# functional equations – Find all functions \$ f: Bbb {R} to Bbb {R} \$ such that for all \$ x, y, z in Bbb {R} \$, \$ f (f (x) + yz) = x + f (y) f (z) \$

Find all functions $$f: Bbb {R} to Bbb {R}$$ as for all $$x, y, z in Bbb {R}$$ , $$f (f (x) + yz) = x + f (y) f (z)$$

I've been told to do it by proving $$f$$ is injective and surjective. I have proved this way: $$y = z = 0$$, so what $$f (f (x)) = x + f ^ 2 (0)$$. For all $$b in Bbb {R}$$, $$x + f ^ 2 (0) = b$$ has a solution, then $$f (f (x)) = b$$ has a solution and follows $$f$$ I surjective. For $$f (x) = f (y)$$, $$f (f (x)) = f (f (y))$$, so $$x + f ^ 2 (0) = y + f ^ 2 (0)$$ , so $$x = y$$. C & # 39; $$f$$ is injective. But how to deduce $$f$$ I do not have the ideal for that.