Find all monic polynomials $f(x)$ in $\mathbb Z[x]$ such that $f(\mathbb Z)$ is closed under multiplication. By Mohsen Jamali
Problem
Source: Iran TST 2007, Day 2
Tags: algebra, polynomial, search, number theory proposed, number theory
harazi
07.05.2007 14:12
Nice problem, but observe that it is an immediate consequence of some problems already discussed on the forum: what are the polynomials whose image contains all numbers $2^{n}$, or $a^{n^{2}}$ and so on. Here, it is clear that once you take some $a\geq 2$ in the image of $f$, all powers of $a$ are in this image and thus you reduced the problem to the mentioned ones. I think however that this should have an absolutely trivial solution, since the constraints are really excessive here.
bilarev
07.05.2007 18:20
Harazi can you give some links to the problems you mentioned?
harazi
08.05.2007 09:36
Here you go for the one with $2^{n}$. http://www.mathlinks.ro/Forum/viewtopic.php?search_id=834109250&t=26076 For my problem with $a^{n^{2}}$ see the files of mathlinks contest.
astrean
04.01.2021 00:05
Clearly, the only constant polynomials satisfying this are $f(x) = 0$, $f(x) = -1$, and $f(x) = 1$. We assume from now on that $f(x)$ is nonconstant. Note that if $f(x)$ is a polynomial satisfying the requirements then $f(x+a)$ must also satisfy the requirements for any integer $a$. Therefore, we can assume WLOG that if $f$ has degree $k$, then the coefficient of the $x^{k-1}$ term for $f$ is $0$. It follows that for sufficiently large $|x|$, $(x-1)^k < f(x) < (x+1)^k$, so if $f(x)$ is a power of $k$ for sufficiently large $k$, $f(x) = x^k$. Now since $f$ is nonconstant, $f(\mathbb Z)$ is an infinite set, and since it is closed under multiplication, then for any $x \in f(\mathbb Z)$, we know that $x^k \in f(\mathbb Z)$ so $f(x) = x^k$ for infinitely many values of $x$. Therefore, $f(x) = x^k$ for all $x$, so it follows that all nonconstant solutions are of the form $f(x) = (x+a)^k$ for some $a$ and $k$. $\Box$
guptaamitu1
05.06.2022 01:29
The following is a classic (known result): If $f \in \mathbb Z[x]$ is non-constant such that $f(\mathbb Z) := \{f(x) : x \in \mathbb Z \}$ contains an infinite geometric progression, then $$ f(x) = (ax + b)^n $$for some $a,b,n \in \mathbb Z_{\ge 0}$ with $a,n \ge 1$ (converse is also true directly). In fact, it came as a problem in Iran TST 2014. We return to our main problem now. Only constant solutions are clearly $f(x) \equiv -1,0,1$. Assume $f$ is non-constant henceforth. As $f(\mathbb Z)$ is closed under multiplication, it is easy to see that it must contain a infinite geometric progression. Using our above result we get $$ f(x) \equiv (ax + b)^n $$for some $a,b,n \in \mathbb Z_{\ge 0}$ with $a,n \ge 1$. As $f$ is monic, hence $a=1$. Thus, $$ f(x) \equiv (x+b)^n $$and it is easy to see all such $f$ indeed work.