( 1119886(119901minus1)2119887(119901+1)2 )120572119901+1 minus 120573119901+1120572 minus 120573

(28)

From Lemma 6 we obtain

119902119901+1 equiv 0 (mod119901) (29)

Thus119898 ∤ 119901 + 1 Also we have

1199022119901+2 = ( 1119886lfloor(2119901+1)2rfloor119887lfloor(2119901+2)2rfloor )1205722119901+2 minus 1205732119901+2120572 minus 120573

= ( 1(119886119887)119901)

(120572119901+1)2 minus (120573119901+1)2120572 minus 120573 equiv 0 (mod119901)


= ( 1(119886119887)119901+1)

(120572119901+1)2 120572 minus (120573119901+1)2 120573120572 minus 120573


= ( 1(119886119887)119901+1 (120572119901+1)

2) 120572 minus 120573120572 minus 120573 equiv(120572119901+1)2(119886119887)119901+1

equiv (120572119901+1)2(minus120572120573)119901+1 equiv

120572119901+1120573119901+1 equiv 1 (mod119901) (30)

Thus from (10)119898 | 2119901 + 1Theorem 8 Let 119901 be a prime let 119898 denote the period of119902119899 (mod119901) and let 1198981015840 denote the period of 119902119899 (mod119901119896) If119898119901119896minus1 is even then 1198981015840 | 119898119901119896minus1 and if 119898119901119896minus1 is odd then1198981015840 | 4119898119901119896minus1Proof We have shown that 120572119898119901119896minus1 equiv 120573119898119901119896minus1 equiv 1 (mod119901119896) inTheorem 3 So that

119902119898119901119896minus1 = ( 1198861minus120585(119898119901119896minus1)(119886119887)lfloor119898119901119896minus12rfloor)

120572119898119901119896minus1 minus 120573119898119901119896minus1120572 minus 120573 equiv 0(mod119901119896)

119902119898119901119896minus1+1 = ( 1198861minus120585(119898119901119896minus1+1)(119886119887)lfloor(119898119901119896minus1+1)2rfloor)

120572119898119901119896minus1+1 minus 120573119898119901119896minus1+1120572 minus 120573equiv 1198861minus120585(119898119901119896minus1+1)(119886119887)lfloor(119898119901119896minus1+1)2rfloor (mod119901119896)

(31)

If119898119901119896minus1 is even then119902119898119901119896minus1+1 equiv 1

(119886119887)119898119901119896minus12 (mod119901119896) (32)

From 120572120573 = minus119886119887 and by usingTheorem 3 it follows that

(119886119887)119898119901119896minus12 = (minus120572120573)119898119901119896minus12 = ((minus120572120573)119898119901119896minus1)12 equiv 1(mod119901119896)

(33)

Then

119902119898119901119896minus1+1 equiv 1 (mod119901119896) (34)

Thus from (10)1198981015840 | 119898119901119896minus1If119898119901119896minus1 is odd then



1199024119898119901119896minus1+1= ( 1198861minus120585(4119898119901119896minus1+1)

(119886119887)lfloor(4119898119901119896minus1+1)2rfloor)1205724119898119901119896minus1+1 minus 1205734119898119901119896minus1+1120572 minus 120573

equiv 1(119886119887)2119898119901119896minus1 (mod119901119896)

(35)

Since 120572120573 = minus119886119887 and 120572119898119901119896minus1 equiv 120573119898119901119896minus1 equiv 1 (mod119901119896) then(119886119887)2119898119901119896minus1 = (minus120572120573)2119898119901119896minus1 = ((minus120572120573)119898119901119896minus1)2 equiv 1

(mod119901119896) 1199024119898119901119896minus1+1 equiv 1 (mod119901119896)

(36)

Thus1198981015840 | 4119898119901119896minus13 Period of Biperiodic Lucas Sequence

In this section we investigate the Biperiodic Lucas sequencemodulo 119901 a prime similar to Biperiodic Fibonacci sequence

Definition 9 The period of the Biperiodic Lucas sequencemodulo a positive integer 119895 is the smallest positive integer 119905such that


