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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?

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P.S. Of course, the sequences p(k) and q(k) are not constant sequences. – Antisha Jan 21 '13 at 14:56
I removed the open-problem tag as this is rather only to be used for (well-)known open problems. – user9072 Jan 21 '13 at 15:06
up vote 8 down vote accepted

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.

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Since $x^2-3x+2=(x-1)(x-2)$, solutions of the recurrence are of the form $a_i=u2^i+v$ for constants $u,v$. – Emil Jeřábek Jan 21 '13 at 15:16
Thank you people, that was helpful. I must ask now does this system have a solution if we set a(1) to have the form: a(1)= 2^(c) + 2^(c-1) + ... + 2 + 1, for some constant c, c is a natural number strictly greater than 2 – Antisha Jan 21 '13 at 15:33
If you consider the description given by EJ and take the difference, you can note that any common divisor will divide $u 2^k$. So you choose $v$ odd and coprime to $u$ for successive terms always coprime. The question is know can you choose $u,v$ coprime (and $v$ odd) so that $u2^k + v$ is (provably) never prime. This is what I am not sure about from the top of my head (though I might overlook something simple). A strange way could be to take $v=1$ and $u=2^{17}$. Then this works if there are no other Fermat primes than the known ones. But perhaps there is a 'better' pair u,v. – user9072 Jan 21 '13 at 16:17
@Antisha: Ronald Graham's 1964 paper "A Fibonacci-like sequence of composite numbers" shows how to construct a linearly recurrent sequence all of whose entries are composite numbers using the Fibonacci recurrence. By modifying his approach it should be possible to find a covering sequence for this recurrence, (although there may be some complications I don't immediately see due to the fact that here the associated polynomial $x^2-3x+2$ factors while for the Fibonacci sequence the associated polynomial $x^2-x-1$ does not factor). – ARupinski Jan 21 '13 at 16:34
According to Section B21 in Richard Guy's Unsolved Problems in Number Theory, John Selfridge (in 1963) "discovered that one of $3,5,7,13,19,37,73$ always divides $78557\cdot2^n+1$." – Barry Cipra Jan 21 '13 at 17:53

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