## Task 5 - 311245

Let $a$ be any real number with $a \geq 2$. Determine all functions that satisfy the following conditions (1), (2), (3) for $a$:

(1) The function $f$ is defined for all non-negative integers $x$; all function values $f(x)$ are real numbers.

(2) For all non-negative integers $x, y$ with $x \geq y$, the following holds:

$$
f(x) \cdot f(y) = f(x+y) + f(x-y)
$$

(3) It holds that $f(1) = a$.

Note: $f$ should be given as an elementary function in closed form, i.e.:

The formulaic specification of the function values $f(x)$ should be achieved by applying arithmetic operations (addition, subtraction, multiplication, division) to $x$ and constants, power functions, exponential functions, trigonometric functions of $x$, or inverse functions of such functions, and this should be done in a number of application steps that is independent of $x$.
Setting $y=0$, one obtains:

$$
f(x) \cdot f(0)=2 f(x)
$$

so that $f(0)=2$. Now, setting $y=1$, one obtains a recursive definition:

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

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

with $f(0)=2$ and $f(1)=a$. This is a linear recursion, so $f(x)$ can be represented as a sum of all functions $f(x)=c^{x}$ that satisfy the recursion:

$$
\begin{gathered}
c^{x+1}=a c^{x}-c^{x-1} \\
c \cdot c^{x}=a c^{x}-\frac{1}{c} c^{x} \\
c=a-\frac{1}{c} \\
c^{2}-a c+1=0 \\
c_{1,2}=\frac{a}{2} \pm \sqrt{\frac{a^{2}}{4}-1}
\end{gathered}
$$

The function is thus

$$
f(x)=k_{1}\left(\frac{a}{2}+\sqrt{\frac{a^{2}}{4}-1}\right)^{x}+k_{2}\left(\frac{a}{2}-\sqrt{\frac{a^{2}}{4}-1}\right)^{x}
$$

For $x=0$, it must hold that:

$$
k_{1}+k_{2}=2
$$

and for $x=1$, it holds that:

$$
\begin{gathered}
k_{1}\left(\frac{a}{2}+\sqrt{\frac{a^{2}}{4}-1}\right)+k_{2}\left(\frac{a}{2}-\sqrt{\frac{a^{2}}{4}-1}\right)=a \\
\left(k_{1}+k_{2}\right) \frac{a}{2}+\left(k_{1}-k_{2}\right) \sqrt{\frac{a^{2}}{4}-1}=a
\end{gathered}
$$

Substituting $k_{1}+k_{2}=2$ into the last equation, one can directly conclude that $k_{1}=k_{2}=1$. Therefore, the function is:

$$
f(x)=\left(\frac{a}{2}+\sqrt{\frac{a^{2}}{4}-1}\right)^{x}+\left(\frac{a}{2}-\sqrt{\frac{a^{2}}{4}-1}\right)^{x}
$$

Note: Since in the quadratic equation, by Vieta's formulas, $c_{1}=\frac{1}{c_{2}}$ holds, the function can also be written as:

$$
f(x)=\left(\frac{a}{2}+\sqrt{\frac{a^{2}}{4}-1}\right)^{x}+\left(\frac{a}{2}+\sqrt{\frac{a^{2}}{4}-1}\right)^{-x}
$$

This can be expressed using the hyperbolic cosine function:

$$
\cosh x=\frac{e^{x}+e^{-x}}{2}
$$

In the present case, this yields:

$$
f(x)=2 \cosh \left(x \cdot \operatorname{arcosh} \frac{a}{2}\right)
$$