Here's a new math puzzle
 

If f(n) is a fabonacci number, p is a prime and p=1or(-1)(mod10)
Show that p|f(p-1)
If you find a correct way, it will be very easy :D

[quote]Show that p|f(p-1) [/quote]

that is...?

Luigi

[quote="ET_"][quote]Show that p|f(p-1) [/quote]

that is...?
[/quote]
p divides the (p-1)th Fibonacci number

There should be a list of math terms somewhere on here for those of us who don't use them alot. I never remember what in the world "mod" means or what it does.

That means p=10k+1 or 10k-1

When p is =+-3(mod 10)
p|f(p+1).....

Well done!
This is [size=18][color=indigo]Lemma 1[/color][/size]
and then…… :rolleyes: :arrow: :arrow: :arrow:

Generally!


A prime number must divide G((p+/-1)/2)
G=Fibonacci, or Lucas number.

Iff p = 1, or 4(mod 5), and p = 1(mod 4) then p must divide F((p-1)/2)
p = 29,41,61,89,101,109,149,181,229,241,269,281,

Iff p = 2, or 3(mod 5) and p= 1(mod 4) then p must divide F((p+1)/2)
p = 13,17,37,53,73,97,113,137,157,173,193,197,233,257,277,293,313,

Iff p = 2,or 3 (mod 5) and p = 3(mod 4) then p must divide L((p+1)/2)
p = 7,23,43,47,67,83,103,107,127,163,167,223,227,263,283,

Iff p = 1, or 4(mod 5), and p = 3(mod 4) then p must divide L((p-1)/2)
p = 11,19,31,59,71,79,131,139,151,179,199,211,239,251,311,

:shock: :surprised: :) :D :D OH!SO IT DOES!!!But how to [size=18]prove[/size]???
PS: I prove my puzzle like this:
As we know,f(n)=(a^n-b^n)/sqrt5,here a=(1+sqrt5)/2,b=(1-sqrt5)/2 are the roots of x^2-x-1=0.
And wpolly said that p|f(p+1) iff p=+-3(mod10)(extend by Binomial theorem,realise that p|[size=9]i[/size]C[size=9]p+1[/size] when i=2,3,...,p-1)(I hide it so it could be more difficult :) )
So we consider f(p+1)f(p-1),we can prove p|f(p+1)f(p-1) in the same way(p|[size=9]i[/size]C[size=9]p+1[/size] when p=1,2,...,p-1,p+1,...,2p-1).Hence p|f(p-1) when p=+-1(mod10)

Another puzzle about Fibonacci numbers(quite easy but interesting):
S=arccot(F(1))+arccot(F(3))+arccot(F(5))+arccot(F(7))+......
Then S=? :?
Even a monkey can get the answer(using a calculator :) ).But prove it is more interesting.

Let U[size=9]i[/size] be a Lucas Sequence with Discriminant d, and p is a prime.

Prove that p|U[size=9]p-(d|p)[/size]. ((d|p) is the Legendre Symbol)

This is just a generalization of the original puzzle.


[Edit:I've made a big mistake as hyh1048576 points it out.]