For the same reasons as the Biperiodic Fibonacci sequencewehave that if 119905 is the period of 119897119899 (mod119895) then

119897119896 equiv 0 (mod119895) 119897119896+1 equiv 1 (mod119895)

119905 | 119896

for any 119896 isin 119885

(38)

Theorem 10 Let 119901 be an odd prime let 119905 denote the period of119897119899 (mod119901) and let 998779 be a nonzero quadratic residue mod119901then 119905 | 119901 minus 1Proof We use Fermatrsquos little theorem to get


so

119897119901minus1 = ( 1119886lfloor(119901minus1)2rfloor119887lfloor1199012rfloor ) (120572119901minus1 + 120573119901minus1) equiv 2(119886119887)(119901minus1)2(mod119901)

(40)

We know (119886119887)(119901minus1)2 equiv 1 (mod119901) thus119897119901minus1 equiv 2 (mod119901) (41)


Also we have

119897119901 = ( 1119886lfloor1199012rfloor119887lfloor(119901+1)2rfloor ) (120572119901 + 120573119901)equiv ( 1119886(119901minus1)2119887(119901+1)2 ) (120572 + 120573) (mod119901)equiv 119886119887(119886119887)(119901minus1)2 equiv

119886119887119887 (mod119901)equiv 119886 (mod119901)

(42)

By using (10) we get 119905 | 119901 minus 1Theorem 11 Let 119901 be an odd prime let 119905 denote the period of119897119899 (mod119901) and let 998779 be a quadratic nonresidue mod119901 then119905 | 2119901 + 2Proof From Lemma 6 and (3) we get

1198972119901+2 = ( 1119886lfloor(2119901+2)2rfloor119887lfloor(2119901+3)2rfloor ) (1205722119901+2 + 1205732119901+2)equiv ( 1

(119886119887)119901+1)((120572119901+1)2 + (120573119901+1)2) (mod119901)

equiv 2 (120573119901+1)2

(119886119887)119901+1 (mod119901)

equiv 2 (120573119901+1)2

120572119901+1120573119901+1 (mod119901)equiv 2120573119901+1120572119901+1 (mod119901)equiv 2 (mod119901)

1198972119901+3 = ( 1119886lfloor(2119901+3)2rfloor119887lfloor(2119901+4)2rfloor ) (1205722119901+3 + 1205732119901+3)= ( 1119886119901+1119887119901+2 ) (1205722119901+3 + 1205732119901+3)= ( 1119886119901+1119887119901+2 ) ((120572119901+1)

2 120572 + (120573119901+1)2 120573)= 1(120572 + 120573) (120572119901+1)2 (119886119887)119901+1 119887

= 1(119886119887) (120572119901+1)2 (119886119887)

119901+1 119887 = 1119886 (120572119901+1)2120572119901+1120573119901+1

= 1119886120572119901+1120573119901+1 = 1119886 (mod119901)

(43)

Thus from (10)119898 | 2119901 + 2Competing Interests

The authors declare that they have no competing interests

References

[1] M Edson S Lewis and O Yayenie ldquoThe k-periodic Fibonaccisequence and extended Binetrsquos formulardquo Integers vol 11 pp739ndash751 2011

[2] J Kramer and V E Hoggatt Special Cases of Fibonacci Period-icity 1972

[3] J Vinson ldquoThe relation of the period modulo m to the rankof apparition of m in the fibonacci sequencerdquo The FibonacciQuarterly pp 37ndash45 1963

[4] M Edson and O Yayenie ldquoA new generalization of Fibonaccisequence and extended Binetrsquos formulardquo Integers ElectronicJournal of Combinatorial Number Theory vol 9 pp 639ndash6542009

[5] S Falcon and A Plaza ldquoOn the Fibonacci k-numbersrdquo ChaosSolitons amp Fractals vol 32 no 5 pp 1615ndash1624 2007

[6] S Falcon and A Plaza ldquok-Fibonacci sequences modulo mrdquoChaos Solitons and Fractals vol 41 no 1 pp 497ndash504 2009

