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?

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.
