Find $f(7)$ given $f(x)f(y)=f(x+y)+f(x-y)$ and $f(1)=3$ 
Let $f:\Bbb R\to\Bbb R$ such that $f(1)=3$ and $f$ satisfies the functional equation
$$f(x)f(y) = f(x+y) + f(x-y)$$
Find the value of $f(7)$.

Attempt:
If $x=1$ and $y=0$, we find
$$f(1) f(0) = 2f(1) \implies f(0) = 2$$
If we fix $y=1$, we get the recurrence relation
$$\begin{cases} f(0) = 2 \\ f(1) = 3 \\ f(x + 1) - 3f(x) + f(x - 1) = 0 & \text{for } x\ge1 \end{cases}$$
From here I can solve for $f(x)$ or sequentially compute $f(2),f(3),f(4),\ldots$ to arrive at $f(7) = 843$.
Question:

Is there a more elegant way of finding $f(7)$ directly from the functional equation?


Well, is not too fancy, but you could do
$f(7)=f(4+3)=f(4)f(3)-f(1),$
and using the fact that
$f(4)=f(3)f(1)-f(2),$
you could compute $f(7)$ by knowing only $f(1),\;f(2),$ and $f(3)$.
