find all polynomials with integer coefficients that $P(\mathbb{Z})= ${$p(a):a\in \mathbb{Z}$} has a Geometric progression.
Problem
Source: iran tst 2014 first exam
Tags: algebra, polynomial, number theory, algebra unsolved
17.04.2014 13:11
sahadian wrote: find all polynomials with integer coefficients that $P(\mathbb{Z})= ${$p(a):a\in \mathbb{Z}$} has a Geometric progression. i think this problem is number theory problem ! am i wrong ?
17.04.2014 18:51
sahadian wrote: find all polynomials with integer coefficients that $P(\mathbb{Z})= \{p(a):a\in \mathbb{Z}\}$ has a Geometric progression. There is one confusion in the wording it is said it contains a geometric progression.Do we say there is a GP $\{g_i\}_{i=1}^{\infty}\subseteq P(\mathbb{Z})$ or $\{P(i)\}_{i\in\mathbb{Z}}$ is a GP? alibez's proof is for the second interpretation but the problem's wording suggests first one. Please clarify.
18.04.2014 08:39
joybangla wrote: sahadian wrote: find all polynomials with integer coefficients that $P(\mathbb{Z})= \{p(a):a\in \mathbb{Z}\}$ has a Geometric progression. There is one confusion in the wording it is said it contains a geometric progression.Do we say there is a GP $\{g_i\}_{i=1}^{\infty}\subseteq P(\mathbb{Z})$ or $\{P(i)\}_{i\in\mathbb{Z}}$ is a GP? alibez's proof is for the second interpretation but the problem's wording suggests first one. Please clarify. my hint it is for first interpretation . i say suppose $P(a_{i})=d^{i}$ !!! I didn't say $P(i)=d^{i}$ !! but i have question is this g.p Infinite ?
21.04.2014 20:41
exist $ {a,e\in\mathbb{R}}$ ,$e>0$$ \forall x\>a :f(x+a)>f(x)$ if isn't correct $f_1=-1*f$ $\forall x\in\mathbb{Z}\ p(x+1)p(x-1)=p(x)^2$ ==> $\forall x\in\mathbb{C}\ p(x+1)p(x-1)=p(x)^2$ if $a $ be root of $P$ $==> P(a)*P(a-2)=P(a-1)^2 ==> p(a-1)=0 ==> p(a-1)*p(a-3)=p(a-2)^2=0 ==> p(a-2)=0 ==>p(a-3)=0 ==> ... $ ==> $p$ have Extreme root
21.04.2014 21:11
arashfree wrote: exist $ {a,e\in\mathbb{R}}$ ,$e>0$$ \forall x\>a :f(x+a)>f(x)$ if isn't correct $f_1=-1*f$ $\forall x\in\mathbb{Z}\ p(x+1)p(x-1)=p(x)^2$ ==> $\forall x\in\mathbb{C}\ p(x+1)p(x-1)=p(x)^2$ if $a $ be root of $P ==> P(a)*P(a-2)=P(a-1)^2 ==> p(a-1)=0 ==> p(a-2)=0 ==> p$ have Extreme root what is it ?
25.05.2014 23:56
This is a consequence of the subspace theorem. http://www.math.jussieu.fr/~miw/articles/pdf/CLMW-DEDA2011.pdf By result 3.1 SE in page 5, let $K=\mathbb{Q}$ and let $f(x)=c^{m-1}P(x)$, where $c$ is the first term of out progression and $m$ is a prime bigger than the polynomial's degree. Clearly $f$ will achieve an $m$-th power an infinite number of times (every $m$-th term of our progression will be an $m$-th power). Since $m$ is a big prime, this means $f$ has at most one complex root. Therefore since $c \neq 0$, $P$ has at most one complex root. So $P(x)=a(x-r)^n$ where $a,n$ are integers and $r$ is complex. It is easy to see that $r$ must be an integer, and that these polynomials satisfy. I have yet to find an easy solution.
31.05.2014 16:45
This is a special case of a problem proposed by V.Senderov in Kvant ( I was too old and I don't remember the issue ) of course proposing it is depends on the proposer's mind....
07.10.2014 06:30
Can anyone give an elaborate explanation of the problem statement?
