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On The Lucas Sequence Equations $$V_{n}=7\square $$ V n = 7 □ and $$V_{n}=7V_{m}\square $$ V n = 7 V m □

On The Lucas Sequence Equations $$V_{n}=7\square $$ V n = 7 □ and $$V_{n}=7V_{m}\square $$ V n =... Let P be a nonzero integer and let $$(U_{n})$$ ( U n ) and $$(V_{n})$$ ( V n ) denote Lucas sequences of first and second kind defined by $$U_{0}=0, U_{1}=1; V_{0}=2, V_{1}=P;$$ U 0 = 0 , U 1 = 1 ; V 0 = 2 , V 1 = P ; and $$U_{n+1}=PU_{n}+U_{n-1},V_{n+1}=PV_{n}+V_{n-1}$$ U n + 1 = P U n + U n - 1 , V n + 1 = P V n + V n - 1 for $$ n\ge 1.$$ n ≥ 1 . In this study, when P is odd, we show that the equation $$ U_{n} =7\square $$ U n = 7 □ has only the solution $$(n,P)=(2,7\square )$$ ( n , P ) = ( 2 , 7 □ ) when 7|P and the equation $$V_{n}=7\square $$ V n = 7 □ has only the solution $$ (n,P)=(1,7\square )$$ ( n , P ) = ( 1 , 7 □ ) when 7|P or $$(n,P)=(4,1)$$ ( n , P ) = ( 4 , 1 ) when $$P^{2}\equiv 1( \text{ mod } 7).$$ P 2 ≡ 1 ( mod 7 ) . In addition, we show that the equation $$V_{n}=7V_{m}\square $$ V n = 7 V m □ has a solution if and only if $$P^{2}=-3+7\square $$ P 2 = - 3 + 7 □ and $$ (n,m)=(3,1).$$ ( n , m ) = ( 3 , 1 ) . Moreover, we show that the equation $$U_{n}=7U_{m} \square $$ U n = 7 U m □ has only the solution $$(n,m,P,\square )=(8,4,1,1)$$ ( n , m , P , □ ) = ( 8 , 4 , 1 , 1 ) when P is odd. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Bulletin of the Malaysian Mathematical Sciences Society Springer Journals

On The Lucas Sequence Equations $$V_{n}=7\square $$ V n = 7 □ and $$V_{n}=7V_{m}\square $$ V n = 7 V m □

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References (15)

Publisher
Springer Journals
Copyright
Copyright © 2016 by Malaysian Mathematical Sciences Society and Universiti Sains Malaysia
Subject
Mathematics; Mathematics, general; Applications of Mathematics
ISSN
0126-6705
eISSN
2180-4206
DOI
10.1007/s40840-015-0295-x
Publisher site
See Article on Publisher Site

Abstract

Let P be a nonzero integer and let $$(U_{n})$$ ( U n ) and $$(V_{n})$$ ( V n ) denote Lucas sequences of first and second kind defined by $$U_{0}=0, U_{1}=1; V_{0}=2, V_{1}=P;$$ U 0 = 0 , U 1 = 1 ; V 0 = 2 , V 1 = P ; and $$U_{n+1}=PU_{n}+U_{n-1},V_{n+1}=PV_{n}+V_{n-1}$$ U n + 1 = P U n + U n - 1 , V n + 1 = P V n + V n - 1 for $$ n\ge 1.$$ n ≥ 1 . In this study, when P is odd, we show that the equation $$ U_{n} =7\square $$ U n = 7 □ has only the solution $$(n,P)=(2,7\square )$$ ( n , P ) = ( 2 , 7 □ ) when 7|P and the equation $$V_{n}=7\square $$ V n = 7 □ has only the solution $$ (n,P)=(1,7\square )$$ ( n , P ) = ( 1 , 7 □ ) when 7|P or $$(n,P)=(4,1)$$ ( n , P ) = ( 4 , 1 ) when $$P^{2}\equiv 1( \text{ mod } 7).$$ P 2 ≡ 1 ( mod 7 ) . In addition, we show that the equation $$V_{n}=7V_{m}\square $$ V n = 7 V m □ has a solution if and only if $$P^{2}=-3+7\square $$ P 2 = - 3 + 7 □ and $$ (n,m)=(3,1).$$ ( n , m ) = ( 3 , 1 ) . Moreover, we show that the equation $$U_{n}=7U_{m} \square $$ U n = 7 U m □ has only the solution $$(n,m,P,\square )=(8,4,1,1)$$ ( n , m , P , □ ) = ( 8 , 4 , 1 , 1 ) when P is odd.

Journal

Bulletin of the Malaysian Mathematical Sciences SocietySpringer Journals

Published: Jan 11, 2016

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