TITLE: $f(550)$ if $f(11)=1$ and $f(a)f(b)=f(a+b)+f(a-b)$
QUESTION [9 upvotes]: $f:\Bbb Z\to\Bbb Z$ is such that $f(11)=1$ and $f(a)f(b)=f(a+b)+f(a-b)$ for all integers $a,b$. What is $f(550)$?

I've first set $f(11)f(0)=2f(11)$, which yields $f(0)=0,2$. Yet I know that 0 won't work because that was a hint. I've also heard that there was a pattern.
The answer is supposed to be $-1$, but I don't know how to get there.

REPLY [4 votes]: We start with $f(11n)f(11)=f(11(n+1))+f(11(n−1))$ for every integer $n$.
and since $f(11)=1$, 
$f(11n)=f(11(n+1))+f(11(n−1))$. Let $a_{n} = f(11n)$ and we have the following relation:
$a_{n+1} = a_{n} - a_{n-1}$ 
From this we get $a_{n+6} = a_{n}$. 
Based on this last relation, we can write $f(550)=a_{50} = a_{6.8+2} = a_{2} = f(22)$
But, $f(11)f(11) = f(22)+f(0)$ and then we get $f(22) = 1 - 2 = -1$. 
We conclude that $f(550) = -1$. As required!