Are you sure that the modulus should surround f^k(a) - a, as the condition given after that is a characteristic of the modulus? And that condition holds no matter what f(x) is, thus making it impossible to determine anything about f(x).

ู^The condition suggest that $f^k(a)\neq a.$ matinyousefi wrote: A polynomial $f(x)$ with integer coefficients is given. We define $d(a,k)=|f^k(a)-a|.$ It is known that for each integer $a$ and natural number $k$, $d(a,k)$ is positive. Prove that for all such $a,k$, $$d(a,k) \geq \frac{k}{3}.$$ Assume FTSOC that there exists $k,l\in\mathbb{N}$ and $a\in\mathbb{Z}$ such that $k>3l$ and $d(a,k)=l.$ Assume that $f^k(a)=a+l$ (the other case can be handled similarly.) Let $f^h(a)=\min\{f^0(a),f^1(a),...,f^k(a)\}$ and $f^g(a)=\max\{f^0(a),f^1(a),...,f^k(a)\}.$ If $g<k,$ then $|f^h(a)-f^g(a)|>|f^{h+1}(a)-f^{g+1}(a)|>0,$ which is a contradiction since $f^h(a)-f^g(a)|f^{h+1}(a)-f^{g+1}(a).$ Thus, $g=k.$ Now, it's easy to see that $f^h(a)\le a+l-k<a-2l.$ (In particular, $h>0.$) Let $a-f^h(a)=j>2l.$ We cah see that for all $i\le k-h,$ we have $j|f^i(a)-f^{i+h}(a)$ and $0<|f^i(a)-f^{i+h}(a)|\le j+l<2j.$ Thus, $|f^i(a)-f^{i+h}(a)|=j$ for all such $i.$ If $2h\le k,$ then $|f^h(a)-f^{2h}(a)|=j\implies f^{2h}(a)=f^h(a)+j=a,$ a contradiction. Thus, $2h>k.$ We can see that $|f^k(a)-f^{k-h}(a)|=j\implies f^{k-h}(a)=a+l-j.$ Thus, $a-f^{k-h}(a)|f^h(a)-f^k(a)\implies j-l|l+j\implies j-l|2l\implies j=3l.$ Finally, $a-f^{k-h}(a)|f^{k-h}(a)-f^{2k-2h}(a)$ and $f^{k-h}(a)-f^{2k-2h}(a)|f^h(a)-f^k(a)$ imply that $f^{k-h}(a)-f^{2k-2h}(a)\in\{2l,4l,-2l,-4l\}.$ Thus, $f^{2k-2h}(a)\in\{a,a+2l,a-4l,a-6l\},$ all cases lead to contradiction.

Let’s prove three statements. 1. $p| |f^d(a)-a|$ => $p | f^{kd}(a)-a|$ it’s obvious 2. Let’s call $d=ord_p(a)$ if $d$ is the smallest positive number such that $p| |f^d(a)-a|$. Then $d=ord_p(a)$ => $d\leq p$. Well, suppose that $d\geq p+1$. Then let’s consider residues mod p of numbers $f(a),f^2(a),..., f^{p+1}(a)$. At least two of them coincide. WLOG $f^k(a)\equiv f^l(a)$ where $l>k$, but when $f^d(a)$ gives a residue coinciding with one of $f^k(a),...,f^{l-1}(a)$, where $l-1\leq p$. So, $d$ isn’t the smallest such number — contradiction. 3. If $k>l$, then $f^{d(l+1)}-f^{dl}| f^{d(k+1)}-f^{dk}$. Let’s note that it would be enough to prove that $f^{d(k+2)}(a)-f^{d(k+1)}(a)| f^{d(k+1)}(a)-f^{dk}(a)$ $f^d(x)=g(x)$. then $f^{dk}(a)=t$, $f^{d(k+1)}(a)=g(t)$. $f^{d(k+2)}(a)-f^{d(k+1)}(a)=g(g(t))-g(t)$. $m= f^{d(k+1)}(a)-f^{dk}(a)$ then $g(t)\equiv t$ mod m. So $g(g(t))\equiv g(t)\equiv t$ mod m and $g(t)\equiv t$. Hence, $f^{d(l+1)}-f^{dl}| f^{d(k+1)}-f^{dk}$. Well, suppose that $|f^m(a)-a|=p<m/3$ $ord_p(a)=d$. So, $m=kd$ and $3p<kd\leq kp$. That means $k>3$. By Besu’s theorem $f^d(a)-a|f^{2d}(a)-f^d(a)$ $f^{2d}(a)-a|f^{3d}(a)-f^{d}(a)$ ... $f^{(k-1)d}(a)-a|f^{kd}(a)-f^{d}(a)$ And let’s note that $f^d(a)-a|f^{kd}(a)-a$. Therefore, $| f^d(a)-a|=p$. First case. $f^d(a)=f^{kd}(a)$. Then $f^{(k-1)d}(f^d(a))-f^d(a)=0$ — contradiction. Second case. $f^{kd}(a)=p+a, f^{d}(a)=-p+a$ So, $f^{(k-1)d}(a)-a|f^{kd}(a)-f^{d}(a)$. That is $f^{(k-1)d}(a)-a|\pm 2p$. Therefore $| f^{(k-1)d}(a)-a|=2p$, otherwise $f^{(k-1)d }(a) $ is equal to either $f^d(a)$ or $f^{kd}(a)$. If $f^{(k-1)d}(a)-a=2p$, then $f^{d(k-1)}-f^{dk}=p$. Thus by (3) $f^{d(l+1)}-f^{dl}=\pm p $ for $l\leq k-1$. But then since $k>3$ at least one of $f^d(a),...,f^{kd}(a)$ is equal to $a$ — contradiction to condition. If $f^{(k-1)d}(a)-a=-2p$, then $f^{d(k-1)}-f^{dk}=-3p$. Thus by (3) $f^{d(l+1)}-f^{dl}=\pm 3p, p $ for $l\leq k-1$. Moreover, if $f^{d(l+1)}-f^{dl}= \pm p $, then for $t<l$ $f^{d(t+1)}-f^{dt}= \pm p $. Well, we have k numbers, the first is $-p+a$, the last is $p+a$. The second can be either $-2p+a$ or $2p+a$. $2p+a$ is impossible because by adding $\pm 3p$ we can’t get $p+a$. The third can’t be $p+a$ since $k>3$. So, the third is $-3p+a$. The fourth can’t be $a$, so the fourth is $-4p+a$. But then we will have at least two coinciding numbers in our sequence — contradiction. To the third case we can apply the same reasoning from the second case.