Consider the polynomial $f(X) \in \mathbb{Z}[x]$ prove the following

Question

Let $f(X) \in \mathbb{Z}[x]$ be a polynomial. Prove the following:

i) For every natural $n>0$ the equation $f(x)=n$ has a solution in $\mathbb{Z}$ then $f(x)=\pm x + c$ with $c$ constant.

ii) If $f(\mathbb{Z})$ contains only prime numbers then $f$ is constant. [Something stronger has been proven here A nonconstant polynomial $q$ with $q(0)>1$ attains infinitely many composite values at integers ... or is it stronger? Here I just need to find one number which isn't prime]

iii) If $f(\mathbb{Z})$ contains only numbers of the set $\{\pm 2^a 3^b; a,b \in \mathbb{N}\}$ then $f$ is constant.

I have proved it, but especially for the last two points my solution (at least, to me) doesn't feel elegant...is there a way to prove it directly? Thank you for the help!

Answer 1

Every number I use, unless specified differently, shall be considered in $\mathbb{Z}$. Also, the generic polynomial shall be $f(x)=a_n x^n + ... + a_1 x + a_0$.

i) Let $deg(f(x))\geq 2$. Consider the derivative $f'(x)$. If the equation $f(x)=n$ has a solution $x_n$ this means we have $f'(x_n)=0$. Thus $f'(x)$ has infinitely many solutions and it is a polynomial. This is absurd $\Rightarrow \ f(x) = a x + c$. If $a=0$ clearly it can't have a solution for every $n$. Then let $|a| \not=1$. Consider $n=|a-1|+c$. The only soution is then $\frac{|a-1|}{a}$ which is not an integer.

ii) We shall prove the converse, namely: "If $f$ is not constant then $f(\mathbb{Z})$ does not contain only prime numbers".

Suppose the absurd "$f(\mathbb{Z})$ contains only prime numbers". Then $f(0)=a_0$ so it's a prime number, which means $a_0>1 \Rightarrow f(a_0)$ is not a prime.

iii) We shall prove the converse, namely: "If $f$ is not constant then $f(\mathbb{Z})$ does not contain only numbers in the form $\pm 2^a 3^b$."

If $a_0=0$ the answer is trivial. If $|a|=1$ then $f(2)$ not in that form. Otherwise $f(a_0) = a_0 (1+a_1 a_0 + ... + a_n a_0^{n-1})$ is not in that form.

Answer 2

i) Suppose that for every $n$ there exists $x_n$ such that $f(x_n)=n$. Then you have $x_{n+1}-x_n |f(x_{n+1})-f(x_n)=1$ which means that $|x_{n+1}-x_n|=1$. This means that $(x_n)$ is a sequence of consecutive integers (increasing or decreasing). So $x_n = \pm n +c$. As a consequence, we know that the polynomial $$ f(x)-(\pm x+c)$$ has infinitely many roots, and is therefore the zero polynomial.

ii) Pick an integer such that $f(k)\neq 0,\pm 1$ and denote $p=f(k)$. Then $p|f(k+l\cdot p)-f(k)$ so $f(k+l\cdot p)$ is also divisible with $p$ for any $l$. Since $f(k+l\cdot p)$ gets arbitrary larger and it is divisible by $p$, it cannot be always a prime number.