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Chukchi
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Fibonacci(i+j-1)

October 17th, 2007, 1:03 am

Prove that, if prime p divides Fibonacci(p-1), then p^2 divides Sum( (Fibonacci(i+j-1))^k, {i,1,p-1}, {j,1,p-1} ) for k = 1, 2, 3, ...
 
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Peniel
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Fibonacci(i+j-1)

October 26th, 2007, 3:30 pm

For k=1, after some calculations, I find that Sum=(Fibo(p) - 1)^2 + (Fibo(p+1) - 1)^2 and I'm stuck with that.Any hint Chukchi?
 
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Vassili
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Fibonacci(i+j-1)

October 27th, 2007, 2:36 pm

I think that for the case k=1 you need to show that, if prime p divides Fibonacci(p-1), the following holds true: if m!=n (mod p) then Fibonacci(m)!=Fibonacci(n) (mod p).
 
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Chukchi
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Fibonacci(i+j-1)

January 4th, 2008, 6:11 am

QuoteOriginally posted by: PenielAny hint Chukchi?Sure, Peniel. For k = 1 there is a nice formula found by Vladeta Javovic:Sum[ Sum[ Fibonacci[i+j-1], {i,1,n} ], {j,1,n} ] = Fibonacci(2*n+3) - 2*Fibonacci(n+3) + 2.Ref. http://www.research.att.com/~njas/sequences/A120297
 
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Peniel
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Fibonacci(i+j-1)

January 5th, 2008, 5:18 pm

Thanks Chukchi but...As I get So I was right and it cheers me up but...I am still stuck with that formula.Another hint?
 
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Chukchi
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Fibonacci(i+j-1)

January 5th, 2008, 7:25 pm

Peniel,May be the following fact would help a little:If prime p divides Fib(p-1) then p divides Fib(p)-1 too, and also if prime p>2 divides Fib(p)-1 then p divides Fib(p-1) too.
 
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Peniel
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Fibonacci(i+j-1)

January 6th, 2008, 4:49 pm

It definitely helps.Seems I'm more comfortable with big summations than with basic arithmetic.Anyway, it's time to tackle the case k>1.
 
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Chukchi
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Fibonacci(i+j-1)

July 24th, 2008, 1:13 am

Here is a hint for k = 2:If I'm not mistaken there is a nice formula for our Sum when k = 2.Something like:Sum( (Fibonacci(i+j-1))^2, {i,1,p-1}, {j,1,p-1} ) = L(2(p-1))*(F(p-1))^2,where L(i) is a Lucas Number L(i) = F(i-1) + F(i+1)