34.13. Find all continuous solutions of the functional equation

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

(Lobachevsky).

## 34.4. Functional equations for differentiable functions
34.13. Let $f(0)=a$ and $f(1)=b$. By setting $x=y=t / 2$ in the original functional equation, we get $f(t / 2)=\sqrt{a f(t)}$, so $f(1 / 2)=\sqrt{a b}=a(a / b)^{1 / 2}=a c^{1 / 2}$, where $c=b / a, f(1 / 4)=\sqrt{a f(1 / 2)}=a c^{1 / 4}$, and generally, $f\left(1 / 2^{n}\right)=a c^{1 / 2^{n}}$ (if $a=0$, we still get $f\left(1 / 2^{n}\right)=0$, since in this case we can avoid dividing by $a$).

Now, let's set $x=\frac{1}{2^{n-1}}$ and $y=\frac{1}{2^{n}}$ in the original functional equation. We then get

$$
f\left(\frac{3}{2^{n}}\right) f\left(\frac{1}{2^{n}}\right)=\left(f\left(\frac{1}{2^{n-1}}\right)\right)^{2}
$$

i.e.,

$$
f\left(\frac{3}{2^{n}}\right)=\left(a c^{1 / 2^{n-1}}\right)^{2}\left(a c^{1 / 2^{n}}\right)^{-1}=a c^{3 / 2^{n}}
$$

It is also clear that if $f\left(\frac{m-1}{2^{n}}\right)=a c^{(m-1) / 2^{n}}$ and $f\left(\frac{m}{2^{n}}\right)=a c^{m / 2^{n}}$, then by setting $x=\frac{m}{2^{n}}$ and $y=\frac{1}{2^{n}}$, we get from the equality

$$
f\left(\frac{m+1}{2^{n}}\right) f\left(\frac{m-1}{2^{n}}\right)=\left(f\left(\frac{m}{2^{n}}\right)\right)^{2}
$$

that $f\left(\frac{m+1}{2^{n}}\right)=a c^{(m+1) / 2^{n}}$. Therefore, $f\left(\frac{m}{2^{n}}\right)=a c^{m / 2^{n}}$ for all natural numbers $m$. Hence, by continuity, $f(x)=a c^{x}$ for all positive $x$.

By setting $x=0$ in the original equation, we get

$$
f(y) f(-y)=(f(0))^{2}=a^{2}
$$

which means, $f(-y)=\frac{a^{2}}{f(y)}=\frac{a^{2}}{a c^{y}}=a c^{-y}$.

It is easy to verify that the function $f(x)=a c^{x}$ satisfies the given functional equation.