Peter wrote: Let the sequence $\{K_{n}\}_{n \ge 1}$ be defined by \[K_{1}=2, K_{2}=8, K_{n+2}=3K_{n+1}-K_{n}+5(-1)^{n}.\] Prove that if $K_{n}$ is prime, then $n$ must be a power of $3$. We have that $K_{n+2}=3K_{n+1}-K_{n}+5(-1)^{n}$ then $K_{n+3}=3K_{n+2}-K_{n+1}+5(-1)^{n+1}$. Summing up those two equations gives $K_{n+3}=2K_{n+2}+2K_{n+1}-K_{n}+5((-1)^{n}+(-1)^{n+1})$ so \[K_{n+3}=2K_{n+2}+2K_{n+1}-K_{n}.\] The roots of the cubic equation ${\lambda}^{3}=2{\lambda}^{2}+2{\lambda}-1$ are \[{\lambda}_{1}=-1,{\lambda}_{2}=\frac{3+\sqrt{5}}{2}=(\frac{1+\sqrt{5}}{2})^{2},{\lambda}_{3}=\frac{3-\sqrt{5}}{2}=(\frac{1-\sqrt{5}}{2})^{2}\] then $K_{n}=c_{1}(-1)^{n}+c_{2}(\frac{1+\sqrt{5}}{2})^{2n}+c_{3}(\frac{1-\sqrt{5}}{2})^{2n}$ for all $n\in{N}$ and constants $c_{1},c_{2}$ and $c_{3}$. Considering that $K_{1}=2,K_{2}=8$ and $K_{3}=3K_{2}-K_{1}-5=17$ we get that $c_{1}=c_{2}=c_{3}=1$ and \[K_{n}=(-1)^{n}+(\frac{1+\sqrt{5}}{2})^{2n}+(\frac{1-\sqrt{5}}{2})^{2n}=\frac{(\frac{1+\sqrt5}{2})^{3n}-(\frac{1-\sqrt5}{2})^{3n}}{(\frac{1+\sqrt5}{2})^{n}-(\frac{1-\sqrt5}{2})^{n}}=\frac{F_{3n}}{F_{n}}\] where $F_{m}=\frac{1}{\sqrt{5}}((\frac{1+\sqrt5}{2})^{m}-(\frac{1-\sqrt5}{2})^{m})$ is the $mth$ term of Fiboncci's sequence. Now let us suppose that $K_{n}$ is a prime number and $n=3^{k}m$, where $\gcd(m;3)=1$ and $k\in{N_{0}}$. Then $F_{3n}=F_{3^{k+1}m}\vdots F_{3^{k+1}}$ and $\gcd(F_{n};F_{3^{k+1}})=\gcd(F_{3^{k}m};F_{3^{k+1}})=F_{3^{k}}$, therefore \[K_{n}=\frac{F_{3n}}{F_{n}}\vdots\frac{F_{3^{k+1}}}{F_{3^{k}}}>1.\] But as we supposed $K_{n}$ is a prime number so $K_{n}=\frac{F_{3n}}{F_{n}}=\frac{F_{3^{k+1}}}{F_{3^{k}}}$. The well-known property of Fibonacci's numbers states that $F_{a+b}=F_{a}F_{b-1}+F_{a+1}F_{b}$. Consequently $F_{3n}=F_{2n}F_{n-1}+F_{2n+1}F_{n}$,$F_{2n}=F_{n}(F_{n-1}+F_{n+1})$ and $F_{2n+1}=F_{n+1}^{2}+F_{n}^{2}$ $\Rightarrow$ $F_{3n}=F_{n}(F_{n+1}^{2}+F_{n}^{2}+F_{n-1}^{2}+F_{n-1}F_{n+1})$. Now it is clear that if $a\geq b$ then $\frac{F_{3a}}{F_{b}}\geq \frac{F_{3b}}{F_{b}}$ and equality holds if and only if $a=b$. We proved above that if $K_{n}$ is a prime number, then $K_{n}=\frac{F_{3n}}{F_{n}}=\frac{F_{3^{k+1}}}{F_{3^{k}}}$ so $n=3^{k}$, as the problem claims.