Hw to find the two second order linear recursive common items? Given two sequences 
$$a(n)=4a(n-1)-a(n-2), \quad a(1)=1，\; a(2)=5;$$
$$b(n)=10b(n-1)-b(n-2), \quad b(1)=1，\; b(2)=11.$$
If some $a(n)=b(m)$, $m,n\in\mathbb{N}$, exist?(Except a(1)=b(1) ) If there are no common items, how to prove it?
The Two solution equation follows from wolfram alpha.
Thanks a lot, I love you master!

Linear difference equations are much like differential equations
To solve
$$
a_n = 4 a_{n-1}-a_{n-2},\: a_1=1,\; a_2 = 5
$$
we assume that $a_n = \alpha^n$ and substituting results in
$$
(1-4\alpha+\alpha^2)\alpha^n = 0
$$
now solving
$$
1-4\alpha+\alpha^2 = 0
$$
gives
$$
a_n = c_1(-2-\sqrt5)^n+c_1(-2+\sqrt5)^n
$$
Analogously for 
$$
b_n = c_3(5-2\sqrt6)^2+c_4(5+\sqrt6)^n
$$
Considering the initial conditions we have
$$
a_n =-\frac{\left(1+\sqrt{3}\right) \left(2 \left(2-\sqrt{3}\right)^n+\sqrt{3} \left(2-\sqrt{3}\right)^n-\left(2+\sqrt{3}\right)^n\right)}{2
   \left(2+\sqrt{3}\right)} = \frac{1}{2} \left(1+\sqrt{3}\right) \left(\left(2+\sqrt{3}\right)^{n-1}-\left(2-\sqrt{3}\right)^n\right)
$$
and
$$
b_n = -\frac{\left(2+\sqrt{6}\right) \left(5 \left(5-2 \sqrt{6}\right)^n+2 \sqrt{6} \left(5-2 \sqrt{6}\right)^n-\left(5+2 \sqrt{6}\right)^n\right)}{4
   \left(5+2 \sqrt{6}\right)} = \frac{1}{4} \left(2+\sqrt{6}\right) \left(\left(5+2 \sqrt{6}\right)^{n-1}-\left(5-2 \sqrt{6}\right)^n\right)
$$
I hope this helps.
NOTE
$$
a_n = \eta_n + \gamma_n\sqrt 3\\
b_n =\xi_n+\mu_n\sqrt6
$$
