Fibonacci numbers. how to prove? [duplicate]












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  • Fibonacci modular results $ F_nmid F_{kn},,$ $, gcd(F_n,F_m) = F_{gcd(n,m)}$

    5 answers




We are give Fibonacci numbers.{fi
| i ∈ N}, where f0 = 0, f1 = 1, fn+2 = fn +fn+1, n∈ N. How to proof with mathematical induction that if n divides by m, then fn divides by fm?
I am having trouble with thinking, what should be the transition.






induction divisibility fibonacci-numbers







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marked as duplicate by anomaly, Crostul, amWhy, Bill Dubuque divisibility
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$begingroup$



This question already has an answer here:




  • Fibonacci modular results $ F_nmid F_{kn},,$ $, gcd(F_n,F_m) = F_{gcd(n,m)}$

    5 answers




We are give Fibonacci numbers.{fi
| i ∈ N}, where f0 = 0, f1 = 1, fn+2 = fn +fn+1, n∈ N. How to proof with mathematical induction that if n divides by m, then fn divides by fm?
I am having trouble with thinking, what should be the transition.






induction divisibility fibonacci-numbers







share|cite|improve this question











$endgroup$



marked as duplicate by anomaly, Crostul, amWhy, Bill Dubuque divisibility
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-2












-2








-2





$begingroup$



This question already has an answer here:




  • Fibonacci modular results $ F_nmid F_{kn},,$ $, gcd(F_n,F_m) = F_{gcd(n,m)}$

    5 answers




We are give Fibonacci numbers.{fi
| i ∈ N}, where f0 = 0, f1 = 1, fn+2 = fn +fn+1, n∈ N. How to proof with mathematical induction that if n divides by m, then fn divides by fm?
I am having trouble with thinking, what should be the transition.






induction divisibility fibonacci-numbers







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Fibonacci modular results $ F_nmid F_{kn},,$ $, gcd(F_n,F_m) = F_{gcd(n,m)}$

    5 answers




We are give Fibonacci numbers.{fi
| i ∈ N}, where f0 = 0, f1 = 1, fn+2 = fn +fn+1, n∈ N. How to proof with mathematical induction that if n divides by m, then fn divides by fm?
I am having trouble with thinking, what should be the transition.





This question already has an answer here:




  • Fibonacci modular results $ F_nmid F_{kn},,$ $, gcd(F_n,F_m) = F_{gcd(n,m)}$

    5 answers







induction divisibility fibonacci-numbers




induction divisibility fibonacci-numbers






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share|cite|improve this question













share|cite|improve this question




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edited Dec 27 '18 at 23:36







Edvards Zakovskis

















asked Dec 27 '18 at 23:28









Edvards ZakovskisEdvards Zakovskis

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marked as duplicate by anomaly, Crostul, amWhy, Bill Dubuque divisibility
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 Dec 27 '18 at 23:57


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    Hi & welcome to MSE. Please show us what you've tried so far & are having difficulty with. Thanks.
    $endgroup$
    – John Omielan
    Dec 27 '18 at 23:29


















  • $begingroup$
    Hi & welcome to MSE. Please show us what you've tried so far & are having difficulty with. Thanks.
    $endgroup$
    – John Omielan
    Dec 27 '18 at 23:29
















$begingroup$
Hi & welcome to MSE. Please show us what you've tried so far & are having difficulty with. Thanks.
$endgroup$
– John Omielan
Dec 27 '18 at 23:29




$begingroup$
Hi & welcome to MSE. Please show us what you've tried so far & are having difficulty with. Thanks.
$endgroup$
– John Omielan
Dec 27 '18 at 23:29










1 Answer
1






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First prove by induction on $m$ that



$F_{m+n}=F_{m+1}F_n+F_mF_{n+1}-F_m F_n$



Then put $m=kn$ and find that if $F_n|F_{kn}$ then $F_n|F_{(k+1)n}$.






share|cite|improve this answer









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  • $begingroup$
    +1 then -1. The battle is on!
    $endgroup$
    – Oscar Lanzi
    Dec 27 '18 at 23:58


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

First prove by induction on $m$ that



$F_{m+n}=F_{m+1}F_n+F_mF_{n+1}-F_m F_n$



Then put $m=kn$ and find that if $F_n|F_{kn}$ then $F_n|F_{(k+1)n}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    +1 then -1. The battle is on!
    $endgroup$
    – Oscar Lanzi
    Dec 27 '18 at 23:58
















0












$begingroup$

First prove by induction on $m$ that



$F_{m+n}=F_{m+1}F_n+F_mF_{n+1}-F_m F_n$



Then put $m=kn$ and find that if $F_n|F_{kn}$ then $F_n|F_{(k+1)n}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    +1 then -1. The battle is on!
    $endgroup$
    – Oscar Lanzi
    Dec 27 '18 at 23:58














0












0








0





$begingroup$

First prove by induction on $m$ that



$F_{m+n}=F_{m+1}F_n+F_mF_{n+1}-F_m F_n$



Then put $m=kn$ and find that if $F_n|F_{kn}$ then $F_n|F_{(k+1)n}$.






share|cite|improve this answer









$endgroup$



First prove by induction on $m$ that



$F_{m+n}=F_{m+1}F_n+F_mF_{n+1}-F_m F_n$



Then put $m=kn$ and find that if $F_n|F_{kn}$ then $F_n|F_{(k+1)n}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 27 '18 at 23:48









Oscar LanziOscar Lanzi

12.9k12136




12.9k12136












  • $begingroup$
    +1 then -1. The battle is on!
    $endgroup$
    – Oscar Lanzi
    Dec 27 '18 at 23:58


















  • $begingroup$
    +1 then -1. The battle is on!
    $endgroup$
    – Oscar Lanzi
    Dec 27 '18 at 23:58
















$begingroup$
+1 then -1. The battle is on!
$endgroup$
– Oscar Lanzi
Dec 27 '18 at 23:58




$begingroup$
+1 then -1. The battle is on!
$endgroup$
– Oscar Lanzi
Dec 27 '18 at 23:58



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