1. Prove f is periodic

Let $$f:\mathbb{R} \to \mathbb{R}$$ such that:
$$\tfrac{{\sqrt 5 + 1}}{2}f\left( x \right) = f\left( {x + 1} \right) + f\left( {x - 1} \right)$$ (1)

a) Prove that f is periodic
b) Prove that there are infinitely many functions such that (1) is true

2. (a)

Edit major mistake in my working so ive removed it, sorry

(b) let f(x) be any function satisfying the condition. consider g(x)= kf(x)

3. This doesn't explain why such a function has to be periodic, but here is an example of one such function (and sure enough it is periodic, with period 10).

Let $f(x) = \sin(\pi x/5)$. Then $f(x\pm1) = \sin(\pi x/5 \pm \pi/5)$, and $f(x+1)+f(x-1) = 2\sin(\pi x/5)\cos(\pi/5) = \frac{\sqrt5+1}2f(x)$ since $\cos(\pi/5) = \frac{\sqrt5+1}4.$

4. Well, I didn't make all the calculus, but:

$x\leftarrow x+1 \Rightarrow \frac{\sqrt{5}+1}{2}f(x+1)=f(x+2)+f(x)$
$\frac{\sqrt{5}+1}{2}f(x+1)=\frac{\sqrt{5}+3}{2}f(x )-\frac{\sqrt{5}+1}{2}f(x-1)$
$\frac{\sqrt{5}+3}{2}f(x)-\frac{\sqrt{5}+1}{2}f(x-1)=f(x+2)+f(x) \Rightarrow f(x+2)=\frac{\sqrt{5}+1}{2}(f(x)-f(x-1))$

Now calculate f(x+3), f(x+4)... and find the period. I am pretty sure this is a way it can be solved.

5. Well i have tried many combinations but with no result. With certainty the period is 10 though. If anyone made it, please post a solution

6. In fact, veileen's method works nicely. Let $\tau = \frac{\sqrt5+1}2$. Then $\tau$ is the golden ratio, and satisfies the quadratic equation $\tau^2-\tau-1=0.$

The function f satisfies $f(x+2) = \tau f(x+1) - f(x)$. Therefore

\begin{aligned}f(x+3) &= \tau f(x+2) - f(x+1) \\ &= \tau \bigl(\tau f(x+1) - f(x)\bigr) - f(x+1) \\ &= (\tau^2-1)f(x+1) - \tau f(x),\end{aligned}

\begin{aligned}f(x+4) &= \tau f(x+3) - f(x+2) \\ &= \tau \bigl((\tau^2-1)f(x+1) - \tau f(x)\bigr) - \bigl(\tau f(x+1) - f(x)\bigr)) \\ &= (\tau^3-2\tau)f(x+1) - (\tau^2-1)f(x),\end{aligned}

\begin{aligned}f(x+5) &= \tau f(x+4) - f(x+3) \\ &= \tau \bigl((\tau^3-2\tau)f(x+1) - (\tau^2-1)f(x)\bigr) - \bigl((\tau^2-1)f(x+1) - \tau f(x)\bigr)) \\ &= (\tau^4-3\tau^2+1)f(x+1) - (\tau^3-2\tau)f(x).\end{aligned}

Now you can check (using the quadratic equation for $\tau$) that $\tau^4-3\tau^2+1 = 0$ and $\tau^3-2\tau = 1$. Thus $f(x+5) = -f(x)$ and so $f(x+10) = f(x).$

Edit. Looking again at veileen's comment, I notice that my solution would be much simpler if I had used the quadratic equation $\tau^2-\tau-1=0$ at each stage of the calculation, rather than saving it up until the end. In fact,

$f(x+3) = (\tau^2-1)f(x+1) - \tau f(x) = \tau\bigl(f(x+1)-f(x)\bigr)$ (because $\tau^2-1=\tau$),

\begin{aligned}f(x+4) = \tau f(x+3) - f(x+2) &= \tau^2\bigl(f(x+1)-f(x)\bigr) - \bigl(\tau f(x+1) - f(x)\bigr) \\ &= (\tau^2-\tau)f(x+1) - (\tau^2-1)f(x) = f(x+1) - \tau f(x),\end{aligned}

$f(x+5) = \tau f(x+4) - f(x+3) = \tau\bigl( f(x+1) - \tau f(x)\bigr) - \tau\bigl(f(x+1)-f(x)\bigr) = (\tau-\tau^2)f(x) = -f(x).$