I was trying to solve one particular problem and to settle that problem I need to solve this problem, and it goes like this (sorry in advance for I still do not know LaTex symbolism so I will try to type it in the old-fashioned way):


If we have system of infinite number of equations that looks like this:


p(3)q(3) - 3p(2)q(2) + 2p(1)q(1)=0


p(4)q(4) - 3p(3)q(3) + 2p(2)q(2)=0


.
.
.


p(k+2)q(k+2) - 3p(k+1)q(k+1) + 2p(k)q(k)=0


.
.
.


,where p(k) and q(k) are natural numbers, not equal to 1 and not equal to 2, for every natural number k.


My question is as in title of the question: Does this system have a solution, or, in other words, do there exist two integer sequences p(k) and q(k), strictly greater than number 2 for every natural number k, such that they are solutions of this system?



Does this system of equations admits a solution?
The $p\_k$ and $q\_k$ always appear together so let us for the moment write $a\_k= p\_kq\_k$ (I will come back to the condition later).


Then your system can be rewritten as $a\_{i+2} = 3a\_{i+1} - 2a\_i$ for $i$ a positive integer. 
Now if you pick any two starting values $0 \lt a\_1 \lt a\_2$ you directly get from these equations a recursively definied sequence of positive integers $a\_i$. 


Now, you want in addition that $a\_i = p\_i q\_i$ with $p\_i,q\_i \ge 3$. To get this notice that if $d$ divides both $a\_{i+1}$ and $a\_i$ then it also divides $a\_{i+2}$. So pick your $a\_1$ and $a\_2$ such that the have a greatest common divisor that is the product of two positive integers at least $3$, say take $a\_1=9$ and $a\_2=18$. Then you can always factor $a\_i$ as $a\_i=p\_i q\_i$ with the condition you want. 


**Added** in view of comments by OP, based mainly on comments by Emil Jeřábek and Barry Cipra (answer now in CW mode):


If one wishes to further analyse the possible solutions, one can notice that the above mentioned recurrence can be solved explicitly and the solutions are of the form $u2^i + v$.


In particular, one can note that successive terms of the sequence $a\_i$ are relatively co-prime if and only if $v$ is odd and co-prime to $u$. 


By a result of Selfridge it is know that for $v=1$ and $u=78557$ the sequence $u2^i + v$ never takes a prime value, and indeed the terms are always divisible by one of $3,5,7,13,19,37,73$. 


Thus the terms of this sequence $78557 \cdot 2^n + 1$, with $i \ge 1$, always admits a decomposition as requested by OP, as a product of two numbers at least $3$.


I guess, but did not check, the proof of Selfridge's result is based on covering congruences, a method alluded to by ARupinski. That is one covers the integers by a finite set of congruence classes and then checks the different finitely many cases individually. 


