mberke wrote: Denote by $d(n)$ be the biggest prime divisor of $|n|>1$. Find all polynomials with integer coefficients satisfy; $$P(n+d(n))=n+d(P(n)) $$for the all $n$ integers such that $|n|>1$ and $P(n)>1$. Infinitely many such polynomials exist. For example $P(x)=-x^2$

Are you sure? $d(n)$ is the biggest prime divisor of $|n|>1$

mberke wrote: Are you sure? $d(n)$ is the biggest prime divisor of $|n|>1$ I am sure. The condition has to be respected only for $n$ such that $|n|>1$ and $P(n)>1$ And so for $P(x)=-x^2$ (and a lot of others), this condition has to be respected by ... no $n$. I think that your teacher gave you this exercise only to teach you to carefully read rhe exact statement. This was just a kind of trap.

Let $P(n)=\sum_{i=1}^ka_in^i$ be such a polynomial of degree $k$. Assume that the problem statement writes $|P(n)|>1$ instead of the current $P(n)>1$. 1. By picking $n$ a large prime, the given equation yields $P(2n)=P(n+d(n))=n+d(P(n)) \ge n$; therefore the leading coefficient $a_k$ is positive. 2. Furthermore, for $n$ a large prime the left hand side $P(2n)$ grows like $a_k(2n)^k$, whereas the right hand side $n+d(P(n))$ grows at most like $n+a_kn^k$. Since the two grows rates are equal, we get $k=1$ for the degree. This yields $P(n)=an+b$ for a positive integer $a$ and an integer $b$. 3. With this the given equation turns into $an+a\,d(n)+b=n+d(an+b)$. For $n$ a large prime, we now get $2an+b=an+a\,d(n)+b=n+d(an+b)\le n+an+b$. This implies $a=1$. 4. With this the given equation turns into $d(n)+b=d(n+b)$. If $b\ge2$, we set $n=b$ and get $d(b)+b=d(2b)\le b$. This implies $d(b)\le0$; a contradiction. For $b=1$, we set $n=3$ and get $d(3)+1=d(4)$, another contradiction. For $b=-1$, we set $n=5$ and get $d(5)-1=d(4)$, another contradiction. If $b\le -2$, we set $n=2|b|$ and get $d(2|b|)=d(|b|)-b$; the left hand side is at most $|b|=-b$, which implies $d(|b|)\le0$; yet another contradiction. 5. This only leaves the case $b=0$ and $P(n)=n$, which clearly is a solution.

mberke wrote: Denote by $d(n)$ be the biggest prime divisor of $|n|>1$. Find all polynomials with integer coefficients satisfy; $$P(n+d(n))=n+d(P(n)) $$for the all $|n|>1$ integers such that $P(n)>1$ and $d(P(n))$ can be defined. Let $P(x)=x^aQ(x)$ suppose that $Q(x)$ is not constan polynomial and $(x,Q(x))=1$ then we can select $p=prime,a$ such that $p|Q(a)$ with $p$ doew not divesi $a$. Then selecting $r=prime$ such that $r=a(modp)$ (such a prime exist by Direclert). For $n=r$ we have: $P(2r)=r+d(P(r))<=r+\frac{P(r)}{p}$ but if $deg(P(x))>1$ we get contradiction. So $deg(P(x))=1$ which means that $P(x)=ax+b$ but again we have:$a2r+b=r+\frac{P(r)}{p}$ which gives contradiction. So $Q(x)=constant$ which gives $P(x)=b*x^a$ take $n=prime$ large enough we get that $P(x)=x$