What are the possible functions satisfying functional equation $f(x) f(y) = f(x + y) + xy$?

Question

I was able to find that $f(0)=1$ by setting $y=0$ to have $f(x)f(0)=f(x)$. DIviding on both sides, I was left with $f(0)=1. I still don't know how I can find all the possible functions and how I can use this.

Answer 1

I'll assume you meant $f:\mathbb{R} \to \mathbb{R}$ $$P(x,y) \implies f(x)f(y)=f(x+y)+xy$$ $$P(x,0) \implies f(x)f(0)=f(x) \implies f(0)=1$$ Since $f(x)=0$ is not a solution, and we can replace $x$ by $a$ such that $f(a) \ne 0$, Now $$P(1,-1) \implies f(1)f(-1)=0$$ Case 1: $f(1)=0$ $$P(x,1) \implies f(x+1)+x=0 \iff f(x)=1-x$$ Case 2: $f(-1)=0$ $$P(x,-1) \implies f(x-1)-x=0 \iff f(x)=x+1$$