Actual Question is to prove that, for a prime p and $a\in F_p$, $a\neq 0$, $f(x)=x^p-x+a$ is irreducible.
This is an exercise Question in Dummit Foote 13.5.5.
Hint : Prove that if $\alpha$ is a root of $f(x)$ then so is $\alpha+1$.
Proof: Suppose $f(\alpha)=0$ i.e., $\alpha^{p}-\alpha+a=0$ Consider $(\alpha+1)^{p}-(\alpha+1)+a=\alpha^p+1-\alpha-1+a=\alpha^{p}-\alpha+a=0$ . So, I proved that if $\alpha$ is a root of $f(x)$ then so is $\alpha+1$.
As further approach I have seen that no element of $F_p$ is a root because if $\alpha\in F_p$ is a root then so is $\alpha+1$ and so is $\alpha+2$ and so would be $0\in F_p$ which is a contradiction as $f(0)=0-0+a=a\neq 0$. So, No element of $F_p$ is a root.
Just by this he wants to say that $f(x)$ is irreducible i.e., if $f(x)$ has no root in $F_p$ then it is irreducible.
I am unable to proceed further in this line. Any help would be appreciated.
Thank You.
P.S : I have seen just now that this question was already answered. I thought of deleting this Question. But Mr.AymanHourieh gave another nice answer. So, i Request Not to close this as this would just give another answer.
Thank You