if $f: \mathbb{Z} \rightarrow \mathbb{Z}$ then find all the possible solutions for $$f(m+n)+f(mn-1)=f(m)f(n)$$
These are all what i have, i really get stucked when there are no hints ,regarding the order of the functional values ,or any values at specific problem.
I tried setting some random values say ,but didnt get to a point,any hints will be appreciated.
Plugging in $m=0$ gives
$$f(n) + f(-1) = f(n)f(0) \Longrightarrow f(n)(1-f(0)) = -f(-1).$$
Letting $m=0=n$ we also have
$$f(0) + f(-1) = f(0)^2 \Longrightarrow f(-1) = f(0)^2 - f(0).$$
Plugging this into the first equation gives
$$f(n)(1-f(0)) = -(f(0)^2-f(0)) = f(0)-f(0)^2 = f(0)(1-f(0)).$$
There are two cases to consider: $f(0) \neq 1$ and $f(0) = 1$. The former is simple. The latter takes a bit more work.
If $f(0) = 1$, then taking $n=0=m$, we get
$$f(-1) = 0.$$
However taking $n=-1$ and $m=0$ we have $f(-1) = 0$ from the above. But then
$$f((-1)+(-1)) + f((-1)(-1) - 1) = f(-1)^2 \Longrightarrow f(-2) + f(0) = f(-1)^2 \Longrightarrow f(-2) = -1.$$
I'll let you continue to explore this to see if you can find any patterns that might emerge.
