Existence of Natural Number n for f(x+n)=f(x) like cos+2pi

[Andrax]

Homework Statement

f(x+2)=f(x-1)f(x+5)
prove that there exists n natural number in which f(x+n)=f(x) like cos+2pi

Homework Equations

anyway i was doing some kind of solution i was pretty onfident that I am going to get the solution i was trying to make for example f(x+6)=f(x+3)f(x+9) we know< that f(x)=f(x-3)f(x+3) so f(x+3)=f(x)/f(x+3) [ignoring that the functoin is from R->R] was hoping that f(x+9)/f(x+3)=f(x+k) then i'll find this k suppose that f(x+6)=f(k)f(x) then f(x) = f(x+6)/f(k) then ill try to find the k from the first equatoin this got me in troubles cause it iddn't work iwas surprised that i didn't work..

The Attempt at a Solution

it seems like the only way anyway final thing I've reached is simplifying
f(x+n)=f(x+n+3)f(x+n-3) n is a natural number

 

[jashua]

As f(x)=f(x+3)f(x-3)=f(x+3)f(x)f(x-6), we have f(x+3)f(x-6)=1 (ignoring that f(x)=0). Then, f(x)f(x-9)=1. Similarly, we find that f(x)f(x+9)=1. As a result, f(x+9)=f(x-9), and so, f(x)=f(x+18). Thus, n=18.

 

[Andrax]

As f(x)=f(x+3)f(x-3)=f(x+3)f(x)f(x-6), we havef( x+3)f(x-6)=1 (ignoring that f(x)=0). Then, f(x)f(x-9)=1. Similarly, we find that f(x)f(x+9)=1. As a result, f(x+9)=f(x-9), and so, f(x)=f(x+18). Thus, n=18.

thanks i already got this one f( x+3)f(x-6)=1 but i didn't think it would help so i totally ignored it