Find all twice differentiable functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying

$$
f(x)^{2}-f(y)^{2}=f(x+y) f(x-y)
$$

for all real numbers $x$ and $y$.
The answer is $f(x)=k x, f(x)=a \sin (c y), f(x)=a \sinh (c y)$, where $a, c \in \mathbb{R}$. The given functional equation is

$$
f(x)^{2}-f(y)^{2}=f(x+y) f(x-y) .
$$

Observe that $x=y=0$ gives $f(0)=0$.

Since $f$ is smooth, we may differentiate with respect to $x$ and obtain

$$
2 f^{\prime}(x) f(x)=f^{\prime}(x+y) f(x-y)+f(x+y) f^{\prime}(x-y) .
$$

Now differentiate this with respect to $y$ to obtain

$$
\begin{aligned}
0 & =\left[f^{\prime \prime}(x+y) f(x-y)-f^{\prime}(x+y) f^{\prime}(x-y)\right] \\
& +\left[f^{\prime}(x+y) f^{\prime}(x-y)-f(x+y) f^{\prime \prime}(x-y)\right] \\
& =f^{\prime \prime}(x+y) f(x-y)-f(x+y) f^{\prime \prime}(x-y)
\end{aligned}
$$

From this we conclude the key relation: for any real numbers $X$ and $Y$ :

$$
f^{\prime \prime}(X) f(Y)=f(X) f^{\prime \prime}(Y) .
$$

Assume $f$ isn't identically zero. Then we deduce there's a constant $k$ such that

$$
f^{\prime \prime}(x)=k f(x)
$$

for all $x$.

This is a standard differential equation with cases on $k$.

- If $k=0$, the solution set is $f(x)=a x+b$. Then $f(0)=0 \Longrightarrow b=0$, and we can check $f(x)=a x$ works.
- If $k0$, the solution set is $f(x)=a \sinh (-k x)+b \cosh (-k x)$. Again $f(0)=$ $0 \Longrightarrow b=0$, and we can check $f(x)=a \sinh (k x)$ works.