Does this system of equations admits a solution? 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?
 A: 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. 