[7] S Falcon and A Plaza ldquoThe k-Fibonacci sequence and thePascal 2-trianglerdquo Chaos Solitons amp Fractals vol 33 no 1 pp38ndash49 2007

[8] S Falcon and A Plaza ldquoOn k-Fibonacci sequences and polyno-mials and their derivativesrdquo Chaos Solutions and Fractals vol39 no 3 pp 1005ndash1019 2009

[9] G Bilgici ldquoTwo generalizations of Lucas sequencerdquo AppliedMathematics and Computation vol 245 pp 526ndash538 2014

[10] G Bilgici ldquoNew generalizations of Fibonacci and Lucassequencesrdquo Applied Mathematical Sciences vol 8 no 29 pp1429ndash1437 2014

[11] S Falcon ldquoOn the k-Lucas numbersrdquo International Journal ofContemporary Mathematical Sciences vol 21 pp 1039ndash10502011

[12] D D Wall ldquoFibonacci series modulomrdquoThe American Mathe-matical Monthly vol 67 no 6 pp 525ndash532 1960

[13] D W Robinson ldquoThe fibonacci matrix modulo mrdquo TheFibonacci Quarterly vol 1 pp 29ndash36 1963

[14] S Gupta P Rockstroh and F E Su ldquoSplitting fields andperiods of Fibonacci sequences modulo primesrdquo MathematicsMagazine vol 85 no 2 pp 130ndash135 2012

[15] M Renault ldquoThe period rank and order of the (a b)minusfibonaccisequence Mod mrdquo Mathematics Magazine vol 86 no 5 pp372ndash380 2013

[16] N Rogers and C W Campbell The period of the fibonaccisequence modulo j [PhD thesis] The University Of ArizonaTucson Ariz USA 2007

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Stochastic AnalysisInternational Journal of


where 120572 and 120573 are defined in (1) Bilgici defined generaliza-tion of Lucas sequence similar to the Biperiodic Fibonaccisequence as follows

1198970 = 21198971 = 119886119897119899 =



(6)

where 119886 and 119887 are two nonzero real numbers We take 119886and 119887 as integers This sequence and other generalizations ofLucas sequence with their properties can be found in [9 10]If we take 119886 = 119887 = 1 we get the Lucas sequence 2 1 3 47 11 18 Also if we take 119886 = 119887 = 119896 we get the 119896ndashLucassequence [11]

119897119899 = ( 119886120585(119899)(119886119887)lfloor(119899+1)2rfloor)120572119899 + 120573119899

= 1119886lfloor1198992rfloor119887lfloor(119899+1)2rfloor (120572119899 + 120573119899) (7)

where 120572 120573 and 120585 are defined in (3)On the other hand several researchers have made signif-

icant studies about the period of the recurrence sequences[2]Wall [12] defined the period-length of the recurring seriesobtained by reducing a Fibonacci series by a modulus 119898 Asan example the Fibonacci sequence mod3 is

0 1 1 2 0 2 2 1 0 1 1 (8)

and has period 8 Vinson [3] and Robinson [13] bothextended Wallrsquos study Moreover they studied the Fibonaccisequence for prime moduli and showed that for primes119901 equiv 1 4 (mod5) the period-length of the Fibonacci sequencemod119901 divides 119901 minus 1 while for primes 119901 equiv 2 3 (mod5) theperiod-length of the Fibonacci sequence mod 119901divides 2(119901+1)

Gupta et al [14] give alternative proofs of this resultsthat also use the Fibonacci matrix They place the roots ofits characteristic polynomial in an appropriate splitting fieldsRenault [15] examined the behaviour of the (119886 119887)-Fibonaccisequence under a modulus

Lucas studied the (119886 119887)-Fibonacci sequence extensivelyHe assigned Δ = 1198862 + 4119887 and deduced that if Δ isquadratic residue (that is a nonzero perfect square) mod119901 then 120572(119901) | 119901 minus 1 If Δ is quadratic nonresidue then120572(119901) | 119901 + 1 Also Rogers and Campbell studied the periodof the Fibonacci sequence mod119895 [16] They investigated theFibonacci sequence modulo 119901 prime and then generalized toprime powers

