Puzzle 789. Almost primes Fibonacci numbers.

(I was lead to that as a 'spin-off' of your puzzle 756). Now I shall prove that 2 is the only Fibonacci number of the form prime-1.

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Let F(n) stand for the nth Fibonacci number, i.e. : F(0) = 0, F(1) = 1 and F(n) = F(n-1)+F(n-2).

I shall prove that F(n)+1 and F(n)-1 factor in two integers (of which some can be +1 or -1 when n is small).

I shall use following abbreviations : w = Sqrt[5] and f = (1+w)/5.

As anybody knows, F(n) equals (f^n - (-1/f)^n)/w = (f^(2n) - e(n))/(w*(f^n)), where e(n) = (-1)^n.

Thus, F(n) + 1 = (f^(2n) + w*f^n - e(n))/(w*f^n).

The numerator of the right hand side of this equation is a quadratic function in f^n and can be written as (f^n - a(n))(f^n - b(n)) where

a(n) = (-w + Sqrt[5+4e(n)])/2 and b(n) = ( -w - Sqrt[5+4e(n)])/2.

Thus, F(n) + 1 = (f^n - a(n)) (f^n - b(n)) / (w*f^n).

If n = 2m is even, we have

a(n) =(-w+3)/2 = (1/f^2) ; b(n) = (-w-3)/2 = -f^2.

So, we can write :

F(2m) + 1 = [1/w]*[(f^(2m+2) - 1)/(f^(m+1))]*[(f^(2m-2) + 1)/(f^(m-1))].

At first sight this is a product of rational numbers. But it is easy to write every factor in the form a + b*f with integer a, b.

When such a factor stays unaltered when w is changed in -w, b will be zero. With that in mind and with some tricky manipulations we can get the following decompositions :

m = 2k => F(4k) + 1 is the product of (f^(4k+2) - 1)/(f^(2k+1)) with the integer (f^(4k-2)+1)/(w*f^(2k-1))

m = 2k+1 => F(4k+2) + 1 is the product of the integer (f^(4k+4) - 1)/(w*f^(2k+2)) with the integer (f^(4k)+1)/(f^(2k)).

If n = 2m+1 is odd, we have :

a(n) =(-w+1)/2 = -1/f ; b(n) = (-w-1)/2 = -f

and thus : F(2m+1) + 1 = [1/f]*[(f^(2m+2) + 1)/(f^(m+1))]*[(f^(2m) + 1)/(f^m)]

and we can get in the same way :

m = 2k => F(4k+1) + 1 is the product of the integer (f^(4k+2) + 1)/(w*f^(2k+1)) with the integer (f^(4k) + 1)/(f^(2k))

m = 2k+1 => F(4k+3) + 1 is the product of the integer (f^(4k+4) + 1)/(f^(2k+2)) with the integer (f^(4k+2) + 1)/(w*f^(2k+1))

For the decomposition of F(n) - 1 the work is analoguous.

Thus, F(n) - 1 = (f^(2n) - w*f^n - e(n))/(w*f^n) = (f^n - a(n))(f^n - b(n))/(w*f^n), where

a(n) = (w + Sqrt[5+4e(n)])/2 and b(n) = ( -w - Sqrt[5+4e(n)])/2.

If n = 2m is even, we have

a(n) =(w+3)/2 = (1/f^2) ; b(n) = (w-3) = -(1/f^2)

so we can write F(2m) - 1 as follows :

F(2m) - 1 = [1/w]*[(f^(2m-2) - 1)/(f^(m-1))]*[(f^(2m+2) + 1)/(f^(m+1))].

m = 2k \[Implies] F(4k) - 1 is the product of the integer (f^(4k+2) + 1)/(w*f^(2k+1)) with the integer (f^(4k-2) - 1)/(f^(2k-1))

m = 2k+1 \[Implies] F(4k+2) - 1 is the product of the integer (f^(4k+4) + 1)/(f^(2k+2)) with the integer (f^(4k) + 1)/(w*f^(2k))

If n = 2m+1 is odd, we have :

a(n) =(w+1)/2 = f ; b(n) = (w-1)/2 = 1/f

and thus : F(2m+1) - 1 = [1/f]*[(f^(2m) - 1)/(f^m)]*[(f^(2m+2) - 1)/(f^(m+1))]

and we can get in the same way :

m = 2k => F(4k+1) - 1 is the product of the integer (f^(4k) - 1)/(w*f^(2k)) with the integer (f^(4k+2) - 1)/(f^(2k+1))

m = 2k+1 => F(4k+3) - 1 is the product of the integer (f^(4k+2) - 1)/(f^(2k+1)) with the integer (f^(4k+4) - 1)/(w*f^(2k+2)).

With some supplementary manipulations we can write the factors in function of the Fibanacci number. This is what I finally found :

F(4n) + 1 = F(2n-1) ( F(2n) + F(2n+2) )

F(4n+1) +1 = F(2n+1) ( F(2n-1) + F(2n+1) )

F(4n+2) +1 = F(2n+2) ( F(2n-1) + F(2n+1) )

F(4n+3) +1 = F(2n+1) ( F(2n+1) + F(2n+3) )

F(4n) - 1 = F(2n+1) ( F(2n-2) + F(2n) )

F(4n+1) - 1 = F(2n) ( F(2n) + F(2n+2) )

F(4n+2) - 1 = F(2n) ( F(2n+1) + F(2n+3) )

F(4n+3) - 1 = F(2n+2) ( F(2n) + F(2n+2) )

Remains only to see for which n the factors are +1 or -1, which give the exceptional near-prime Fibonacci numbers.

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The last formulas can also be 'verified' with a computer.

(No doubt they must have been discovered already during the centuries in which mankind studied these remarkable Fibonacci numbers.

So, for secutity and to avoid being accused of plagiarism, I do not claim them).

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The question of puzzle 789 has been posed in :

http://math.stackexchange.com/questions/16877/fibonaccin-1-is-always-composite-for-n6-why

As can be seen there, the eight last relations I submitted are indeed known (Lucas numbers can be expressed as sums of Fibonacci numbers).

So, my feeling was right.