In the standard way, we compute the explicit formulas for each sequence: \begin{align*} a_n &= \frac{11 \cdot 3^n + 21(-1)^n}{12} \\ b_n &= \frac{9 \cdot 4^n + 16 (-1)^n }{20}\\ \end{align*}Thus it is enough to show that the equation \[ 55 \cdot 3^n + 105(-1)^n = 27 \cdot 4^m + 48(-1)^m \]has no solutions over the integers. Assume otherwise for the sake of contradiction. One can easily check that there is no $b_i=10=a_2$. Henceforth we assume $n > 2$. Then taken mod 27, we obtain \[ 105 \equiv \pm 48 \pmod{27} \]which is false.

I think you need to prove $$55\cdot 3^m +105(-1)^m=27\cdot 4^n+48(-1)^n$$has no solutions for $m,n$, because you have to prove that $a_i\neq b_j\forall i,j>1$, not that $a_n\neq b_n$ @below great solution

$a_n \ (\text{mod}\ 9)$ is $(1,10,5,4,5,4,5,4,...)$ $b_n\ (\text{mod}\ 9)$ is $(1,8,1,8,1,8,1,8,...)$ Now you have to observe that $n\ge 3\implies b_n> a_1$

Math-wiz wrote: I think you need to prove $$55\cdot 3^m +105(-1)^m=27\cdot 4^n+48(-1)^n$$has no solutions for $m,n$, because you have to prove that $a_i\neq b_j\forall i,j>1$, not that $a_n\neq b_n$ Yes this is true. The solution did this actually; it was just a typo in the equation. The actual proof is correct, and I've edited the equation statement. Thanks!

basically the same as 1973 USAMO Problem 2...

Generating fuction is perfect solution for this problem

Perfect-Square wrote: Generating fuction is perfect solution for this problem Can we see this perfect solution?

How i continue from here to find the form of the first one?$a_{n+1}+a_n=3(a_n+a_{n-1})$ and this gives $a_{n+1}+a_n=11\cdot 3^{n-1}$

We can just solve the characteristic equation : $a^{2}-2 a - 3 = 0$ ,and the roots $\alpha = 3,\beta =-1$ We know that ,$a_n = A\alpha ^{n} + B\beta ^{n}$ , Putting $a_1=1$ and $a_2=10$ ,we get $A=11/12 , B=7/4$ So we have $a_n = \frac{11 \cdot 3^n + 21(-1)^n}{12}$

any links to read about this method ?

@above! http://nms.lu.lv/wp-content/uploads/2016/04/21-linear-recurrences.pdf

$$mod3^k$$