2 Period of Biperiodic Fibonacci Sequence

In this section we investigate the Biperiodic Fibonaccisequence modulo 119901 prime and then generalize to primepowers

Definition 1 Theperiod of the Biperiodic Fibonacci sequencemodulo a positive integer 119895 is the smallest positive integer 119898such that


By the definition above the only members that can possiblycome back to the starting point are multiples of 119898This canbe summed up in the statement that if 119898 is the period of119902119899 (mod119895) then

119902119896 equiv 0 (mod119895)119902119896+1 equiv 1 (mod119895)

119898 | 119896

for any 119896 isin 119885

(10)

Theorem 2 Let 119901 be a prime and let 119899 be a positive integer If119886 equiv 1 (mod119901) (11)

then

119886119901119899 equiv 1 (mod119901119899+1) (12)

We remark that the proof of the Theorem 2 can be seen in[16]

Theorem 3 Let 119901 be a prime let 119896 be a positive integer and let120572 and 120573 be the fundamental roots of the Biperiodic Fibonaccisequence If119898 is the period of 119902119899 (mod119901)


Proof For 119886 = 0 and 119901 is a prime integer we have

119902119898 = ((1198861minus120585(119898))

(119886119887)lfloor1198982rfloor)120572119898 minus 120573119898120572 minus 120573 equiv 0 (mod119901) (14)

and then

120572119898 equiv 120573119898 (mod119901) (15)

Also we obtain119902119898 equiv 119902119898+1 minus 1199021 (mod119901)equiv ( 1198861minus120585(119898+1)

(119886119887)lfloor(119898+1)2rfloor)120572119898+1 minus 120573119898+1120572 minus 120573 minus 120572 minus 120573120572 minus 120573

(mod119901)equiv 0 (mod119901)

(16)

If119898 is even then

119902119898 equiv ( 1(119886119887)1198982)

120572 (120572119898 minus 1) minus 120573 (120573119898 minus 1)120572 minus 120573

(mod119901) (17)



(119886119887)1198982) (120572119898 minus 1) equiv 0 (mod119901) (18)



If119898 is odd

119902119898 equiv ( 119886(119886119887)(119898+1)2)(


(20)


(119886119887)(119898+1)2) (120572119898 minus 1) equiv 0 (mod119901) (21)





Thus we have





(119886119887)(119901minus1)2(mod119901)

(24)


and then

119902119901 equiv 1 (mod119901) (26)





= ( 1(119886119887)(119901minus1)2)



equiv ( 1(119886119887)(119901minus1)2)

(radic998779)119901radic998779 (mod119901)

equiv ( 1(119886119887)(119901minus1)2)




(28)


119902119901+1 equiv 0 (mod119901) (29)

Thus119898 ∤ 119901 + 1 Also we have


= ( 1(119886119887)119901)



= ( 1(119886119887)119901+1)

(120572119901+1)2 120572 minus (120573119901+1)2 120573120572 minus 120573


= ( 1(119886119887)119901+1 (120572119901+1)



120572119901+1120573119901+1 equiv 1 (mod119901) (30)






(31)


(119886119887)119898119901119896minus12 (mod119901119896) (32)



(33)

Then

119902119898119901119896minus1+1 equiv 1 (mod119901119896) (34)






equiv 1(119886119887)2119898119901119896minus1 (mod119901119896)

(35)



(36)







119905 | 119896

for any 119896 isin 119885

(38)



so


(40)



Also we have


119886119887119887 (mod119901)equiv 119886 (mod119901)

(42)



(119886119887)119901+1)((120572119901+1)2 + (120573119901+1)2) (mod119901)

equiv 2 (120573119901+1)2

(119886119887)119901+1 (mod119901)

equiv 2 (120573119901+1)2



2 120572 + (120573119901+1)2 120573)= 1(120572 + 120573) (120572119901+1)2 (119886119887)119901+1 119887

= 1(119886119887) (120572119901+1)2 (119886119887)

119901+1 119887 = 1119886 (120572119901+1)2120572119901+1120573119901+1

= 1119886120572119901+1120573119901+1 = 1119886 (mod119901)

(43)



