
> 
> 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. $1 < d\_{n} < P(n)$), such that the sequence $d\_{1},d\_{2}, d\_{3},\ldots$ is increasing. Prove that either:  
> 
>  i) $P(n)$ is the product of two linear polynomials with integer coefficients, or  
> 
>  ii) all the values of $P(n)$, for positive integers $n$, are divisible by the same integer $m>1$.
> 
> 
> 


This Russian problem was posted some time ago at [math.SE](https://math.stackexchange.com/questions/1274380/increasing-sequence-of-divisors-of-a-quadratic-trinomial). 
A bounty was placed on it, but no solution was received.
Any suggestion is appreciated. Thanks in advance.



Divisors of a quadratic trinomial
Assume the contrary. 


At first, $P(n+2)-3P(n+1)+3P(n)-P(n-1)=0$, thus if the same number $m>1$ divides three consecutive values of $P$, it divides all values of $P$ and case ii) takes place.


Denote $P(n)=d\_na\_n$, $\ell(n)=P(n+1)-P(n)$ is a linear function. If $\ell(x)$ divides $P(x)$ as a polynomial, case i) takes place, if not, then $a(x)P(x)+b(x)\ell(x)=C\ne 0$ for some $a(x),b(x)\in \mathbb{Z}[x]$, hence ${\rm gcd}\,(P(n),\ell(n))\leq C$. Thus equality $d\_{n+1}=d\_n$ implies $d\_n\leq C$, so, this may hold only for finitely many $n$ (otherwise $d\_n=m$ for large enough $m$ and case ii) takes place.)


Now assume that $d\_n\geq 2\ell(20n)$ (and $n$ is large enough). Then $a\_{k+1}\leq P(k+1)/d\_k=a\_k+\ell(k)/d\_k<a\_k+1$ for $k=n,n+1,\dots,20n$. Thus either $a\_k$ takes the same value three times in a row, see above why it implies case ii), or $a\_{k+2}\leq a\_k-1$ for $k=n,\dots,20n$. That is, $0<a\_{20n}<a\_n-9n$, i.e. $a\_n>9n$, $P(n)=a\_nd\_n>9n\cdot 2\ell(20 n)$, this is false for large enough $n$. 


So, $d\_n=O(n)$ (and of course $d\_n=\Omega(n)$ as $d\_n$ strictly increases from some place; $a\_n=P(n)/d\_n$ tends to infinity, thus $a\_{n+1}\ne a\_n$ for large $n$ (if $a\_n=a\_{n+1}=m$, $m$ divides above constant $C$.) For some $A>0$ there is a set $U$ of positive integers with positive upper density so that $d\_{n+1}=d\_n+A$ for $n\in U$. We have $a\_{n+1}-a\_n=P(n+1)/(d\_n+A)-P(n)/d\_n=O(1)$. It means that $a\_{n+1}-a\_n=A$ for some constant $A$ for all $n$ from some smaller set $
\tilde{U}\subset U$ of positive density.


Well, this is quadratic equation for $d\_n$ if we fix $n$:
$\ell(n)=Ad\_n+DP(n)/d\_n+AD$, $Ad\_n^2-(\ell(n)-AD)d\_n+DP(n)=0$. Its discriminant $(\ell(n)-AD)^2-4ADP(n)$ is (at most) quadratic trinomial in $n$. Three cases:


1) discriminant is a square of a linear function. Then $d\_n$ is linear in $n$, and since $P(n)$ is divisible by $d\_n$ for infinitely many $n$ we see that $P$ factorizes into two linear factors.


2) discriminant is a non-square quadratic trinomial. It may be a perfect square, but quite rarely. That is, for fixed $A,D$ only a null density of $n$ are ok (by Pell's type equation and blah-blah.)


3) discriminant is a linear function in $n$. The same as 2), but without Pell type argument.


