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!
 A: 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
$$
a_n = \eta_n + \gamma_n\sqrt 3\\
b_n =\xi_n+\mu_n\sqrt6
$$
A: I do not think this is elementary. For your sequence called $a_n \; ,$ you are giving the $x$ values in the Pell type
$$  x^2 - 3 y^2 = -2.  $$
For your sequence called $b_n \; ,$ you are giving the $x$ values in the Pell type
$$  2x^2 - 3 z^2 = -1.  $$
Multiply and subtract, whenever there is a common $x$ value, 
$$  z^2 - 2 y^2 = -1.  $$
There are results, for example in Mordell's book, on pairs of Pell type equations. I will see if I can find anything. This question has some references (but no final resolution) A System of Simultaneous Pell Equations
Alright, SZALAY gives an algorithm for completely solving this sort of thing. 
A: I did not realize that this problem moves naturally into my area, ternary quadratic forms. The outcome is that the overlap you have already found is the only one. I found out this simple technique from reading Szalay. Oh, it all reduces to confirming that there are only trivial integer solutions to a quartic, I put the information and a request for confirmation at 
Quartic: using a Thue tool truly
Alright, we have
$$ x^2 - 3 y^2 = -2$$
and $$ 2x^2 - 3 z^2 = -1$$
Multiply these by $-1$ and $2$ to get
$$ -x^2 + 3 y^2 = 2 \; , $$
$$ 4 x^2 - 6 z^2 = -2 \; , $$
so
$$ 3x^2 + 3 y^2 - 6 z^2 = 0 \; ,  $$
finally
$$  x^2 + y^2 = 2 z^2. $$
Since we can see from the original Pell type equations that $z$ must be odd, meanwhile $\gcd(x,y,z) = 1.$
Here is the part related to Pythagorean triples, but adjusted for $x^2 + y^2 = 2 z^2$: all (primitive) triples can be formed by
$$  x = r^2 + 2rs - s^2 \; , $$
$$  y = r^2 - 2rs - s^2 \; , $$
$$  z = r^2 + s^2.  $$
Actually, for any of the three variables you can choose $\pm \; .$  See Parametric characterization for $x^2 + y^2 = 2z^2$
Plugging back in, either to $x^2 - 3 y^2 = -2$ or $2 x^2 - 3 z^2 = -1,$ we get the (integer) equation
$$ r^4 - 8 r^3 s + 2 r^2 s^2 + 8 r s^3 + s^4 = 1.  $$
This is called a Thue equation. I ran it in gp-Pari, the only solutions are
$$ (1,0) \; ; \; (-1,0) \; ; \; (0,1) \; ; \; (0,-1)  $$
so that the $x,y,z$ triples can only be made up of $\pm 1 \; .$
I also ran it in the online MAGMA calculator. The syntax in June 2018 has updated compared with published articles.

