Find all functions f:R->R such that for all x,y∈R
f(xy + 1) = f(x+y) + f(x)f(y)
(In reply to
Solution by FrankM)
From f(0) + f(0)^2 = 0, it is certainly possible that f(0) = 0. But f(0) = -1 also satisfies the relation.
Now how does the f(0) = -1 branch turn out?