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 ?

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.

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 ?

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

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 ?

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.

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....

Can anyone give an elaborate explanation of the problem statement?

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.

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?

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 $r=a+1$)