67th Romanian Mathematical Olympiad

a) It is natural to look for solutions of the exponential type $f(x)=a^x$, where $a>0$. If this is a solution, we get $a+a^{-1}=\sqrt{5}$, and thus $a=\frac{\sqrt{5}+1}{2}$. It is easy to check that the functions $f_1:\mathbb{R}\to \mathbb{R}$, $f_1(x)=(\frac{\sqrt{5}-1}{2}{)}^x$ and $f_2:\mathbb{R}\to \mathbb{R}$, $f_2(x)=(\frac{\sqrt{5}+1}{2}{)}^x$ satisfy the required conditions.

b) Let $g$ be a function which satisfies the required condition. Then, $g(x+2)+g(x)=\sqrt{3}g(x+1)$, and thus $g(x+2)+g(x)=\sqrt{3}(\sqrt{3}g(x)-g(x-1))$, from where we obtain that $g(x+2)=2g(x)-\sqrt{3}g(x-1)$. Next $g(x+3)=2g(x+1)-\sqrt{3}g(x)=2(\sqrt{3}g(x)-g(x-1))-\sqrt{3}g(x)$, and we get $g(x+3)=\sqrt{3}g(x)-2g(x-1)$. Similarly $g(x+4)=\sqrt{3}g(x+1)-2g(x)=\sqrt{3}(\sqrt{3}g(x)-g(x-1))-2g(x)$, to get $g(x+4)=g(x)-\sqrt{3}g(x-1)$. Finally, $g(x+5)=g(x+1)-\sqrt{3}g(x)$, and thus $g(x+5)=-g(x-1)$, to obtain $g(x+6)=-g(x)$. To get the conclusion we have to notice that $g(x+12)=-g(x+6)=g(x)$.