EST-

Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ for which

$$
f(x+y)=f(x-y)+f(f(1-x y))
$$

holds for all real numbers $x$ and $y$.

Answer: $f(x) \equiv 0$.
Substituting $y=0$ gives $f(x)=f(x)+f(f(1))$, hence $f(f(1))=0$. Using this after substituting $x=0$ into the original equation gives $f(y)=f(-y)$ for all $y$, i.e., $f$ is even.

Substituting $x=1$ into the original equation gives $f(1+y)=f(1-y)+$ $f(f(1-y))$. By $f$ being even, also $f(-1-y)=f(1-y)+f(f(1-y))$. Hence $f(f(1-y))=f(1-y-2)-f(1-y)$. As $1-y$ covers all real values, one can conclude that

$$
f(f(z))=f(z-2)-f(z)
$$

for all real numbers $z$.

Substituting $-z$ for $z$ into (4) and simplifying the terms by using that $f$ is even, one obtains $f(f(z))=f(z+2)-f(z)$. Together with (4), this implies

$$
f(z+2)=f(z-2)
$$

for all real numbers $z$.

Now taking $y=2$ in the original equation followed by applying (5) leads to $f(f(1-2 x))=0$ for all real $x$. As $1-2 x$ covers all real values, one can conclude that

$$
f(f(z))=0
$$

for all real numbers $z$. Thus the original equation reduces to

$$
f(x+y)=f(x-y) .
$$

Taking $x=y$ here gives $f(2 x)=f(0)$, i.e., $f$ is constant, as $2 x$ covers all real numbers. As 0 must be among the values of $f$ by (6), $f(x) \equiv 0$ is the only possibility.

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