Prove no two numbers besides 1 in the sequences are identical Question:
Given that   {$a_n$} and   {$b_n$} are two sequences of integers defined by
$$a_1=1,a_2=10,a_{n+1}=2a_n+3a_{n-1}$$ 
$$b_1=1,b_2=8,b_{n+1}=3b_n+4b_{n-1}$$ 
for $n=2,3,4,...$
Prove that, besides the number '1' , no two numbers in the sequences are identical. 
What have I done so far
I have tried prove by contradiction, parity and the usual way to solve recursive problem but failed, does anyone can tell me some hints on how to approach this problem?
 A: John Omielan's answer suggests looking modulo $9$.
Without solving explicitly for $a_n$ and $b_n$, 
can you show that $a_{2n-1}\equiv5$, $a_{2n}\equiv4$, $b_{2n-1}\equiv1,$ and $b_{2n}\equiv8\pmod 9,$ for $n\ge2$?
A: Using the method of characteristic equation and roots, given in Linear difference equation, 
gives for the first sequence the characteristic equation of
$$\begin{equation}\begin{aligned}
\lambda^{2} & = 2\lambda + 3 \\
\lambda^{2} - 2\lambda - 3 & = 0 \\
(\lambda - 3)(\lambda + 1) & = 0
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Thus, $\lambda = 3, -1$, giving the equation of
$$a_t = c_1(3)^{t} + c_2(-1)^{t} \tag{2}\label{eq2A}$$
Using $a_1 = 1$ and $a_2 = 10$ gives
$$1 = 3c_1 - c_2 \tag{3}\label{eq3A}$$
$$10 = 9c_1 + c_2 \tag{4}\label{eq4A}$$
Adding these $2$ equations gives $11 = 12c_1 \implies c_1 = \frac{11}{12}$. Using this in \eqref{eq3A} gives $c_2 = 3c_1 - 1 = \frac{21}{12}$. Thus, you have for all $n \ge 1$ that
$$a_n = \frac{11(3^{n}) + 21(-1)^{n}}{12} \tag{5}\label{eq5A}$$
For the second sequence, using a similar procedure (I'll leave this up to you to do), you will get for all $n \ge 1$ that
$$b_n = \frac{9(4^{n}) + 16(-1)^{n}}{20} \tag{6}\label{eq6A}$$
Since both sequences are strictly increasing, no $2$ values among the same sequence can be equal to each other. Instead, for some $i \gt 1$ and $j \gt 1$, assume $a_i = b_j$. You then get
$$\begin{equation}\begin{aligned}
\frac{11(3^{i}) + 21(-1)^{i}}{12} & = \frac{9(4^{j}) + 16(-1)^{j}}{20} \\
\frac{11(3^{i-1}) + 7(-1)^{i}}{4} & = \frac{9(4^{j-1}) + 4(-1)^{j}}{5} \\
5(11(3^{i-1}) + 7(-1)^{i}) & = 4(9(4^{j-1}) + 4(-1)^{j})
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
Since $a_2 = 10$ and all $b_j \gt 10$ for $j \ge 2$, then consider $i \gt 2$. Then $9 \mid 3^{i-1}$, giving that \eqref{eq7A} becomes, modulo $9$,
$$\begin{equation}\begin{aligned}
5(11(3^{i-1}) + 7(-1)^{i}) & \equiv 4(9(4^{j-1}) + 4(-1)^{j}) \pmod 9 \\
35(-1)^{i} & \equiv 16(-1)^{j} \pmod 9 \\
(-1)(-1)^{i} & \equiv (-2)(-1)^{j} \pmod 9 \\
(-1)^{i+1} & \equiv (-2)(-1)^{j} \pmod 9
\end{aligned}\end{equation}\tag{8}\label{eq8A}$$
The LHS is either $-1$ or $1$ while the RHS is either $-2$ or $2$, so they are never equal to each other. This shows there are never $2$ values, apart from $a_1 = b_1 = 1$, of the sequences which are equal to each other.
