Increasing sequence of divisors of a quadratic trinomial This question is from a Russian contest, the 2011 Tuymaada Olympiad. It's the fourth question on day two for the problems at grade level 2.

Let $P(n)$ be a quadratic trinomial with integer coefficients. For each positive integer $n$, the number $P(n)$ has a proper divisor $d_{n}$, i.e., $1 &lt; d_{n} &lt; P(n)$, such that the sequence $d_{1},d_{2},d_{3},\ldots$ is increasing. Prove that either

$P(n)$ is the product of two linear polynomials with integer coefficients, or
all the values of $P(n)$, for positive integers $n$, are divisible by the same integer $m &gt; 1$.


Part (2) of the last sentence says that if $P(n)=an^2+bn+c$, where $a,b,c$ are integers such that $b^2-4ac$ is not a perfect square, then the sequence $d_{1},d_{2}, \ldots$ is increasing only if there is a positive integer $m&gt;1$ which divides all $P(n)$.
I tried to analyze two different cases: one when $b^2-4ac$ is negative
and one when $b^2-4ac$ is positive and not a perfect square,
but I couldn't go anywhere.
Any suggestion would be appreciated.
This is a hint really but long for a comment I wanted. Conditions {(i) or (ii)} are deny by {not (i) and not (ii)} i.e. $p(n)=an^2+bn+c$  is irreducible and for all m > 1 there is $n_m$ such that  p($n_m$) and m are coprimes.

NOTE.- When p is irreducible quadratic  (or of a greater degree) and it is not constantly multiple of a number (such as $x^2$ + x + 2 which is always even) it was conjectured by Bouniakowsky that p(n) is infinitely many times a prime (this is so far not proven and partial advances have been made by Sierpinski, Garrison, Forman  and probably  other people). The fact here is that just “not (i)” by itself (without “not (ii)”) in general produce some prime with which the condition 1 &lt; $d_n$ &lt; p(n) would be deny. The irreducible binomial $x^{12}$ +488669 is a known example  which is composite for 0 &lt; x &lt; 616980 giving a first prime at x = 616980 but for irreducible quadratic  often it is not hard to produce a prime.

It seems to be plausible try to find an answer without the NOTE but always denying the condition (i) and (ii) so get a proof by Reductio Ad Absurdum.
This is a hint really but long for a comment I wanted. Conditions {(i) or (ii)} are deny by {not (i) and not (ii)} i.e. $p(n)=an^2+bn+c$  is irreducible and for all m > 1 there is $n_m$ such that  p($n_m$) and m are coprimes.

NOTE.- When p is irreducible quadratic  (or of a greater degree) and it is not constantly multiple of a number (such as $x^2$ + x + 2 which is always even) it was conjectured by Bouniakowsky that p(n) is infinitely many times a prime (this is so far not proven and partial advances have been made by Sierpinski, Garrison, Forman  and probably  other people). The fact here is that just “not (i)” by itself (without “not (ii)”) in general produce some prime with which the condition 1 &lt; $d_n$ &lt; p(n) would be deny. The irreducible binomial $x^{12}$ +488669 is a known example  which is composite for 0 &lt; x &lt; 616980 giving a first prime at x = 616980 but for irreducible quadratic  often it is not hard to produce a prime.

It seems to be plausible try to find an answer without the NOTE but always denying the condition (i) and (ii) so get a proof by Reductio Ad Absurdum.