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The sum of any ten consecutive numbers from a fibonacci sequence is divisible by 11 [on hold]

Fibonacci and Lucas identityWhy is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?Fibonacci sequence divisible by 7?How to prove gcd of consecutive Fibonacci numbers is 1?Fibonacci numbers divisible by $9$The sum of $n$ consecutive Fibonacci numbers.Product of consecutive Fibonacci numbers divisibilityWhy is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?Fibonacci sequence divisible by 3?Any non-negative integer can be written as the sum of distinct and non-consecutive Fibonacci numbersThe sum of 8 consecutive Fibonacci numbers is not a Fibonacci numberFibonacci sequence from natural numbers

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How do I prove that any ten consecutive numbers of a fibonacci sequence is divisible by 11?

divisibility fibonacci-numbers

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asked Apr 1 at 14:10

AbigailAbigail

171

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put on hold as off-topic by Javi, Mike Pierce, John Douma, YiFan, heropup Apr 2 at 2:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Javi, Mike Pierce, John Douma, YiFan, heropup

If this question can be reworded to fit the rules in the help center, please edit the question.

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  I mean, just bash? Use the recursive formula to show it.
  $endgroup$
  – Don Thousand
  Apr 1 at 14:12
  

  
  
  
  


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  See also here: math.stackexchange.com/questions/60049/…
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  – Jonas Lenz
  Apr 1 at 14:12
  

  
  
  
  


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  How about using artofproblemsolving.com/wiki/index.php?title=Binet%27s_Formula
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  – lab bhattacharjee
  Apr 1 at 14:21
  

  
  
  
  


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  $begingroup$
  Possible duplicate of Why is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?
  $endgroup$
  – Quuxplusone
  Apr 1 at 21:25
  

  
  
  
  

add a comment | 

2

2

$begingroup$

How do I prove that any ten consecutive numbers of a fibonacci sequence is divisible by 11?

divisibility fibonacci-numbers

share|cite|improve this question

asked Apr 1 at 14:10

AbigailAbigail

171

New contributor

Abigail is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

put on hold as off-topic by Javi, Mike Pierce, John Douma, YiFan, heropup Apr 2 at 2:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Javi, Mike Pierce, John Douma, YiFan, heropup

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I mean, just bash? Use the recursive formula to show it.
  $endgroup$
  – Don Thousand
  Apr 1 at 14:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  See also here: math.stackexchange.com/questions/60049/…
  $endgroup$
  – Jonas Lenz
  Apr 1 at 14:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How about using artofproblemsolving.com/wiki/index.php?title=Binet%27s_Formula
  $endgroup$
  – lab bhattacharjee
  Apr 1 at 14:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Why is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?
  $endgroup$
  – Quuxplusone
  Apr 1 at 21:25
  

  
  
  
  

add a comment | 

$begingroup$

How do I prove that any ten consecutive numbers of a fibonacci sequence is divisible by 11?

divisibility fibonacci-numbers

share|cite|improve this question

asked Apr 1 at 14:10

AbigailAbigail

171

New contributor

Abigail is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked Apr 1 at 14:10

AbigailAbigail

171

New contributor

Abigail is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked Apr 1 at 14:10

AbigailAbigail

171

New contributor

Abigail is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Apr 1 at 14:10

AbigailAbigail

171

New contributor

Abigail is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Apr 1 at 14:10

AbigailAbigail

171

2 Answers
2

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16

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Suggestion:
If the first two numbers in the sequences are $a, b$, then you can use the Fibonacci recursion to generate the next $8$ numbers. Then add them all up and see what you get. (Your answer will be in terms of $a$ and $b$.)

share|cite|improve this answer

answered Apr 1 at 14:16

paw88789paw88789

29.6k12351

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  For the lazy, the sum is 55a + 88b. ;)
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  – Mukul Gupta
  Apr 1 at 19:14
  

  
  
  
  

add a comment | 

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If you use a Binet-like formula for the Fibonacci sequence $mod 11$, you'll notice that $$F_nequiv 8cdot (4^n-(-3)^n)pmod11$$

And thus $$sum_k=0^9 F_n+kequiv8cdot 4^ncdot(-7)cdot(4^10-1)-8cdot3^ncdot8cdot((-3)^10-1)pmod11$$

By Fermat's theorem, $a^10equiv1pmod 11$ for all $anotequiv0pmod11$.

More generally, the sum of $p-1$ consecutive Fibonacci numbers is divisible by the prime $p$ as soon as the polynomial $x^2-x-1$ is reducible in $Bbb F_p[x]$ (and $1$ is not a root, which can never occur).

share|cite|improve this answer

edited Apr 2 at 2:06

answered Apr 1 at 14:46

Saucy O'PathSaucy O'Path

6,3621627

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  I can't help but read that as "for all values of 11 which do not divide a".
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  – hobbs
  Apr 1 at 20:41
  

  
  
  
  


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  @hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
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  – J.G.
  Apr 1 at 20:59
  
  

  
  
  
  


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  @J.G. ... or $anotequiv0pmod11$, for that matter.
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  – Saucy O'Path
  Apr 1 at 21:00
  

  
  
  
  


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  @SaucyO'Path It'd certainly be more consistent with the rest of your formalism. You may as well edit it in, then.
  $endgroup$
  – J.G.
  Apr 1 at 21:01
  

  
  
  
  

add a comment | 

16

$begingroup$

Suggestion:
If the first two numbers in the sequences are $a, b$, then you can use the Fibonacci recursion to generate the next $8$ numbers. Then add them all up and see what you get. (Your answer will be in terms of $a$ and $b$.)

