Mersenne composite using fibonacci

Let p= 1 mod 4

If Mp, does not divide F(Mp-1), then Mp is composite.


For example.
2^13466917 divides F(2^13466917 -2)

Oops,
2^13466917 -1 divides F(2^13466917 -2)

And so may be prime!
sure enough it is.

What would take longer?

(a)The execution of the Lucas-Lehmer test on 2^13466917 -1
or,
(b)Dividing F(2^13466917 -2) by 2^13466917 -1, make sure it is not compostie
?

[svempasnake]

Originally Posted by TTn

What would take longer?

(a)The execution of the Lucas-Lehmer test on 2^13466917 -1
or,
(b)Dividing F(2^13466917 -2) by 2^13466917 -1, make sure it is not compostie
?

I hope (a), but I think (b). My fantasy fails when trying to figure out any way of dealing with that number. F(2^13466917 -2) just looks like an astronomically long number to me. :(

Well, Fibonacci numbers are smaller than the currently used Lehmer test numbers, but the algorithm lay undiscovered.

For example Lucas sequence 2,1, 3 ,4, 7 ,11,18,29...
L(2^n) =
3, then 3^2 -2 =7, then 7^2-2=47, and so on, just like Lehmer test but with a smaller starting number of three.



I found more !


Let p be a prime>7 satisfying the following conditions:
1. p= 2,4(mod 5)
2. 2^[p+1] -3, is also prime
Then (2^[p+1]-3) | F(2^p-1)

Let p be a prime>5 satisfying the following conditions:
1. p = 4 (mod5)
2. 2^[p+1]-1 is also prime
Then(2^[p+1]-1) | L(2^p-1)

[toferc]

Never mind.