Let sequence $\{x_n\}$ satisfy the relation $x_{n+2}=x_{n+1}+2x_n$ for $n\ge 1$ where $x_1=1$ and $x_2=3$.
Let sequence $\{y_n\}$ satisfy the relation $y_{n+2}=2y_{n+1}+3y_n$ for $n\ge 1$ where $y_1=7$ and $y_2=17$.
Show that these two sequences do not share any common term.
We are going to show that both $\{x_n \pmod{8}\}$ and $\{y_n\pmod{8}\}$ are cyclic after certain terms.
For $\{x_n\}$, we have $x_2\equiv 3\pmod{8}$ and $x_3 = 3+2\times 1\equiv 5\pmod{5}$. For $n\ge 4$, we have $$\left\{ \begin{array}{rllll} x_{n+2} &= x_{n+1} + 2x_n&\equiv 5 + 2\times 3&\equiv 3&\pmod{8}\\ x_{n+3} &= x_{n+2} + 2x_{n+1}&\equiv 3 + 2\times 5&\equiv 5&\pmod{8} \end{array}\right.$$
Therefore, we conclude $\{x_n\pmod{8}\} =\{1, 3, 5, 3, 5, 3, 5,\cdots \}$.
Meanwhile, for $\{y_n\}$, we have $y_1 \equiv 7\pmod{8}$ and $y_2\equiv 1\pmod{8}$. When $n\ge 3$, we have $$\left\{ \begin{array}{rllll} y_{n+2} &= 2y_{n+1} + 3y_n&\equiv 2\times 7 + 3\times 1&\equiv 1&\pmod{8}\\ y_{n+3} &= 2y_{n+2} + 3y_{n+1}&\equiv 2\times 1 + 3\times 7&\equiv 7&\pmod{8} \end{array}\right.$$
Therefore, we conclude $\{y_n\pmod{8}\} =\{7, 1, 7, 1, 7, 1,\cdots \}$.
Clearly, $\{x_n\}$ and $\{y_n\}$ do not share some initial terms. Therefore no terms will be equal in these two sequences.