References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











(119886119887)1198982) (120572119898 minus 1) equiv 0 (mod119901) (18)



If119898 is odd

119902119898 equiv ( 119886(119886119887)(119898+1)2)(


(20)


(119886119887)(119898+1)2) (120572119898 minus 1) equiv 0 (mod119901) (21)





Thus we have





(119886119887)(119901minus1)2(mod119901)

(24)


and then

119902119901 equiv 1 (mod119901) (26)





= ( 1(119886119887)(119901minus1)2)



equiv ( 1(119886119887)(119901minus1)2)

(radic998779)119901radic998779 (mod119901)

equiv ( 1(119886119887)(119901minus1)2)




(28)


119902119901+1 equiv 0 (mod119901) (29)

Thus119898 ∤ 119901 + 1 Also we have


= ( 1(119886119887)119901)



= ( 1(119886119887)119901+1)

(120572119901+1)2 120572 minus (120573119901+1)2 120573120572 minus 120573


= ( 1(119886119887)119901+1 (120572119901+1)



120572119901+1120573119901+1 equiv 1 (mod119901) (30)






(31)


(119886119887)119898119901119896minus12 (mod119901119896) (32)



(33)

Then

119902119898119901119896minus1+1 equiv 1 (mod119901119896) (34)






equiv 1(119886119887)2119898119901119896minus1 (mod119901119896)

(35)



(36)







119905 | 119896

for any 119896 isin 119885

(38)



so


(40)



Also we have


119886119887119887 (mod119901)equiv 119886 (mod119901)

(42)



(119886119887)119901+1)((120572119901+1)2 + (120573119901+1)2) (mod119901)

equiv 2 (120573119901+1)2

(119886119887)119901+1 (mod119901)

equiv 2 (120573119901+1)2



2 120572 + (120573119901+1)2 120573)= 1(120572 + 120573) (120572119901+1)2 (119886119887)119901+1 119887

= 1(119886119887) (120572119901+1)2 (119886119887)

119901+1 119887 = 1119886 (120572119901+1)2120572119901+1120573119901+1

= 1119886120572119901+1120573119901+1 = 1119886 (mod119901)

(43)



References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










= ( 1(119886119887)119901+1 (120572119901+1)



120572119901+1120573119901+1 equiv 1 (mod119901) (30)






(31)


(119886119887)119898119901119896minus12 (mod119901119896) (32)



(33)

Then

119902119898119901119896minus1+1 equiv 1 (mod119901119896) (34)






equiv 1(119886119887)2119898119901119896minus1 (mod119901119896)

(35)



(36)







119905 | 119896

for any 119896 isin 119885

(38)



so


(40)



Also we have


119886119887119887 (mod119901)equiv 119886 (mod119901)

(42)



(119886119887)119901+1)((120572119901+1)2 + (120573119901+1)2) (mod119901)

equiv 2 (120573119901+1)2

(119886119887)119901+1 (mod119901)

equiv 2 (120573119901+1)2



2 120572 + (120573119901+1)2 120573)= 1(120572 + 120573) (120572119901+1)2 (119886119887)119901+1 119887

= 1(119886119887) (120572119901+1)2 (119886119887)

119901+1 119887 = 1119886 (120572119901+1)2120572119901+1120573119901+1

= 1119886120572119901+1120573119901+1 = 1119886 (mod119901)

(43)



References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Also we have


119886119887119887 (mod119901)equiv 119886 (mod119901)

(42)



(119886119887)119901+1)((120572119901+1)2 + (120573119901+1)2) (mod119901)

equiv 2 (120573119901+1)2

(119886119887)119901+1 (mod119901)

equiv 2 (120573119901+1)2



2 120572 + (120573119901+1)2 120573)= 1(120572 + 120573) (120572119901+1)2 (119886119887)119901+1 119887

= 1(119886119887) (120572119901+1)2 (119886119887)

119901+1 119887 = 1119886 (120572119901+1)2120572119901+1120573119901+1

= 1119886120572119901+1120573119901+1 = 1119886 (mod119901)

(43)



References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









research article on the periods of biperiodic fibonacci and biperiodic lucas...

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