Conjecture
Max Alekseyev
maxale at gmail.com
Sun Dec 9 01:08:30 CET 2007
There is a stronger statement:
If Fibonacci(n) is divisible by a prime p of the form 4k+3 then n is even.
To prove this statement it is enough to show that
(1+sqrt(5))/(1-sqrt(5)) is never a square modulo such p (which is a
straightforward exercise).
In particular, primes of the form 4k+3 never divide Fibonacci(q) for
prime q that proves your conjecture.
Max
On Dec 8, 2007 12:39 PM, Artur <grafix at csl.pl> wrote:
> Who is able to proove my conjecture
> "Prime factors of **A050937(n)*
> <http://www.research.att.com/%7Enjas/sequences/A050937> are product of
> *A134852 <http://www.research.att.com/%7Enjas/sequences/A134852>(n)* sum
> of squares"
> Exmple:
> *4181 = 37*113 = (1^2+6^2)(7^2+8^2)
> 1346269=557*2417 = (14^2+19^2)(4^2+49^2)
> etc.
>
> Another words all of prime factors have to be congruent 1 mod 4
>
> *ARTUR
>
>
> *
>
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