Factors of Fibonacci numbers

RMLabrador · 2020-10-09, 06:52

Greetings.
The factors of Fibonacci numbers:
For prime p:
$F(p)=\prod_{i=1}^{m} (k_{i}\cdot p\pm 1)$
For an any n
$n=p_{1}^{t1}\cdot p_{2}^{t2}\cdot...\cdot p_{k}^{tk}$

$F(n)=\prod_{i=1}^{m} (k_{i}\cdot p_{1}\pm 1)\cdot (l_{i}\cdot p_{2}\pm 1)\cdot ...\cdot(z_{i}\cdot p_{k}\pm 1)$

Example p=103
F(103)=1500520536206896083277=519121*5644193*512119709
(519121-1)/103=5040
(5644193+1)/103=54798
(512119709-1)/103=4972036

P.S.
Of course,m and coefficients k both are function of p or n)

RMLabrador · 2020-10-10, 08:31

Interesting. It is not so hard to write some simple code, like this (Pari GP)

forprime(n=11,900,x=Mod(factorint(fibonacci(n)),n);print(n,lift(x)))

and see, that posted above is true for all prime, at least for those, that in the scope of your patience))

Taking into account the level of your obsession with factorization 2 ^ n + 1 or so, my next post will be about it)))

Dr Sardonicus · 2020-10-11, 00:27

Quote:

Originally Posted by RMLabrador

Greetings.
The factors of Fibonacci numbers:
For prime p:
$F(p)=\prod_{i=1}^{m} (k_{i}\cdot p\pm 1)$
For an any n
$n=p_{1}^{t1}\cdot p_{2}^{t2}\cdot...\cdot p_{k}^{tk}$

I note that F₅ = 5 does not conform to your prescription.

The following is well known. It is an exercise for beginners in the subject of Fibonacci and Lucas numbers.

Let p be a prime number.

1) If (5/p) = +1 [p == 1 or 4 (mod 5)], then p divides F_p-1. Corollary: If (5/p) = +1, and p divides F_q with q prime, then q divides p-1, i.e. p == 1 (mod q).

2) If (5/p) = -1 [p == 2 or 3 (mod 5)], then p divides F_p+1. Corollary: If (5/p) = -1, and p divides F_q with q prime, then q divides p + 1, i.e. p == -1 (mod q).

RMLabrador · 2020-10-11, 16:30

i guess, "well known" is not true))
here, no mention about
http://www.maths.surrey.ac.uk/hosted....html#section2

and i can find 10+ links like this, and 0 link with similar formulae, like mine.
I just state the global form of factors of Fibonacci numbers, and not properties of F(n) itself.
Besides, 2,3,5 and 7 in my opinion are not prime.
They uber prime, they posses the rule of 1 and themselves division, and the broke the many other, i.e. 2 is even, 3 and 5 and 7 are too monadic) remember 6k+/-1? 6=2*3 and 6=5+1 and 6=7-1.
Forget this for a while.

RMLabrador · 2020-10-11, 16:58

Just run this in some other than Pari/GP system
x = 251;
c = Fibonacci[x];
(*Solve[(((a*x+c-1)/((a*x-1)*x))^(-1))-(1/b)==0&&(x*a-1)*(b*x-1)
==c,{a,b},Integers]*)
Solve[((a + b)^2 + 4*a*b*(c - 1) - z^2 == 0) && (x*a - 1)*(b*x - 1) -
c == 0 && b > a && a > 0 && z > 0, {a, b, z}, Integers]

2020-10-09, 06:52	#1
RMLabrador "Roman V. Makarchuk" Aug 2020 Ukraine 34₁₀ Posts	Factors of Fibonacci numbers Greetings. The factors of Fibonacci numbers: For prime p: $F(p)=\prod_{i=1}^{m} (k_{i}\cdot p\pm 1)$ For an any n $n=p_{1}^{t1}\cdot p_{2}^{t2}\cdot...\cdot p_{k}^{tk}$ $F(n)=\prod_{i=1}^{m} (k_{i}\cdot p_{1}\pm 1)\cdot (l_{i}\cdot p_{2}\pm 1)\cdot ...\cdot(z_{i}\cdot p_{k}\pm 1)$ Example p=103 F(103)=1500520536206896083277=5191215644193512119709 (519121-1)/103=5040 (5644193+1)/103=54798 (512119709-1)/103=4972036 P.S. Of course,m and coefficients k both are function of p or n)

2020-10-11, 16:58	#5
RMLabrador "Roman V. Makarchuk" Aug 2020 Ukraine 2×17 Posts	Just run this in some other than Pari/GP system x = 251; c = Fibonacci[x]; (Solve[(((ax+c-1)/((ax-1)x))^(-1))-(1/b)==0&&(xa-1)(bx-1) ==c,{a,b},Integers]) Solve[((a + b)^2 + 4ab(c - 1) - z^2 == 0) && (xa - 1)(bx - 1) - c == 0 && b > a && a > 0 && z > 0, {a, b, z}, Integers] Last fiddled with by RMLabrador on 2020-10-11 at 17:03 Reason: clear mind

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2020-10-10, 08:31	#2
RMLabrador "Roman V. Makarchuk" Aug 2020 Ukraine 42₈ Posts	Interesting. It is not so hard to write some simple code, like this (Pari GP) forprime(n=11,900,x=Mod(factorint(fibonacci(n)),n);print(n,lift(x))) and see, that posted above is true for all prime, at least for those, that in the scope of your patience)) Taking into account the level of your obsession with factorization 2 ^ n + 1 or so, my next post will be about it)))

2020-10-11, 16:30	#4
RMLabrador "Roman V. Makarchuk" Aug 2020 Ukraine 2×17 Posts	i guess, "well known" is not true)) here, no mention about http://www.maths.surrey.ac.uk/hosted....html#section2 and i can find 10+ links like this, and 0 link with similar formulae, like mine. I just state the global form of factors of Fibonacci numbers, and not properties of F(n) itself. Besides, 2,3,5 and 7 in my opinion are not prime. They uber prime, they posses the rule of 1 and themselves division, and the broke the many other, i.e. 2 is even, 3 and 5 and 7 are too monadic) remember 6k+/-1? 6=2*3 and 6=5+1 and 6=7-1. Forget this for a while.