Let $x_{n}$ be a sequence such that $x_{0}=9,x_{1}=89$,and such $$x_{n+2}=10x_{n+1}-x_{n}\forall n\ge 0$$
Show that $$x_{n}\equiv 0\pmod {27}\Longleftrightarrow n\equiv 3\pmod 9$$
Is there a very simple way to prove that a sequence has a period of 9 in the sense of mod 27?
Note the corresponding characteristic equation is $$ \lambda^2-10\lambda+1=0 $$ which has two roots $r_1=5+2\sqrt6,r_2=5-2\sqrt6$. Therefore $x_n$ can be expressed as $$ x_n=C_1r_1^n+C_2r_2^n. $$ Using $x_0=9,x_1=89$, it is easy to see $C_1=\frac92+\frac{11}{\sqrt6},C_2=\frac92-\frac{11}{\sqrt6}.$ So \begin{eqnarray} x_n&=&\bigg(\frac92+\frac{11}{\sqrt6}\bigg)(5+2\sqrt6)^n+\bigg(\frac92-\frac{11}{\sqrt6}\bigg)(5-2\sqrt6)^n\\ &=&\frac92\bigg[(5+2\sqrt6)^n+(5-2\sqrt6)^n\bigg]+\frac{11}{\sqrt6}\bigg[(5+2\sqrt6)^n-(5-2\sqrt6)^n\bigg]\\ &=&9\sum_{2k\le n}\binom{n}{2k}5^{n-2k}6^k+22\sum_{2k-1\le n}\binom{n}{2k-1}5^{n-2k+1}6^{k-1}\\ &\equiv&9\cdot5^n+22\bigg[\binom{n}{1}5^{n-1}+\binom{n}{3}5^{n-3}6\bigg]\mod27\\ &\equiv&5^{n-3}\bigg[9\cdot5^3+22n\cdot5^2+22n(n-1)(n-2)\bigg]\mod27\\ &\equiv&-9+10n-5n(n-1)(n-2)\mod27 \end{eqnarray} Clearly if $x_n\equiv0\mod27$, then $3|n$ since $3|n(n-1)(n-2)$. Let $n=3k$. Then \begin{eqnarray} x_n&\equiv&-9+10n-5n(n-1)(n-2)\mod27\\ &\equiv&-9+30k-15k(3k-1)(3k-2)\mod27\\ &\equiv&3\bigg[-3-15k(9k^2-9k+4)\bigg]\mod27\\ &\equiv&3(-3-60k)\mod27\\ &=&-9(20k+1)\mod27\\ \end{eqnarray} Thus if $x_n\equiv0\mod27$, then $3|(20k+1)$ which implies that $k=3l+1$ or $$ n\equiv3\mod 9$$ Conversely if $ n\equiv3\mod 9$, it is not hard to show $x_n\equiv0\mod27$.