Question
Find all functions $f:\mathbb{R}\to \mathbb{R}$ such that $f(f(x)+yz)=x+f(y)f(z).$
My doubt:
In the hint of this they write show that $f(0)=0$ or $f(0)^2=2$. I was not able to see why this thing works after a lot of substitution tries.
Any hints ?
Set $x=y=z=0$, to get $f(f(0))=f^2(0)$.
Set $x=z=0$, to get $f(f(0))=f(y)f(0)$. Therefore, $f(y)f(0)=f^2(0)$, and we must have either $f(y)=f(0)$ or $f(0)=0$. The constant function $f(y)=f(0)=c$ can be easily shown not to hold since plugging in the original equation, we get $c=x+c^2, \forall x$ which cannot hold. Hence, we must have $f(0)=0$.
Now set $z=0$ to get $f(f(x))=x$.
Set $x=0$ to get $f(yz)=f(y)f(z)$.
These are both well-known functional equations and I will let you do the remaining work.
Clearly there exists such $c$ that $f(c)=0$ (just write $x=-f(y)f(z)$).
Let $z=c$ and we get $$f(f(x)+cy) = x$$ for all $x$, so $f$ is bijective function. Now let $x=0$ and we get from $(*)$ $$f(f(0)+cy) =0\underbrace{\implies}_{injectivety} f(0)+cy=c$$ which must be true for all $y$, so if we put $y=1$ we get $\boxed{f(0)=0}$. So if $x=0$ we get $$f(yz) = f(y)f(z)$$ for all $y,z$, so for $y=z=1$ we get $$f(1)= f(1)^2\implies f(1)=1$$ and if $y=z$ we have $$f(y^2)= f(y)^2\geq 0$$, so $f$ is positive on positve numbers. If we set $y=0$ in starting equation we get also $$f(f(x))=x$$ for all $x$. Finally set $x=f(t)$ and $z=1$ in starting eqiuation and we get $$f(t+y)= f(t)+f(y)$$ which means $f$ is a solution of Cauchy equation which is bounded on positive numbers, so $f(x)=x$.