share|cite|improve this answer

answered Apr 1 at 14:16

paw88789paw88789

29.6k12351

$endgroup$

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  $begingroup$
  For the lazy, the sum is 55a + 88b. ;)
  $endgroup$
  – Mukul Gupta
  Apr 1 at 19:14
  

  
  
  
  

add a comment | 

16

$begingroup$

Suggestion:
If the first two numbers in the sequences are $a, b$, then you can use the Fibonacci recursion to generate the next $8$ numbers. Then add them all up and see what you get. (Your answer will be in terms of $a$ and $b$.)

share|cite|improve this answer

answered Apr 1 at 14:16

paw88789paw88789

29.6k12351

$endgroup$

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  $begingroup$
  For the lazy, the sum is 55a + 88b. ;)
  $endgroup$
  – Mukul Gupta
  Apr 1 at 19:14
  

  
  
  
  

add a comment | 

$begingroup$

Suggestion:
If the first two numbers in the sequences are $a, b$, then you can use the Fibonacci recursion to generate the next $8$ numbers. Then add them all up and see what you get. (Your answer will be in terms of $a$ and $b$.)

share|cite|improve this answer

answered Apr 1 at 14:16

paw88789paw88789

29.6k12351

$endgroup$

share|cite|improve this answer

answered Apr 1 at 14:16

paw88789paw88789

29.6k12351

answered Apr 1 at 14:16

paw88789paw88789

29.6k12351

answered Apr 1 at 14:16

paw88789paw88789

29.6k12351

3

$begingroup$

If you use a Binet-like formula for the Fibonacci sequence $mod 11$, you'll notice that $$F_nequiv 8cdot (4^n-(-3)^n)pmod11$$

And thus $$sum_k=0^9 F_n+kequiv8cdot 4^ncdot(-7)cdot(4^10-1)-8cdot3^ncdot8cdot((-3)^10-1)pmod11$$

By Fermat's theorem, $a^10equiv1pmod 11$ for all $anotequiv0pmod11$.

More generally, the sum of $p-1$ consecutive Fibonacci numbers is divisible by the prime $p$ as soon as the polynomial $x^2-x-1$ is reducible in $Bbb F_p[x]$ (and $1$ is not a root, which can never occur).

share|cite|improve this answer

edited Apr 2 at 2:06

answered Apr 1 at 14:46

Saucy O'PathSaucy O'Path

6,3621627

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  I can't help but read that as "for all values of 11 which do not divide a".
  $endgroup$
  – hobbs
  Apr 1 at 20:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
  $endgroup$
  – J.G.
  Apr 1 at 20:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.G. ... or $anotequiv0pmod11$, for that matter.
  $endgroup$
  – Saucy O'Path
  Apr 1 at 21:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SaucyO'Path It'd certainly be more consistent with the rest of your formalism. You may as well edit it in, then.
  $endgroup$
  – J.G.
  Apr 1 at 21:01
  

  
  
  
  

add a comment | 

3

$begingroup$

If you use a Binet-like formula for the Fibonacci sequence $mod 11$, you'll notice that $$F_nequiv 8cdot (4^n-(-3)^n)pmod11$$

And thus $$sum_k=0^9 F_n+kequiv8cdot 4^ncdot(-7)cdot(4^10-1)-8cdot3^ncdot8cdot((-3)^10-1)pmod11$$

By Fermat's theorem, $a^10equiv1pmod 11$ for all $anotequiv0pmod11$.

More generally, the sum of $p-1$ consecutive Fibonacci numbers is divisible by the prime $p$ as soon as the polynomial $x^2-x-1$ is reducible in $Bbb F_p[x]$ (and $1$ is not a root, which can never occur).

share|cite|improve this answer

edited Apr 2 at 2:06

answered Apr 1 at 14:46

Saucy O'PathSaucy O'Path

6,3621627

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  I can't help but read that as "for all values of 11 which do not divide a".
  $endgroup$
  – hobbs
  Apr 1 at 20:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
  $endgroup$
  – J.G.
  Apr 1 at 20:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.G. ... or $anotequiv0pmod11$, for that matter.
  $endgroup$
  – Saucy O'Path
  Apr 1 at 21:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SaucyO'Path It'd certainly be more consistent with the rest of your formalism. You may as well edit it in, then.
  $endgroup$
  – J.G.
  Apr 1 at 21:01
  

  
  
  
  

add a comment | 

$begingroup$

If you use a Binet-like formula for the Fibonacci sequence $mod 11$, you'll notice that $$F_nequiv 8cdot (4^n-(-3)^n)pmod11$$

And thus $$sum_k=0^9 F_n+kequiv8cdot 4^ncdot(-7)cdot(4^10-1)-8cdot3^ncdot8cdot((-3)^10-1)pmod11$$

By Fermat's theorem, $a^10equiv1pmod 11$ for all $anotequiv0pmod11$.

More generally, the sum of $p-1$ consecutive Fibonacci numbers is divisible by the prime $p$ as soon as the polynomial $x^2-x-1$ is reducible in $Bbb F_p[x]$ (and $1$ is not a root, which can never occur).

share|cite|improve this answer

edited Apr 2 at 2:06

answered Apr 1 at 14:46

Saucy O'PathSaucy O'Path

6,3621627

$endgroup$

share|cite|improve this answer

edited Apr 2 at 2:06

answered Apr 1 at 14:46

Saucy O'PathSaucy O'Path

6,3621627

answered Apr 1 at 14:46

Saucy O'PathSaucy O'Path

6,3621627

answered Apr 1 at 14:46

Saucy O'PathSaucy O'Path

6,3621627