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

I was told to do this by proving $f$ is injective and surjective. I have proved it this way : setting $y=z=0 $, and then $f(f(x))=x+f^2(0)$. For any $b \in \Bbb{R}$, $x+f^2(0)=b$ has a solution ,then $f(f(x))=b$ has a solution and it follows that $f$ is surjective. For $f(x)=f(y)$, $f(f(x))=f(f(y))$, so $x+f^2(0)=y+f^2(0)$ , so $x=y$. That's $f$ is injective. But how to find $f$ , I have no idea.

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Substitute $(x,y,z)=(0,0,1),$ then $x=z=0$. We obtain $f(f(0))=f(0)f(1)$ and $f(f(0))=f(y)f(0)$. Hence for $f(0)\neq 0$, $f(y)=f(1)$ for all $y$. Let $f(1)=c$ since it's constant. Substituting this into our original equation shows $c=x+c^2$ which is obviously not true for all $x$. Hence $f(0)=0$.

Substituting $x=0$ we find $f(yz)=f(y)f(z)$ so that $f$ is multiplicative. Now substitute $y=z=0$ to show $f(f(x))=x$. Hence if we substitute $x\to f(x)$ we find $$f(x+yz)=f(x)+f(y)f(z)=f(x)+f(yz)$$ so that $f$ is both multiplicative and additive. Thus $f(x)=x$ or $f(x)=0$. Checking both of these shows that the only such function satisfiying the given equation is $f(x)=x$.

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