Let $f (x)= x^7 -105x + 12$ then which of the following options is correct?

Question

Let $f (x)= x^7 - 105 x +12$ then

1. $f (x)$ is reducible over $ \mathbb {Q} $

2. There exists an integer $m$ such that $f (m)=105$

3. There exists an integer $m$ such that $f (m)=2$

4. $f (m)$ is not a prime number for any integer $m$

By Eisenstein 1 option is false. For option 2, if possible let there be an integer such that $f (m)=105$ then

$ m^7 - 105 m +12= 105$ implies $ m^7 -105 m -93=0$, $m$ can be $1, -1, 3, -3, 31, -31, 93, -93$ so clearly option 2 is false and in a similar way option 3 is also false. I have no idea about option 4.

So my request is to check whether my reasoning of option 1, 2 and 3 are alright and to give me a hint for option 4.

Thank you in advance.

Answer 1

Hint for 4). Note that for $n\in\mathbb{N}$, and $x\in\mathbb{Z}$, $x^n\equiv x \pmod{2}$, therefore $$f(x)=x^7-105x+12\equiv x-x+0=0\pmod{2}$$ which means that for any integer $x$, $f(x)$ is divisible by $2$.

Since $2$ is the only even prime it remains to show that there is no integer $x$ such that $f(x)=2$. That is, the equation $$x^7-105x+10=0$$ has no integer solution. Can you take it from here? Recall the Rational Root Theorem.

P.S. BTW option 2 is false because 105 is odd.

Answer 2

Look at $f(x)$ modulo $2$. $$x^7-105x+12 \equiv x^7-x \pmod{2}$$ This polynomial evaluates $0$ for $x\equiv 0$ and for $x \equiv 1$: this suffices to prove that $f(x)$ is always divisible by $2$.