When is a sum of products of positive powers of 2 and 3 divisible by $2^b-3^n$?

Question

Here we have a really tough exercise.

Find all natural solution:

$$\frac{\sum\limits_{k=1}^n 2^{a_k} 3^{n-k}}{c}+3^n=2^{b} ,\quad b\geq a_n; \quad a_k, b, c ,n\in \mathbb N $$

Any ideas, hints?

Answer 1

$$2^a3^{n-1}+2^{2a}3^{n-2}+\cdots+2^{an}3^0=c(2^b-3^n)$$ $$3^n\left(\left(\frac{2^a}3\right)^1+\left(\frac{2^a}3\right)^2+\left(\frac{2^a}3\right)^3+\cdots+\left(\frac{2^a}3\right)^n\right)=c(2^b-3^n)$$ $$3^n\left(\frac{\frac{2^a}3((2^a/3)^n-1)}{2^a/3-1}\right)=c(2^b-3^n)$$ $$2^a\left(\frac{2^a-3^n}{2^a-3}\right)=c(2^b-3^n)$$ $$2^a(2^a-3^n)=c(2^a-3)(2^b-3^n)$$ I hope you can solve now.