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.

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 $f(n) = 2^n + 2^{-n}, \forall n\in\mathbb{Z}$.