31.05.2015 20:35
This is, in fact, an analysis question. It seems like the Romanians are not the only ones who propose analysis problems for TSTs. Let $P(x)=a_mx^m+a_{m-1}x^{m-1}+...+a_1x+a_0$, let $(x_n)_{n\ge 0}$ be a sequence for which $(P(x_n))_{n\ge 0}$ is a geometric progression and let $r$ be its ratio. We can choose the progression with $r>0$ (otherwise we look at $r^2$). Furthermore, for every positive integer $N$ there are at most $2N-1$ terms with absolute value less than $N$ in the sequence (since they are distinct), so $\displaystyle\lim_{n\rightarrow \infty}|x_n|=+\infty$. We have that $r^m=\dfrac{P(x_{m+n})}{P(x_n)}=\dfrac{x_{m+n}^m}{x_n^m}\cdot \dfrac{a_m+a_{m-1}/x_{m+n}+...+a_0/x_{m+n}^m}{a_m+a_{m-1}/x_n+...+a_0/x_n^m}$, and using the relation above we deduce that $\lim_{n\rightarrow \infty} \left( \dfrac{x_{n+m}}{x_n} \right)=r$. The next step in our proof is to show that $\lim_{n\rightarrow \infty}(x_{n+m}-rx_n)<\infty$, which will almost finish the problem. Notice that: $$P(x_{n+m})-P(rx_n)=(x_{m+n}-rx_n) \cdot \left( a_m(x_{m+n}^{m-1}+rx_n x_{m+n}^{m-2}+...+r^{m-1}x_n^{m-1})+a_{m-1}(x_{m+n}^{m-2}+rx_n x_{m+n}^{m-3}+...+r^{m-2}x_n^{m-2})+...+a_1 \right).$$ But we also have: $$P(x_{n+m})-P(rx_n)=r^mP(x_n)-P(rx_n)=a_{m-1}x_n^{m-1}(r^m-r^{m-1})+a_{m-2}x_n^{m-2}(r^m-r^{m-2})+...+a_1x_n(r^m-r).$$ Combining these two relations and denoting by $z_n=x_{m+n}/x_n$, we get that: $$\lim_{n\rightarrow \infty}(x_{n+m}-rx_n)=\lim_{n\rightarrow \infty}\left( \dfrac{a_{m-1}r^{m-1}(r-1)+(stuff\ with\ low\ degree)}{a_m(z_n^{m-1}+rz_n^{m-2}+...+r^{m-1}+(stuff\ with\ low\ degree)}\right) =\dfrac{a_{m-1}r^{m-1}(r-1)}{ma_m r^{m-1}}=\dfrac{a_{m-1}(r-1)}{ma_m}.$$ Hence the sequence $(x_{m+n}-rx_n)_{n\ge 0}$ converges, and since its terms are integers, the limit is an integer $l$ and the sequence becomes stationary. So for all $n\ge N_0$ we have $x_{m+n}=rx_n+l$. Define now $q=l/(r+1)$ and $y_k=x_{N_0+km}-q$ to obtain $y_n=ry_{n-1}$. Now $P(x_{N_0+km})$ becomes $Q(y_k)$ for some polynomial $Q\in \Bbb{Q}[X]$. Define $R(x)=\dfrac{Q(y_0)x^m}{y_0^m}$. It follows that $Q(y_n)=R(y_n)$ for all positive integers $n$, therefore $Q(x)=cx^m$ for some rational number $c$. Putting back in our original variables we conclude that $P(x)=a(x-s)^m$ for some rational numbers $a$ and $s$. It is quite obvious that $a$ and $s$ must be integers and the problem is finished.
31.01.2017 19:40
alibez wrote: sahadian wrote: find all polynomials with integer coefficients that $P(\mathbb{Z})= ${$p(a):a\in \mathbb{Z}$} has a Geometric progression. i think this problem is number theory problem ! am i wrong ?
Could someone explain ths more please?
23.01.2022 09:06
We translate and multiply $p$ by a factor such that $Cp(x)=a_nx^n+a_{n-2}x^{n-2}+\cdots+a_0$ for integers $a_n, a_{n-2},\cdots,a_0$. From now, we can assume $[x^{n-1}]p(x)=0$ and $p(x)\in \mathbb{Z}[x]$. Let $r$ be the common ratio. Then $\frac{p(x_{j+n})}{p(x_j)} = r^n$. For $\nu_p$ reasons, $r\in \mathbb{Z}$ We know $|x_j|$ becomes large when $j$ is large. I contend that $x_{j+n} = \pm rx_j$. Assume not, then note $p(rx_j)-r^np(x_j) = O(x_j^{n-2})$ but $|p(rx_j\pm 1) - p(rx_j)|=O(x_j^{n-1})$, contradiction. (If $\deg p$ is even, we also need to consider $p(-rx_j)$, but the same conclusions hold) This implies that $p(rx)=r^np(x)$ for infinitely many $x$, so it is an identity. Thus $p(x)=cx^n$. Now, translating back, $p(x)=c(ax+b)^n$ for some $a,b,c\in \mathbb{Z}$ (Actually, $ax+b$ works because we can take $r=a+1$)