20201009, 06:52  #1 
"Roman V. Makarchuk"
Aug 2020
Ukraine
42_{8} Posts 
Factors of Fibonacci numbers
Greetings.
The factors of Fibonacci numbers: For prime p: For an any n Example p=103 F(103)=1500520536206896083277=519121*5644193*512119709 (5191211)/103=5040 (5644193+1)/103=54798 (5121197091)/103=4972036 P.S. Of course,m and coefficients k both are function of p or n) 
20201010, 08:31  #2 
"Roman V. Makarchuk"
Aug 2020
Ukraine
22_{16} 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))) 
20201011, 00:27  #3  
Feb 2017
Nowhere
10114_{8} Posts 
Quote:
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_{p1}. Corollary: If (5/p) = +1, and p divides F_{q} with q prime, then q divides p1, 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). 

20201011, 16:30  #4 
"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=71. Forget this for a while. 
20201011, 16:58  #5 
"Roman V. Makarchuk"
Aug 2020
Ukraine
22_{16} Posts 
Just run this in some other than Pari/GP system
x = 251; c = Fibonacci[x]; (*Solve[(((a*x+c1)/((a*x1)*x))^(1))(1/b)==0&&(x*a1)*(b*x1) ==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] Last fiddled with by RMLabrador on 20201011 at 17:03 Reason: clear mind 
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