A Popular Function, or Constant?

My intuitive solution:

Let $f^{\circ n}(x)$ stand for the function $f$ applied $n$ times, then this equation implies that $f^{\circ n+2}(x) = f^{\circ n+1}(x) + f^{\circ n}(x),$ which means that the sequence $f^{\circ n}(x)$ has the Fibonacci property for every real number $x$ . Since Fibonacci-type sequences can be generated as the linear combinations of two exponential functions with bases $\tfrac12(1\pm\sqrt 5)$ , and only positive bases result in continuous functions defined on all of $\mathbb R$ , I figured that $f$ simply involved multiplication by $\tfrac12(1+\sqrt 5)$ . Thus $f(2015) = \tfrac12(1+\sqrt 5)\cdot 2015 = \boxed{3260.3}.$

My boring solution:

If $f$ is polynomial of degree one or higher, then applying $f$ apparently does not increase the degree of the polynomial. Thus I choose the linear function $f(x) = ax + b$ . Now $f(f(x)) \equiv f(x) + x \\ a(ax + b)+b \equiv ax + b + x \\ a^2x + ab \equiv (a+1)x \\ a^2 = a+1\ \wedge\ ab = 0 \\ a = \tfrac12(1\pm\sqrt 5)\ \wedge b = 0,$ so that $f(x) = \tfrac12(1+\sqrt 5) x$ . (The negative value of $a$ cannot be made into a continuous function.)

this functional equation does not have one unique solution; in fact $\begin{cases} x=C_1(t)\left(\dfrac{1+\sqrt{5}}{2}\right)^t+C_2(t)\left(\dfrac{1-\sqrt{5}}{2}\right)^t \\ f=C_1(t)\left(\dfrac{1+\sqrt{5}}{2}\right)^{t+1}+C_2(t)\left(\dfrac{1-\sqrt{5}}{2}\right)^{t+1} \end{cases}$

where $C_1(t)$ and $C_2(t)$ are arbitrary periodic functions with unit period.

first, $deg(f(f(x)))=deg(f(x))^2$ obviously the degree is not changing but the polynomial is gaining an x.it cannot have degree 0. and only 1.so $f(x)=ax+b$ $f(f(x))=a^2x+b(a-1)$ $ax+b+x=a^2x+b(a-1)$ $\begin{cases} a^2=a+1\rightarrow a=\phi(the-negative-solution-neglected)\\ b=b(a-1)\rightarrow b=0(a-cannot-be-2) \end{cases}$ $f(x)=\phi x$ $f(2015)=\boxed{3260.34}(2 d.p)$

Let $f(x) = kx$ , then

$f(f(x)) = f(x)+x$

$f(kx) = kx+x$

$k^{2}x = x(k+1)$

$x(k^2-k-1) = 0$

$k^2-k-1 = 0$

$k = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$

Since $k > 0$ , $k=\frac{1 + \sqrt{5}}{2}$

$f(2015) = \frac{1 + \sqrt{5}}{2} \times 2015 = 3260.33$