Find all functions $f:\Bbb{Z}\to\Bbb{R}$ such that $f(m)\cdot f(n)=f(m-n)+f(m+n)$ 
Find all functions from the integers to the real numbers such that $$f(m)\cdot f(n)=f(m-n)+f(m+n)$$ and $f(1)=\frac{5}{2}$.

I think the answer should be $f(m)=2^m+2^{-m}$ by calculating small values, but I can’t prove that.
 A: First note that the function is even:
$$f(m)f(-n) = f(m + n) + f(m - n) = f(m)f(n)$$
This implies $f = 0$, which is not a solution, or $f(-n) = f(n), \forall n \in\mathbb{N}$.
Set $m = 0$ in the original equation. We get:
$$f(0)f(n) = f(-n) + f(n) = 2f(n) \implies f(0) = 2$$ 
By setting $m = n$ we get:
$$f(n)^2 = f(0) + f(2n) = 2 + f(2n)$$
From here we get $f(2n) = f(n)^2 - 2$.
By setting $m = n + 1$ we get:
$$f(n+1)f(n) = f(1) + f(2n+1) = \frac{5}{2} + f(2n+1)$$
From here we get $f(2n+1) = f(n+1)f(n) - \frac{5}{2}$.
Now notice that $f$ is defined recursively on $\mathbb{N}$: if we know $f(n)$ and $f(n + 1)$, we can calculate $f(2n)$ and $f(2n+1)$.
Using this definition we can show by induction that one of the solutions is $f(n) = 2^n + 2^{-n}, \forall n\in\mathbb{N}$ and using the fact that $f$ is even we get $f(n) = 2^n + 2^{-n}, \forall n\in\mathbb{Z}$.
A: First of all, we have that ($m=n=0$) $$f(0)^2=2f(0)\implies f(0)=0 \:\mbox{or}\: f(0)=2.$$
Assume $f(0)=0.$ Then $(m=0)$
$$f(n)+f(-n)=0$$ and $(m=n)$
$$f(n)^2=f(2n).$$ Thus
$$f(-2n)=f(-n)^2=f(n)^2=f(2n)\\f(-2n)=f(2n)$$ gives $f\equiv 0.$ This is not a valid solution since $f(1)=0\ne \frac52.$
Assume $f(0)=2.$ Then $(m=0)$ $$f(n)+f(-n)=2f(n)\implies f(-n)=f(n).$$ So we only need to know $f$ for positive $n.$ But with $m=k-1,n=1$ we have
$$f(k)=f(1)f(k-1)-f(k-2)=\dfrac52 f(k-1)-f(k-2).$$ This recurrence equation should be easy to solve. (And you'll get your asnwer confirmed.)
